buch

A Course in Model Theory 1

Katrin Tent and Martin Ziegler

Münster and Freiburg 2010

1v0.8-53-g4c86431, Thu Nov 18 14:57:21 2010 +0100

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1 The Basics 6

1 Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Language . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Elementary extensions and compactness 25

4 Elementary substructures . . . . . . . . . . . . . . . . . . . . . . 25

5 The Compactness Theorem . . . . . . . . . . . . . . . . . . . . . 28

6 The Löwenheim–Skolem Theorem . . . . . . . . . . . . . . . . . . 34

3 Quantifier elimination 36

7 Preservation theorems . . . . . . . . . . . . . . . . . . . . . . . . 36

8 Quantifier elimination . . . . . . . . . . . . . . . . . . . . . . . . 42

9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4 Countable models 58

10 The Omitting Types Theorem . . . . . . . . . . . . . . . . . . . . 58

11 The space of types . . . . . . . . . . . . . . . . . . . . . . . . . . 60

12 ℵ0–categorical theories . . . . . . . . . . . . . . . . . . . . . . . . 64

13 The amalgamation method . . . . . . . . . . . . . . . . . . . . . 69

14 Prime models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

1

CONTENTS 2

5 ℵ1–categorical theories 76

15 Indiscernibles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

16 ω–stable Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

17 Prime extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

18 Lachlan’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 88

19 Vaughtian Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

20 Algebraic formulas . . . . . . . . . . . . . . . . . . . . . . . . . . 94

21 Strongly minimal sets . . . . . . . . . . . . . . . . . . . . . . . . 97

22 The Baldwin-Lachlan theorem . . . . . . . . . . . . . . . . . . . . 103

6 Morley Rank 104

23 Saturated models and the Monster . . . . . . . . . . . . . . . . . 104

24 Morley Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

25 Countable models of ℵ1-categorical theories . . . . . . . . . . . . 117

26 Computation of Morley Rank . . . . . . . . . . . . . . . . . . . . 121

7 Simple theories 125

27 Dividing and Forking . . . . . . . . . . . . . . . . . . . . . . . . . 125

28 Simplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

29 The Independence Theorem . . . . . . . . . . . . . . . . . . . . . 136

30 Lascar strong types . . . . . . . . . . . . . . . . . . . . . . . . . . 142

31 Example: Pseudo-finite fields . . . . . . . . . . . . . . . . . . . . 145

8 Stable Theories 148

32 Heirs and Coheirs . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

33 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

34 Definable types . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

35 Elimination of imaginaries and T eq . . . . . . . . . . . . . . . . . 159

36 Properties of forking in stable theories . . . . . . . . . . . . . . . 166

37 SU-rank and the stability spectrum . . . . . . . . . . . . . . . . . 173

9 Prime extensions 178

CONTENTS 3

38 Indiscernibles in stable theories . . . . . . . . . . . . . . . . . . . 178

39 Totally transcendental theories . . . . . . . . . . . . . . . . . . . 181

40 Countable stable theories . . . . . . . . . . . . . . . . . . . . . . 185

10 The fine structure of ℵ1–categorical theories 187

41 Internal types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

42 Analysable types . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

43 Locally modular strongly minimal sets . . . . . . . . . . . . . . . 195

44 Hrushovski’s Examples . . . . . . . . . . . . . . . . . . . . . . . . 199

A Set Theory 208

1 Sets and classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

2 Cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

3 Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

B Fields 216

4 Ordered fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

5 Differential Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

6 Separable and regular field extensions . . . . . . . . . . . . . . . 224

7 Pseudo-finite fields and profinite groups . . . . . . . . . . . . . . 228

C Combinatorics 232

8 Pregeometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

9 The Erdös–Makkai Theorem . . . . . . . . . . . . . . . . . . . . . 238

10 The Erdös–Rado Theorem . . . . . . . . . . . . . . . . . . . . . . 239

D Solutions of Exercises 240

Bibliography 258

Index 262

CONTENTS 4

1 Introduction

This book aims to be an introduction to model theory which can be used with-out any background in logic. We start from scratch, introducing first-orderlogic, structures, languages etc. but move on fairly quickly to the fundamentalresults in model theory and stability theory. We also decided to cover simpletheories and Hrushovski constructions, which over the last decade developedinto an important subject. We try to give the necessary background in algebra,combinatorics and set theory either in the course of the text or in the corre-sponding section of the appendices. The exercises form an integral part of thebook. Some of them are used later on, others complement the text or presentaspects of the theory that we felt should not be completely omitted. For themost important exercises (and the more difficult ones) we include (hints for)solutions at the end of the book.

The book falls into four parts. The first three chapters introduce the basicsas would be contained in a course giving a general introduction to model theory.This first part ends with Chapter 4 which introduces and explores the notion ofa type, the topology on the space of types and a way to make sure that a certaintype will not be realized in a model to be constructed. The chapter ends withFraïssé’s amalgamation method, a simple but powerful tool for constructingmodels.

Chapter 5 is devoted to Morley’s famous theorem that a theory with a uniquemodel in some uncountable cardinality has a unique model in every uncountablecardinality. To prove this theorem, we describe the analysis of uncountablycategorical theories due to Baldwin and Lachlan in terms of strongly minimalsets. These are in some sense the easiest examples of stable theories and serveas an introduction to the topic. This chapter forms a unit with Chapter 6 inwhich the Morley rank is studied in a bit more detail.

For the road to more general stable theories we decided to go via simplicity.The notion of a simple theory was introduced by Shelah in [49]. Such theoriesallow for a notion of independence which is presented in Chapter 7. Funda-mental examples such as pseudo-finite fields make simple theories an importantgeneralisation of the stable ones. We specialize this notion of independence inChapter 8 to characterize forking in stable theories.

In Chapters 9 and 10 we go back to more classical topics of stability theorysuch as existence and uniqueness of prime extensions and their analysis in theuncountably categorical case due to Hrushovski. We end the exposition byexplaining a variant of Hrushovski’s construction of a strongly minimal set.

Model theory does not exist independently of set theory or other areas ofmathematics. Many proofs require a knowledge of certain principles of infinitecombinatorics which we were hesitant to assume as universally known. Similarly,to study theories of fields we felt it necessary to explain a certain amount of

CONTENTS 5

algebra. In the three appendices we try to give the background on set theoryand algebra necessary to be able to follow the exposition in the text.

Other books, some emphasising particular aspects of the theory, some gen-eral introductions that we recommend for further reading include the following[39],[37], [35], [11], [19],[40], [41], [12], [53].

We would like to thank Manuel Bleichner, Juan-Diego Caycedo, PhilippDoebler, Antongiulio Fornasiero, Nina Frohn, Zaniar Ghadernezhad, GuntramHainke, Immanuel Halupczok Franziska Jahnke, Alexander Kraut, Moritz Müller,Alexandra Omar Aziz, Lars Scheele and Nina Schwarze for carefully reading ear-lier versions of the manuscript. We also thank Andreas Baudisch for trying outthe book in a seminar and Bernhard Herwig, who translated early parts of thelecture notes from which parts of this book evolved.

Chapter 1

The Basics

1 Structures

In this section we start at the very beginning, by introducing the prerequisitesfor the objects of study. We deal with first-order logic and its structures. Tothis end we first introduce the languages. These will be chosen in different waysfor the different mathematical structures that one wants to study.

Definition 1.1. A language L is a set of constants, function symbols andrelation symbols1.

Function symbols and relation symbols have an arity ≥ 1. One can think ofconstants as 0–ary function symbols2. This allows us to omit the constantsymbol case in many proofs.

The language per se has no inherent meaning. However, the choice of languagewill reflect the nature of the intended objects. Here are some standard examples:

L∅ = ∅ The empty language.LAG = {0,+,−} The language of abelian groups.LR = LAG ∪ {1, ·} The language of rings.LG = {e, ◦,−1} The language of groups.LO = {<} The language of orders.LOR = LR ∪ LO The language of ordered rings.LN = {0,S,+, ·, <} The language of the natural numbers.Lsets = {ε} The language of set theory.

The symbols are1 we also use predicate for relation symbol.2By an unfortunate convention 0–ary relation symbols are not considered.

6

CHAPTER 1. THE BASICS 7

constants: 0, 1, eunary function symbols: −, −1, Sbinary function symbols: +, ·, ◦binary relation symbols: <, ε.

The languages obtain their meaning only when interpreted in an appropriatestructure:

Definition 1.2. Let L be a language. An L–structure is a pair A =(A, (ZA)Z∈L

),

where

A is a non–empty set, the domain or universe of A,ZA ∈ A if Z is a constant,ZA : An −→ A if Z is an n–ary function symbol, andZA ⊂ An if Z is an n–ary relation symbol.

ZA is called the A–interpretation of Z.

The requirement on A to be not empty is merely a (sometimes annoying)convention. The cardinality of a structure is the cardinality of its universe. Wewrite |A| or |A| for the cardinality of A.

Definition 1.3. Let A and B be L–structures. A map h : A → B is called ahomomorphism if for all a1, . . . , an ∈ A

h(cA) = cB

h(fA(a1, . . . , an)) = fB(h(a1), . . . , h(an))

RA(a1, . . . , an) ⇒ RB(h(a1), . . . , h(an))

for all constants c, n-ary function symbols f and relation symbols R from L.We denote this by

h : A → B.

If in addition h is injective and

RA(a1, . . . , an) ⇔ RB(h(a1), . . . , h(an))

for all a1, . . . , an ∈ A, then h is called an (isomorphic) embedding . An isomor-phism is a surjective embedding. We denote isomorphisms by

h : A∼→ B.

If there is an isomorphism between A and B, the two structures are calledisomorphic and we write

A ∼= B.

It is easy to see that being isomorphic is an equivalence relation between struc-tures and that bijections can be used to transfer the structures between sets.


Definition 1.4. An automorphism of A is an isomorphism A∼→ A. The set of

automorphisms Aut(A) forms a group under composition.

Definition 1.5. A is called substructure of B if A ⊂ B and if the inclusionmap is an embedding from A to B. We denote this by

A ⊂ B.

B is an extension of A if A is a substructure of B.

Remark 1.6. If B is an L–structure and A a non–empty subset of B, thenA is the universe of a (uniquely determined) substructure A if and only if Acontains all cB and A is closed under all operations fB. In particular, if L doesnot contain any constants or function symbols, then any non–empty subset ofan L–structure is again an L–structure. Also, if h : A → B, then h(A) is theuniverse of a substructure of B.

It is also clear that for any family Ai of substructures of B, the intersectionof the Ai is empty or a substructure of B. Therefore, if S is any non–emptysubset of B, then there exists a smallest substructure A = 〈S〉B which containsS. We call A the substructure generated by S. If S is finite, then A is said tobe finitely generated.

If L contains a constant, then the intersection of all substructures of B isnot empty as it contains the B–interpretation of this constant. Thus B has asmallest substructure 〈∅〉B. If L has no constants, we set 〈∅〉B = ∅

Lemma 1.7. If A is generated by S, then every homomorphism h : A → B isdetermined by its values on S.

Proof. If h′ : A → B is another homomorphism, then C = {b | h(b) = h′(b)} isempty or a substructure. If h and h′ coincide on S, then S is a subset of C, andtherefore C = A.

Lemma 1.8. Let h : A∼→ A′ be an isomorphism and B an extension of A. Then

there exists an extension B′ of A′ and an isomorphism g : B∼→ B′ extending

h.

Proof. First extend the bijection h : A → A′ to a bijection g : B → B′ and useg to define an L–structure on B′.

Definition 1.9. Let (I,≤) be a directed partial order. This means that for alli, j ∈ I there exists a k ∈ I such that i ≤ k and j ≤ k. A family (Ai)i∈I ofL–structures is called directed if

i ≤ j ⇒ Ai ⊂ Aj .


If I is linearly ordered, we call (Ai)i∈I a chain. If, for example, a structureA1 is isomorphic to a substructure A0 of itself,

h0 : A0∼→ A1,

then Lemma 1.8 gives an extension

h1 : A1∼→ A2.

Continuing in this way, we obtain a chain A0 ⊂ A1 ⊂ A2 . . . and an increasingsequence hi : Ai

∼→ Ai+1 of isomorphisms.

Lemma 1.10. Let (Ai)i∈I be a directed family of L–structures. Then A =⋃i∈I Ai is the universe of a (uniquely determined) L–structure

A =⋃i∈I

Ai,

which is an extension of all Ai.

Proof. Let R be an n–ary relation symbol and a1, . . . , an ∈ A. As I is directed,there exists k ∈ I such that all ai are in Ak. We define (and this is the onlypossibility)

RA(a1, . . . , an) ⇔ RAk(a1, . . . , an).

Constants and function symbols are treated similarly.

A subset K of L is called a sublanguage. An L–structure becomes a K–structure if we forget the interpretations of the symbols from L \K:

A � K =(A, (ZA)Z∈K

)We call this process restriction. Conversely we call A an expansion of A � K.Here are some examples:

Let A be an L–structure.

a) Let R be an n–ary relation on A. We introduce a new relation symbol Rand we denote by

(A, R)

the expansion of A to an L ∪ {R}–structure in which R is interpreted by R.

b) For given elements a1, . . . , an we may introduce new constants a1, . . . , an andconsider the L ∪ {a1, . . . , an}–structure

(A, a1, . . . , an) .


c) Let B be a subset of A. By considering every element of B as a new constant,we obtain the new language

L(B) = L ∪B

and the L(B)–structureAB = (A, b)b∈B .

Note that Aut(AB) is the group of automorphisms of A fixing B elementwise.We denote this group by Aut(A/B).

Similarly, if C is a set of new constants, we write L(C) for the languageL ∪ C.

Many–sorted structures

Without much effort, the concepts introduced here can be extended to many–sorted languages and structures, which we shall need to consider later on.

Let S be a set, which we call the set of sorts. An S-sorted language L is givenby a set of constants for each sort in S, and typed function and relation symbolswhich carry the information about their arity and the sorts of their domain andrange. More precisely, for any tuple (s1, . . . sn) and (s1, . . . , sn, t) there is a setof relation symbols and function symbols, respectively. An S-sorted structureis a pair A =

(A, (ZA)Z∈L

), where

A is a family (As)s∈S of non–empty sets.ZA ∈ As if Z is a constant of sort s ∈ S,ZA : As1 × . . .×Asn −→ At if Z is a function symbol of type (s1, . . . , sn, t),ZA ⊂ As1 × . . .×Asn if Z is a relation symbol of type (s1, . . . , sn).

It should be clear how to define homomorphisms between many–sorted struc-tures A and B: they are given by maps taking As to Bs for s ∈ S and behavingas before with respect to constants, function and relation symbols.

Example:Consider the two-sorted language LPG for permutation groups with a sort x forthe set and a sort g for the group. The constants and function symbols for LPG

are those of LG restricted to the sort g and an additional function symbol ϕof type (x, g, x). Thus, an LPG-structure (X,G) is given by a set X and anLG-structure G together with a function X ×G −→ X.

Exercise 1.1 (Direct products).Let A1,A2 be L–structures. Define an L–structure A1×A2 with universe A1×A2

such that the natural epimorphisms πi : A1 × A2 −→ Ai for i = 1, 2 satisfy thefollowing universal property: Given any L–structure D and homomorphismsϕi : D −→ Ai, i = 1, 2 there is a unique homomorphism ψ : D −→ A1 ×A2 suchthat πi ◦ψ = ϕi, i = 1, 2, i.e., this is the product in the category of L–structureswith homomorphisms.


Exercise 1.2 (Afshordel).Let f : A → A be an embedding. Then there is an extension A ⊂ B and anextension of f to an automorphism g of B. We can find B as the union of thechain A ⊂ g−1(A) ⊂ g−2(A) ⊂ · · · . The pair (B, g) is uniquely determined bythat property.


2 Language

Starting from the inventory of the languages defined in Section 1 we now describethe grammar which allows us to build well-formed terms and formulas whichwill again be interpreted in the according structures.

Definition 2.1. L–terms are words (sequences of symbols) built from con-stants, the function symbols of L and the variables v0, v1, . . . according to thefollowing rules:

1. Every variable vi and every constant c is an L–term.

2. If f is an n–ary function symbol and t1, . . . , tn are L–terms, then alsoft1 . . . tn is an L–term.

The number of function symbols occurring in a term is called its complexity.This will be used in induction arguments.

We often write f(t1, . . . , tn) instead of ft1 . . . tn for better readability anduse the usual conventions for some particular function symbols. For example

(x+ y) · (z + w)

stands for·+ xy + zw

and (x ◦ y)−1 for −1 ◦ xy.

Let A be an L–structure and ~b = (b0, b1, . . .) a sequence of elements whichwe consider as assignments to the variables v0, v1, . . .. If we replace in t eachvariable vi by ai, the term t determines an element tA[~b] of A in an obvious way:

Definition 2.2. For an L–term t, an L–structure A and an assignment ~b wedefine the interpretation tA[~b] by

vAi [~b] = bi

cA[~b] = cA

ft1 . . . tAn [~b] = fA(tA1 [

~b], . . . , tAn [~b])

This (recursive) definition is possible because every term has a unique decom-position into its constituents: if ft1 . . . tn = ft′1 . . . t

′n, then t1 = t′1, . . . , tn = t′n.

This as well as the following lemma are easy to prove using induction on thecomplexity of the terms involved.

Lemma 2.3. The interpretation tA[~b] depends on bi only if vi occurs in t.


If the variables x1, . . . , xn are pairwise distinct3 and if no other variablesoccur in t, we write

t = t(x1, . . . , xn).

According to the previous lemma, if ~b is an assignment for the variables whichassigns ai to xi, we can define

tA[a1, . . . , an] = tA[~b].

If t1, . . . , tn are terms, we can substitute t1, . . . , tn for the variables x1, . . . , xn.The resulting term is denoted by

t(t1, . . . , tn).

One easily proves:

Lemma 2.4 (Substitution Lemma).

t(t1, . . . , tn)A[~b] = tA

[tA1 [~b], . . . , tAn [

~b]]

If we expand A to the L(A)–structure AA, we get as a special case

t(a1, . . . , an)AA = tA[a1, . . . , an].

Lemma 2.5. Let h : A → B be a homomorphism and t(x1, . . . , xn) a term.Then we have for all a1, . . . , an from A

tB[h(a1), . . . , h(an)] = h(tA[a1, . . . , an]

)Proof. Induction on the complexity of t.

Lemma 2.6. Let S be a subset of the L–structure A. Then

〈S〉A ={tA[s1, . . . , sn] | t(x1, . . . , xn) L–term, s1, . . . , sn ∈ S

}.

Proof. We may assume that S is not empty or that L contains a constant sinceotherwise both sides of the equation are empty. It follows from Lemma 2.5that the universe of a substructure is closed under interpretations of termstA[−, . . . ,−]. Thus the right hand side is contained in 〈S〉A. For the conversewe have to show that the right hand side is closed under the operations fA. Theassertion now follows using Remark 1.6.

3Remember that xi ∈ {v0, v1, . . .}


A constant term is a term without variables. As a special case of Lemma 2.6we thus have

〈∅〉A ={tA | t constant L–term

}.

The previous lemma implies:

Corollary 2.7.|〈S〉A| ≤ max{|S|, |L|,ℵ0}

Proof. There are at most max{|L|,ℵ0} many L–terms and for every term t atmost max{|S|,ℵ0} many assignments of elements of S to the variables of t.

We still need to define L–formulas . These are sequences of symbols whichare built from the symbols of L, the parentheses “(” and “)” as auxiliary symbolsand the following logical symbols:

variables v0, v1, . . .equality symbol .

=negation symbol ¬conjunction symbol ∧existential quantifier ∃

Definition 2.8. L–formulas are

1. t1.= t2 where t1, t2 are L–terms,

2. Rt1 . . . , tn where R is an n–ary relation symbol from L andt1, . . . , tn are L–terms,

3. ¬ψ where ψ is an L–formula,

4. (ψ1 ∧ ψ2) where ψ1 and ψ2 are L–formulas,

5. ∃x ψ where ψ is an L–formula and x a variable.

Formulas of the form t1.= t2 or Rt1 . . . , tn are called atomic.

As with terms, we define the complexity of a formula as the number ofoccurrences of ¬,∃ and ∧. This allows us to do induction on (the complexityof) formulas.

We use the following abbreviations:

(ψ1 ∨ ψ2) = ¬(¬ψ1 ∧ ¬ψ2)

(ψ1 → ψ2) = ¬(ψ1 ∧ ¬ψ2)

(ψ1 ↔ ψ2) = ((ψ1 → ψ2) ∧ (ψ2 → ψ1))

∀x ψ = ¬∃x¬ψ

for disjunction, implication, equivalence and universal quantifier.


Sometimes we write t1Rt2 for Rt1t2, ∃x1 . . . xn for ∃x1 . . .∃xn and ∀x1 . . . xnfor ∀x1 . . . ∀xn. To increase readability we might use superfluous parentheses.On the other hand we might omit parentheses with an implicit understandingof the binding strength of logical symbols: ¬, ∃, and ∀ bind more stronglythan ∧ which in turn binds more strongly than ∨. Finally → and ↔ have theleast binding strength. For example ¬ψ1 ∧ ψ2 → ψ3 is understood to mean((¬ψ1 ∧ ψ2) → ψ3).

Given an L–structure A and an L–formula ϕ(x1, . . . xn) it should now beclear what it means for ϕ to hold for ~b. Here is the formal definition:

Definition 2.9. Let A be an L–structure. For L–formulas ϕ and all assignments~b we define the relation

A |= ϕ[~b]

recursively over ϕ:

A |= t1.= t2 [~b] ⇔ tA1 [

~b] = tA2 [~b]

A |= Rt1 . . . tn [~b] ⇔ RA(tA1 [~b], . . . , tAn [

~b])

A |= ¬ψ [~b] ⇔ A 6|= ψ [~b]

A |= (ψ1 ∧ ψ2) [~b] ⇔ A |= ψ1 [~b] and A |= ψ2 [~b]

A |= ∃xψ [~b] ⇔ there exists a ∈ A such that A |= ψ[~ba

x

]Here we use the notation

~ba

x= (b0, . . . , bi−1, a, bi+1,...) if x = vi.

If A |= ϕ[~b] holds we say ‘ϕ holds in A for ~b ’ or ‘~b satisfies ϕ (in A)’.

For this definition to work one has to check that every formula has a uniquedecomposition4 into subformulas: ifRt1, . . . , tn = Rt′1, . . . , t

′n, then t1 = t′1, . . . , tn =

t′n; and (ψ1 ∧ ψ2) = (ψ′1 ∧ ψ′

2) implies ψ1 = ψ′1 and ψ2 = ψ′

2.

It should be clear that our abbreviations have the intended meaning, e.g.

A |= (ψ1 → ψ2)[~b] if and only if (A |= ψ1[~b] implies A |= ψ2[~b]).

Whether ϕ holds in A for ~b depends only on the free variables of ϕ:

Definition 2.10. The variable x occurs freely in the formula ϕ if it occurs at aplace, which is not within the scope of a quantifier ∃x. Otherwise its occurrence

4It is precisely for this uniqueness that brackets were introduced when defining formulas.


is called bound. Here is the formal definition (recursive in ϕ):

x free in t1.= t2 ⇔ x occurs in t1 or in t2.

x free in Rt1 . . . tn ⇔ x occurs in one of the ti.x free in ¬ψ ⇔ x free in ψ

x free in (ψ1 ∧ ψ2) ⇔ x free in ψ1 or x free in ψ2

x free in ∃y ψ ⇔ x 6= y and x free in ψ

For example the variable v0 does not occur freely in ∀v0(∃v1R(v0, v1) ∧P (v1)); v1 occurs both freely and bound5.

Lemma 2.11. Suppose ~b and ~c agree on all variables which are free in ϕ. Then

A |= ϕ[~b] ⇔ A |= ϕ[~c].

Proof. By induction on the complexity of ϕ.

If we write a formula in the form ϕ(x1, . . . , xn), we mean:

• the xi are pairwise distinct,

• all free variables in ϕ are among {x1, . . . , xn}.

If furthermore a1, . . . , an are elements of the structure A, we define

A |= ϕ[a1, . . . , an]

by A |= ϕ[~b], where ~b is an assignment satisfying ~b(xi) = ai. Because ofLemma 2.11 this is well defined.

Thus ϕ(x1, . . . , xn) defines an n–ary relation

ϕ(A) = {a | A |= ϕ[a]}

on A, the realisation set of ϕ. Such realisation sets are called 0–definablesubsets of An, or 0–definable relations.

Let B be a subset of A. A B–definable subset of A is a set of the form ϕ(A)for an L(B)–formula ϕ(x). We also say that ϕ (and ϕ(A)) are defined over B andthat the set ϕ(A) is defined by ϕ. Often we don’t explicitly specify a parameterset B and just talk about definable subsets. 0–definable sets are definable overthe empty set. We call two formulas equivalent if in every structure they definethe same set.

5However, we usually make sure that no variable occurs both freely and bound. This canbe done by renaming the free occurrence by an unused variable.


Definition 2.12. A formula ϕ without free variables is called a sentence. Wewrite A |= ϕ if A |= ϕ[~b] for some (all) ~b.

In that case A is called a model of ϕ. We also say ϕ holds in A. If Σ is a setof sentences, then A is a model of Σ if all sentences of Σ hold in A. We denotethis by

A |= Σ.

Let ϕ = ϕ(x1, . . . , xn) and let t1, . . . , tn be terms. The formula

ϕ(t1, . . . , tn)

is the formula we obtain by first renaming all bound variables by variables whichdo not occur in the ti and then replacing every free occurrence of xi by ti.

Lemma 2.13 (Substitution lemma).

A |= ϕ(t1, . . . , tn)[~b] ⇐⇒ A |= ϕ[tA1 [~b], . . . , tAn [

~b]]

Proof. The proof is an easy induction on ϕ.

Also note this (trivial) special case:

AA |= ϕ(a1, . . . , an) ⇐⇒ A |= ϕ[a1, . . . , an].

Frequently we suppress the subscript and simply write

A |= ϕ(a1, . . . , an).

Atomic formulas and their negation are called basic. Formulas without quan-tifiers (or: quantifier free formulas) are Boolean combinations of basic formulas,i.e. they can be built from basic formulas by successively applying ¬ and ∧. Theconjunction of formulas πi is denoted by

∧i<m πi and

∨i<m πi denotes their dis-

junction. By convention∧i<1 πi =

∨i<1 πi = π0. It is convenient to allow the

empty conjunction and the empty disjunction. For that we introduce two newformulas: the formula >, which is always true, and the formula ⊥, which isalways false. We define ∧

i<0

πi = >∨i<0

πi =⊥

A formula is in negation normal form if it is built from basic formulas using∧,∨,∃,∀.


Lemma 2.14. Every formula can be transformed into an equivalent formulawhich is in negation normal form.

Proof. Let ∼ denote equivalence of formulas. We consider formulas which arebuilt using ∧,∨,∃,∀ and ¬ and move the negation symbols in front of atomicformulas using

¬(ϕ ∧ ψ) ∼ (¬ϕ ∨ ¬ψ)¬(ϕ ∨ ψ) ∼ (¬ϕ ∧ ¬ψ)

¬∃xϕ ∼ ∀x¬ϕ¬∀xϕ ∼ ∃x¬ϕ¬¬ϕ ∼ ϕ

Definition 2.15. A formula in negation normal form which does not containany existential quantifier is called universal. Formulas in negation normal formwithout universal quantifiers are called existential.

Clearly an isomorphism h : A → B preserves the validity of every formula:

A |= ϕ[a1, . . . , an] ⇐⇒ B |= ϕ[h(a1), . . . , h(an)].

Embeddings preserve the validity of existential formulas:

Lemma 2.16. Let h : A → B be an embedding. Then for all existential formulasϕ(x1, . . . , xn) and all a1, . . . , an ∈ A we have

A |= ϕ[a1, . . . , an] =⇒ B |= ϕ[h(a1), . . . , h(an)].

For universal ϕ, the dual holds:

B |= ϕ[h(a1), . . . , h(an)] =⇒ A |= ϕ[a1, . . . , an].

Proof. By an easy induction on ϕ: for basic formulas the assertion follows fromthe definition of an embedding and Lemma 2.5. The inductive step is trivial forthe cases ∧ and ∨. Let finally ϕ(x) be ∃y ψ(x, y). If A |= ϕ[a], there exists ana ∈ A such that A |= ψ[a, a]. By induction we have B |= ψ[h(a), h(a)]. ThusB |= ϕ[h(a)].

Let A be an L–structure. The atomic diagram of A is

Diag(A) ={ϕ basic L(A)–sentence

∣∣ AA |= ϕ},

the set of all basic sentences with parameters from A which hold in A.

Lemma 2.17. The models of Diag(A) are precisely the structures (B, h(a))a∈Afor embeddings h : A → B.


Proof. The structures (B, h(a))a∈A are models of the atomic diagram by Lemma 2.16.For the converse note that a map h is an embedding if and only if it preservesthe validity of all formulas of the form

(¬)x1.= x2

c.= x1

f(x1, . . . , xn).= x0

(¬)R(x1, . . . , xn).

Many–sorted languages

In a many–sorted language with sorts in S, terms and formulas are builtwith respect to the sorts. For each sort s ∈ S we have variables vs0, vs1, . . . fromwhich we build the following terms of sort s.

Every variable vsi is an L–term of sort s.

Every constant c of sort s is an L–term of sort s.

If f is a function symbol of type (s1, . . . , sn, s) and ti is an L–term of sortsi for i = 1, . . . n, then ft1 . . . tn is an L–term of sort s.

L-formulas are constructed as before with the following adjustments:

t1.= t2 where t1, t2 are L–terms of the same sort,

Rt1 . . . , tn where R is a relation symbol from L of type (s1, . . . , sn)and ti is an L–term of sort si,

∃x ψ where ψ is an L–formula and x a variable (of some sorts).

It should be clear how to extend the definitions of this section to the many–sorted situation and that the results presented here continue to hold withoutchange. In the sequel we will not deal separately with many–sorted languagesuntil we meet them again in Section 35.

Exercise 2.1.Let L be a language and P be a new n–ary relation symbol. Let ϕ = ϕ(P )be an L(P ) = L ∪ {P}–sentence and π(x1, . . . , xn) an L–formula. Now replaceevery occurrence of P in ϕ by π. More precisely, every subformula of the formPt1 . . . tn is replaced by π(t1 . . . tn). We denote the resulting L–formula by ϕ(π).Show

A |= ϕ(π) if and only if(A, π(A)) |= ϕ(P ).


Exercise 2.2.Every quantifier free formula is equivalent to a formula of the form∧

i<m

∨j<mi

πij

and to a formula of the form ∨i<m

∧j<mi

πij

where the πij are basic formulas. The first form is called conjunctive normalform, the second disjunctive normal form.

Exercise 2.3.Every formula is equivalent to a formula in prenex normal form:

Q1x1 . . . Qnxnϕ

The Qi are quantifiers (∃ or ∀) and ϕ is quantifier free.

Exercise 2.4 (Ultraproducts and Łos’s Theorem).A filter on a set I is a non–empty set F ⊂ P(I) which does not contain theempty set and is closed under intersections and supersets, i.e. for A,B ∈ F , wehave A ∩ B ∈ F and if A ∈ F and A ⊆ C ⊆ I we have C ∈ F . A filter F iscalled ultrafilter if for every A ∈ P we have A ∈ F or I \ A ∈ F . (By Zorn’sLemma, any filter can be extended to an ultrafilter.)

For a family (Ai : i ∈ I) of L–structures and F an ultrafilter on I we definethe ultraproduct Πi∈IAi/F as follows. On the cartesian product Πi∈IAi, theultrafilter F defines an equivalence relation ∼F by

(ai)i∈I ∼F (bi)i∈I ⇔ {i ∈ I : ai = bi} ∈ F

On the set of equivalence classes (ai)F we define an L–structure Πi∈IAi/F .

• For constants c ∈ L, put cΠFAi = (cAi)F .

• For n-ary function symbols f ∈ L put

fΠFAi((a1i )F , . . . , (ani )F )) = (fAi(a1i , . . . , a

ni ))F .

• For n-ary relation symbols R ∈ L put

RΠFAi((a1i )F , . . . , (ani )F )) ⇔ {i ∈ I : RAi(a1i , . . . , a

ni )} ∈ F .

1. Show that the ultraproduct Πi∈IAi/F is well-defined.

2. Prove Łos’s Theorem: For any L–formula ϕ we have

Πi∈IAi/F |= ϕ((a1i )F , . . . , (ani )F ) ⇔ {i ∈ I : Ai |= ϕ(a1i , . . . , a

ni )} ∈ F .


3 Theories

Having defined a language, we can now take a closer look at which formulashold in a given structure. Conversely, we can start with a set of sentences andconsider those structures in which they hold. In this way, these sentences serveas a set of axioms for a theory.

Definition 3.1. An L–theory T is a set of L–sentences.

A theory which has a model is a consistent theory. More generally, we calla set Σ of L-formulas consistent if there is an L–structure A and an assignment~b such that A |= ϕ(~b) for all ϕ ∈ Σ. Σ is consistent with T if T ∪Σ is consistent.

Lemma 3.2. Let T be an L–theory and L′ be an extension of L. Then T isconsistent as an L–theory if and only if T is consistent as a L′–theory.

Proof. This follows from the (trivial) fact, that every L–structure is expandableto an L′–structure.

Examples:To keep algebraic expressions readable we will write 0 and 1 for the symbols 0and 1 in the following examples. We will omit the point for the multiplicationand brackets if they are implied by the order of operations rule.

TAG, the theory of abelian groups, has the axioms

• ∀x, y, z (x+ y) + z.= x+ (y + z)

• ∀x 0 + x.= x

• ∀x (−x) + x.= 0

• ∀x, y x+ y.= y + x

TR, the theory of commutative rings:

• TAG

• ∀x, y, z (xy)z .= x(yz)

• ∀x 1x.= x

• ∀x, y xy .= yx

• ∀x, y, z x(y + z).= xy + xz

TF, the theory of fields:

• TR

• ¬0 .= 1

• ∀x (¬x .= 0 → ∃y xy .

= 1)


Definition 3.3. If a sentence ϕ holds in all models of T , we say that ϕ followsfrom T (or that T proves ϕ) and write6

T ` ϕ.

By Lemma 3.2 this relation is independent of the language. Sentences whichfollow from the empty theory ∅ are called valid .

The most important properties of ` are:

Lemma 3.4.

1. If T ` ϕ and T ` (ϕ→ ψ), then T ` ψ.

2. If T ` ϕ(c1, . . . , cn) and the constants c1, . . . , cn occur neither in T nor inϕ(x1, . . . , xn), then T ` ∀x1 . . . xnϕ(x1, . . . , xn).

Proof. We prove 2: Let L′ = L \ {c1, . . . , cn}. If the L′–structure A is a modelof T and a1, . . . , an are arbitrary elements, then (A, a1, . . . , an) |= ϕ(c1, . . . , cn).That means A |= ∀x1 . . . xnϕ(x1, . . . , xn). Thus T ` ∀x1 . . . xnϕ(x1, . . . , xn).

We generalise this relation to theories S: we write T ` S if all models of Tare models of S. S and T are called equivalent, S ≡ T , if S and T have thesame models.

Definition 3.5. A consistent L–theory T is called complete if for all L–sentences ϕ

T ` ϕ or T ` ¬ϕ.

This notion clearly depends on L. If T is complete and L′ is an extension ofL, T will in general not be complete as a L′–theory.

Definition 3.6. For a complete theory T we define

|T | = max(|L|,ℵ0).

The typical (and, as we will see below, only) example of a complete theoryis the theory of a structure A

Th(A) = {ϕ | A |= ϕ}.

Lemma 3.7. A consistent theory T is complete if and only if Ded(T ) = {ϕ |T ` ϕ} is maximal consistent, or equivalently, if T ≡ S for every consistentextension S of T .

6Note that sometimes this relation is denoted T |= ϕ to distinguish this notion from themore syntactic notion of logical inference, see [50, section 2.6]


Proof. We call ϕ independent from T if neither ϕ nor ¬ϕ follows from T . So ϕis independent from T exactly if T ∪{ϕ} is a proper (in the sense of `) consistentextension of T . From this the lemma follows directly.

Definition 3.8. Two L–structures A and B are called elementarily equivalent,

A ≡ B,

if they have the same theory, that is, if for all L–sentences ϕ

A |= ϕ ⇐⇒ B |= ϕ.

Isomorphic structures are always elementarily equivalent. The converseholds only for finite structures, see Exercise 3.3 and Theorem 6.1.

Lemma 3.9. Let T be a consistent theory. Then the following are equivalent

a) T is complete.

b) All models of T are elementarily equivalent.

c) There exists a structure A with T ≡ Th(A).

Proof. a⇒ c: Let A be a model of T . If ϕ holds in A, then T 6` ¬ϕ and thusT ` ϕ. So Ded(T ) = Th(A) holds.

c ⇒b: If B |= T , B |= Th(A) and therefore B ≡ A. Note that ≡ is anequivalence relation.

b⇒ a: Let A be a model of T . If ϕ holds in A, then ϕ holds in all models ofT , i.e. T ` ϕ. Otherwise, ¬ϕ holds in A and we have T ` ¬ϕ.

For an L-theory T , we letMod(T ) denote the class of all models of T . A classof L–structures forms an elementary class if it is the class of models of someL-theory T . By Łos’s Theorem (Exercise 2.4), elementary classes are closedunder ultraproducts (see Exercise 4.2).

By the previous examples, the class of all abelian groups (commutative rings,fields, respectively) is an elementary class as is the subclass of elementary abelianp-groups for some prime p. However, the class of all finite abelian p-groups doesnot form an elementary class (see Exercise 3.4).

Exercise 3.1.Write down the axioms for the theory TDLO of dense linear orders withoutendpoint in the language LO of orders and the axioms for the theory TACF ofalgebraically closed fields in LR. Is TACF complete?

Exercise 3.2.


1. For a prime number p, let Zp∞ denote the p-Prüfer group, i.e. the groupof all pk-th roots of unity for all k ∈ N. Show that the groups Zmp∞ andZnp∞ are not elementarily equivalent for m 6= n.

2. Show that Zn 6≡ Zm in the language of groups if n 6= m.

Exercise 3.3.Show that if A is a finite L–structure and B is elementarily equivalent to A,then they are isomorphic. (Show this first for finite L.)

Exercise 3.4.Show that the class of all finite groups (all torsion groups, all nilpotent groups,respectively) does not form an elementary class.

Chapter 2

Elementary extensions andcompactness

4 Elementary substructures

As in other fields of mathematics, we need to compare structures and considermaps from one structure to another. For this to make sense we consider a fixedlanguage L. Maps and extensions are then required to respect this language:

Let A and B be two L–structures. A map h : A → B is called elementaryif it preserves the validity of arbitrary formulas ϕ(x1, . . . , xn)1. More precisely,for all a1, . . . , an ∈ A we have:

A |= ϕ[a1, . . . , an] ⇐⇒ B |= ϕ[h(a1), . . . , h(an)].

In particular, h preserves quantifier free formulas and is therefore an embedding.Hence h is also called elementary embedding. We write

h : A≺−→ B

The following lemma is clear.

Lemma 4.1. The models of Th(AA) are exactly the structures of the form(B, h(a))a∈A for elementary embeddings h : A

≺−→ B.

Th(AA) is called the elementary diagram of A.

A substructure A of B is called elementary if the inclusion map is elementary,i.e. if

A |= ϕ[a1, . . . , an] ⇐⇒ B |= ϕ[a1, . . . , an]

1This only means that formulas which hold in A also hold in B. But taking negations, theconverse follows.

25

CHAPTER 2. ELEMENTARY EXTENSIONS AND COMPACTNESS 26

for all a1, . . . , an ∈ A. In this case we write

A ≺ B

and B is called an elementary extension of A.

Theorem 4.2 (Tarski’s Test). Let B be an L–structure and A a subset of B. Ais the universe of an elementary substructure if and only if every L(A)–formulaϕ(x) which is satisfiable in B can be satisfied by an element of A.

Proof. If A ≺ B and B |= ∃xϕ(x), we also have A |= ∃xϕ(x) and there existsa ∈ A such that A |= ϕ(a). Thus B |= ϕ(a).

Conversely, suppose that the condition of Tarski’s test is satisfied. First weshow that A is the universe of a substructure A. The L(A)–formula x .

= x issatisfiable in B, so A is not empty. If f ∈ L is an n–ary function symbol (n ≥ 0)and a1, . . . , an from A, we consider the formula

ϕ(x) = f(a1, . . . , an).= x.

Since ϕ(x) is always satisfied by an element of A, A is closed under fB.Now we show, by induction on ψ, that

A |= ψ ⇐⇒ B |= ψ

for all L(A)–sentences ψ. This is clear for atomic sentences. The inductionsteps for ψ = ¬ϕ and ψ = (ϕ1 ∧ ϕ2) are trivial.It remains to consider the case ψ = ∃xϕ(x). If ψ holds in A, there exists a ∈ Asuch that A |= ϕ(a). The inductive hypothesis yields B |= ϕ(a), thus B |= ψ.For the converse suppose ψ holds in B. Then ϕ(x) is satisfiable in B and byTarski’s test we find a ∈ A such that B |= ϕ(a). By induction A |= ϕ(a) andA |= ψ holds.

We use Tarski’s Test to construct small elementary substructures:

Corollary 4.3. Suppose S is a subset of the L–structure B. Then B has anelementary substructure A containing S and of cardinality at most

max{|S|, |L|,ℵ0}.

Proof. We construct A as the union of an ascending sequence S0 ⊂ S1 ⊂ . . .of subsets of B. We start with S0 = S. If Si is already defined, we choosean element aϕ ∈ B for every L(Si)–formula ϕ(x) which is satisfiable in B anddefine Si+1 to be Si together with these aϕ. It is clear that A is the universe ofan elementary substructure. It remains to prove the bound on the cardinalityof A.

An L–formula is a finite sequence of symbols from L, auxiliary symbols andlogical symbols. These are |L|+ℵ0 = max(|L|,ℵ0) many symbols and therefore


there are exactly max(|L|,ℵ0) many L–formulas (see Appendix A.2.6).Let κ = max(|S|, |L|,ℵ0). There are κ many L(S)–formulas, thus |S1| ≤ κ.Inductively it follows for every i that |Si| ≤ κ. Finally we have |A| ≤ κ · ℵ0 =κ.

A directed family (Ai)i∈I of structures is elementary if Ai ≺ Aj for alli ≤ j. The following lemma is called Chain Lemma as it is mainly applied toelementary chains.

Theorem 4.4 (Tarski’s Chain Lemma). The union of an elementary directedfamily is an elementary extension of all its members.

Proof. Let A =⋃i∈I (Ai)i∈I . We prove by induction over ϕ(x) that for all i

and a ∈ AiAi |= ϕ(a) ⇐⇒ A |= ϕ(a).

If ϕ is atomic, nothing is to be proved. If ϕ is a negation or a conjunction, theclaim follows directly from the inductive hypothesis.If ϕ(x) = ∃yψ(x, y), ϕ(a) holds in A exactly if there exists b ∈ A with A |=ψ(a, b). As the family is directed, there always exists a j ≥ i with b ∈ Aj . Bythe inductive hypothesis we have A |= ψ(a, b) ⇐⇒ Aj |= ψ(a, b). Thus ϕ(a)holds in A exactly if it holds in an Aj (j ≥ i). Now the claim follows fromAi ≺ Aj .

Exercise 4.1.Let A be an L–structure and ES(A) = {B : B � A}. Show that if (Ai)i∈I is achain of elements in ES(A) and B their union, then B ∈ ES(A).

Exercise 4.2.Show that a class C of L–structure is an elementary class if and only if C isclosed under ultraproducts and elementary equivalence.


5 The Compactness Theorem

In this section we prove the Compactness Theorem, one of the fundamentalresults in first-order logic. It states that a first-order theory has a model ifevery finite part of it does. Its name is motivated by the results in Section 11which associate to each theory a certain compact topological space.

We call a theory T finitely satisfiable if every finite subset of T is consistent.

Theorem 5.1 (Compactness Theorem). Finitely satisfiable theories are consis-tent.

Let L be a language and C a set of new constants. An L(C)–theory T ′ iscalled a Henkin–theory if for every L(C)–formula ϕ(x) there is a constant c ∈ Csuch that

∃xϕ(x) → ϕ(c) ∈ T ′.

The elements of C are called Henkin–constants of T ′.

Let us call an L–theory T finitely complete if it is finitely satisfiable and ifevery L–sentence ϕ satisfies ϕ ∈ T or ¬ϕ ∈ T . This terminology is only prelim-inary (and not standard): by the compactness theorem a theory is equivalentto a finitely complete one if and only if it is complete.

The compactness theorem follows from the following two lemmas.

Lemma 5.2. Every finitely satisfiable L–theory T can be extended to a finitelycomplete Henkin–theory T ∗.

Note that conversely the lemma follows directly from the compactness theo-rem: choose a model A of T . Then Th(AA) is a finitely complete Henkin–theorywith A as a set of Henkin–constants.

Proof. We define an increasing sequence ∅ = C0 ⊂ C1 . . . of new constants byassigning to every L(Ci)–formula ϕ(x) a constant cϕ(x) and

Ci+1 ={cϕ(x)

∣∣ ϕ(x) L(Ci)–formula}.

Let C be the union of the Ci and TH the set of all Henkin–axioms

∃xϕ(x) → ϕ(cϕ(x))

for L(C)–formulas ϕ(x). It is easy to see that one can expand every L–structureto a model of TH . Hence T ∪ TH is a finitely satisfiable Henkin–theory. Usingthat the union of a chain of finitely satisfiable theories is again finitely satisfiable,we can apply Zorn’s Lemma and get a maximal finitely satisfiable L(C)–theoryT ∗ which contains T∪TH . As in Lemma 3.7 we show that T ∗ is finitely complete:if neither ϕ nor ¬ϕ belongs to T ∗, neither T ∗ ∪ {ϕ} nor T ∗ ∪ {¬ϕ} would be


finitely satisfiable. Hence there would be a finite subset ∆ of T ∗ which wouldbe consistent neither with ϕ nor with ¬ϕ. Then ∆ itself would be inconsistentand T ∗ would not be finitely satisfiable. This proves the lemma.

Lemma 5.3. Every finitely complete Henkin–theory T ∗ has a model A (uniqueup to isomorphism) consisting of constants; i.e.,

(A, ac)c∈C |= T ∗

with A = {ac | c ∈ C}.

Proof. Let us first note that since T ∗ is finitely complete, every sentence whichfollows from a finite subset of T ∗ belongs to T ∗. Otherwise the negation of thatsentence would belong to T ∗, but would also be inconsistent together with afinite part of T ∗.

We define for c, d ∈ C

c ' d ⇐⇒ c.= d ∈ T ∗

As c .= c is valid, and d

.= c follows from c

.= d, and c

.= e follows from c

.= d

and d .= e, ' is an equivalence relation. We denote the equivalence class of c by

ac and setA = {ac | c ∈ C}.

We expand A to an L–structure A by defining

RA(ac1 , . . . , acn) ⇐⇒ R(c1, . . . , cn) ∈ T ∗(1)fA(ac1 , . . . , acn) = ac0 ⇐⇒ f(c1, . . . , cn)

.= c0 ∈ T ∗(2)

for relation symbols R and function symbols f (n ≥ 0–ary) from L.

We have to show that this is well-defined: For (1) we have to show that

ac1 = ad1 , . . . , acn = adn , R(c1, . . . , cn) ∈ T ∗

implies R(d1, . . . , dn) ∈ T ∗. But clearly R(d1, . . . , dn) holds in any structuresatisfying

c1.= d1, . . . , cn

.= dn, R(c1, . . . , cn).

Similarly for (2) we first notice that ac0 = ad0 follows from

ac1 = ad1 , . . . , acn = adn , f(c1, . . . , cn).= c0 ∈ T ∗, f(d1, . . . , dn)

.= d0 ∈ T ∗.

For (2) we also have to show that for all c1, . . . , cn there exists c0 with f(c1, . . . , cn).=

c0 ∈ T ∗. As T ∗ is a Henkin–theory, there exists c0 with

∃xf(c1, . . . , cn).= x→ f(c1, . . . , cn)

.= c0 ∈ T ∗.


Now the valid sentence ∃xf(c1, . . . , cn).= x belongs to T ∗, so f(c1, . . . , cn)

.= c0

belongs to T ∗. This shows that everything is well defined.

Let A∗ be the L(C)–structure (A, ac)c∈C . We show by induction on thecomplexity of ϕ that for every L(C)-sentence ϕ

A∗ |= ϕ⇐⇒ ϕ ∈ T ∗.

There are four cases:

a) ϕ is atomic.

If ϕ has the form c.= d or R(c1, . . . , cn), the statement follows from the

construction of A∗. Other atomic sentences contain function symbols f orconstants (which we consider in this proof as 0–ary function symbols) fromL. We inductively reduce the number of such occurrences and apply theprevious case. Suppose ϕ contains a function symbol from L. Then ϕ canbe written as

ϕ = ψ(f(c1, . . . , cn))

for a function symbol f ∈ L and an L(C)–formula ψ(x). Choose c0 satisfyingf(c1, . . . , cn)

.= c0 ∈ T ∗. By construction f(c1, . . . , cn)

.= c0 holds in A∗.

Thus A∗ |= ϕ ⇐⇒ A∗ |= ψ(c0) and ϕ ∈ T ∗ ⇐⇒ ψ(c0) ∈ T ∗. By inductionhypothesis on the number of occurrences we have A∗ |= ψ(c0) ⇐⇒ ψ(c0) ∈T ∗. This suffices.

b) ϕ = ¬ψ.

As T ∗ is finitely complete, ϕ ∈ T ∗ ⇐⇒ ψ 6∈ T ∗ holds and by the inductionhypothesis we have

A∗ |= ϕ⇐⇒ A∗ 6|= ψ ⇐⇒ ψ 6∈ T ∗ ⇐⇒ ϕ ∈ T ∗.

c) ϕ = (ψ1 ∧ ψ2).

As T ∗ contains all sentences which follow from a finite subset of T ∗, ϕ belongsto T ∗ if and only if ψ1 and ψ2 belong to T ∗. Thus

A∗ |= ϕ⇐⇒ A∗ |= ψi (i = 1, 2) ⇐⇒ ψi ∈ T ∗ (i = 1, 2) ⇐⇒ ϕ ∈ T ∗.

d) ϕ = ∃xψ(x)We have

A∗ |= ϕ⇔ A∗ |= ψ(c) for some c ∈ C ⇔ ψ(c) ∈ T ∗ for some c ∈ C ⇔ ϕ ∈ T ∗.

The second equivalence is the induction hypothesis and for the third we argueas follows: if ϕ ∈ T ∗, we choose c satisfying ϕ → ψ(c) ∈ T ∗. As ϕ ∈ T ∗ wealso have ψ(c) ∈ T ∗.


Corollary 5.4. T ` ϕ if and only if ∆ ` ϕ for a finite subset ∆ of T .

Proof. The formula ϕ follows from T if and only if T ∪{¬ϕ} is inconsistent.

Corollary 5.5. A set of formulas Σ(x1, . . . , xn) is consistent with T if and onlyif every finite subset of Σ is consistent with T .

Proof. Introduce new constants c1, . . . , cn. Then Σ is consistent with T if andonly if T ∪ Σ(c1, . . . , cn) is consistent. Now apply Theorem 5.1.

Definition 5.6. Let A be an L–structure and B ⊆ A. Then a ∈ A realises aset of L(B)–formulas Σ(x) (containing at most the free variable x), if a satisfiesall formulas from Σ(x). We write

A |= Σ(a).

Σ(x) is called finitely satisfiable in A if every finite subset of Σ is realised in A.

Lemma 5.7. Σ(x) is finitely satisfiable in A if and only if there is an elementaryextension of A in which Σ(x) is realised.

Proof. By Lemma 4.1, Σ is realised in an elementary extension of A if and onlyif Σ is consistent with Th(AA). So the lemma follows from the easy observationthat a finite set of L(A)–formulas is consistent with Th(AA) if and only if it isrealised in A.

Definition 5.8. Let A be an L–structure and B a subset of A. A set p(x) ofL(B)–formulas is a type over B if p(x) is maximal finitely satisfiable in A. Wecall B the domain of p. Let

S(B) = SA(B)

denote the set of types over B.

Every element a of A determines a type

tp(a/B) = tpA(a/B) = {ϕ(x) | A |= ϕ(a), ϕ an L(B)–formula}.

So an element a realises the type p ∈ S(B) exactly if p = tp(a/B). Note that ifA′ is an elementary extension of A, then

SA(B) = SA′(B) and tpA

′(a/B) = tpA(a/B).

We will use the notation tp(a) for tp(a/∅).

Similarly, maximal finitely satisfiable sets of formulas in x1, . . . , xn are calledn–types and

Sn(B) = SAn (B)


denotes the set of n–types over B. For an n–tuple a from A, there is an obviousdefinition of tpA(a/B) ∈ SAn (B). Very much in the same way, we can define thetype tp(C/B) of an arbitrary set C over B. This will be convenient in laterchapters. In order to do this properly we allow free variables xc indexed byc ∈ C and define

tp(C/B) = {ϕ(xc1 , . . . , xcn) | A |= ϕ(c1, . . . , cn), ϕ an L(B)–formula}.

Corollary 5.9. Every structure A has an elementary extension B in which alltypes over A are realised.

Proof. We choose for every p ∈ S(A) a new constant cp. We have to find amodel of

Th(AA) ∪⋃

p∈S(A)

p(cp).

It is easy to see that this theory is finitely satisfiable using that every p is finitelysatisfiable in A. The compactness theorem now shows that the model exists.

We give a second proof which only uses Lemma 5.7: Let (pα)α<λ be anenumeration of S(A), where λ is an ordinal number (see Appendix A.3). UsingTheorem A.3.1, we construct an elementary chain

A = A0 ≺ A1 ≺ . . .Aβ . . . (β ≤ λ)

such that each pα is realised in Aα+1.

Let us suppose that the elementary chain (Aα′)α′<β is already constructed. Ifβ is a limit ordinal, we let Aβ =

⋃α<β Aα.2 The longer chain (Aα′)α′≤β is

elementary because of Lemma 4.4. If β = α + 1 we first note that pα is alsofinitely satisfiable in Aα. Therefore we can realise pα in a suitable elementaryextension Aβ � Aα. Then B = Aλ is the model we were looking for.

Exercise 5.1.Prove the Compactness Theorem using ultraproducts (see Exercise 2.4).

Exercise 5.2.A class C of L–structure is finitely axiomatisable if it is the class of models of afinite theory. Show that C is finitely axiomatisable if and only if both C and itscomplement form an elementary class.

Exercise 5.3.Show that the class of connected graphs is not an elementary class. A graph(V,R) is a set V with a symmetric, irreflexive binary relation. It is connectedif for any x, y ∈ V there is a sequence of elements x0 = x, . . . xk = y such that(xi−1, xi) ∈ R for i = 1, . . . n.

2 We call a chain (Aα) indexed by ordinal numbers continuous if Aβ =⋃

α<β Aα for alllimit ordinals β.


Exercise 5.4.Let A = (R, 0, <, fA) where f is a unary function symbol. Call an elementx ∈ A∗ � A infinitesimal if −r < x < r for all r ∈ A>0. Show: If fA(0) = 0,then fA is continuous in 0 if and only if for any elementary extension A∗ of Athe map fA

∗takes infinitesimal elements to infinitesimal elements.

Exercise 5.5.Let T be an LR-theory containing the theory of fields TF. Show:

1. If T has models of arbitrary large characteristic, then it has a model ofcharacteristic 0.

2. The theory of fields of characteristic 0 is not finitely axiomatisable.


6 The Löwenheim–Skolem Theorem

One of the consequences of the Compactness Theorem is the fact that a first-order theory cannot pin down the size of an infinite structure. This is thecontent of the following theorem:

Theorem 6.1 (Löwenheim–Skolem). Let B be an L–structure, S a subset ofB and κ an infinite cardinal.

1. Ifmax{|S|, |L|} ≤ κ ≤ |B|,

then B has an elementary substructure of cardinality κ containing S.

2. If B is infinite andmax{|B|, |L|} ≤ κ,

then B has an elementary extension of cardinality κ.

Proof. 1: Choose a set S ⊂ S′ ⊂ B of cardinality κ and apply Corollary 4.3.

2: We first construct an elementary extension B′ of cardinality at least κ.Choose a set C of new constants of cardinality κ. As B is infinite, the theory

Th(BB) ∪ {¬ c .= d | c, d ∈ C, c 6= d}

is finitely satisfiable (even in B: just interpret the finitely many new constantsin a finite subset by elements of B). By Lemma 4.1, any model (B′

B , bc)c∈C isan elementary extension of B with κ many different elements (bc).Finally we apply the first part of the theorem to B′ and S = B.

Corollary 6.2. A theory which has an infinite model has a model in everycardinality κ ≥ max(|L|,ℵ0).

Thus, no theory with an infinite model can describe this model up to isomor-phism. So the best we can hope for is a unique model for a given cardinality:

Definition 6.3 (preliminary, cf. 6.5). Let κ be an infinite cardinal. A theoryT is called κ–categorical if all models of T of cardinality κ are isomorphic.

Theorem 6.4 (Vaught’s Test). A κ–categorical theory T is complete if in ad-dition the following conditions are satisfied:

a) T consistent,

b) T has no finite model,

c) |L| ≤ κ.


Proof. We have to show that all models A and B of T are elementarily equiv-alent. As A and B are infinite, Th(A) and Th(B) have models A′ and B′ ofcardinality κ. By assumption A′ and B′ are isomorphic, and follows that

A ≡ A′ ≡ B′ ≡ B.

Examples:

1. (Theorem of Cantor, see Exercise 3.1) The theory TDLO of dense linearorders without endpoints is ℵ0–categorical and by Vaught’s test complete.To see this let A and B be countable dense linear orders without end-points, and let A = {ai : i ∈ ω}, B = {bi : i ∈ ω}. We inductivelydefine sequences (ci)i<ω, (di)i<ω exhausting A and B, respectively, andsuch that the assignment ci 7→ di is the required isomorphism. Assumethat (ci)i<m, (di)i<m have been defined so that ci 7→ di, i < m is an orderisomorphism. If m = 2k, let cm = aj where aj is the element with minimalindex in {ai : i ∈ ω} not occurring in (ci)i<m. Since B is a dense linearorder without endpoints, there is some element dm ∈ {bi : i ∈ ω} such that(ci)i≤m and (di)i≤m are order isomorphic. If m = 2k + 1 we interchangethe roles of A and B.

2. For any prime p or p = 0, the theory TACFp of algebraically closed fieldsof characteristic p is κ-categorical for any κ > ℵ0. Any two algebraicallyclosed field of the same characteristic and of cardinality κ > ℵ0 havetranscendence bases (over the algebraic closure of the prime field) of car-dinality κ, see Appendix 2.6 and Appendix 8. Any bijection between thesetranscendence bases induces an isomorphism of the fields. It follows thatTACFp is complete.

Considering Theorem 6.4 we strengthen our definition:

Definition 6.5. Let κ be an infinite cardinal. A theory T is called κ–categoricalif it is complete, |T | ≤ κ and, up to isomorphy, has exactly one model ofcardinality κ.

Exercise 6.1.1. The theory of K–vector spaces TM(K) (see p. 49) is κ-categorical forκ > |K|.

2. Is TACFp ℵ0–categorical?

Exercise 6.2.Show that an ∀∃–sentence, which holds in all finite fields, is true in all alge-braically closed fields.

Chapter 3

Quantifier elimination

7 Preservation theorems

In general, it can be difficult to tell which sentences belong to a given theory orwhich extensions are consistent. Therefore it is helpful to know that in certaintheories one can restrict attention to sentences of a certain class, e.g., quantifier-free or, say, existential formulas. We consider a fixed language L and first provesome separation results:

Lemma 7.1 (Separation Lemma). Let T1 and T2 be two theories. Assume His a set of sentences which is closed under ∧ and ∨ and contains > and ⊥ (trueand false). Then the following are equivalent:

a) There is a sentence ϕ ∈ H which separates T1 from T2. This means

T1 ` ϕ and T2 ` ¬ϕ.

b) All models A1 of T1 can be separated from all models A2 of T2 by asentence ϕ ∈ H. This means

A1 |= ϕ and A2 |= ¬ϕ.

Proof. a) ⇒ b):If ϕ separates T1 from T2, it separates all models of T1 from all models of T2.

b) ⇒ a):For any model A1 of T1 let HA1 be the set of all sentences from H which aretrue in A1. b) implies that HA1 and T2 cannot have a common model. By theCompactness Theorem there is a finite conjunction ϕA1 of sentences from HA1

inconsistent with T2. Clearly,

T1 ∪ {¬ϕA1 | A1 |= T1}

36

CHAPTER 3. QUANTIFIER ELIMINATION 37

is inconsistent since any model A1 of T1 satisfies ϕA1 . Again by compactness T1implies a disjunction ϕ of finitely many of the ϕA1 . This ϕ is in H and separatesT1 from T2.

For structures A,B and a map f : A→ B preserving all formulas from a setof formulas ∆, we use the notation

f : A →∆ B.

We also writeA ⇒∆ B

to express that all sentences from ∆ true in A are also true in B.

Lemma 7.2. Let T be a theory, A a structure and ∆ a set of formulas, closedunder existential quantification, conjunction and substitution of variables. Thenthe following are equivalent:

a) All sentences ϕ ∈ ∆ which are true in A are consistent with T .

b) There is a model B |= T and a map f : A →∆ B.

Proof. b) ⇒ a):Assume f : A →∆ B |= T . If ϕ ∈ ∆ is true in A it is also true in B and thereforeis consistent with T .

a) ⇒ b):Consider Th∆(AA), the set set of all sentences δ(a), (δ(x) ∈ ∆), which are true inAA. The models (B, f(a)a∈A) of this theory correspond to maps f : A →∆ B.This means that we have to find a model of T ∪ Th∆(AA). To show finitesatisfiability it is enough to show that T ∪D is consistent for every finite subsetD of Th∆(AA). Let δ(a) be the conjunction of the elements of D. Then A is amodel of ϕ = ∃x δ(x), so by assumption T has a model B which is also a modelof ϕ. This means that there is a tuple b such that (B, b) |= δ(a).

Note that Lemma 7.2 applied to T = Th(B) shows that A ⇒∆ B if and only ifthere exist a map f and a structure B′ ≡ B such that f : A →∆ B′.

Theorem 7.3. Let T1 and T2 be two theories. Then the following are equivalent:

a) There is a universal sentence which separates T1 from T2.

b) No model of T2 is a substructure of a model of T1.

Proof. a) ⇒ b):Let ϕ be a universal sentence which separates T1 from T2. Let A1 be a modelof T1 and A2 a substructure of A1. Since A1 is a model of ϕ, A2 is also modelof ϕ by Lemma 2.16. Therefore A2 cannot be a model of T2.


b) ⇒ a):We use the Separation Lemma. Let A1 and A2 be models of T1 and T2, whichcannot be separated by a universal sentence. This can be denoted by

A2 ⇒∃ A1.

Now Lemma 7.2 implies that A2 has an extension A′1 ≡ A1. Then A′

1 is again amodel of T1 contradicting b).

Definition 7.4. For any L-theory T , the formulas ϕ(x), ψ(x) are said to beequivalent modulo T (or relative to T ) if T ` ∀x(ϕ(x) ↔ ψ(x)).

Corollary 7.5. Let T be a theory.

1. Consider a formula ϕ(x1, . . . , xn). The following are equivalent:

a) ϕ(x1, . . . , xn) is, modulo T , equivalent to a universal formula.

b) If A ⊂ B are models of T and a1, . . . , an ∈ A, then B |= ϕ(a1, . . . , an)implies A |= ϕ(a1, . . . , an).

2. We call a theory which consists of universal sentences universal. T isequivalent to a universal theory if and only if all substructures of modelsof T are again models of T .

Proof. 1): Assume b). We extend L by an n–tuple c of new constants c1, . . . , cnand consider the theory

T1 = T ∪ {ϕ(c)} and T2 = T ∪ {¬ϕ(c)}.

b) says that substructures of models of T1 cannot be models of T2. By Theo-rem 7.3 T1 and T2 can be separated by a universal L(c)–sentence ψ(c). T1 ` ψ(c)implies by Lemma 3.4.2

T ` ∀x (ϕ(x) → ψ(x))

and from T2 ` ¬ψ(c) we see

T ` ∀x (¬ϕ(x) → ¬ψ(x)).

2): It is clear that substructures of models of a universal theory are modelsagain. Now suppose that a theory T has this property. Let ϕ be an axiom ofT . If A is a substructure of B, it is not possible that B is a model of T and Ais a model of ¬ϕ at the same time. By 7.3 there is a universal sentence ψ withT ` ψ and ¬ϕ ` ¬ψ. Hence all axioms of T follow from

T∀ = {ψ | T ` ψ, ψ universal}.


An ∀∃–formula is of the form

∀x1 . . . xnψ

where ψ is existential (see p. 18). The following is clear:

Lemma 7.6. Suppose ϕ is an ∀∃–sentence, (Ai)i∈I is a directed family of modelsof ϕ and B the union of the Ai. Then B is also a model of ϕ.

Proof. Writeϕ = ∀x ψ(x),

where ψ is existential. For any a ∈ B there is an Ai containing a. Since Ai |= ϕ,clearly ψ(a) holds in Ai. As ψ(a) is existential it must also hold in B.

Definition 7.7. We call a theory T inductive if the union of any directed familyof models of T is again a model.

Theorem 7.8. Let T1 and T2 be two theories. Then the following are equivalent:

a) There is an ∀∃–sentence which separates T1 from T2.

b) No model of T2 is the union of a chain (or of a directed family) of modelsof T1.

Proof. a) ⇒ b):Assume ϕ is a ∀∃–sentence which separates T1 from T2, (Ai)i∈I is a directedfamily of models of T1 and B the union of the Ai. Since the Ai are models ofϕ, B is also a model of ϕ by Lemma 7.6. Since B |= ϕ, B cannot be a modelof T2.

¬a) ⇒ ¬b):If a) is not true, the Separation Lemma gives us models A and B0 of T1 and T2which cannot be separated by an ∀∃–sentence. Since ∃∀–formulas are equivalentto negated ∀∃–formulas, we have

B0 ⇒∃∀ A.

By Lemma 7.2 there is a map

f : B0 →∀ A0

with A0 ≡ A. We can assume that B0 ⊂ A0 and f is the inclusion map. Then

A0B ⇒∃ B0

B .

Applying Lemma 7.2 again, we obtain an extension B1B of A0

B with B1B ≡ B0

B ,i.e. B0 ≺ B1.


B0 ⊂ A0 ⊂ B1

≺

The same procedure applied to A and B1 gives us two extensions A1 ⊂ B2 withA1 ≡ A and B1 ≺ B2. This results in an infinite chain

B0 ⊂ A0 ⊂ B1

≺

⊂ A1 ⊂ B2

≺

⊂ . . .

≺

with Ai ≡ A and Bi ≺ Bi+1. Let B be the union of the Ai. Since B is also theunion of the elementary chain of the Bi, B is an elementary extension of B0

and hence a model of T2. But the Ai are models of T1, so b) does not hold.

Corollary 7.9. Let T be a theory.

1. For each sentence ϕ the following are equivalent:

a) ϕ is, modulo T , equivalent to an ∀∃–sentence.

b) IfA0 ⊂ A1 ⊂ . . .

and their union B are models of T , then ϕ holds in B if it is true inall the Ai.

2. T is inductive if and only if T can be axiomatised by ∀∃–sentences.

Proof. 1): Theorem 7.8 shows that ∀∃–formulas are preserved by unions ofchains. Hence a) ⇒ b). For the converse consider the theories

T1 = T ∪ {ϕ} and T2 = T ∪ {¬ϕ}.

b) says that the union of a chain of models of T1 cannot be a model of T2. ByTheorem 7.8 we can separate T1 and T2 by an ∀∃–sentence ψ. T1 ` ψ impliesT ` ϕ→ ψ and T2 ` ¬ψ implies T ` ¬ϕ→ ¬ψ.

2): Clearly ∀∃–axiomatised theories are inductive. For the converse assume thatT is inductive and ϕ an axiom of T . If B is a union of models of T , it cannotbe a model of ¬ϕ. By Theorem 7.8 there is an ∀∃–sentence ψ with T ` ψ and¬ϕ ` ¬ψ. Hence all axioms of T follow from

T∀∃ = {ψ | T ` ψ, ψ ∀∃–formula}.


Exercise 7.1.Let X be a topological space, Y1 and Y2 quasi-compact1 subsets, and H a setof clopen subsets. Then the following are equivalent:

a) There is a positive Boolean combination B of elements from H such thatY1 ⊂ B and Y2 ∩B = ∅.

b) For all y1 ∈ Y1 and y2 ∈ Y2 there is an H ∈ H such that y1 ∈ H and y2 6∈ H.

(This, in fact, is a generalisation of the Separation Lemma 7.1.)

1i.e. compact but not necessarily Hausdorff


8 Quantifier elimination

Having quantifier elimination in a reasonable language is a property which makesa theory ‘tame’. In this section we will collect some criteria and extensions of thisconcept. They will be applied in Section 9 to show that a number of interestingtheories do have quantifier elimination in the appropriate language.

Definition 8.1. A theory T has quantifier elimination if every L–formulaϕ(x1, . . . , xn) is equivalent modulo T to some quantifier free formulaρ(x1, . . . , xn).

For n = 0 this means that modulo T every sentence is equivalent to aquantifier free sentence. If L has no constants, > and ⊥ are the only quantifierfree sentences. T is then either inconsistent or complete.

Note that it is easy to transform any theory T into a theory with quantifierelimination if one is willing to expand the language: just enlarge L by adding ann–place relation symbol Rϕ for every L–formula ϕ(x1, . . . , xn) and T by addingall axioms

∀x1, . . . , xn(Rϕ(x1, . . . , xn) ↔ ϕ(x1, . . . , xn)).

The resulting theory, the Morleyisation Tm of T , has quantifier elimination.Many other properties of a theory are not affected by Morleyisation. So T iscomplete if and only if Tm is, similarly for κ–categoricity and other propertieswe will define in later chapters.

A prime structure of T is a structure which embeds into all models of T .The following is clear:

Lemma 8.2. A consistent theory T with quantifier elimination which possessesa prime structure is complete.

Definition 8.3. A simple existential formula has the form

ϕ = ∃y ρ

for a quantifier free formula ρ. If ρ is a conjunction of basic formulas, ϕ is calledprimitive existential.

Lemma 8.4. T has quantifier elimination if and only if every primitive exis-tential formula is, modulo T , equivalent to a quantifier free formula.

Proof. We can write every simple existential formula in the form ∃y∨i<n ρi

for ρi which are conjunctions of basic formulas. This shows that every simpleexistential formula is equivalent to a disjunction of primitive existential formulas,namely to

∨i<n(∃y ρi). We can therefore assume that every simple existential

formula is, modulo T , equivalent to a quantifier free formula.


We are now able to eliminate the quantifiers in arbitrary formulas in prenexnormal form (see Exercise 2.3)

Q1x1 . . . Qnxnρ.

If Qn = ∃, we choose a quantifier free formula ρ0 which, modulo T , is equivalentto ∃xnρ. Then we proceed with the formula Q1x1 . . . Qn−1xn−1ρ0. If Qn = ∀,we find a quantifier free ρ1 which is, modulo T , equivalent to ∃xn¬ρ and proceedwith Q1x1 . . . Qn−1xn−1¬ρ1.

The following theorem gives useful criteria for quantifier elimination.

Theorem 8.5. For a theory T the following are equivalent:

a) T has quantifier elimination.

b) For all models M1 and M2 of T with a common substructure A we have

M1A ≡ M2

A.

c) For all models M1 and M2 of T with a common substructure A and forall primitive existential formulas ϕ(x1, . . . , xn) and parameters a1, . . . , anfrom A

M1 |= ϕ(a1, . . . , an) ⇒ M2 |= ϕ(a1, . . . , an).

If L has no constants, A is allowed to be the empty “structure”.

Proof. a) ⇒ b):Let ϕ(a) be an L(A)–sentence which holds in M1. Choose a quantifier free ρ(x)which is, modulo T , equivalent to ϕ(x). Then

M1 |= ϕ(a) ⇒ M1 |= ρ(a)⇒ A |= ρ(a) ⇒

M2 |= ρ(a) ⇒ M2 |= ϕ(a).

b) ⇒ c):Clear.

c) ⇒ a):Let ϕ(x) be a primitive existential formula. In order to show that ϕ(x) isequivalent, modulo T , to a quantifier free formula ρ(x) we extend L by a n–tuplec of new constants c1, . . . , cn. We have to show that we can separate T ∪{ϕ(c)}and T ∪ {¬ϕ(c)} by a quantifier free sentence ρ(c). We apply the SeparationLemma. Let M1 and M2 be two models of T with two distinguished n–tuplesa1 and a2. Suppose that (M1, a1) and (M2, a2) satisfy the same quantifier freeL(c)–sentences. We have to show that

(1) M1 |= ϕ(a1) ⇒ M2 |= ϕ(a2).


Consider the substructures Ai = 〈ai〉Mi

, generated by ai. If we can show thatthere is an isomorphism

f : A1 → A2

taking a1 to a2, we may assume that A1 = A2 = A and a1 = a2 = a. Then (1)follows directly from c).

Every element of A1 has the form tM1

[a1] for an L–term t(x), (2.6). The iso-morphism f to be constructed must satisfy:

f(tM

1

[a1])= tM

2

[a2].

We now define f by this equation, and we have to check that f is well–definedand injective. Assume

sM1

[a1] = tM1

[a1].

Then s(c).= t(c) holds in (M1, a1), and by our assumption also in (M2, a2),

which meanssM

2

[a2] = tM2

[a2].

This shows that f is well–defined. Swapping the two sides yields injectivity.

That f is surjective is clear. It remains to show that f commutes with theinterpretation of the relation symbols.

M1 |= R[tM

1

1 [a1], . . . , tM1

m [a1]],

is equivalent to (M1, a1) |= R(t1(c), . . . , tm(c)), which is equivalent to (M2, a2) |=R(t1(c), . . . , tm(c)), which in turn is equivalent to

M2 |= R[tM

2

1 [a2], . . . , tM2

m [a2]].

Note that part b) of Theorem 8.5 is saying that T is substructure complete ;i.e., for any model M |= T and substructure A ⊂ M the theory T ∪ Diag(A) iscomplete.

Definition 8.6. T is model complete if for all models M1 and M2 of T

M1 ⊂ M2 ⇒ M1 ≺ M2.

Note that T is model complete if and only if for any M |= T the theoryT ∪Diag(M) is complete.

Clearly, by 8.5.b) applied to A = M1 all theories with quantifier eliminationare model complete.

Lemma 8.7 (Robinson’s Test). Let T be a theory. Then the following areequivalent:


a) T is model complete.

b) For all models M1 ⊂ M2 of T and all existential sentences ϕ from L(M1)

M2 |= ϕ ⇒ M1 |= ϕ.

c) Each formula is, modulo T , equivalent to a universal formula.

Proof. a)⇒b) is trivial.a)⇔c) follows from 7.5.1.b) implies that every existential formula is, modulo T , equivalent to a universalformula. As in the proof of 8.4 this implies c).

If M1 ⊂ M2 satisfies b), we call M1 existentially closed in M2. We denotethis by

M1 ≺1 M2.

Definition 8.8. Let T be a theory. A theory T ∗ is a model companion of T ifthe following three conditions are satisfied.

a) Each model of T can be extended to a model of T ∗.

b) Each model of T ∗ can be extended to a model of T .

c) T ∗ is model complete.

Theorem 8.9. A theory T has, up to equivalence, at most one model companionT ∗.

Proof. If T+ is another model companion of T , every model of T+ is containedin a model of T ∗ and conversely. Let A0 be a model of T+. A0 can be embeddedin a model B0 of T ∗. B0 again is contained in a model A1 of T+. In this waywe find two elementary chains, (Ai) and (Bi), which have a common union C.A0 ≺ C and B0 ≺ C implies A0 ≡ B0. Thus A0 is a model of T ∗. InterchangingT ∗ and T+ yields that every model of T ∗ is a model of T+.

Digression: Existentially closed structures and the Kaiser hull

Let T be an L–theory. It follows from 7.2 that the models of T ∀ are the substruc-tures of models of T . The conditions a) and b) in the definition of “model companion”can therefore be expressed as

T ∀ = T ∗∀.

Hence the model companion of a theory T depends only on T ∀.

Definition 8.10. An L–structure A is called T–existentially closed (or T -e.c), if


a) A can be embedded in a model of T .

b) A is existentially closed in every extension which is a model of T .

A structure A is T–e.c exactly if it is T ∀–e.c. This is clear for condition a) since everymodel B of T ∀ can be embedded in a model M of T . For b) this follows from the factthat A ⊂ B ⊂ M and A ≺1 M implies A ≺1 B.

Lemma 8.11. Every model of a theory T can be embedded in a T–e.c structure.

Proof. Let A be a model of T ∀. We choose an enumeration (ϕα)α<κ of all existentialL(A)–sentences and construct an ascending chain (Aα)α≤κ of models of T ∀. We beginwith A0 = A. Let Aα be constructed. If ϕα holds in an extension of Aα which is amodel of T , we let Aα+1 be such a model. Otherwise we set Aα+1 = Aα. For limitordinals λ we define Aλ to be the union of all Aα, (α < λ). Note that Aλ is again amodel of T ∀.The structure A1 = Aκ has the following property: Every existential L(A)–sentencewhich holds in an extension of A1 which is a model of T holds in A1. Now, in the samemanner, we construct A2 from A1, etc. The union M of the chain A0 ⊂ A1 ⊂ A2 . . . isthe desired T–e.c structure.

The structure M constructed in the proof can be very big. On the other hand, itis easily seen that every elementary substructure N of a T–e.c structure M is againT–e.c: let N ⊂ A be a model of T . Since MN ⇒∃ AN , there is an embedding of M inan elementary extension B of A which is the identity on N . Since M is existentiallyclosed in B, N is existentially closed in B and therefore also in A.

N

A M

B

��7

SS

SSo

��7

SS

SSo

≺1 ≺

≺ ≺1

Lemma 8.12 ([28]). Let T be a theory. Then there is a biggest inductive theory TKH

with T ∀ = TKH∀ . TKH is called the Kaiser hull of T .

Proof. Let T 1 and T 2 be two inductive theories with T 1∀ = T 2

∀ = T ∀. We have to showthat (T 1 ∪ T 2) ∀ = T∀. Let M be a model of T . As in the proof of 8.9 we extend Mby a chain A0 ⊂ B0 ⊂ A1 ⊂ B1 . . . of models of T 1 and T 2. The union of this chainis a model of T 1 ∪ T 2.

Lemma 8.13. TKH is the ∀∃–part of the theory of all T -e.c structures.

Proof. Let T ∗ be the ∀∃–part of the theory of all T -e.c structures. Since T–e.c struc-tures are models of T ∀ , we have T ∀ ⊂ T ∗

∀. It follows from 8.11 that T ∗∀ ⊂ T ∀. Hence

T ∗ is contained in the Kaiser hull.


It remains to show that every T–e.c structure M is a model of the Kaiser hull. Choosea model N of TKH which contains M. Then M ≺1 N. This implies N ⇒∀∃ M andtherefore M |= TKH.

The previous lemma implies immediately that T–e.c structures are models of T∀∃.

Theorem 8.14. For any theory T the following are equivalent:

a) T has a model companion T ∗

b) All models of TKH are T–e.c.

c) The T–e.c structures form an elementary class.

If T ∗ exists, we have

T ∗ = TKH = theory of all T–e.c structures.

Proof. a) ⇒ b):Let T ∗ be the model companion of T . As a model complete theory, T ∗ is inductive.So T ∗ is contained in the Kaiser hull and it suffices to show that every model M of T ∗

is T–e.c. Let A be a model of T which extends M. A can be embedded in a model Nof T ∗. Now M ≺ N implies M ≺1 A.

b) ⇒ c):By the last lemma all T–e.c structures are models of TKH. Thus b) implies that T–e.cstructures are exactly the models of TKH.

c) ⇒ a):Assume that the T -e.c structures are exactly the models of the theory T+. By 8.11 wehave T ∀ = T+

∀ . Criterion 8.7 implies, that T+ is model complete. So T+ is the modelcompanion of T .

The last assertion of the theorem follows easily from the proof.

Exercise 8.1.A theory T with quantifier elimination is axiomatisable by sentences of the form

∀x1 . . . xnψ

where ψ is primitive existential formula.

Exercise 8.2.Let L be the language containing an unary function f and a binary relation symbol Rand consider the L–theory T = {∀x∀y(R(x, y) → (R(x, f(y))}. Show the following

1. For any T–e.c structure M and a, b ∈ M with b /∈ {a, fM(a), (fM)2(a), . . .} wehave M |= ∃z(R(z, a) ∧ ¬R(z, b)).

2. Let M be a model of T and a an element ofM such that {a, fM(a), (fM)2(a), . . .}is infinite. Then in an elementary extension M′ there is an element b withM′ |= ∀z(R(z, a) → R(z, b)).

3. The class of T–e.c structures is not elementary, so T does not have a modelcompanion.


9 Examples

In this section we present a number of theories with quantifier elimination,or at least elimination down to some well-understood formulas, among themthe theories of vector spaces and of algebraically, differentially, and real closedfields. Since such theories are comparatively easy to understand, they form acore inventory of the working model theorist.

9.1 Infinite sets

The models of the theory T∞ of infinite sets are all infinite sets without addi-tional structure. The language L∅ is empty, the axioms are (for n = 1, 2, . . .)

• ∃x0 . . . xn−1

∧i<j<n ¬xi

.= xj

Theorem 9.1. The theory T∞ of infinite sets has quantifier elimination andis complete.

Proof. Clear.

9.2 Dense linear orderings

Theorem 9.2. TDLO has quantifier elimination.

Proof. Let A be a finite common substructure of the two models O1 and O2.We choose an ascending enumeration A = {a1, . . . , an}. Let ∃y ρ(y) be a simpleexistential L(A)–sentence, which is true in O1 and assume O1 |= ρ(b1). Wewant to extend the order preserving map ai 7→ ai to an order preserving mapA∪{b1} → O2. For this we have to find an image b2 of b1. There are four cases

i) b1 ∈ A. We set b2 = b1.

ii) b1 lies between ai and ai+1. We choose b2 in O2 with the same property.

iii) b1 is smaller than all elements of A. We choose a b2 ∈ O2 of the same kind.

iv) b1 is bigger that all ai. Choose b2 in the same manner.

This defines an isomorphism A ∪ {b1} → A ∪ {b2}, which shows that O2 |=ρ(b2).

Since LO has no constants, we have another proof that TDLO is complete.


9.3 Modules

Let R be a (possibly non–commutative) ring with 1. An R–module

M = (M, 0,+,−, r)r∈R

is an abelian group (M, 0,+,−) together with operations r : M → M for everyring element r ∈ R which satisfies certain axioms. We formulate the axiomsin the language LM(R) = LAG ∪ {r | r ∈ R}. The theory TM(R) of R–modulesconsists of

• TAG

• ∀x, y r(x+ y).= rx+ ry

• ∀x (r + s)x.= rx+ sx

• ∀x (rs)x.= r(sx)

• ∀x 1x.= x

for all r, s ∈ R. Then T∞ ∪ TM(R) is the theory of all infinite R–modules.

We start with the case where the ring is a field K. Of course, a K–moduleis just a vector space over K.

Theorem 9.3. Let K be a field. Then the theory of all infinite K–vector spaceshas quantifier elimination and is complete.

Proof. Let A be a common finitely generated substructure (i.e. a subspace) ofthe two infinite K–vector spaces V1 and V2. Let ∃y ρ(y) be a simple existentialL(A)–sentence which holds in V1. Choose a b1 from V1 which satisfies ρ(y). Ifb1 belongs to A, we are finished since then V2 |= ρ(b1). If not, we choose ab2 ∈ V2 \ A. Possibly we have to replace V2 by an elementary extension. Thevector spaces A +Kb1 and A +Kb2 are isomorphic by an isomorphism whichmaps b1 to b2 and fixes A elementwise. Hence V2 |= ρ(b2).

The theory is complete since a quantifier free sentence is true in a vector spaceif and only if it is true in the zero–vector space.

For arbitrary rings R, we only get a relative elimination result down topositive primitive formulas:

Definition 9.4. An equation is a LM(R)–formula γ(x) of the form

r1x1 + r2x2 + ...+ rmxm = 0.

A positive primitive formula (pp–formula) is of the form

∃y(γ1 ∧ . . . ∧ γn)

where the γi(xy) are equations.


Theorem 9.5. For every ring R and any R-module M , every LM(R)-formulais equivalent (modulo the theory of M) to a Boolean combination of positiveprimitive formulas.

Remark 9.6. 1. We assume the class of positive primitive formulas to beclosed under ∧.

2. A pp–formula ϕ(x1, . . . , xn) defines a subgroup ϕ(Mn) of Mn:

M |= ϕ(0) and M |= ϕ(x) ∧ ϕ(y) → ϕ(x− y).

Lemma 9.7. Let ϕ(x, y) be a pp–formula and a ∈ M . Then ϕ(M,a) is emptyor a coset of ϕ(M, 0).

Proof. M |= ϕ(x, a) → (ϕ(y, 0) ↔ ϕ(x+ y, a)).

Corollary 9.8. Let a, b ∈ M , ϕ(x, y) a pp–formula. Then (in M) ϕ(x, a) andϕ(x, b) are equivalent or contradictory.

For the proof of Theorem 9.5 we need two further lemmas:

Lemma 9.9 (B.H. Neumann). Let Hi denote subgroups of some abelian group.If H0+a0 ⊂

⋃ni=1Hi+ai and H0/(H0∩Hi) is infinite for i > k, then H0+a0 ⊂⋃k

i=1Hi + ai.

For a proof see Exercise 23.15. The following is an easy calculation:

Lemma 9.10. Let Ai, i ≤ k, be any sets. If A0 is finite, then A0 ⊂⋃ki=1Ai if

and only if ∑∆⊂{1,...k}

(−1)|∆||A0 ∩⋂i∈∆

Ai| = 0.

P roof of 9.5. Fix M . It is enough to show that if ψ(x, y) is in M equivalentto a Boolean combination of pp–formulas, then so is ∀xψ. Since pp–formulasare closed under conjunction, ψ is M -equivalent to a conjunction of formulasϕ0(x, y) → ϕ1(x, y) ∨ . . . ∨ ϕn(x, y) where the ϕi(x, y) are pp–formulas.

We may assume that ψ itself is of this form. Let Hi = ϕi(M, 0), so theϕi(M,y) are empty or cosets of Hi. (Think of y as being fixed in M .) LetH0/(H0 ∩Hi) be finite for i = 1, . . . , k and infinite for i = k + 1, . . . , n, k ≥ 0.By Neumann’s Lemma we have

M |= ∀xψ ↔ ∀x (ϕ0(x, y) → ϕ1(x, y) ∨ . . . ∨ ϕk(x, y)) .

We apply Lemma 9.10 to the sets Ai = ϕi(M,y)/(H0 ∩ . . . ∩ Hk): Soϕ(M,y) ∩

⋂i∈∆ ϕi(M,y) is empty or consists of N∆ cosets of H0 ∩ . . . ∩ Hk

where

N∆ =

∣∣∣∣∣H0 ∩⋂i∈∆

Hi/(H0 ∩ . . . ∩Hk)

∣∣∣∣∣ .


WhenceM |= ∀xψ ↔

∑∆∈N

(−1)|∆|N∆ = 0.

where

N =

{∆ ⊂ {1, . . . k} : ∃x

(ϕ0(x, y) ∧

∧i∈∆

ϕi(x, y)

)}.

9.4 Algebraically closed fields

As the next group of examples we consider fields:

Theorem 9.11 (Tarski). The theory TACF of algebraically closed fields hasquantifier elimination.

Proof. Let K1 and K2 be two algebraically closed fields and R a common sub-ring. Let ∃y ρ(y) be a simple existential sentence with parameters in R whichholds in K1. We have to show that ∃y ρ(y) is also true in K2.

Let F1 and F2 be the quotient fields of R in K1 and K2, respectively, andlet f : F1 → F2 be an isomorphism which is the identity on R (see e.g. [31],Ch. II.4). Then f extends to an isomorphism g : G1 → G2 between the relativealgebraic closures Gi of Fi in Ki, (i = 1, 2) (see e.g. [31], Ch. V.2) .

R R-id

F1 F2-f

G1 G2-g

G1(b1)

K1 K2

Choose an element b1 of K1 which satisfies ρ(y). There are two cases:

Case 1: b1 ∈ G1

Then b2 = g(b1) satisfies the formula ρ(y) in K2.

Case 2: b1 6∈ G1

Then b1 is transcendental over G1 and the field extension G1(b1) is isomorphicto the rational function field G1(X). If K2 is a proper extension of G2, we


choose any element from K2 \ G2 for b2. Then g extends to an isomorphismbetween G1(b1) and G2(b2) which maps b1 to b2. Hence b2 satisfies ρ(y) in K2.In case that K2 = G2 we take a proper elementary extension K ′

2 of K2. (K ′2

exists by 6.1.2 since K2 is infinite.) Then (for the same reason) ∃y ρ(y) holdsin K ′

2 and therefore in K2.

Corollary 9.12. TACF is model complete.

Obviously, TACF is not complete:

For prime numbers p let

TACFp = TACF ∪ {p · 1 .= 1 + . . .+ 1︸︷︷︸

p− times

.= 0}

be the theory of algebraically closed fields of characteristic p and

TACF0 = TACF ∪ {¬n · 1 = 1 + . . .+ 1︸︷︷︸n− times

.= 0 | n = 1, 2, . . .}

the theory of algebraically closed fields of characteristic 0.

Corollary 9.13. The theories TACFp and TACF0 are complete.

Proof. This follows from 8.2 since the prime fields are prime structures for thesetheories.

Corollary 9.14 (Hilbert’s Nullstellensatz). Let K be a field. Then any properideal I in K[X1, . . . , Xn] has a zero in the algebraic closure acl(K).

Proof. As a proper ideal, I is contained in a maximal ideal P . Then L =K[X1, . . . , Xn]/P is an extension field of K in which the cosets of the Xi are azero of I. If I is generated by f0, . . . , fk−1 and

ϕ = ∃x1, . . . xn∧i<k

fi(x1, . . . , xn).= 0,

then ϕ holds in L and therefore in acl(L). We can assume that acl(K) lies inacl(L). Since acl(K) ≺ acl(L), ϕ holds in acl(K).

9.5 Real closed fields

The theory of real closed fields, TRCF, will be discussed in section B.4. It isaxiomatised in the language LOR of ordered rings.

Theorem 9.15 (Tarski-Seidenberg). TRCF has quantifier elimination and iscomplete.


Proof. Let (K1, <) and (K2, <) be two real closed field with a common subringR. Consider an LOR(R)–sentence ∃y ρ(y) (for a quantifier free ρ) which holdsin (K1, <). We have to show ∃y ρ(y) also holds in (K2, <).

We build first the quotient fields F1 and F2 of R in K1 and K2. By B.4.1 thereis an isomorphism f : (F1, <) → (F2, <) which fixes R. The relative algebraicclosure Gi of Fi in Ki is a real closure of (Fi, <), (i = 1, 2). By B.4.5 f extendsto an isomorphism g : (G1, <) → (G2, <).

Let b1 be an element of K1 which satisfies ρ(y). There are two cases:

Case 1: b1 ∈ G1

Then b2 = g(b1) satisfies ρ(y) in K2.

Case 2: b1 6∈ G1:Then b1 is transcendental over G1 and the field extension G1(b1) is isomorphic tothe rational function field G1(X). Let G`1 be the set of all elements of G1 whichare smaller than b1, Gr1 the set of all elements of G1 which are larger than b1.Then all elements of G`2 = g(G`1) are smaller than all elements of Gr2 = g(Gr1).Since fields are densely ordered, we find in an elementary extension (K ′

2, <) of(K2, <) an element b2 which lies between the elements of G`2 and the elementsof Gr2. Since b2 is not in G2, it is transcendental over G2. Hence g extends toan isomorphism h : G1(b1) → G2(b2) which maps b1 to b2.

In order to show that h is order preserving it suffices to show that h is orderpreserving on G1[b1] (B.4.1). Let p(b1) be an element of G1[b1]. Corollary 4.8gives us a decomposition

p(X) = ε∏i<m

(X − ai)∏j<n

((X − cj)

2 + dj)

with positive dj . The sign of p(b1) depends only on the signs of the factors ε,b1 − a0, . . . , b1 − am−1. The sign of h(p(b1)) depends in the same way on thesigns of g(ε), b2 − g(a0), . . . , b2 − g(am−1). But b2 was chosen in such a waythat

b1 < ai ⇔ b2 < g(ai).

Hence p(b1) is positive if and only if h(p(b1)) is positive.

Finally we have

(K1, <) |= ρ(b1) ⇒ (G1(b1), <) |= ρ(b1) ⇒ (G2(b2), <) |= ρ(b2) ⇒⇒ (K ′

2, <) |= ∃y ρ(y) ⇒ (K2, <) |= ∃y ρ(y),

which proves quantifier elimination.

TRCF is complete since the ordered field of the rationals is a prime structure.

Corollary 9.16 (17. Hilbert’s Problem). Let (K,<) be a real closed field. Apolynomial f ∈ K[X1, . . . , Xn] is a sum of squares

f = g21 + . . .+ g2k


of rational functions gi ∈ K(X1, . . . , Xn) if and only if

f(a1, . . . , an) ≥ 0

for all a1, . . . , an ∈ K.

Proof. Clearly a sum of squares cannot have negative values. For the converseassume that f is not a sum of squares. Then, by Corollary 4.3, K(X1, . . . , Xn)has an ordering in which f is negative. Since in K the positive elements aresquares, this ordering, which we denote also by <, extends the ordering of K.Let (L,<) be the real closure of

(K(X1, . . . , Xn), <

). In (L,<) the sentence

∃x1, . . . , xn f(x1, . . . , xn) < 0

is true. Hence it is also true in (K,<).

9.6 Separably closed fields

A field is separably closed if every non–constant separable polynomial has a zero.Clearly, separably closed fields which are also perfect are algebraically closed.

For any field K of characteristic p > 0, Kp = {ap | a ∈ K} is a subfield ofK. If the degree [K : Kp] is finite, it has the form pe and e is called the degreeof imperfection of K. If the degree [K : Kp] is infinite, then we say that K hasinfinite degree of imperfection. See also page 234.

For any natural number e we denote by SCFp,e the theory of separably closedfields with degree of imperfection e. SCFp,∞ is the theory of separably closedclosed field of characteristic p with infinite degree of imperfection. We will provebelow that SCFp,e is complete. A proof of the completeness of SCFp,∞ can befound in [15].

To study SCFp,e we consider an expansion of it: SCFp(c1, · · · , ce), the theoryof separably closed fields of characteristic p in the language L(c1, . . . , ce) of ringswith constants c1, . . . , ce for a distinguished finite p–basis. We show

Proposition 9.17. SCFp(c1, · · · , ce) is model complete.

For the proof we need the following lemma:

Lemma 9.18. Let K and L be extensions of F . Assume that K/F is sepa-rable and that L is separably closed. Then K embeds over F in an elementaryextension of L.

Proof. Since L is infinite, it has arbitrarily large elementary extensions. Sowe may assume that tr.deg(L/F ) is infinite. Let K ′ be a finitely generatedsubfield of K over F . By compactness it suffices to show that all such K ′/F


can be embedded into L/F . By LemmaB 6.12 K ′/F has a transcendence basisx1, . . . , xn so that K ′/F (x1, . . . , xn) is separably algebraic. F (x1, . . . , xn)/Fcan be embedded into L/F . Since L is separably closed this embedding extendto K ′.

Proof of Proposition 9.17. Let (F, b1, . . . , be) ⊂ (K, b1, . . . , be) be an extensionof models of SCFp(c1, · · · , ce). Since F and K have the same p–basis, K isseparable over F by Remark B6.9. By Lemma 9.18, K embeds over F into anelementary extension of F , showing that F is existentially closed in K. Nowthe claim follows by Robinson’s Test (Lemma 8.7).

Corollary 9.19 (Ershov). SCFp,e is complete.

Proof. Consider the polynomial ring R = Fp[x1, . . . , xe]. It is easy to see thatx1, . . . , xe is a p–basis of R in the sense of Lemma B 6.11. The same lemmaimplies that x1, . . . , xe is a p–basis of F = Fp(x1, . . . , xe)sep (where Lsep denotesthe separabel algebraic closure of the field L). So (F, x1, . . . , xe) is a model ofSCFp(c1, · · · , ce). If (L, b1, . . . , be) is another model of SCFp(c1, · · · , ce), thenby Lemma B 6.10 the bi are algebraically independent over Fp, so we can embed(F, x1, . . . , xe) into (L, b1, . . . , be). This embedding is elementary by Proposi-tion 9.17, so L is elementarily equivalent to F .

Let K be a field with p–basis b1, . . . , be. The λ-functions λν : K → K aredefined by

x =∑

λν(x)pbν ,

where the ν are multi-indices (ν1, . . . , νe) with 0 ≤ νi < p and bν denotesbν11 · · · bνee .

Theorem 9.20 (Delon). SCFp(c1, · · · , ce) has quantifier elimination in the lan-guage L(c1, . . . , ce, λν)ν∈pe .

It can be shown that SCFp(c1, · · · , ce) has quantifier elimination already in thelanguage L(λν)ν∈pe without naming a p–basis.

Proof. Let K = (K, b1, . . . , be) and L = (L, b1, . . . , be) be models ofSCFp(c1, · · · , ce) and let R be a common subring which contains the bi andis closed under the λ-functions of K and L.2 Since R is closed under the λ-functions, the bi form a p–basis of R in the sense of Lemma B 6.11. Let Fbe the separable closure of the quotient field of R. By Lemma B 6.11 the bialso form a p–basis of F . So F is an elementary subfield of K and of L byProposition 9.17, and hence KR and LR are elementarily equivalent. Now theclaim follows from Theorem 8.5.

2Note that the λ-functions of K and L agree automatically on R.


9.7 Differentially closed fields

We next consider fields with a derivation in the language of fields expandedby a function symbol d. Differential fields are introduced and discussed inSection B.5.

Definition 9.21. The theory of differentially closed fields, TDCF0 , is the theoryof differential fields (K, d) in characteristic 0 satisfying the following property:

For f ∈ K[x0, . . . , xn] \K[x0, . . . , xn−1] and g ∈ K[x0, . . . , xn−1], g 6= 0, thereis some a ∈ K such that f(a, da, . . . , dna) = 0 and g(a, da, . . . , dn−1a) 6= 0.

Clearly, models of TDCF0 are algebraically closed.

Theorem 9.22.

1. Any differential field can be extended to a model of TDCF0 .

2. TDCF0 is complete and has quantifier elimination.

Proof. Let (K, d) be a differential field and f and g as in the definition. Wemay assume that f is irreducible. Then f determines a field extension F =K(t0, . . . , tn−1, b) where the ti are algebraically independent over K and

f(t0, . . . , tn−1, b) = 0

(see Remark B 6.7). By Lemma B.5.2 and B.5.3 there is an extension ofthe derivation to F with dti = ti+1 and dtn−1 = b. For a = t0 we havef(a, da, . . . , dna) = 0 and g(a, da, . . . , dn−1a) 6= 0. Let K0 denote the differen-tial field that we obtain from K by doing this for all pairs f, g as above withcoefficients from K. Inductively, we define differential fields Ki+1 satisfying therequired condition for polynomials with coefficients in Ki. Their union

⋃i<ωKi

is a model of TDCF0 .

Since the rational numbers with trivial derivation are a prime structure forTDCF0 , by Lemma 8.2 it suffices for part 2. to prove quantifier elimination.For this, let K be a differential field with two extensions F1 and F2 which aremodels of TDCF0 . Let a be an element of F1 and let K{a} = K(a, da, d2a, . . . )be the differential field generated by K and a. We have to show that K{a} canbe embedded over K into an elementary extension of F2. We distinguish twocases.

1. The derivatives a, da, d2a, . . . are algebraically independent over K: SinceF2 is a model of TDCF0 , there is an element b in some elementary extensionsuch that g(a, da, . . . , dn−1a) 6= 0 for all n and all g ∈ K[x0, . . . , xn−1] \ 0.The isomorphism from K{a} to K{b} defined by dia 7→ dib is the requiredembedding.


2. Let dna be algebraic over K(a, da, . . . , dn−1a) and n minimal: Choose anirreducible f ∈ K[x0, . . . , xn] such that f(a, da, . . . , dna) = 0 (cf. Remark B 6.7).We may find some b with f(b, db, . . . , dnb) = 0 and g(b, ba, . . . , dn−1b) 6= 0 for allg ∈ K[x0, . . . , xn−1]\0 in an elementary extension of F2. The field isomorphismfrom K1 = K(a, . . . , dna) to K2 = K(b, . . . , dnb) fixing K and taking dia to dibtakes the derivation of F1 restricted to K(a, . . . , dn−1a) to the derivation of F2

restricted to K(b, . . . , dn−1b). The uniqueness part of Lemma B.5.3 implies thatK1 and K2 are closed under the respective derivations, and that K1 and K2 areisomorphic over K as differential fields.

Exercise 9.1.The theory TRG of the random graph is axiomatised by the following axiomscheme:

∀x0, . . . xm−1, y1, . . . yn−1

(∧i 6=j

¬xi.= yj →

∃z(∧i<m

zRxi ∧ ¬z .= xi) ∧ (

∧j<n

¬zRyj ∧ ¬z .= yj)

)Show that this theory has quantifier elimination.

Exercise 9.2.In models of TACF ,TRCF and TDCF0 , the model theoretic algebraic closure ofa set A coincides with the algebraic closure of the (differential) field generatedby A.

Exercise 9.3.Show that the following is true in any algebraically closed fieldK: every injectivepolynomial map of a definable subset of Kn in itself is surjective.

In fact, more is true: use Exercise 23.13 to show that the previous statementholds for every injective definable map.

Chapter 4

Countable models

10 The Omitting Types Theorem

As we have seen in Corollary 5.9, it is not hard to realise a given type or in factany number of them. But as Sacks [44] pointed out, it needs a model theoristto avoid realizing a given type.

Definition 10.1. Let T be an L–theory and Σ(x) a set of L–formulas. A modelA of T not realizing Σ(x) is said to omit Σ(x). A formula ϕ(x) isolates Σ(x) if

a) ϕ(x) is consistent with T .

b) T ` ∀x (ϕ(x) → σ(x)) for all σ(x) in Σ(x).

A set of formulas is often called a partial type. This explains the name ofthe following theorem.

Theorem 10.2 (Omitting Types). If T is countable1 and consistent and ifΣ(x) is not isolated in T , then T has a model which omits Σ(x).

If Σ(x) is isolated by ϕ(x) and A is a model of T , then Σ(x) is realised inA by all realisations of ϕ(x). Therefore the converse of the theorem is true forcomplete theories T : if Σ(x) is isolated in T , then it is realised in every modelof T .

Proof. We choose a countable set C of new constants and extend T to a theoryT ∗ with the following properties:

1An L–theory is countable if L is at most countable.

58

CHAPTER 4. COUNTABLE MODELS 59

a) T ∗ is a Henkin–theory: for all L(C)–formulas ψ(x) there exists a constantc ∈ C with ∃xψ(x) → ψ(c) ∈ T ∗.

b) For all c ∈ C there is a σ(x) ∈ Σ(x) with ¬σ(c) ∈ T ∗.

We construct T ∗ inductively as the union of an ascending chain

T = T0 ⊂ T1 ⊂ . . .

of consistent extensions of T by finitely many axioms from L(C), in each stepmaking an instance of a) or b) true.

Enumerate C = {ci : i < ω} and let {ψi(x) : i < ω} be an enumeration ofthe L(C)-formulas.

Assume that T2i is already constructed. Choose some c ∈ C which does notoccur in T2i and set T2i+1 = T2i∪{∃xψi(x) → ψi(c)}. Clearly T2i+1 is consistent

Up to equivalence T2i+1 has the form T ∪ {δ(ci, c)} for an L–formula δ(x, y)and a tuple c ∈ C which does not contain ci. Since ∃y δ(x, y) does not isolateΣ(x), for some σ ∈ Σ the formula ∃y δ(x, y)∧¬σ(x) is consistent with T . Thus,T2i+2 = T2i+1 ∪ {¬σ(ci)} consistent.

Take a model (A′, ac)c∈C of T ∗. Since T ∗ is a Henkin theory, Tarski’s Test 4.2shows that A = {ac | c ∈ C} is the universe of an elementary substructure A(cf. Lemma 5.3). By property b), Σ(x) is omitted in A.

Corollary 10.3. Let T be countable and consistent and let

Σ0(x1, . . . , xn0), Σ1(x1, . . . , xn1), . . .

be a sequence of partial types. If all Σi are not isolated, then T has a modelwhich omits all Σi.

Proof. Generalise the proof of the Omitting Types Theorem.

Exercise 10.1.Prove Corollary 10.3.


11 The space of types

We now endow the set of types of a given theory with a topology. The Com-pactness Theorem 5.1 then translates into the statement that this topology iscompact, whence its name.

Fix a theory T . An n–type is a maximal set of formulas p (x1, . . . , xn) con-sistent with T . We denote by Sn(T ) the set of all n–types of T . We also writeS(T ) for S1(T ).2

If B is a subset of an L–structure A, we recover SAn (B) (see p. 31) asSn(Th(AB)). In particular, if T is complete and A is any model of T , wehave SA(∅) = S(T ).

For any L–formula ϕ(x1, . . . , xn), let [ϕ] denote the set of all types contain-ing ϕ.

Lemma 11.1.

1. [ϕ] = [ψ] if and only if ϕ and ψ are equivalent modulo T .

2. The sets [ϕ] are closed under Boolean operations. In fact [ϕ]∩[ψ] = [ϕ∧ψ],[ϕ] ∪ [ψ] = [ϕ ∨ ψ], Sn(T ) \ [ϕ] = [¬ϕ], Sn(T ) = [>] and ∅ = [⊥].

Proof. For the first part just notice that if ϕ and ψ are not equivalent moduloT , then ϕ ∧ ¬ψ or ¬ϕ ∧ ψ is consistent with T and hence [ϕ] 6= [ψ]. The rest isclear.

It follows that the collection of sets of the form [ϕ] is closed under finiteintersections and includes Sn(T ). So these sets form a basis of a topology onSn(T ).

Lemma 11.2. Sn(T ) is a 0–dimensional compact space.

Proof. Being 0–dimensional means having a basis of clopen sets. Our basic opensets are clopen since their complements are also basic open.

If p and q are two different types, there is a formula ϕ contained in p butnot in q. It follows that [ϕ] and [¬ϕ] are open sets which separate p and q. Thisshows that Sn(T ) is Hausdorff.

To show compactness consider a family [ϕi], (i ∈ I), with the finite intersec-tion property. This means that all ϕi1 ∧ · · · ∧ ϕik are consistent with T . So, byCorollary 5.5, {ϕi | i ∈ I} is consistent with T and can be extended to a typep, which then belongs to all [ϕi].

Lemma 11.3. All clopen subsets of Sn(T ) have the form [ϕ].2S0(T ) can be considered as the set of all complete extensions of T , up to equivalence.


Proof. It follows from Exercise 7.1 that we can separate any two disjoint closedsubsets of Sn(T ) by a basic open set.

Remark. The Stone duality theorem asserts that the map

X 7→ {C | C clopen subset of X}

yields an equivalence between the category of 0–dimensional compact spacesand the category of Boolean algebras. The inverse map assigns to every Booleanalgebra B its Stone space S(B), the set of all ultrafilters (see Exercise 2.4) of B.For more on Boolean algebras see [17].

Definition 11.4. A map f from a subset of a structure A to a structure B iselementary if it preserves the truth of formulas; i.e., f : A0 → B is elementaryif for every formula ϕ(x1, . . . , xn) and a ∈ A0 we have

A |= ϕ(a) ⇒ B |= ϕ(f(a)).

Note that the empty map is elementary if and only if A and B are elementarilyequivalent. An elementary embedding of A is an elementary map which isdefined on all of A.

Lemma 11.5. Let A and B be L–structures, A0 and B0 subsets of A and B,respectively. Any elementary map A0 → B0 induces a continuous surjectivemap Sn(B0) → Sn(A0).

Proof. If q(x) ∈ Sn(B0), we define

S(f)(q) = {ϕ(x1, . . . , xn, a) | a ∈ A0, ϕ(x1, . . . , xn, fa) ∈ q}.

It is easy to see that S(f) defines a map from Sn(B0) to Sn(A0). S(f) is surjectivesince {ϕ(x1, . . . , xn, fa) | ϕ(x1, . . . , xn, a) ∈ p} is finitely satisfiable for everyp ∈ Sn(A0). And S(f) is continuous since [ϕ(x1, . . . , xn, fa)] is the preimage of[ϕ(x1, . . . , xn, a)] under S(f).

There are two main cases:

• An elementary bijection f : A0 → B0 defines a homeomorphism Sn(A0) →Sn(B0). We write f(p) for the image of p.

• If A = B andA0 ⊂ B0, the inclusion map induces the restriction Sn(B0) →Sn(A0). We write q � A0 for the restriction of q to A0. We call q an ex-tension of q � A0.

We leave the following lemma as an exercise (see Exercise 11.1).


Lemma 11.6. A type p is isolated in T if and only if p is an isolated point inSn(T ). If fact, ϕ isolates p if and only if [ϕ] = {p}. I.e. [ϕ] is an atom in theBoolean algebra of clopen subsets of Sn(T ).

We call a formula ϕ(x) complete if

{ψ(x) | T ` ∀x (ϕ(x) → ψ(x))}

is a type. We have shown:

Corollary 11.7. A formula isolates a type if and only if it is complete.

Exercise 11.1.Show that a type p is isolated if and only if it is isolated as an element in theStone space.

Exercise 11.2.

a) Closed subsets of Sn(T ) have the form {p ∈ Sn(T ) | Σ ⊂ p}, where Σ is anyset of formulas.

b) Let T be countable and consistent. Then any meager3 subset X of Sn(T )can be omitted, i.e. there is model which omits all p ∈ X.

Exercise 11.3.Consider the space Sω(T ) of all complete types in variables v0, v1, . . .. Note thatSω(T ) is again a compact space and therefore not meager by Baire’s theorem.

1. Show that {tp(a0, a2, . . .) | the ai enumerate a model of T} is comeagerin Sω(T ).

2. Use this to give a purely topological proof the Omitting Types Theorem(10.3).

Exercise 11.4.Let L ⊂ L′, T an L–theory, T ′ an L′–theory and T ⊂ T ′. Show that there is anatural continuous map Sn(T

′) → Sn(T ). This map is surjective if and only ifT ′/T is a conservative extension, i.e. if T ′ and T prove the same L–sentences.

Exercise 11.5.Let B be a subset of A. Show that the restriction map Sm+n(B) → Sn(B) isopen, continuous and surjective. Let a be an n–tuple in A. Show that the fibreover tp(a/B) is canonically homeomorphic to Sm(aB).

Exercise 11.6.A theory T has quantifier elimination if and only if every type is implied by itsquantifier free part.

3A subset of a topological space is nowhere dense if its closure has no interior. A countableunion of nowhere dense sets is meager.


Exercise 11.7.Consider the structure M = (Q, <). Determine all types in S1(Q). Which ofthese types are realised in R? Which extensions does a type over Q have to atype over R?


12 ℵ0–categorical theories

In this section, we consider theories with a unique countable model (up toisomorphism, of course). These theories can be characterised by the fact thatthey have only finite many n-types for each n, see Exercise 12.3. We show thefollowing equivalent statement:

Theorem 12.1 (Ryll–Nardzewski). Let T be a countable complete theory withinfinite models. Then T is ℵ0–categorical if and only if for every n there areonly finitely many formulas ϕ(x1, . . . , xn) up to equivalence relative to T .

The proof will make use of the following notion:

Definition 12.2. An L–structure A is ω–saturated if all types over finite subsetsof A are realised in A.

The types in the definition are meant to be 1–types. On the other hand, itis not hard to see that an ω-saturated structure realises all n-types over finitesets (cp. Exercise 12.9), for all n ≥ 1. The following lemma is a generalisationof the ℵ0–categoricity of TDLO. The proof is essentially the same, see p. 35

Lemma 12.3. Two elementarily equivalent countable and ω–saturated struc-tures are isomorphic.

Proof. Suppose A and B are as in the lemma. We choose enumerations A ={a0, a1, . . . } and B = {b0, b1, . . .}. Then we construct an ascending sequencef0 ⊂ f1 ⊂ . . . of finite elementary maps

fi : Ai → Bi

between finite subset of A and B. We will choose the fi in such a way that Ais the union of the Ai and B the union of the Bi. The union of the fi is thenthe desired isomorphism between A and B.

The empty map f0 = ∅ is elementary since A and B are elementarily equiv-alent. Assume that fi is already constructed. There are two cases:

i = 2n:We will extend fi to Ai+1 = Ai ∪ {an}.Consider the type

p(x) = tp(an/Ai).

Since fi is elementary, fi(p)(x) is in B a type over Bi. Since B is ω–saturated,there is a realisation b′ of this type. So for a ∈ Ai

A |= ϕ(an, a) ⇒ B |= ϕ(b′, fi(a)).

This shows that fi+1(an) = b′ defines an elementary extension of fi.


i = 2n+ 1:We exchange A and B: since A is ω–saturated, we find an elementary map fi+1

with image Bi+1 = Bi ∪ {bn}.

Proof of Theorem 12.1. Assume that there are only finitely many ϕ(x1, . . . , xn)relative to T for every n. By Lemma 12.3 it suffices to show that all models ofT are ω–saturated. Let M be a model of T and A an n–element subset. If thereare only N many formulas, up to equivalence, in the variables x1, . . . , xn+1,there are, up to equivalence in M, at most N many L(A)–formulas ϕ(x). Thus,each type p(x) ∈ S(A) is isolated (with respect to Th(MA)) by a “smallest”formula ϕp(x). Each element of M which realises ϕp(x) also realises p(x), so Mis ω–saturated.

Conversely, if there are infinitely many ϕ(x1, . . . , xn) modulo T for some n,then – as the type space Sn(T ) is compact – there must be some non-isolatedtype p. By the Omitting Types Theorem (10.2) there is a countable model ofT in which this type is not realised. On the other hand, there also exists acountable model of T realizing this type. So T is not ℵ0–categorical.

Remark 12.4. The proof shows that a countable complete theory with infinitemodels is ℵ0–categorical if and only if all countable models are ω–saturated.

In Theorem 16.10 this characterisation will be extended to theories categor-ical in uncountable cardinalities.

Remark 12.5. The proof of Lemma 12.3 also shows that ω-saturated modelsare ω–homogeneous in the following sense:

Definition 12.6. An L–structure M is ω–homogeneous if for every elementarymap f0 defined on a finite subset A of M and for any a ∈ M there is someb ∈M such that

f = f0 ∪ {〈a, b〉}is elementary.

Note that f = f0∪{〈a, b〉} is elementary if and only if b realises f0(tp(a/A)).

Corollary 12.7. Let A be a structure and a1, . . . , an elements of A. ThenTh(A) is ℵ0–categorical if and only if Th(A, a1, . . . , an) is ℵ0–categorical.

Examples:The following theories are ℵ0–categorical:

T∞, the theory of infinite sets.

For every finite field Fq, the theory of infinite Fq–vector spaces. (Indeed, thistheory is categorical in all infinite cardinals. This follows directly from thefact that vector spaces over the same field and of the same dimension areisomorphic.)


The theory TRG of the random graph (see Exercise 9.1): This follows fromTheorem 12.1 since TRG has quantifier elimination and for any n there areonly finitely many graphs on n elements.

The theory TDLO of dense linear orders without endpoints. This follows fromTheorem 12.1 since TDLO has quantifier elimination: For every n there are onlyfinitely many (say Nn) ways to order n (not necessarily distinct) elements.For n = 2 for example there are the three possibilities a1 < a2, a1 = a2and a2 < a1. Each of these possibilities corresponds to a complete formulaψ(x1, . . . , xn). Hence there are, up to equivalence, exactly 2Nn many formulasϕ(x1, . . . , xn).

Next we study the existence of countable ω–saturated structures:

Definition 12.8. A theory T is small if Sn(T ) are at most countable for alln < ω.

A countable theory with at most countably many non–isomorphic at mostcountable models is always small. The converse is not true.

Lemma 12.9. A countable4 complete theory is small if and only if it has acountable ω–saturated model.

Proof. If T has a finite model A, T is small and A is ω–saturated. So we mayassume that T has infinite models.

If all types can be realised in a single countable model, there can be at mostcountably many types.

If conversely all Sn+1(T ) are at most countable, then over any n–elementsubset of a model of T there are at most countably many types. We constructan elementary chain

A0 ≺ A1 ≺ . . .

of models of T . For A0 we take any countable model. If Ai is already con-structed, we use, Corollary 5.9 and Theorem 6.1.1 to construct a countablemodel Ai+1 in such a way that all types over finite subsets of Ai are realised inAi+1. This can be done since there are only countable many such types. Theunion A =

⋃i∈ω Ai is countable and ω–saturated since every type over a finite

subset B of A is realised in Ai+1 if B ⊆ Ai.

Theorem 12.10 (Vaught). A countable complete theory cannot have exactlytwo countable models.

Proof. We can assume that T is small and not ℵ0–categorical. We will show thatT has at least three non–isomorphic countable models. T has an ω–saturated

4The statement is true even for uncountable L.


countable model A and there is a non–isolated type p(x), which can be omittedin a countable model B. Let p(x) be realised in A by a. Since Th(A, a) is notℵ0–categorical, Th(A, a) has a countable model (C, c) which is not ω–saturated.C is not ω–saturated and therefore not isomorphic to A. But C realises p(x) andis therefore not isomorphic to B.

Exercise 12.5 shows, that for any n 6= 2, n ≤ ω there is a countable com-plete theory with exactly n countable models. Vaught’s Conjecture statesthat if a complete countable theory has fewer than continuum many countablenon-isomorphic models, the number of countable models is at most countable(see [43] for a survey on what is currently known).

Exercise 12.1. 1. If T is ℵ0–categorical, then in any model M the algebraicclosure of a finite set is finite (see Definition p. 94). In particular, M islocally finite, i.e., any substructure generated by a finite subset is finite.(In many–sorted structures we mean that in each sort the trace of thealgebraic closure is finite.)

2. There is no ℵ0–categorical theory of fields, i.e., if T is a complete LR–theory containing TF, then T is not ℵ0–categorical.

Exercise 12.2.A theory T is small exactly if T has at most countably many completions, eachof which is small.

Exercise 12.3.Show that T is ℵ0–categorical if and only if Sn(T ) is finite for all n.

Exercise 12.4.Write down a theory with exactly two countable models.

Exercise 12.5.Show that for every n > 2 there is a countable complete theory with exactly ncountable models. (Consider (Q, <, P0, . . . , Pn−2, c0, c1, . . .), where the Pi forma partition of Q into dense subsets and the ci are an increasing sequence ofelements of P0.)

Exercise 12.6.Give an example of an uncountable complete theory with exactly one countablemodel which does not satisfy the conclusion of Theorem 12.1.

Exercise 12.7.Suppose M is countable and ℵ0–categorical. Show that if X ⊆Mn is invariantunder all automorphisms of M, then X is definable.

Exercise 12.8.Let M be a structure and assume that for some n only finitely many n-typesare realised in M. Then any structure elementarily equivalent to M satisfiesexactly the same n-types.


Exercise 12.9.If A is ω–saturated, all n–types over finite sets are realised. More generallyprove the following: If A is κ–saturated i.e. if all 1-types over sets of cardinalityless than κ are realised in A, then the same is true for all n-types. See alsoExercise 23.5.

Exercise 12.10.Show:

1. The theory of (R, 0,+) has exactly two 1-types but ℵ0 many 2-types.

2. The theory of (R, 0,+, <) has exactly three 1-types but 2ℵ0 many 2-types.


13 The amalgamation method

In this section we will present one of the main methods to construct new andinteresting examples of first order structures. It goes back to Fraïssé, but hasmore recently been modified by Hrushovski [24]. We here focus mainly on theℵ0–categorical examples and return to the fancier version in Section 44.

Definition 13.1. For any language L, the skeleton K5 of an L–structure Mis the class of all finitely generated L–structures which are isomorphic to asubstructure of M. We say that an L–structure M is K–saturated if its skeletonis K and if for all A, B in K and all embeddings f0 : A → M and f1 : A → Bthere is an embedding g1 : B → M with f0 = g1 ◦ f1.

Theorem 13.2. Let L be a countable language. Any two countable K-saturatedstructures are isomorphic.

Proof. Let M and N be countable L–structures with the same skeleton K, andassume that M and N are K–saturated. As in the proof of Lemma 12.3 weconstruct an isomorphism between M and N as the union of an ascending se-quence of isomorphisms between finitely generated substructures of M and N .This can be done because if f1 : A→ N is an embedding of a finitely generatedsubstructure A of M into N, and a is an element of M , then by K–saturationf1 can be extended to an embedding g1 : A

′ → N where A′ = 〈Aa〉M . Nowinterchange the roles of M and N.

Remark 13.3. The proof shows that any countable K-saturated structure Mis ultrahomogeneous i.e. any isomorphism between finitely generated substruc-tures extends to an automorphism of M.

Theorem 13.4. Let L be a countable language and K a countable class offinitely generated L–structures. There is a countable K-saturated L–structureM if and only if

a) (Heredity) If A0 belongs to K, then all elements of the skeleton of A0 alsobelong to K.

b) (Joint Embedding) For B0,B1 ∈ K there are some D ∈ K and embeddingsgi : Bi → D.

c) (Amalgamation) If A,B0,B1 ∈ K and fi : A → Bi, (i = 0, 1) are embed-dings, there is some D ∈ K and two embeddings gi : Bi → D such thatg0 ◦ f0 = g1 ◦ f1.

5This is also called the age of M.


A

B0 B1

D

��7

SS

SSo

��7

SS

SSo

f0 f1

g0 g1

In this case, M is unique up to isomorphism and is called the Fraïssé limitof K.

Proof. Let K be the skeleton of a countable K-saturated structure M. Clearly,K has the Hereditary Property. To see that K has the Amalgamation Propertylet A, B0, B1, f0 and f1 be as in c). We may assume that B0 ⊂ M andf0 is the inclusion map. Furthermore we can assume A ⊂ B1 and that f1 isthe inclusion map. Now the embedding g1 : B1 → M is the extension of theisomorphism f0 : A → f0(A) to B1 and satisfies f0 = g1 ◦ f1. For D we choosea finitely generated substructure of M which contains B0 and the image of g1.For g0 : B0 → D take the inclusion map. The Joint Embedding Property isproved similarly.

For the converse assume that K has properties a), b), and c). Choose anenumeration (Bi)i∈ω of all isomorphism types in K. We construct M as theunion of an ascending chain

∅ = M0 ⊂ M1 ⊂ . . .M

of elements of K. Suppose that Mi is already constructed. If i = 2n is even, wechoose Mi+1 as the top of the diagram

Mi Bn

Mi+1

��7

SS

SSog0 g1

where we can assume that g0 is the inclusion map. If i = 2n + 1 is odd, let Aand B from K and two embeddings f0 : A → Mi and f1 : A → B be given. Weconstruct Mi+1 using the diagram


A

Mi B

Mi+1

��7

SS

SSo

��7

SS

SSo

f0 f1

g0 g1

To ensure that M is K-saturated we do the following. Assume that wehave A, B ∈ K and embeddings f0 : A → M and f1 : A → B. For large j theimage of f0 will be contained in Mj . During the construction of the the Mi, inorder to guarantee the K-saturation of M, we have to ensure that eventually,for some odd i ≥ j, the embeddings f0 : A → Mi and f1 : A → B were usedin the construction of Mi+1. This can be done since for each j there are – upto isomorphism – at most countably many possibilities. Thus there exists anembedding g1 : B → Mi+1 with f0 = g1 ◦ f1.

Clearly, K is the skeleton of M: the finitely generated substructures of M arethe substructures of the Mi. Since the Mi belong to K, their finitely generatedsubstructures also belong to K. On the other hand each Bn is isomorphic to asubstructure of M2n+1.

Uniqueness follows from Theorem 13.2

For finite relational languages L, any non-empty finite subset is itself a (fi-nitely generated) substructure. For such languages, the construction yields ℵ0–categorical structures. We now take a closer look at ℵ0–categorical theories withquantifier elimination in a finite relational language.

Remark 13.5. A complete theory T in a finite relational language with quan-tifier elimination is ℵ0–categorical. So all its models are ω–homogeneous byRemarks 12.4 and 12.5.

Proof. For every n, there is only a finite number of non–equivalent quantifierfree formulas ρ(x1, . . . , xn). If T has quantifier elimination, this number is alsothe number of all formulas ϕ(x1, . . . , xn) modulo T and so T is ℵ0–categoricalby Theorem 12.1.

Clearly, if a theory has quantifier elimination, any isomorphism betweensubstructures is elementary. For relational languages we can say more:

Lemma 13.6. Let T be a complete theory in a finite relational language andM an infinite model of T . The following are equivalent:

a) T has quantifier elimination.


b) Any isomorphism between finite substructures is elementary.

c) The domain of any isomorphism between finite substructures can be extendedto any further element.

Proof. a)→ b) is clear.

b)→ a): If any isomorphism between finite substructures of M is elementary,all n–tuples a which satisfy in M the same quantifier free type

tpqf(a) = {ρ(x) | M |= ρ(a), ρ(x) quantifier free}

satisfy the same simple existential formulas. We will show from this that everysimple existential formula ϕ(x1, . . . , xn) = ∃y ρ(x1, . . . , xn, y) is, modulo T ,equivalent to a quantifier free formula. Let r1(x), . . . , rk−1(x) be the quanti-fier free types of all n–tuples in M which satisfy ϕ(x). Let ρi(x) be equivalentto the conjunction of all formulas from ri(x). Then

T ` ∀x(ϕ(x) ↔

∨i<k

ρi(x)).

a) → c): T is ℵ0–categorical and hence all models are ω–homogeneous. Sinceany isomorphism between finite substructures is elementary by the equivalenceof a) and b) the claim follows.

c) → b): If the domain of any finite isomorphism can be extended to anyfurther element, it is easy to see that every finite isomorphism is elementary.

We have thus established the following:

Theorem 13.7. Let L be a finite relational language and K a class of finiteL–structures. If the Fraïssé limit of K exists, its theory is ℵ0–categorical andhas quantifier elimination.

Example:The class of finite linear orders obviously has the Amalgamation Property. TheirFraïssé limit is the dense linear order without endpoints.

Exercise 13.1.Prove Remark 13.3.

Exercise 13.2.Let K be the class of finite graphs. Show that its Fraïssé limit is the countablerandom graph. This yields another proof that the theory of the random graphhas quantifier elimination.


14 Prime models

Some, but not all, theories have models which are smallest in the sense that theyelementarily embed into any other model of the theory. For countable completetheories these are the models realizing only the ‘necessary’ types. If they exist,they are unique and ω-homogeneous.

In this section - unless explicitly stated otherwise - we let T be acountable complete theory with infinite models.

Definition 14.1. Let T be a countable theory with infinite models, not neces-sarily complete.

1. A0 is a prime model of T if A0 can be elementarily embedded into allmodels of T .

2. A structure A is called atomic if all n–tuples a of elements of A are atomic.This means that the types tp(a) = tp(a/∅) are isolated in SAn (∅).

Prime models need not exist, see the example on p. 74. By Corollary 11.7,a tuple a is atomic if and only if it satisfies a complete formula. For the termi-nology cf. Lemma 11.6.

Since T has countable models, prime models must be countable and sincenon–isolated types can be omitted in suitable models by Theorem 10.2, onlyisolated types can be realised in prime models. Thus, one direction of thefollowing theorem is clear:

Theorem 14.2. A model of T is prime if and only if it is countable and atomic.

Proof. As just noted, a prime model has to be countable and atomic. For theconverse let M0 be a countable and atomic model of T and M any model of T .We construct an elementary embedding of M0 to M as a union of an ascendingsequence of elementary maps

f : A→ B

between finite subsets A of M0 and B of M . We start with the empty map,which is elementary since M0 and M are elementarily equivalent.It is enough to show that every f can be extended to any given A∪{a}. Let p(x)be the type of a over A and f(p) the image of p under f (cf. Lemma 11.5). Wewill show that f(p) has a realisation b ∈M . Then f ∪ {〈a, b〉} is an elementaryextension of f .

Let a be a tuple which enumerates the elements of A and ϕ(x, x) an L–formula which isolates the type of aa. Then p is isolated by ϕ(x, a): clearlyϕ(x, a) ∈ tp(a/a) and if ρ(x, a) ∈ tp(a/a), we have ρ(x, y) ∈ tp(a, a). Thisimplies M0 |= ∀x, y (ϕ(x, y) → ρ(x, y)) and M |= ∀x (ϕ(x, a) → ρ(x, a)).Thus f(p) is isolated by ϕ(x, f(a)) and, since ϕ(x, f(a)) can be realised in M,so can be f(p).


Theorem 14.3. All prime models of T are isomorphic.

Proof. Let M0 and M′0 be two prime models. Since prime models are atomic,

elementary maps between finite subsets of M0 and M′0 can be extended to all

finite extensions. Since M0 and M′0 are countable, it follows exactly as in the

proof of Lemma 12.3 that M0 and M′0 are isomorphic.

The previous proof also shows the following:

Corollary 14.4. Prime models are ω-homogeneous.

Proof. Let M0 be prime and a any tuple of elements fromM0. By Theorem 14.2,(M0, a) is a prime model of its theory. The claim follows now from Theorem 14.3.

Definition 14.5. The isolated types are dense in T if every consistent L–formula ψ(x1, . . . , xn) belongs to an isolated type p(x1, . . . , xn) ∈ Sn(T ).

Remark 14.6. By Corollary 11.7 this definition is equivalent to asking thatevery consistent L–formula ψ(x1, . . . , xn) contains a complete formula ϕ(x1, . . . , xn):

T ` ∀x (ϕ(x) → ψ(x))

Theorem 14.7. T has a prime model if and only if the isolated types are dense.

Proof. Suppose T has a prime model M (so M is atomic by 14.2). Since consis-tent formulas ψ(x) are realised in all models of T , ψ(x) is realised by an atomictuple a and ψ(x) belongs to the isolated type tp(a).

For the other direction notice that a structure M0 is atomic if and only if forall n the set

Σn(x1, . . . , xn) = {¬ϕ(x1, . . . , xn) | ϕ(x1, . . . , xn) complete}

is not realised in M0. Hence, by Corollary 10.3, it is enough to show that theΣn are not isolated in T . This is the case if and only if for every consistentψ(x1, . . . , xn) there is a complete formula ϕ(x1, . . . , xn) with T 6` ∀x (ψ(x) →¬ϕ(x)). Since ϕ(x) is complete, this is equivalent to T ` ∀x (ϕ(x) → ψ(x)). Weconclude that Σn is not isolated if and only if the isolated n–types are dense.

Notice that the last part shows in fact the equivalence directly. (Because if Σnis isolated for some n, then it is realised in every model and no atomic modelcan exist.)

Example:Let L be the language having a unary predicate Ps for every finite 0 –1–

sequence s ∈ <ω2. The axioms of TTree say that the Ps, s ∈ <ω2, form a binarydecomposition of the universe:


• ∀xP∅(x)

• ∃xPs(x)

• ∀x ((Ps0(x) ∨ Ps1(x)) ↔ Ps(x))

• ∀x ¬(Ps0(x) ∧ Ps1(x))

TTree is complete and has quantifier elimination. There are no complete formulasand no prime model.

Definition 14.8. A family of formulas ϕs(x), s ∈ <ω2, is a binary tree if forall s ∈ <ω2 the following holds:

a) T ` ∀x((ϕs0(x) ∨ ϕs1(x)) → ϕs(x)

)b) T ` ∀x ¬

(ϕs0(x) ∧ ϕs1(x)

)Theorem 14.9. Let T be a complete theory.

1. If T is small, it has no binary tree of consistent formulas. If T is count-able, the converse holds as well.

2. If T has no binary tree of consistent formulas, the isolated types are dense.

Proof. 1. Let(ϕs(x1, . . . , xn)

)be a binary tree of consistent formulas. Then,

for all η ∈ ω2, the set {ϕs(x)

∣∣ s ⊂ η}

is consistent and therefore is contained in some type pη(x) ∈ Sn(T ). The pη(x)are all different showing that T is not small. We leave the converse as Exer-cise 14.1.

2. If the isolated types are not dense, there is a consistent ϕ(x1, . . . , xn) whichdoes not contain a complete formula. Call such a formula perfect. Since perfectformulas are not complete, they can be decomposed into disjoint6 consistentformulas, which again have to be perfect. This allows us to construct a binarytree of perfect formulas.

Exercise 14.1.Countable theories without a binary tree of consistent formulas are small.

Exercise 14.2.Show that the isolated types being dense is equivalent to the isolated typesbeing (topologically) dense in the Stone space Sn(T ).

Exercise 14.3.Let T be the theory of (R, <,Q) where Q is a predicate for the rational numbers.Does T have a prime model?

6We call two formulas disjoint if their conjunction is not consistent with T .

Chapter 5

ℵ1–categorical theories

We have already seen examples of ℵ0–categorical theories (e.g. the theory ofdense linear orderings without endpoints) and of theories categorical in all in-finite κ (e.g. the theory of infinite dimensional vector spaces over finite fields)and all uncountably infinite κ (e.g. the theory of algebraically closed fields offixed characteristic), respectively.

The aim of this chapter is to understand the structure of ℵ1-categorical the-ories and to prove, in Corollary 22.2, Morley’s theorem that a countable theorycategorical in some uncountable cardinality is categorical in all uncountablecardinalities (but not necessarily countably categorical).

As in the case of ℵ0-categorical theories, we will see that the number ofcomplete types in an ℵ1-categorical theory is rather small (the theory is ω-stable) albeit not always finite. We will define a geometry associated to astrongly minimal set whose dimension determines the isomorphism type of amodel of such a theory. This then implies Morley’s theorem.

15 Indiscernibles

In this section we begin with a few facts about ‘indiscernible’ elements. We willsee that structures generated by them realise only few types.

Definition 15.1. Let I be a linear order and A an L–structure. A family (ai)i∈Iof elements1 of A is called a sequence of indiscernibles if for all L–formulasϕ(x1, . . . , xn) and all i1 < . . . < in and j1 < . . . < jn from I

A |= ϕ(ai1 , . . . , ain) ↔ ϕ(aj1 , . . . , ajn).

1or, more generally, of tuples of elements, all of the same length

76

CHAPTER 5. ℵ1–CATEGORICAL THEORIES 77

If two of the ai are equal, all ai are the same. Therefore it is often assumedthat the ai are distinct.

Sometimes sequences of indiscernibles are also called order indiscernible todistinguish them from totally indiscernible sequences in which the ordering of theindex set does not matter. However, in stable theories, these notions coincide.So if nothing else is said, indiscernible elements will always be order indiscerniblein the sense just defined.

Definition 15.2. Let I be an infinite linear order and I = (ai)i∈I a sequenceof k-tuples in M, A ⊆ M . The Ehrenfeucht-Mostowski-type EM(I/A) of Iover A is the set of L(A)–formulas ϕ(x1, . . . , xn) with M |= ϕ(ai1 , . . . , ain) forall i1 < . . . < in ∈ I, n < ω.

Lemma 15.3 (The Standard Lemma). Let I and J be two infinite linear ordersand I = (ai)i∈I a sequence of elements of a structure M. Then there is a struc-ture N ≡ M with an indiscernible sequence (bj)j∈J realizing the Ehrenfeucht-Mostowski-type EM(I) of I.

Corollary 15.4. Assume that T has an infinite model. Then, for any linearorder I, T has a model with a sequence (ai)i∈I of distinct indiscernibles.

For the proof of the Standard Lemma we need Ramsey’s Theorem. Let [A]ndenote the set of all n–element subsets of A.

Theorem 15.5 (Ramsey). Let A be infinite and n ∈ ω. Partition the set ofn-element subsets [A]n into subsets C1, . . . Ck. Then there is an infinite subsetof A whose n–element subsets all belong to the same subset Ci.

Proof. Thinking of the partition as a colouring on [A]n, we are looking for aninfinite subset B of A such that [B]n is monochromatic. We prove the theoremby induction on n. For n = 1, the statement is evident from the pigeonholeprinciple. Assuming the theorem is true for n, we now prove it for n + 1.Let a0 ∈ A. Then any colouring of [A]n+1 induces a colouring of the n-elementsubsets of A′ = A\{a0}: just colour x ∈ [A′]n by the colour of {a0}∪x ∈ [A]n+1.By the induction hypothesis, there exists an infinite monochromatic subset B1 ofA′ in the induced colouring. Thus, all the (n+1)-element subsets of A consistingof a0 and n elements of B1 have the same colour. Now pick any a1 ∈ B1. By thesame argument we obtain an infinite subset B2 of B1 with the same properties.Inductively, we thus construct an infinite sequence A = B0 ⊃ B1 ⊃ B2 ⊃ . . .,and elements ai ∈ Bi \Bi+1 such that the colour of each (n+1)-element subset{ai(0), ai(2), . . . , ai(n)} with i(0) < i(2) < . . . < i(n) depends only on the valueof i(0). Again by the pigeonhole principle there are infinitely many values ofi(0) for which this colour will be the same. These ai(0)’s then yield the desiredmonochromatic set.


Proof of 15.3. Choose a set C of new constants with an ordering isomorphic toJ . Consider the theories

T ′ = {ϕ(c) | ϕ(x) ∈ EM(I)}(1)

T ′′ = {ϕ(c) ↔ ϕ(d) | c, d ∈ C}(2)

Here the ϕ(x) are L–formulas and c, d tuples in increasing order. We have toshow that T ∪ T ′ ∪ T ′′ is consistent. It is enough to show that

TC0,∆ = T ∪ {ϕ(c) ∈ T ′ | c ∈ C0} ∪{ϕ(c) ↔ ϕ(d)

∣∣ ϕ(x) ∈ ∆, c, d ∈ C0

}is consistent for finite sets C0 and ∆. We can assume that the elements of ∆are formulas with free variables x1, . . . , xn and that all tuples c and d have thesame length n.

For notational simplicity we assume that all ai are different. So we mayconsider A = {ai |∈ I} as an ordered set. We define an equivalence relation on[A]n by

a ∼ b ⇔ M |= ϕ(a) ↔ ϕ(b) for all ϕ(x1, . . . , xn) ∈ ∆

where a, b are tuples in increasing order. Since this equivalence relation has atmost 2|∆| many classes, by Ramsey’s Theorem there is an infinite subset B ofA with all n–element subsets in the same equivalence class. We interpret theconstants c ∈ C0 by elements bc in B ordered in the same way as the c. Then(M, bc)c∈C0 is a model of TC0,∆.

Lemma 15.6. Assume L is countable. If the L–structure M is generated bya well–ordered sequence (ai) of indiscernibles, then M realises only countablymany types over every countable subset of M .

Proof. If A = {ai | i ∈ I}, then every element b of M has the form b = t(a),where t is an L–term and a is a tuple from A.Consider a countable subset S of M . Write

S = {tMn (an) | n ∈ ω}.

Let A0 = {ai | i ∈ I0} be the (countable) set of elements of A which occur inthe an. Then every type tp(b/S) is determined by tp(b/A0) since every L(S)–formula

ϕ(x, tMn1

(an1), . . .)

can be replaced by the L(A0)–formula ϕ(x, tn1(an1), . . .).

Now the type of b = t(a) over A0 depends only on t(x) (countably many pos-sibilities) and the type tp(a/A0). Write a = ai for a tuple i from I. Since theai are indiscernible, the type depends only on the quantifier free type tpqf (i/I0)in the structure (I,<). This type again depends on tpqf (i) (finitely many pos-sibilities) and on the types p(x) = tpqf(i/I0) of the elements i of i. There arethree kinds of such types:


1. i is bigger than all elements of I0

2. i is an element i0 of I0

3. For some i0 ∈ I0, i is smaller than i0 but bigger than all elements of{j ∈ I0 | j < i0}.

There is only one type in the first case, in the other cases the type is determinedby i0. This results in countably many possibilities for each component of i.

Definition 15.7. Let L be a language. A Skolem theory TSk(L) is a theory ina bigger language LSk with the following properties:

a) TSk(L) has quantifier elimination.

b) TSk(L) is universal.

c) Every L–structure can be expanded to a model of TSk(L).

d) |LSk| ≤ max{|L|,ℵ0}

Theorem 15.8. Every L has a Skolem theory.

Proof. We define an ascending sequence of languages

L = L0 ⊂ L1 ⊂ L2 ⊂ . . . ,

by introducing for every quantifier free Li–formula ϕ(x1, . . . , xn, y) a new n–place Skolem function2 fϕ and defining Li+1 as the union of Li and the set ofthese function symbols. LSk is the union of all Li. We set

TSk ={∀x(∃y ϕ(x, y) → ϕ(x, fϕ(x))

) ∣∣∣ ϕ(x, y) q.f. LSk–formula}.

Corollary 15.9. Let T be a countable theory with an infinite model and let κbe an infinite cardinal. Then T has a model of cardinality κ which realises onlycountably many types over every countable subset.

Proof. Consider the theory T ∗ = T ∪ TSk(L). T ∗ is countable, has an infinitemodel and quantifier elimination.

Claim: T ∗ is equivalent to a universal theory.

Proof: Modulo TSk(L) every axiom ϕ of T is equivalent to a quantifier freeLSk–sentence ϕ∗. Therefore T ∗ is equivalent to the universal theory {ϕ∗ | ϕ ∈T} ∪ TSk(L).

2If n = 0, fϕ is a constant.


Let I be a well-ordering of cardinality κ and N∗ a model of T ∗ with indis-cernibles (ai)i∈I . The claim implies that the substructure M∗ generated by theai is a model of T ∗. M∗ has cardinality κ. Since T ∗ has quantifier elimina-tion, M∗ is an elementary substructure of N∗ and (ai) is indiscernible in M∗.By Lemma 15.6, there are only countably many types over every countable setrealised in M∗. The same is then true for the reduct M = M∗ � L.

Exercise 15.1.A sequence of elements in (Q, <) is indiscernible if and only if it is either con-stant, strictly increasing or strictly decreasing.

Exercise 15.2.Prove Ramsey’s Theorem 15.5 by induction on n similarly to the proof of C 10.2using a non-principal ultrafilter on A. (For ultrafilters see Exercise 2.4.)


16 ω–stable Theories

In this section we fix a complete theory T with infinite models. In the previ-ous section we saw that we may add indiscernible elements to a model withoutchanging the number of realised types. We will now use this to show that ℵ1-categorical theories have a small number of types, i.e. they are ω-stable. Con-versely, with few types it is easier to be saturated and since saturated structuresare unique we find the connection to categorical theories.

Definition 16.1. Let κ be an infinite cardinal. T is κ–stable if in each modelof T , over every set of parameters of size at most κ, there are at most κ many1-types, i.e.

|A| ≤ κ ⇒ |S(A)| ≤ κ.

Example 16.2 (Algebraically closed fields). The theories TACFp for p a primeor 0 are κ–stable for all κ.

Note that by Theorem 16.5 below it would suffice to prove that the theoriesTACFp are ω–stable. The converse holds as well: any infinite ω-stable field is infact algebraically closed (see [34]).

Proof. Let K be a subfield of an algebraically closed field. By quantifier elimi-nation the type of an element a over K is determined by the isomorphism typeof the extension K[a]/K. If a is transcendental over K, K[a] is isomorphic tothe polynomial ring K[X]. If a is algebraic with minimal polynomial f ∈ K[X],K[a] is isomorphic to K[X]/(f). So there is one more 1–type over K than thereare irreducible polynomials.

Exercise 16.5 shows that in κ–stable theories there are only κ many n–typesover a set of cardinality κ. This has a nice algebraic reason in algebraically closedfields: the isomorphism type of K[a1, . . . , an]/K is determined by the vanishingideal P of a1, . . . , an (see Lemma B6.6). By Hilbert’s Basis Theorem, P isfinitely generated. So, ifK has cardinality κ, the polynomial ringK[X1, . . . , Xn]has only κ many ideals.

Theorem 16.3. A countable theory T which is categorical in an uncountablecardinal κ is ω–stable 3.

Proof. Let N be a model and A ⊂ N countable with S(A) uncountable. Let(bi)i∈I be a sequence of ℵ1 many elements with pairwise distinct types overA. (Note that we can assume that all types over A are realised in N.) Wechoose first an elementary substructure M0 of cardinality ℵ1 which contains Aand all bi. Then we choose an elementary extension M of M0 of cardinality κ.M is a model of cardinality κ which realises uncountably many types over the

3ω–stable and ℵ0–stable are synonymous.


countable set A. By Corollary 15.9, T has another model in which this is notthe case. So T cannot be κ–categorical.

Definition 16.4. T is totally transcendental if T has no model M with a binarytree of consistent L(M)–formulas.

Theorem 16.5.

1. ω–stable theories are totally transcendental.

2. Totally transcendental theories are κ–stable for all κ ≥ |T |.

It follows that a countable theory T is ω–stable if and only if it is totallytranscendental.

Proof. 1. Let M be a model with a binary tree of consistent L(M)–formulas.The set A of parameters which occur in the tree’s formulas is countable butS(A) has cardinality 2ℵ0 . (cf. Theorem 14.9).

2. Assume that there are more than κ many types over some set A of cardinal-ity κ. Let us call an L(A)–formula big if it belongs to more than κ many typesover A and thin otherwise.

By assumption x .= x is big. If we can show that each big formula decomposes

into two big formulas, we can construct a binary tree of big formulas, whichfinishes the proof.So assume that ϕ is big. Since each thin formula belongs to at most κ manytypes and since there are at most κ many formulas, there are at most κ manytypes which contain thin formulas. Therefore ϕ belongs to two distinct typesp and q which contain only big formulas. If we separate p and q by ψ ∈ p and¬ψ ∈ q, we decompose ϕ into the big formulas ϕ ∧ ψ and ϕ ∧ ¬ψ.

The following definition generalises the notion of ω–saturation.

Definition 16.6. Let κ be an infinite cardinal. An L–structure A is κ–saturatedif in A all types over sets of cardinality less than κ are realised. An infinitestructure A is saturated if it is |A|–saturated.

Even though we ask for saturation only that 1-types are realised, as in theω-saturated case this easily implies that all n-types are realised as well (cp.Exercise 12.9).

Lemma 12.3 generalises to sets

Lemma 16.7. Elementarily equivalent saturated structures of the same cardi-nality are isomorphic.


Proof. Let A and B be elementarily equivalent saturated structures both ofcardinality κ. We choose enumerations (aα)α<κ and (bα)α<κ of A and B andconstruct an increasing sequence of elementary maps fα : Aα → Bα. Assumethat the fβ are constructed for all β < α. The union of the fβ is an elementarymap f∗α : A∗

α → B∗α. The construction will imply that A∗

α and B∗α have at most

cardinality |α|, which is smaller than κ.We write α = λ+ n (as in p. 214) and distinguish two cases:

n = 2i:In this case we consider p(x) = tp(aλ+i/A

∗α). Realise f∗α(p) by b ∈ B and define

fα = f∗α ∪ {〈aλ+i, b〉}.

n = 2i+ 1:Similarly. We find an extension

fα = f∗α ∪ {〈a, bλ+i〉}.

Then⋃α<κ fα is the desired isomorphism between A and B.

Lemma 16.8. If T is κ–stable, then for all regular λ ≤ κ there is a model ofcardinality κ which is λ–saturated.

Proof. We construct a continuous elementary chain

M0 ≺ M1 . . . ≺ Mα ≺ . . . (α < λ),

of models of T with cardinality κ such that all p ∈ S(Mα) are realised in Mα+1.This is possible since by κ–stability S(Mα) has cardinality κ. Finally let M bethe union of this chain. Then M is λ–saturated. In fact, if |A| < λ and if a ∈ Ais contained in Mα(a) then Λ =

⋃a∈A α(a) is an initial segment of λ of smaller

cardinality than λ. So Λ has an upper bound µ < λ. It follows that A ⊂ Mµ

and all types over A are realised in Mµ+1.

Remark 16.9. If T is κ-stable for a regular cardinal κ, the previous lemmayields a saturated model of cardinality κ. Harnik [18] showed that this holds infact for arbitrary κ. See also Corollary 23.3 for more general constructions.

Theorem 16.10. A countable theory T is κ–categorical if and only if all modelsof cardinality κ are saturated.

Proof. If all models of cardinality κ are saturated, it follows from Lemma 16.7that T is κ–categorical.

Assume, for the converse, that T is κ–categorical. For κ = ℵ0 the theoremfollows from (the proof of) Theorem 12.1. So we may assume that κ is uncount-able. Then T is totally transcendental by Theorems 16.3 and 16.5 and thereforeκ–stable by Theorem 16.5.


By Lemma 16.8, all models of T of cardinality κ are µ+–saturated for all µ < κ.i.e. κ–saturated.

Exercise 16.1.Use Exercise 33.8 to show that a theory with an infinite definable linear ordering(like TDLO and TRCF) cannot be κ-stable for any κ.

Exercise 16.2.Show that the theory of an equivalence relation with two infinite classes hasquantifier elimination and is ω–stable. Is it ℵ1–categorical?

Exercise 16.3.If A is κ-saturated, then all definable subsets are either finite or have cardinalityat least κ.

Exercise 16.4.An L–theory T is totally transcendental if and only if T � L0 is ω-stable for allat most countable L0 ⊂ L.

Exercise 16.5.If T is κ–stable, then in each model of T , over every set of parameters with atmost κ many elements, there are at most κ many n–types.


17 Prime extensions

As with prime models, prime extensions are the smallest ones in the sense ofelementary embeddings. We will see here (and in Sections 39 and 40) thatprime extensions, if they exist, share a number of important properties withprime models.

Definition 17.1. Let M be a model of T and A ⊂M .

1. M is a prime extension of A (or: prime over A) if every elementary mapA→ N extends to an elementary map M → N.

A

M N

6

��

��

-

idA

2. B ⊆M is constructible over A if B has an enumeration

B = {bα | α < λ},

where each bα is atomic over A ∪Bα, with Bα = {bµ | µ < α}.

So M is a prime extension of A if and only if MA is a prime model ofTh(MA).

Notice the following:

Lemma 17.2. If a model M is constructible over A, then M is prime over A.

Proof. Let (mα)α<λ an enumeration of M , such that each mα is atomic overA ∪ Mα. Let f : A → N be an elementary map. We define inductively anincreasing sequence of elementary maps fα : A ∪ Mα+1 → N with f0 = f .Assume that fβ is defined for all β < α. The union of these fβ is an elementarymap f ′α : A∪Mα → N. Since p(x) = tp(aα/A∪Mα) is isolated, f ′α(p) ∈ S(f ′α(A∪Mα)) is also isolated and has a realisation b in N. We set fα = f ′α ∪ {〈aα, b〉}.Finally, the union of all fα (α < λ) is an elementary embedding M → N.

We will see below that in totally transcendental theories prime extensionsare atomic.

Theorem 17.3. If T is totally transcendental, every subset of a model of Thas a constructible prime extension.

We will see in Section 40 that in totally transcendental theories prime extensionsare unique up to isomorphism (cf. Theorem 14.3).

For the proof we need the following lemma which generalises Theorem 14.7:


Lemma 17.4. If T is totally transcendental, the isolated types are dense overevery subset of any model.

Proof. Consider a subset A of a model M. Then Th(MA) has no binary tree ofconsistent formulas (in one variable). By Theorem 14.9, the isolated 1–types inTh(MA) are dense.

We can now prove Theorem 17.3:

Proof. By Lemma 17.2 it suffices to construct an elementary substructure M0 ≺M which contains A and is contructible over A. An application of Zorn’s Lemmagives us a maximal construction (aα)α<λ, which cannot be prolonged by anelement aλ ∈ M \ Aλ. Clearly A is contained in Aλ. We show that Aλ is theuniverse of an elementary substructure M0 using Tarski’s Test. So assume thatϕ(x) is an L(Aλ)–formula and M |= ∃xϕ(x). Since isolated types over Aλ aredense by Lemma 17.4, there is an isolated p(x) ∈ S(Aλ) containing ϕ(x). Let bbe a realisation of p(x) in M. We can prolong our construction by aλ = b, thusb ∈ Aλ by maximality and ϕ(x) is realised in Aλ.

To prove that in totally transcendental theories prime extensions are atomic,we need the following:

Lemma 17.5. Let a and b be two finite tuples of elements of a structure M.Then tp(ab) is atomic if and only if tp(a/b) and tp(b) are atomic.

Proof. First assume that ϕ(x, y) isolates tp(a, b). As in the proof of Theo-rem 14.2, ϕ(x, b) isolates tp(a/b) and we claim that ∃xϕ(x, y) isolates p(y) =tp(b): we have ∃xϕ(x, y) ∈ p(y) and if σ(y) ∈ p(y), then M |= ∀x, y (ϕ(x, y) →σ(y)). This means that M |= ∀y (∃xϕ(x, y) → σ(y)).

Now, conversely, assume that ρ(x, b) isolates tp(a/b) and that σ(y) isolatesp(y) = tp(b). Then ρ(x, y)∧σ(y) isolates tp(a, b). For, clearly, we have ρ(x, y)∧σ(y) ∈ tp(a, b). If, on the other hand, ϕ(x, y) ∈ tp(a, b), then ϕ(x, b) belongs totp(a/b) and

M |= ∀x (ρ(x, b) → ϕ(x, b)).

Hence∀x (ρ(x, y) → ϕ(x, y)) ∈ p(y)

and it follows that

M |= ∀y (σ(y) → ∀x (ρ(x, y) → ϕ(x, y))).

Thus M |= ∀x, y (ρ(x, y) ∧ σ(y) → ϕ(x, y)).

Corollary 17.6. Constructible extensions are atomic.


Proof. Let M0 be a constructible extension of A and let a be a tuple fromM0. We have to show that a is atomic over A. We can clearly assume thatthe elements of a are pairwise distinct and do not belong to A. We can alsopermute the elements of a so that

a = aαb

for some tuple b ∈ Aα. Let ϕ(x, c) be an L(Aα)–formula which is complete overAα and satisfied by aα. Then aα is also atomic over A∪ {bc}. Using induction,we know that bc is atomic over A. By Lemma 17.5 applied to (M0)A, aαbc isatomic over A, which implies that a = aαb is atomic over A.

Corollary 17.7. If T is totally transcendental, prime extensions are atomic.

Proof. Let M be a model of T and A ⊂M . Since A has at least one constructibleextension M0 and since all prime extensions of A are contained in M0

4, all primeextensions are atomic.

A structure M is called a minimal extension of the subset A if M has noproper elementary substructure which contains A.

Lemma 17.8. Let M be a model of T and A ⊂ M . If A has a prime ex-tension and a minimal extension, they are isomorphic over A, i.e., there is anisomorphism fixing A elementwise.

Proof. A prime extension embeds elementarily in the minimal extension. Thisembedding must be surjective by minimality.

Exercise 17.1.For every (countable) theory T the following are equivalent (cf. Theorem 14.7):

a) Every parameter set has a prime extension. (We say: T has prime exten-sions.)

b) Over every parameter set A the isolated types are dense in the type spaceS1(A).

c) Over every countable parameter set A the isolated types are dense in S1(A).

Exercise 17.2.Lemma 17.5 follows from Exercise 11.5 and the following observation: Letπ : X → Y be a continuous open map between topological spaces. Then apoint x ∈ X is isolated if and only if π(x) is isolated in Y and x is isolated inπ−1(π(x)).

4More precisely, they are isomorphic over A to elementary substructures of M0.


18 Lachlan’s theorem

Using the fact (established in Section 16) that uncountably categorical theoriesare totally transcendental, we will prove the downward direction of Morley’stheorem. We use Lachlan’s result that in totally transcendental theories modelshave arbitrary large elementary extensions realizing few types:

Theorem 18.1 (Lachlan). ([1] Lemma 10) Let T be totally transcendental andM an uncountable model of T . Then M has arbitrarily large elementary ex-tensions which omit every countable set of L(M)–formulas which is omitted inM.

Proof. For the proof, we call an L(M)–formula ϕ(x) large if its realisation setϕ(M) is uncountable. Since there is no infinite binary tree of large formulas,there exists a minimal large formula ϕ0(x). This means that for every L(M)–formula ψ(x) either ϕ0(x) ∧ ψ(x) or ϕ0(x) ∧ ¬ψ(x) is at most countable. Nowit is easy to see that

p(x) = {ψ(x) | ϕ0(x) ∧ ψ(x) large}

is a type in S(M).

Clearly p(x) contains no formula of the form x.= a for a ∈ M , so p(x) is

not realised in M . On the other hand, every countable subset Π(x) of p(x) isrealised in M: since ϕ0(M) \ψ(M) is at most countable for every ψ(x) ∈ Π(x),the elements of ϕ0(M) which do not belong to the union of these sets realiseΠ(x).

Let a be a realisation of p(x) in a (proper) elementary extension N. ByTheorem 17.3 we can assume that N is atomic over M ∪ {a}.

Fix b ∈ N . We have to show that every countable subset Σ(y) of tp(b/M)is realised in M .

Let χ(x, y) be an L(M)–formula such that χ(a, y) isolates q(y) = tp(b/M ∪{a}). If b realises an L(M)–formula σ(y), we have N |= ∀y (χ(a, y) → σ(y)).Hence the formula

σ∗(x) = ∀y (χ(x, y) → σ(y))

belongs to p(x). Note that ∃y χ(x, y) belongs also to p(x).

Choose an element a′ ∈M which satisfies

{σ∗(x) | σ ∈ Σ} ∪ {∃y χ(x, y)}

and choose b′ ∈ M with M |= χ(a′, b′). Since σ∗(a′) is true in M, σ(b′) is truein M. So b′ realises Σ(y).

We have shown that M has a proper elementary extension which realises nonew countable set of L(M)–formulas. By iteration we obtain arbitrarily longchains of elementary extensions with the same property.


The corollary is the downwards part of Morley’s Theorem, p. 76:

Corollary 18.2. A countable theory which is κ–categorical for some uncount-able κ, is ℵ1–categorical.

Proof. Let T be κ–categorical and assume that T is not ℵ1–categorical. Then Thas a model M of cardinality ℵ1 which is not saturated. So there is a type p overa countable subset of M which is not realised in M. By Theorems 16.3 and 16.5T is totally transcendental. Theorem 18.1 gives an elementary extension N ofM of cardinality κ which omits all countable sets of formulas which are omittedin M. Thus also p is omitted. Since N is not saturated, T is not κ–categorical,a contradiction.

Exercise 18.1 (Afshordel).Prove in a similar way: If a countable theory T is κ–categorical for some un-countable κ, it is λ–categorical for every uncountable λ ≤ κ.


19 Vaughtian Pairs

A crucial fact about uncountably categorical theories is the absence of definablesets whose size is independent of the size of the model in which they live (cap-tured in the notion of a Vaughtian pair). In fact, in an uncountably categoricaltheory each model is prime over any infinite definable subset. This will allow usin Section 21 to attach a dimension to the models of an uncountably categori-cal theory. In this section let T be a countable complete theory withinfinite models.

Definition 19.1. T has a Vaughtian pair if there are two models M ≺ N andan L(M)–formula ϕ(x) such that

a) M 6= N,

b) ϕ(M) is infinite,

c) ϕ(M) = ϕ(N)

If ϕ(x) does not contain parameters, we say that T has a Vaughtian pair forϕ(x).

Remark. Notice that T does not have a Vaughtian pair if and only if everymodel M is a minimal extension of ϕ(M) ∪ A for any formula ϕ(x) with para-meters in A ⊂M which defines an infinite set in M.

Let N be a model of T where ϕ(N) is infinite but has smaller cardinalitythan N. The Löwenheim–Skolem Theorem yields an elementary substructure Mof N which contains ϕ(N) and has the same cardinality as ϕ(N). Then M ≺ Nis a Vaughtian pair for ϕ(x). The next theorem shows that a converse of thisobservation is also true.

Theorem 19.2 (Vaught’s Two-cardinal Theorem). If T has a Vaughtian pair,it has a model M of cardinality ℵ1 with ϕ(M) countable for some formulaϕ(x) ∈ L(M).

For the proof of Theorem 19.2 we need the following:

Lemma 19.3. Let T be complete, countable, and with infinite models.

1. Every countable model of T has a countable ω–homogeneous elementaryextension.

2. The union of an elementary chain of ω–homogeneous models is ω–homo-geneous.

3. Two ω–homogeneous countable models of T realizing the same n–types forall n < ω are isomorphic.


Proof. 1. Let M0 be a countable model of T . We realise the countably manytypes

{f(tp(a/A)) | a,A ⊂M0, A finite, f : A→M0 elementary}

in a countable elementary extension M1. By iterating this process we obtain anelementary chain

M0 ≺ M1 ≺ . . . ,

whose union is ω–homogeneous.

2. Clear.

3. Suppose A and B are ω–homogeneous, countable and realise the same n-types. We show that we can extend any finite elementary map f : {a1, . . . ai} →{b1, . . . bi}; aj 7→ bj to any a ∈ A\Ai. Realise the type tp(a1, . . . , ai, a) by sometuple b′ = b′1, . . . , b

′i+1 in B. Using the ω-homogeneity of B we may extend the

finite partial isomorphism g = {(b′j , bj) : 1 ≤ j ≤ i} by (b′i+1, b) for some b ∈ B.Then fi+1 = fi∪{(a, b)} is the required extension. Reversing the roles of B andA we construct the desired isomorphism.

Proof. (of Theorem 19.2) Suppose that the Vaughtian pair is witnessed (incertain models) by some formula ϕ(x). For simplicity we assume that ϕ(x) doesnot contain parameters (see Exercise 19.4). Let P be a new unary predicate. Itis easy to find an L(P )–theory TVP whose models (N,M) consist of a model Nof T and a subset M defined by the new predicate P which is the universe ofan elementary substructure M which together with N forms a Vaughtian pairfor ϕ(x). The Löwenheim Skolem Theorem applied to TVP yields a Vaughtianpair M0 ≺ N0 for ϕ(x) with M0,N0 countable.

We first construct an elementary chain

(N0,M0) ≺ (N1,M1) ≺ . . .

of countable Vaughtian pairs, with the aim that both components of the unionpair

(N,M)

are ω–homogeneous and realise the same n–types. If (Ni,Mi) is given, we firstchoose a countable elementary extension (N′,M ′) such that M′ realises all n–types which are realised in Ni. Then we choose as in the proof of Lemma 19.3(1) a countable elementary extension (Ni+1,Mi+1) of (N′,M ′) for which Ni+1

and Mi+1 are ω–homogeneous.

It follows from Lemma 19.3 (3) that M and N are isomorphic.

Next we construct a continuous elementary chain

M0 ≺ M1 ≺ . . . ≺ Mα ≺ . . . (α < ω1)

with (Mα+1,Mα) ∼= (N,M) for all α. We start with M0 = M. If Mα is con-structed, we choose an isomorphism M → Mα and extend it to an isomorphism


N → Mα+1 (see Lemma 1.8). For a countable limit ordinal λ, Mλ is the unionof the Mα (α < λ). Mλ is isomorphic to M by Lemma 19.3.2 and 19.3.3.

Finally we setM =

⋃α<ω1

Mα.

Since the Mα are growing, M has cardinality ℵ1 while ϕ(M) = ϕ(Mα) = ϕ(M0)is countable.

Corollary 19.4. If T is categorical in an uncountable cardinality, it does nothave a Vaughtian pair.

Proof. If T has a Vaughtian pair, then by Theorem 19.2 it has a model M ofcardinality ℵ1 such that for some ϕ(x) ∈ L(M) the set ϕ(M) is countable. Onthe other hand, if T is categorical in an uncountable cardinal, it is ℵ1-categoricalby Corollary 18.2 and by Theorem 16.10, all models of T of cardinality ℵ1 aresaturated. In particular, each formula is either satisfied by a finite number orby ℵ1 many elements, a contradiction.

Corollary 19.5. Let T be categorical in an uncountable cardinal, M a model,ϕ(M) infinite and definable over A ⊂ M . Then M is – the unique – primeextension of A ∪ ϕ(M).

Proof. By Corollary 19.4, T does not have a Vaughtian pair, so M is minimalover A ∪ ϕ(M). If N is a prime extension over A ∪ ϕ(M), which exists byTheorem 17.3, N is isomorphic to M over A ∪ ϕ(M) by Lemma 17.8.

Definition 19.6. We say that T eliminates the quantifier ∃∞x, there are in-finitely many x, if for every L–formula ϕ(x, y) there is a finite bound nϕ suchthat in all models M of T and for all parameters a ∈M ,

ϕ(M, a)

is either infinite or has or at most nϕ elements

Remark. This means that for all ϕ(x, y) there is a ψ(y) such that in all modelsM of T and for all a ∈M

M |= ∃∞x ϕ(x, a) ⇐⇒ M |= ψ(a).

We denote this byT ` ∀y

(∃∞x ϕ(x, y) ↔ ψ(y)

).

If nϕ exists, we can use ψ(y) = ∃>nϕx ϕ(x, y) (there are more than nϕ manyx such that ϕ(x, y)). If, conversely, ψ(y) is a formula which is implied by∃∞x ϕ(x, y), a compactness argument shows that there must be a bound nϕsuch that

T ` ∃>nϕxϕ(x, y) → ψ(y).


Lemma 19.7. A theory T without Vaughtian pair eliminates the quantifier∃∞x.

Proof. Let P be a new unary predicate and c1, . . . , cn new constants. Let T ∗

be the theory of all L ∪ {P, c1, . . . , cn}–structures

(M, N, a1, . . . , an),

where M is a model of T , N the universe of a proper elementary substructure,a1, . . . , an elements of N and ϕ(M, a) ⊂ N . Suppose that the bound nϕ doesnot exist. Then, for any n, there is a model N of T and a ∈ N such that ϕ(N, a)is finite, but has more than n elements. Let M be a proper elementary extensionof N. Then ϕ(M, a) = ϕ(N, a) and the pair (M, N, a) is a model of T ∗. Thisshows that the theory

T ∗ ∪ {∃>nxϕ(x, c) | n = 1, 2, . . .}

is finitely satisfiable. A model of this theory gives a Vaughtian pair for T .

Exercise 19.1.If T is totally transcendental and has a Vaughtian pair for ϕ(x), T has, for alluncountable κ, a model of cardinality κ with countable ϕ(M). Prove Corol-lary 19.4 from this. (Use Theorem 18.1.)

Exercise 19.2.Show directly (without using Lemma 16.8) that a theory T which is categoricalin some uncountable cardinality, has a model M of cardinality ℵ1 in which eachL(M)-formula is either satisfied by a finite number or by ℵ1 many elements.

Exercise 19.3.Show that the theory TRG of the random graph has a Vaughtian pair.

Exercise 19.4.Let T be a theory, M a model of T and a ⊂M a finite tuple of parameters. Letq(x) be the type of a in M. Then for new constants c, the L(c)-theory

T (q) = Th(M, a) = T ∪ {ϕ(c)| ϕ(x) ∈ q(x)}

is complete. Show that T is λ-stable (or without Vaughtian pair etc.) if andonly if T (q) is. For countable languages, this implies that T is categorical insome uncountable cardinal if and only if T (q) is.

Exercise 19.5.Assume that T eliminates the quantifier ∃∞. Then for every formula ϕ(x1, . . . , xn, y)there is a formula θ(y) such that in all models M of T a tuple b satisfies θ if andonly if M has an elementary extension M′ with elements a1, . . . , an ∈ M ′ \Msuch that M′ |= ϕ(a1. . . . , an, b).

Exercise 19.6.Let T1 and T2 be two model complete theories in disjoint languages L1 and L2.Assume that both theories eliminate ∃∞. Then T1∪T2 has a model companion.


20 Algebraic formulas

Formulas defining a finite set are called algebraic. In this section we collect afew facts and a bit of terminology around this concept which will be crucial inthe following sections.

Definition 20.1. Let M be a structure and A a subset of M . A formulaϕ(x) ∈ L(A) is called algebraic if ϕ(M) is finite. An element a ∈M is algebraicover A if there is a finite A–definable set which contains a. We call an elementalgebraic if it is algebraic over the empty set. The algebraic closure of A, acl(A),is the set of all elements of M algebraic over A, and A is called algebraicallyclosed if it equals its algebraic closure.

Remark. Note that the algebraic closure of A does not grow in elementaryextensions of M because an L(A)–formula which defines a finite set in M definesthe same set in every elementary extension. As a special case we have thatelementary substructures are algebraically closed.

It is easy to see that

(3) | acl(A)| ≤ max{|T |, |A|}

(cf. Theorem 6.1).

In algebraically closed fields, an element a is algebraic over A precisely if ais algebraic (in the field theoretical sense) over the field generated by A. Thisfollows easily from quantifier elimination in TACF .

We call a type p(x) ∈ S(A) algebraic if (and only if) p contains an algebraicformula. Any algebraic type p is isolated by an algebraic formula ϕ(x) ∈ L(A),namely by any ϕ ∈ p having the minimal number of solutions in M. Thisnumber is called the degree deg(p) (or multiplicity) of p. As isolated types arerealised in every model, the algebraic types over A are exactly those of the formtp(a/A) where a is algebraic over A. The degree of a over A deg(a/A) is thedegree of tp(a/A).

Lemma 20.2. Let p ∈ S(A) be non–algebraic and A ⊂ B. Then p has anon–algebraic extension q ∈ S(B).Proof.

q0(x) = p(x) ∪ {¬ψ(x) | ψ(x) algebraic L(B)–formula}

is finitely satisfiable. For otherwise there are ϕ(x) ∈ p(x) and algebraic L(B)–formulas ψ1(x) . . . ψn(x) with

M |= ∀x (ϕ(x) → ψ1(x) ∨ . . . ∨ ψn(x)).

But then ϕ(x) and hence p(x) is algebraic. So we can take for q any typecontaining q0.


Remark 20.3. Since algebraic types are isolated by algebraic formulas, an easycompactness argument shows that a type p ∈ S(A) is algebraic if and only ifp has only finitely many realisations (namely deg(p) many) in all elementaryextensions of M.

Lemma 20.4. Let M and N be two structures and f : A → B an elementarybijection between two subsets. Then f extends to an elementary bijection betweenacl(A) and acl(B).

Proof. Let g : A′ → B′ a maximal extension of f to two subsets of acl(A) andacl(B). Let a be an element of acl(A). Since a is algebraic over A′, a is atomicover A′. We can therefore realise the type g(tp(a/A′)) in N – by an elementb ∈ acl(B) – and obtain an extension g ∪ {〈a, b〉} of g. It follows a ∈ A′. Sog is defined on the whole of acl(A). Interchanging A and B shows that g issurjective. (See Lemma 23.9 for an alternative proof.)

The algebraic closure operation will be used to study models of ℵ1-categoricaltheories in further detail.

Definition 20.5. A pregeometry 5 (or matroid) (X, cl) is a set X with a closureoperator cl : P(X) → P(X) such that for all A ⊂ X and a, b ∈ X

a) A ⊂ cl(A)

b) cl(A) is the union of all cl(A′), where the A′ range over all finite subsetsof A.

c) cl(cl(A)) = cl(A)

d) exchange property

a ∈ cl(Ab) \ cl(A) ⇒ b ∈ cl(Aa)

A set A is called closed (or cl-closed) if A = cl(A). Note that the closurecl(A) of A is the smallest cl–closed set containing A. So a pregeometry is givenby the system of cl–closed subsets.

The operator cl(A) = A for all A ⊆ X is a trivial example of a pregeometry.The three standard examples from algebra are vector spaces with the linearclosure operator, for a field K with prime field F , the relative algebraic closurecl(A) = F (A)alg ∩K,6 and for a field K of characteristic p, the p-closure givenby cl(A) = Kp(A) (see Appendix C, Remark 8.1).

Lemma 20.6. If X is the universe of a structure, acl has properties a), b) andc) of a pregeometry.

5Pregeometries were introduced by van der Waerden (1930) and Whitney (1934).6Lalg denotes the algebraic closure of the field L.


Proof. Properties a) and b) are clear. For c) assume that c is algebraic overb1, . . . , bn and the bi are algebraic over A. We have to show that c is algebraicover A. Choose an algebraic formula ϕ(x, b1, . . . , bn) statisfied by c and algebraicL(A)–formulas ϕi(y) satisfied by the bi. Let ϕ(x, b1, . . . , bn) be satisfied byexactly k elements. Then the L(A)–formula

∃ y1 . . . yn(ϕ1(y1) ∧ . . . ∧ ϕn(yn) ∧ ∃≤kzϕ(z, y1, . . . , yn) ∧ ϕ(x, y1, . . . , yn))

is algebraic and realised by c.

Exercise 20.1.If A ⊂ M and M is |A|+–saturated, then p ∈ S(A) is algebraic if and only ifp(M) is finite.

Exercise 20.2 (P.M. Neumann).Let A and B be subsets of M and (c0 . . . cn) a sequence of elements which arenot algebraic over A. If M is sufficiently saturated, the type tp(c0, . . . cn/A)has a realisation which is disjoint from B. (Hint: Use induction on n. Distinguishbetween whether or not ci is algebraic over Acn for some i < n.)


21 Strongly minimal sets

Strongly minimal theories defined below turn out to be uncountably categorical:the isomorphism type of the model is determined by the dimension of an asso-ciated geometry. While this appears to be a very special case of such theories,we will see in the next section that we can always essentially reduce to thissituation.

Throughout this section we fix a complete theory T with infinitemodels. For Corollary 21.9 we have to assume T countable.

Definition 21.1. Let M be a model of T and ϕ(x) a non–algebraicL(M)–formula, not necessarily in one variable.

1. The set ϕ(M) is called minimal in M if for all L(M)–formulas ψ(x) theintersection ϕ(M) ∧ ψ(M) is either finite or cofinite in ϕ(M).

2. The formula ϕ(x) is strongly minimal if ϕ(x) defines a minimal set inall elementary extensions of M. In this case, we also call the definableset ϕ(M) strongly minimal. A non-algebraic type containing a stronglyminimal formula is called strongly minimal.

3. A theory T is strongly minimal if the formula x .= x is strongly minimal.

Clearly, strong minimality is preserved under definable bijections; i.e., if Aand B are definable subsets of Mk,Mm defined by ϕ and ψ, respectively, suchthat there is a definable bijection between A and B, then if ϕ is strongly minimalso is ψ.

Examples:

1. The following theories are strongly minimal, which is easily seen in eachcase using quantifier elimination.

• T∞, the theory of infinite sets. The sets which are definable over aparameter set A in a model M are the finite subsets S of A and theircomplements M \ S.

• For a field K, the theory of infinite K–vector spaces. The sets defin-able over a set A are the finite subsets of the subspace spanned by Aand their complements.

• The theories TACFp of algebraically closed fields of fixed character-istic. The definable sets of any model K are Boolean combinationsof zero–sets

{a ∈ K | f(a) = 0}

of polynomials f(X) ∈ K[X]. Zero–sets are finite or, if f = 0, all ofK.


2. If K is a model TACFp , for any a, b ∈ K, the formula ax1+b = x2 definingan affine line A in K2 is strongly minimal as there is a definable bijectionbetween A and K.

3. For any strongly minimal formula ϕ(x), the induced theory T �ϕ is stronglyminimal. Here, for any model M of T the induced theory is the theory ofϕ(M) with the structure given by all intersections of 0–definable subsetsof Mn with ϕ(M)n for all n ∈ ω. This theory depends only on T and ϕ,not on M.

Whether ϕ(x, a) is strongly minimal depends only on the type of the parametertuple a and not on the actual model: observe that ϕ(x, a) is strongly minimalif and only if for all L–formulas ψ(x, z) the set

Σψ(z, a) ={∃>nx(ϕ(x, a) ∧ ψ(x, z)) ∧ ∃>nx(ϕ(x, a) ∧ ¬ψ(x, z))

∣∣ n = 1, 2, . . .}

cannot be realised in any elementary extension. This means that for all ψ(x, z)there is a bound nψ such that

M |= ∀z(∃≤nψx(ϕ(x, a) ∧ ψ(x, z)) ∨ ∃≤nψx(ϕ(x, a) ∧ ¬ψ(x, z))

).

This is an elementary property of a, i.e. expressible by a first-order formula. Soit makes sense to call ϕ(x, a) a strongly minimal formula without specifying amodel.

Lemma 21.2. If M is ω–saturated, or if T eliminates the quantifier ∃∞, anyminimal formula is strongly minimal. If T is totally transcendental, every infi-nite definable subset of M contains a minimal set ϕ(M).

Proof. If M is ω-saturated and ϕ(x, a) not strongly minimal, then for someL–formula ψ(x, z) the set Σψ(z, a) is realised in M, so ϕ is not minimal.

If on the other hand ϕ(x, a) is minimal and T eliminates the quantifier ∃∞,then for all L–formulas ψ(x, z)

¬(∃∞x(ϕ(x, a) ∧ ψ(x, z)) ∧ ∃∞x(ϕ(x, a) ∧ ¬ψ(x, z))

)is an elementary property of z.

If ϕ0(M) does not contain a minimal set, one can construct from ϕ0(x) abinary tree of L(M)–formulas defining infinite subsets of M. This contradictsω–stability.

Lemma 21.3. ϕ(M) is minimal if and only if there is a unique non-algebraictype p ∈ S(M) containing ϕ(x).

Proof. If ϕ(M) is minimal, then clearly

p = {ψ | ψ(x) ∈ L(M) such that ϕ ∧ ¬ψ is algebraic}


is the unique non-algebraic type in S(M) containing it. If ϕ(M) is not minimal,there is some L–formula ψ with both ϕ ∧ ψ and ϕ ∧ ¬ψ non-algebraic. ByLemma 20.2 there are at least two non-algebraic types in S(M) containingϕ.

Corollary 21.4. A strongly minimal type p ∈ S(A) has a unique non-algebraicextension to all supersets B of A in elementary extensions of M. Consequently,the type of n realisations a1, . . . an of p with ai /∈ acl(a1, . . . ai−1A), i = 1, . . . n,is uniquely determined.

Proof. Existence of non-algebraic extensions follows from Lemma 20.2, whichalso allows us to assume that B is a model. Uniqueness now follows fromLemma 21.3 applied to any strongly minimal formula of p. The last sentencefollows by induction.

Theorem 21.5. If ϕ(x) is a strongly minimal formula in M without parameters,the operation

cl : P(ϕ(M)) → P(ϕ(M))

defined bycl(A) = aclM (A) ∩ ϕ(M)

is a pregeometry (ϕ(M), cl).

Proof. We have to verify the exchange property. For notational simplicity weassume A = ∅. Let a ∈ ϕ(M) be not algebraic over ∅ and b ∈ ϕ(M) not algebraicover a. By Corollary 21.4, all such pairs a, b have the same type p(x, y). LetA′ be an infinite set of non–algebraic elements realising ϕ (which exists in anelementary extension of M) and b′ non–algebraic over A′. Since all a′ ∈ A′

have the same type p(x, b′) over b′, no a′ is algebraic over b′. Thus also a is notalgebraic over b.

The same proof shows that algebraic closure defines a pregeometry on the setof realizations of a minimal type, i.e. a non-algebraic type p ∈ S1(A) having aunique non-algebraic extension to all supersets B of A in elementary extensionsof M. Here is an example to show that a minimal type need not be stronglyminimal:

Let T be the theory of M = (M,Pi)i<ω in which the Pi form a properdescending sequence of subsets. The type p = {x ∈ Pi : i < ω} ∈ S1(∅) isminimal. If all Pi+1 are coinfinite in Pi, then p does not contain a minimalformula and is not strongly minimal.

In pregeometries there is a natural notion of independence and dimension(see p. 233, Def. 8.2), so in light of Theorem 21.5 we may define the following:


If ϕ(x) is strongly minimal without parameters, the ϕ–dimension of a modelM of T is the dimension of the pregeometry (ϕ(M), cl)

dimϕ(M).

If M is the model of a strongly minimal theory, we just write dim(M).

If ϕ(x) is defined over A0 ⊂ M , the closure operator of the pregeometryϕ(MA0) is given by

cl(A) = aclM (A0 ∪A) ∩ ϕ(M)

anddimϕ(M/A0) := dimϕ(MA0)

is called the ϕ–dimension of M over A0.

Lemma 21.6. Let ϕ(x) be defined over A0 and strongly minimal, and let M andN be models containing A0. Then there exists an A0–elementary map betweenϕ(M) and ϕ(N) if and only if M and N have the same ϕ–dimension over A0.

Proof. An A0–elementary map between ϕ(M) and ϕ(N) maps bases to bases,so one direction is clear.

For the other direction we use Corollary 21.4: if ϕ(M) and ϕ(N) have thesame dimension over A0, let U and V be bases of ϕ(M) and ϕ(N), respectively,and let f : U → V be a bijection. By Corollary 21.4, f is A0–elementary andby Lemma 20.4 f extends to an elementary bijection g : acl(A0U) → acl(A0V ).Thus, g � ϕ(M) is an A0–elementary map from ϕ(M) to ϕ(N).

We now turn to showing that strongly minimal theories are categorical inall uncountable cardinals. For reference we first note the following special casesof the preceding lemmas:

Corollary 21.7. 1. A theory T is strongly minimal if and only if over everyparameter set there is exactly one non–algebraic type.

2. In models of a strongly minimal theory the algebraic closure defines apregeometry.

3. Bijections between independent subsets of models of a strongly minimaltheory are elementary. In particular, the type of n independent elementsis uniquely determined.

If T is strongly minimal, by the preceding we have

|S(A)| ≤ | acl(A)|+ 1.


Strongly minimal theories are therefore λ–stable for all λ ≥ |T |. It follows thatevery restriction to a countable language is ω–stable, so T is totally transcenden-tal (cf. Exercise 16.4). If ϕ(M) is cofinite and N a proper elementary extensionof M, then ϕ(N) is a proper extension of ϕ(M). Thus strongly minimal theorieshave no Vaughtian pairs.

Theorem 21.8. Let T be strongly minimal. Models of T are uniquely deter-mined by their dimensions. The set of possible dimensions is an end segmentof the cardinals. A model M is ω–saturated if and only if dim(M) ≥ ℵ0. Allmodels are ω–homogeneous.

Proof. Let M0,M1 be models of the same dimension, and let B0, B1 be bases forM0 and M1, respectively. Then any bijection f : B0 → B1 is an elementary mapby Corollary 21.7, which extends to an isomorphism of the algebraic closuresM0 and M1 by Lemma 20.4.

The next claim implies that the possible dimensions form an end segment ofthe cardinals.

Claim: Every infinite algebraically closed subset S of M is the universe of anelementary substructure.Proof: By Theorem 4.2 it suffices to show that every consistent L(S)–formulaϕ(x) can be realised in S. If ϕ(M) is finite, all realisations are algebraic over Sand belong therefore to S. If ϕ(M) is cofinite, ϕ(M) meets all infinite sets.

Let A be a finite subset of M and p the non–algebraic type in S(A). Thus,p is realised in M exactly if M 6= acl(A), i.e. if and only if dim(M) > dim(A).Since all algebraic types over A are always realised in M, this shows that M isω–saturated if and only if M has infinite dimension.

It remains to show that all models are ω–homogeneous. Let f : A→ B be anelementary bijection between two finite subsets ofM . By Lemma 20.4, f extendsto an elementary bijection between acl(A) and acl(B). If a ∈ M \ acl(A), thenp = tp(a/A) is the unique non–algebraic type over A and f(p) is the unique non–algebraic type over B. Since dim(A) = dim(B), the argument in the previousparagraph shows that f(p) is realised in M.

Corollary 21.9. If T is countable and strongly minimal, it is categorical in alluncountable cardinalities.

Proof. Let M1 and M2 be two models of cardinality κ > ℵ0. Choose two basesB1 and B2 of M1 and M2 respectively. By p. 94, equation (3), B1 and B2 bothhave cardinality κ. Then any bijection f : B1 → B2 is an elementary map byCorollary 21.7, which extends to an isomorphism of the algebraic closures M1

and M2 by Lemma 20.4.

Exercise 21.1.If M is minimal and ω–saturated, then Th(M) is strongly minimal.


Exercise 21.2.Show directly that strongly minimal theories eliminate ∃∞ (without using Corol-laries 19.4 and 21.9.)

Exercise 21.3.A type is minimal if and only if its set of realisations in any model is minimal(i.e. has no infinite and coinfinite relatively definable subsets).

Exercise 21.4.Show that acl defines a pregeometry on µ(M) if µ(M) is minimal. In fact thefollowing is true: If b ∈ µ(M), a ∈ acl(Ab), b 6∈ acl(Aa), then a ∈ acl(A).Furthermore we have deg(a/A) = deg(a/Ab).


22 The Baldwin-Lachlan theorem

In this section, we present the characterisation of uncountably categorical theo-ries due to Baldwin and Lachlan [1]. Since this characterisation is independentof the uncountable cardinal, it implies Morley’s Theorem. The crucial point isthe existence of a strongly minimal formula ϕ in a totally transcendental the-ory. By Corollary 19.5, each model M is prime over the set of realisations ϕ(M)whose dimension determines the isomorphism type of the model.

Theorem 22.1 (Baldwin–Lachlan). Let κ be an uncountable cardinal. A count-able theory T is κ–categorical if and only if T is ω–stable and has no Vaughtianpair.

Proof. If T is categorical in some uncountable cardinal, then T is ω-stable byTheorem 16.3 and has no Vaughtian pair by Corollary 19.4.

For the other direction we first obtain a strongly minimal formula: since Tis totally transcendental, it has a prime model M0. (This follows from Theo-rems 14.7 and 14.9 or from Theorem 17.3.) Let ϕ(x) be a minimal formula inL(M0), which exists by Lemma 21.2. Since T has no Vaughtian pairs, ∃∞ can beeliminated by Lemma 19.7 and hence ϕ(x) is strongly minimal by Lemma 21.2.

Let M1,M2 be models of cardinality κ. We may assume that M0 is anelementary submodel of both M1 and M2. Since T has no Vaughtian pair, Mi

is a minimal extension of M0∪ϕ(Mi), i = 1, 2. Therefore, ϕ(Mi) has cardinalityκ and hence (since κ is uncountable) we conclude that dimϕ(M1/M0) = κ =dimϕ(M2/M0). By Lemma 21.6 there exists anM0-elementary map from ϕ(M0)to ϕ(M1), which by Lemma 17.8 extends to an isomorphism from M1 to M2.

Corollary 22.2 (Morley). Let κ be an uncountable cardinal. T is ℵ1–categoricalif and only if T is κ-categorical.

Notice that the proof of Theorem 22.1 shows in fact the following:

Corollary 22.3. Suppose T is ℵ1–categorical, M1,M2 are models of T , ai ∈ Mi

and ϕ(x, ai) strongly minimal, i = 1, 2, with tp(a1) = tp(a2). If M1 and M2

have the same respective ϕ-dimension, then they are isomorphic.

For uncountable models, the ϕ-dimension equals the cardinality of the model,so clearly does not depend on the realisation of tp(ai). We will show in Section 25that the converse to Corollary 22.3 holds also for countable models, i.e., ifϕ(x, a0) is strongly minimal, then the ϕ-dimension of M is the same for all arealizing tp(a0). Thus, also the countable models of an uncountably categoricaltheory are in one-to-one correspondence with the possible ϕ-dimensions.

Chapter 6

Morley Rank

In this chapter we collect a number of further results about totally transcen-dental theories, in particular we will introduce Morley rank. For convenience,we will introduce the ‘monster model’ (for arbitrary theories), a very large, verysaturated, very homogeneous model. From now on, all models we consider willbe elementary submodels of this monster model. We then finish the analysis ofthe countable models of uncountably categorical theories.

23 Saturated models and the Monster

The importance of saturated structures was already visible in Section 16 wherewe showed that saturated structures of fixed cardinality are unique up to iso-morphism. Saturated structures need not exist (think about why not), butby considering special models instead, we can preserve many of the importantproperties - and prove their existence.

Definition 23.1. A structure M of cardinality κ ≥ ω is special if M is theunion of an elementary chain Mλ where λ runs over all cardinals less than κand each Mλ is λ+–saturated.

We call (Mλ) a specialising chain.

Remark. Saturated structures are special. If |M| is regular, the converse istrue.

Lemma 23.2. Let λ be an infinite cardinal ≥ |L|. Then every L–structure Mof cardinality 2λ has a λ+–saturated elementary extension of cardinality 2λ.

Proof. Every set of cardinality 2λ has 2λ many subsets of cardinality at most

104

CHAPTER 6. MORLEY RANK 105

λ. This allows us to construct a continuous elementary chain

M = M0 ≺ M1 . . . ≺ Mα ≺ . . . (α < λ+)

of structures of cardinality 2λ such that all p ∈ S(A), for A ⊂Mα, |A| ≤ λ, arerealised in Mα+1. The union of this chain has the desired properties.

Corollary 23.3. Let κ > |L| be an uncountable cardinal. Assume that

(1) λ < κ ⇒ 2λ ≤ κ

Then every infinite L–structure M of cardinality smaller than κ has a specialextension of cardinality κ.

Let α be a limit ordinal. Then for any cardinal µ, κ = iα(µ) satisfies (1)and we have cf(κ) = cf(α) (see p. 214).

The following is a generalisation of Lemma 16.7

Theorem 23.4. Two elementarily equivalent special structures of the samecardinality are isomorphic.

Proof. The proof is a refined version of the proof of Lemma 16.7. Let A and Bbe two elementarily equivalent special structures of cardinality κ. Let A be theunion of a chain of elementary substructures Aλ, B the union of Bλ, where λruns over all cardinals less than κ and Aλ and Bλ are λ+–saturated. Lemma 3.2of the appendix shows that A and B have enumerations (aα)α<κ and (bα)α<κsuch that aα ∈ A|α| and bα ∈ B|α|. We construct an increasing sequence ofelementary maps fα : Aα → Bα such that |Aα| ≤ |α|, Aα ⊂ A|α|, |Bα| ≤ |α|,Bα ⊂ B|α|.

Definition 23.5. A structure M is

• κ–universal if every structure of cardinality < κ which is elementarilyequivalent to M can be elementarily embedded into M.

• κ–homogeneous if for every subset A of M of cardinality smaller than κand for every a ∈ M , every elementary map A → M can be extended toan elementary map A ∪ {a} →M .

• strongly κ–homogeneous if for every subset A ofM of cardinality less thanκ, every elementary map A→M can be extended to an automorphism ofM.

Theorem 23.6. Special structures of cardinality κ are κ+–universal and stronglycf(κ)–homogeneous.


Proof. Let M be a special structure of cardinality κ. The κ+–universality ofM can be proved in the same way as Theorem 23.4. Let A be a subset of Mof cardinality less than cf(κ) and let f : A → M an elementary map. Fix aspecialising sequence (Mλ). For λ0 sufficiently large, Mλ0 contains A. Thesequence

M∗λ =

{(Mλ, a)a∈A, if λ0 ≤ λ

(Mλ0 , a)a∈A, if λ < λ0

is then a specialising sequence of (M, a)a∈A. For the same reason (M, f(a))a∈Ais special. By Theorem 23.4 these two structures are isomorphic under an au-tomorphism of M which extends f .

The monster model

Let T be a complete theory with infinite models. For convenience, we wouldlike to work in a very large saturated structure, large enough so that any modelof T can be considered as an elementary substructure. If T is totally transcen-dental, by Remark 16.9 we can choose such a monster model as a saturatedmodel of cardinality κ where κ is a regular cardinal greater than all the mod-els we ever consider otherwise. Using Exercise 33.7, this also works for stabletheories and regular κ with κ|T | = κ. κ = (λ|T |)+ has this properties for anyinfinite λ.

In order to construct the monster model C for an arbitrary theory T wewill work in BGC (Bernays–Gödel + Global Choice). This is a conservativeextension of ZFC (cf. Appendix, Section A) which adds classes to ZFC. Thenbeing a model of T is interpreted as being the union of an elementary chain of(set-size) models of T . The universe of our monster model C will be a properclass:

Theorem 23.7 (BGC). There is a class-size model C of T such that all typesover all subsets of C are realised in C. C is uniquely determined up to isomor-phism.

Proof. Global choice allows us to construct a long continuous elementary chain(Mα)α∈On of models of T such that all types over Mα are realised in Mα+1. LetC be the union of this chain. The uniqueness is proved as in Lemma 16.7.

We call C the monster model of T . Note that Global Choice implies that Ccan be well-ordered.

Corollary 23.8.

• C is κ-saturated for all cardinals κ.

• Any model of T is elementarily embeddable in C

• Any elementary bijection between two subsets of C can be extended to anautomorphism of C.


We say that two elements are conjugate over some parameter set A if thereis an automorphism of C fixing A elementwise and taking one to the other. Notethat a, b ∈ C are conjugate over A if and only if they have the same type over A.We call types p ∈ S(A), q ∈ S(B) conjugate over D if there is an automorphism fof C fixing D and taking A to B and such that q = {ϕ(x, f(a)) | ϕ(x, a) ∈ p}.Note that strictly speaking Aut(C) is not an object in Bernays-Gödel Set theorybut we will nevertheless use this term as a way of talking about automorphisms.

Readers who mistrust set theory can fix a regular cardinal γ bigger than thecardinality of all models and parameter sets they want to consider. For C theymay then use a special model of cardinality κ = iγ(ℵ0). This is κ+–universaland strongly γ–homogeneous.

We will use the following convention throughout the rest of thisbook:

• Any model of T is an elementary substructure of C. We identify modelswith their universes and denote them by M , N ,. . . .

• Parameter sets A, B, . . . are subsets of C

• Formulas ϕ(x) with parameters define a subclass F = ϕ(C) of C. Twoformulas are equivalent if they define the same class.

• We write |= ϕ for C |= ϕ.

• A set of formulas with parameters from C is consistent if it is realised inC.

• A global type is a type over C.

This convention changes the flavour of quite a number of proofs. As anexample look at the following

Lemma 23.9. An elementary bijection f : A → B extends to an elementarybijection between acl(A) and acl(B).

Proof. Extend f to an automorphism f ′ of C. Clearly f ′ maps acl(A) to acl(B).

This implies Lemma 20.4 and the second claim in the proof of Theorem 21.6.Note that by the remark on p. 94 for any model M and any A ⊂M the algebraicclosure of A in the sense of M equals the algebraic closure in the sense of C.

Lemma 23.10. Let D be a definable class and A a set of parameters. Then thefollowing are equivalent:


a) D is definable over A.

b) D is invariant under all automorphisms of C which fix A pointwise.

Proof. Let D be defined by ϕ, defined over B ⊃ A. Consider the maps

Cτ−→ S(B)

π−→ S(A),

where τ(c) = tp(c/B) and π is the restriction map. Let Y be the image of D inS(A). Since Y = π[ϕ], Y is closed.

Assume that D is invariant under all automorphisms of C which fix A point-wise. Since elements which have the same type over A are conjugate by anautomorphism of C, this means that D–membership depends only on the typeover A i.e. D = (πτ)−1(Y ).

This implies that [ϕ] = π−1(Y ), or S(A) \ Y = π[¬ϕ], hence S(A) \ Y is alsoclosed and we conclude that Y is clopen. By Lemma 11.3 Y = [ψ] for someL(A)–formula ψ. This ψ defines D.

Definition 23.11. The definable closure dcl(A) of A is the set of elements c forwhich there is an L(A)-formula ϕ(x) such that c is the unique element satisfyingϕ. Elements or tuples a and b are said to be interdefinable if a ∈ dcl(b) andb ∈ dcl(a).

Corollary 23.12.

1. a ∈ dcl(A) if and only if a has only one conjugate over A.

2. a ∈ acl(A) if and only if a has finitely many conjugates over A.

Proof. 1. is clear, since a ∈ dcl(A) means that {a} is A–definable. 2. followsfrom Remark 20.3, since the realisations of tp(a/A) are exactly the conjugatesof a over A.

Exercise 23.1.Finite structures are saturated.

Exercise 23.2.acl(A) is the intersection of all models which contain A.

Exercise 23.3.Prove Robinson’s Joint Consistency Lemma: Extend the complete L–theory Tto an L1–theory T1 and an L2–theory T2 such that L = L1 ∩ L2. If T1 and T2are both consistent, show that T1 ∪ T2 is consistent.

Exercise 23.4.Prove Beth’s Interpolation Theorem: If ` ϕ1 → ϕ2 for Li–sentences ϕi, thereis an L = L1 ∩ L2–sentence θ such that ` ϕ1 → θ and ` θ → ϕ2.


Exercise 23.5.If M is κ–saturated, then over every set of cardinality smaller than κ every typein κ many variables is realised in M .

Exercise 23.6.Let κ be an infinite cardinal, not smaller than the cardinality of L and M anL–structure. Show that the following are equivalent

a) M is κ–saturated

b) M is κ+–universal and κ–homogeneous

If max(|L|,ℵ0) < κ this is also equivalent to

c) M is κ–universal and κ–homogeneous.

Exercise 23.7.Let κ be a an uncountable regular cardinal > |L|. Show that

1. Every L–structure M of cardinality 2<κ has a κ–saturated elementaryextension of cardinality 2<κ.

2. Assume that λ < κ implies 2λ ≤ κ. Then every L–structure M of cardi-nality κ has a saturated elementary extension of cardinality κ.

Exercise 23.8.Let Pω(N) be the set of all finite subsets of N. Show that the theory of(Pω(N),⊂) has a saturated model of cardinality κ if and only if κ is regularand λ < κ implies 2λ ≤ κ.

Exercise 23.9.A type-definable class is the class of all realisations of a set of formulas. Showthat a type-definable class is invariant under all automorphisms of C which fixA pointwise if and only if it can be defined by a set of L(A)–formulas. (UseExercise 11.2.a and the proof of Lemma 23.10.)

Exercise 23.10.Let A be contained in B. Show that the following are equivalent:

1. B ⊂ dcl(A).

2. Every type over A extends uniquely to B.

Exercise 23.11.1. b is in the definable closure of a if and only if there is a 0–definable class

D containing a and a 0–definable map D → C which maps a to b.

2. a and b are interdefinable if and only if a and b are contained in 0–definableclasses D and E and there is a 0–definable bijection between D and E whichmaps a to b.


Exercise 23.12.Let K be a model of TACF0 , TACFp for p > 0, of TRCF or of TDCF0 and let Abe a (differential) subfield of K. Prove that the definable closure of A is

1. (TACF0) A itself,

2. (TACFp) the perfect hull (see Definition B6.8) of A,

3. (TRCF) the relative algebraic closure of A,

4. (TDCF0) A itself.

Exercise 23.13.Use Exercise 23.12 to show the following:

1. Let K be a model of TACF0 and f : Kn → K a definable function. ThenKn can be decomposed into a finite number of definable subsets Xi suchthat, on each Xi, f is given by a rational function.

2. The same is true for models of TACFp , p > 0. But on each Xi, f is of theform hp

−m, for some rational function h.

3. In models of TDCF0 , f is given on each Xi by a differential rational func-tion.

Exercise 23.14 (P. Neumann).Let X be an infinite set, G ≤ Sym(X), and B ⊂ X finite. Suppose that theorbit of each of the elements c0, . . . cn ∈ X under G is infinite. Then there issome g ∈ G with g(ci) /∈ B for i = 0, . . . n. (Hint: Proceed as in Exercise 20.2. Infact, Exercise 20.2 follows from this by compactness.)

Exercise 23.15 (B. Neumann).Let G be a group (not necessarily abelian), H0, . . . ,Hn, subgroups of infinite in-dex. Show that G is not a finite union of cosets of the Hi (but see Exercise 23.16,which is easy). Deduce Lemma 9.9 from this.

Exercise 23.16 (B. & P. Neumann).Deduce Exercise 23.14 from 23.15 and conversely.


24 Morley Rank

The Morley Rank is a rather natural notion of dimension on the formulas of atheory T or, equivalently, on the definable subsets of the monster model, definedinductively very much like the dimension of algebraic sets. It is ordinal valuedfor all consistent formulas if and only if T is totally transcendental. In thissection, we let T be a complete (possibly uncountable) theory.

We now define the Morley rank MR for formulas ϕ(x) with parameters inthe monster model. We begin by defining1 the relation MRϕ ≥ α by inductionon the ordinal α.

MRϕ ≥ 0 if ϕ is consistent;

MRϕ ≥ β + 1 if there is an infinite family (ϕi(x)) of formulas (in thesame variables x) which imply ϕ, are pairwise incon-sistent and such that MRϕi ≥ β for all i;

MRϕ ≥ λ (for a limit ordinal λ) if MRϕ ≥ β for all β < λ.

Definition 24.1. For a given formula ϕ there are three cases

1. If there is no α with MRϕ ≥ α, we put MRϕ = −∞.

2. MRϕ ≥ α for all α, we put MRϕ = ∞.

3. Otherwise, by the definition of MRϕ ≥ λ for limit ordinals λ, there is amaximal α with MRϕ ≥ α, and we set MRϕ = max{α | MRϕ ≥ α}.

It is easy to see by induction on α that the relation MRϕ ≥ α implies therelation MRϕ ≥ β for β ≤ α. It follows from this that indeed the Morley rankof ϕ is at least α if and only if the relation MRϕ ≥ α holds.2

Note that

MRϕ = −∞ ⇔ ϕ is inconsistentMRϕ = 0 ⇔ ϕ is consistent and algebraic.

If a formula has ordinal valued Morley rank, we also say that this formula hasMorley rank. The Morley rank MR(T ) of T is the Morley rank of the formulax.= x. The Morley rank of a formula ϕ(x, a) only depends on ϕ(x, y) and the

type of a. It follows that if a formula has Morley rank, then it is less than1Set-theoretically we define a function which maps each α to a class of formulas. In BG

one cannot in general define functions from ordinals to classes by a recursion scheme. Themore conscienteous reader should therefore use a different definition: Define for each set Athe relation MRA ϕ ≥ α, where only formulas with parameters from A are used. And sayMRϕ ≥ α if MRA ϕ ≥ α for some (large enough) A. By Exercise 24.3, if ϕ is defined over Mand M is ω–saturated, we have MRϕ = MRM ϕ.

2Here, of course −∞ is considered as being smaller than all ordinals and ∞ to be biggerthan all ordinals.


(2|T |)+. We will see in Exercise 24.6 that in fact all ordinal ranks are smallerthat |T |+.

Remark. Clearly, if ϕ implies ψ, then MRϕ ≤ MRψ. It is also clear fromthe definition that if ϕ has rank α <∞, then for every β < α there is a formulaψ which implies ϕ and has rank β.

The next lemma expresses the fact that the formulas of rank less than αform an ideal in the Boolean algebra of equivalence classes of formulas.

Lemma 24.2.MR(ϕ ∨ ψ) = max(MRϕ,MRψ)

Proof. By the previous remark, we have MR(ϕ ∨ ψ) ≥ max(MRϕ,MRψ). Forthe other inequality we show by induction on α that

MR(ϕ ∨ ψ) ≥ α+ 1 implies max(MRϕ,MRψ) ≥ α+ 1.

Let MR(ϕ ∨ ψ) ≥ α + 1. Then there is an infinite family of formulas (ϕi)which imply ϕ ∨ ψ, are pairwise inconsistent and such that MRϕi ≥ α. Byinduction hypothesis we have for all i MR(ϕi ∧ ϕ) ≥ α or MR(ϕi ∧ ψ) ≥ α.If the first case holds for infinitely many i, then MRϕ ≥ α + 1. OtherwiseMRψ ≥ α+ 1.

We call ϕ and ψ α–equivalent,

ϕ ∼α ψ,

if their symmetric difference ϕ 4 ψ has rank less than α. By our previousconsiderations it is clear that α-equivalence is in fact an equivalence relation.

We call a formula ϕ α–strongly minimal if it has rank α and for any formulaψ implying ϕ either ψ or ϕ ∧ ¬ψ, has rank less than α, (equivalently, if everyψ ⊂ ϕ is α–equivalent to ∅ or to ϕ). In particular, ϕ is 0–strongly minimal ifand only if ϕ is realised by a single element and ϕ is 1–strongly minimal if andonly if ϕ is strongly minimal.

Lemma 24.3. Each formula ϕ of rank α < ∞ is equivalent to a disjunctionof finitely many pairwise disjoint α–strongly minimal formulas ϕ1, . . . , ϕn, theα–strongly minimal components (or just components) of ϕ. The componentsare uniquely determined up to α–equivalence.

Proof. Suppose ϕ is a formula of rank α without such a decomposition. Then ϕcan be written as the disjoint disjunction of a formula ϕ1 of rank α and anotherformula ψ1 of rank α not having such a decomposition. Inductively there areformulas ϕ = ϕ0, ϕ1 . . . of rank α and ψ1, ψ2, . . . so that ϕi is the disjoint unionof ϕi+1 and ψi+1. But then the rank of ϕ would be greater than α.


To see the uniqueness of this decomposition, let ψ be an α–strongly minimalformula implying ϕ and let ϕ1, . . . , ϕn be the α–strongly minimal components.Then ψ can be decomposed as ψ ∧ϕi, with, say, ψ ∧ϕi0 , α–equivalent to ψ. Soup to α–equivalence the components of ϕ are exactly the α–strongly minimalformulas implying ϕ.

Definition 24.4. For a formula ϕ of Morley rank α, the Morley degree MD(ϕ)is the number of its α–strongly minimal components.

The Morley degree is not defined for inconsistent formulas or formulas nothaving Morley rank. The Morley degree of an algebraic formula equals itsdegree as defined on p. 94. Strongly minimal formulas are exactly the formulasof Morley rank and Morley degree 1. As with strongly minimal formulas it iseasy to see that Morley rank and degree are preserved under definable bijections.

Defining MDα(ϕ) as the Morley degree for formulas ϕ of rank α, as 0 forformulas of smaller rank and as ∞ for formulas ϕ of higher rank, we obtain thefollowing:

Lemma 24.5. If ϕ is the disjoint union of ψ1 and ψ2, then

MDα(ϕ) = MDα(ψ1) +MDα(ψ2).

Theorem 24.6. T is totally transcendental if and only if each formula hasMorley rank.

Proof. Since there are not arbitrarily large ordinal Morley ranks, each formulawithout Morley rank can be decomposed into two disjoint formulas withoutMorley rank, yielding a binary tree of consistent formulas, a contradiction.

For the other direction let (ϕs)s∈2<ω be a binary tree of consistent formulas.Choose ϕs so that its rank α is minimal and (among the formulas of rank α)of minimal degree. Then both ϕs0 and ϕs1 have rank α and therefore smallerdegree than ϕ, a contradiction.

A group is said to have the the descending chain condition (dcc) on definablesubgroups, if there is no infinite properly descending chain H0 ⊃ H1 ⊃ H2 ⊃ . . .of definable subgroups.

Remark 24.7. A totally transcendental group has the descending chain con-dition on definable subgroups.

Proof. If H is a definable proper subgroup of a totally transcendental group G,then either the Morley rank or the Morley degree of H must be less than that ofG since any coset of H has the same Morley rank and degree as H. Thereforethe claim follows from the fact that the ordinals are well-ordered.


The previous remark is a crucial tool in the theory of totally transcendentalgroups. For example, it immediately implies that (Z,+) is not totally transcen-dental.

Corollary 24.8. The theory of separably closed fields of degree of imperfectione > 0 is not totally transcendental.

Proof. The subfields K ⊃ Kp ⊃ Kp2 ⊃ Kp3 ⊃ . . . form an infinite definablechain of properly descending (additive) subgroups. In fact we will see later thatthe proof shows that that separably closed fields are not even superstable (seeExercise 37.8.)

Definition 24.9. The Morley rank MR(p) of a type p is the minimal rank ofany formula in p. If MR(p) is an ordinal α, then its Morley degree MD(p) of pis the minimal degree of a formula of p having rank α. If p = tp(a/A) we alsowrite MR(a/A) and MD(a/A).

Algebraic types have Morley rank 0 and

MD(p) = deg(p).

Strongly minimal types are exactly the types of Morley rank and Morley de-gree 1.

Let p ∈ S(A) have Morley rank α and Morley degree d. Then by definitionthere is some ϕ ∈ p of rank α and degree d. Clearly, ϕ is uniquely determinedup to α–equivalence since for all ψ we have MR(ϕ ∧ ¬ψ) < α if and only ifψ ∈ p. Thus, p is uniquely determined by ϕ:

(2) p = {ψ(x) | ψ L(A)–formula, MR(ϕ ∧ ¬ψ) < α} .

Obviously, α–equivalent formulas determine the same type (cf. Lemma 21.3).

Thus ϕ ∈ L(A) belongs to a unique type of rank α if and only if ϕ is α–minimal over A; i.e. if ϕ has rank α and cannot be decomposed as the union oftwo L(A)-formulas of rank α.

Lemma 24.10. Let ϕ be a consistent L(A)–formula.

1.MRϕ = max{MR(p) | ϕ ∈ p ∈ S(A)}

2. Let MRϕ = α. Then

MDϕ =∑(

MD(p)∣∣ ϕ ∈ p ∈ S(A), MR(p) = α

).

Proof. 1: If MRϕ = ∞, then {ϕ} ∪ {¬ψ | ψ L(A)–formula, MRψ < ∞} isconsistent. Any type over A containing this set of formulas has rank ∞.


If MRϕ = α, there is a decomposition of ϕ in L(A)–formulas ϕ1, . . . , ϕk, α–minimal over A. (Note that k is bounded by MDϕ.) By (2), the ϕi determinetypes pi of rank α.

2: The pi are exactly the types of rank α containing ϕ. Furthermore,

MDϕi = MD(pi).

Corollary 24.11. If p ∈ S(A) has Morley rank and A ⊂ B, then

MD(p) =∑(

MD(q)∣∣ p ⊂ q ∈ S(B), MR(p) = MR(q)

)

Corollary 24.12. Let p ∈ S(A) have Morley rank and A ⊂ B. Then p ∈ S(A)has at least one and at most MD(p) many extensions to B of the same rank.

We will later show that extensions of the same Morley rank are a specialcase of the non-forking extensions studied in Chapters 7 and 8. For types withMorley rank those of the same Morley rank are exactly the non-forking ones.

Exercise 24.1.Let ϕ be a formula of Morley rank α <∞ and ψ0, ψ1, . . . an infinite sequence offormulas. Assume that there is a number k such that the conjunction any k ofthe ψi has Morley rank smaller than α. Then MR(ϕ ∧ ψi) < α for almost all i.

Exercise 24.2.We only defined Morley rank for formulas and types in one variable, but itgeneralises to formulas and types in any fixed number of variables x1, . . . , xn inthe obvious way. Prove the following:

1. If T is ω–stable, there are only countably many n-types over any countableset.

2. If T is totally transcendental, then any formula in n variables has Morleyrank.

Exercise 24.3.Let ϕ be a formula with parameters in the ω–saturated modelM . If MR(ϕ) > α,show that there is an infinite family of formulas with parameters in M whicheach imply ϕ, are pairwise inconsistent and have Morley rank ≥ α.

Exercise 24.4.Show that a totally transcendental group G has a connected component, i.e. asmallest definable subgroup G0 of finite index. Show also that any finite normalsubgroup of G0 lies in the center of G0.


Exercise 24.5.If T is totally transcendental, then all types over ω–saturated models haveMorley degree 1. (We will see later that this is true without assuming ω–saturation (36.12).)

Exercise 24.6 (Lachlan).If a type p has Morley rank, then MR(p) < |T |+. Hence, if T is totally tran-scendental, we have MR(T ) < |T |+.

Exercise 24.7.For a topological space X we define recursively on ordinals X0 := X,Xα+1 :=Xα \ {x : x isolated in Xα} and Xλ :=

⋂α<λX

α if λ is a limit ordinal. TheCantor-Bendixson rank of x ∈ X is equal to α if α is maximal with x ∈ Xα.

Show that on S(C) the Morley rank equals the Cantor-Bendixson rank. Notethat S(C) is not even a class. So, for this exercise we have to ignore set–theoreticsubtleties.

Exercise 24.8.If p is a type over acl(A), then p and p � A have the same Morley rank.


25 Countable models of ℵ1-categorical theories

Uncountable models of ℵ1-categorical theories are determined up to isomor-phism by their cardinality. Section 22 showed that this cardinality coincideswith the dimension of a strongly minimal formula. We here extend this analysisin order to classify also the countable models of ℵ1-categorical theories and showin Theorem 25.7 that for each possible dimension there is a unique model of thetheory.

Throughout this section we fix a countable ℵ1–categorical theory T .For models M ≺ N of T and ϕ(x) ∈ L(M) a strongly minimal formula, we writedimϕ(N/M) for the ϕ–dimension of N over M .

Theorem 25.1. Let T be a countable ℵ1-categorical theory, M ≺ N be modelsof T , A ⊂M and ϕ(x) ∈ L(A) a strongly minimal formula.

1. If b1, . . . bn ∈ ϕ(N) are independent over M and N is prime over M ∪{b1, . . . bn}, then

dimϕ(N/M) = n, and

2.dimϕ(N) = dimϕ(M) + dimϕ(N/M)

Proof. For ease of notation we assume A = ∅.

1:Let c ∈ ϕ(N). We want to show that c is algebraic over M ∪ {b1, . . . , bn}.Assume the contrary. Then p(x) = tp(c/M ∪ {b1, . . . , bn}) is strongly minimaland is axiomatised by

{ϕ(x)} ∪ {¬ϕi(x) | i ∈ I},

where the ϕi range over all algebraic formulas defined over M ∪ {b1, . . . , bn}.Since ϕ(M) is infinite, any finite subset of p(x) is realised by an element of M .Thus, p(x) is not isolated. But all elements of the prime extension N are atomicover M ∪ {b1, . . . , bn} by Corollary 17.7, a contradiction.

2:This follows from Remark C 8.8, if we can show that a basis of ϕ(N) over ϕ(M)is also a basis of ϕ(N) overM . So the proof is complete once we have establishedthe following lemma.

Lemma 25.2. Let T be ω–stable, M ≺ N models of T , ϕ(x) strongly minimaland bi ∈ ϕ(N). If the bi are independent over ϕ(M), they are independent overM .

Proof. Assume that b = b1 . . . bn are algebraically independent over ϕ(M) butdependent over a ∈ M . An argument as in the proof of Theorem 19.2 showsthat we may assume that M is ω–saturated. Let p be the type of b over M .


We choose a sequence b0, b1, . . . in ϕ(M) such that b2i is an n-tuple of elementsalgebraically independent over b0 . . . b2i−1 and b2i+1 realises p � ab0 . . . b2i. Letq be the type of a over the set B of elements of (bi). Since the sequence (bi)is indiscernible, every permutation π of ω defines a type π(q) over B. If {i |π(2i) even} 6= {i | π′(2i) even}, we have π(q) 6= π′(q). So there are uncountablymany types over B and T is not ω–stable.

Corollary 25.3. The dimension

dim(N/M) = dimϕ(N/M)

of N over M does not depend on ϕ: it is the maximal length of an elementarychain

M = N0 � N1 � . . . � Nn = N

between M and N .

Proof. This follows from the previous theorem since T has no Vaughtian pairs.

For the remainder of this section, we let M0 denote the primemodel of T . We also fix a strongly minimal formula ϕ(x, a0) ∈ L(M0)and put p0(x) = tp(a0).

Note that the type p0(x) of a0 is isolated by Theorem 14.2, hence realisedin every model of T . For any model M and realisation a of p0 in M , letdimϕ(x,a)(M) denote the ϕ(x, a)–dimension of M over a. To simplify notationwe assume that a0 is some element a0 rather than a tuple.

Since M0 is ω–homogeneous by Corollary 14.4, the dimension

m0 = dimϕ(x,a)(M0)

does not depend on the realisation a of p0 in M0. We will show in Lemma 25.6that the same is true for any model of T .

Lemma 25.4. A countable model M is saturated if and only if its ϕ(x, a)–dimension is ω. Hence in this case, the dimension is independent of the reali-sation of p0(x) in M . In particular, T is ℵ0–categorical if and only if m0 = ω.

Proof. In a saturated model the ϕ(x, a)–dimension is infinite. Since there existsa unique countable saturated model by Lemmas 16.7 and 16.8, the first claimfollows. This obviously does not depend on the realisation of p0. The lastsentence now follows from Theorem 16.10.

We need the following observation:

Lemma 25.5. If M is prime over a finite set and m0 < ω, then dimϕ(x,a)(M)is finite.


Proof. Suppose M is prime over the finite set C. Let B be a basis of ϕ(M,a)over M0. Since M is the minimal prime extension of M0 ∪B, C is atomic overM0 ∪ B. Thus there exists a finite subset B0 of B such that C is contained inthe prime extension N of M0 ∪B0. As M is prime over Ca, it suffices to showthat dimϕ(x,a)(N) is finite and this follows by Theorem 25.1 from

dimϕ(x,a)(N) = m0 + |B0| .

The crucial lemma and promised converse to Corollary 22.3 is the following:

Lemma 25.6. The dimension dimϕ(x,a)(M) does not depend on the realisationa of p0 in M .

Proof. The lemma is clear if M is uncountable and also if M is countable withinfinite ϕ-dimension by Lemma 25.4. Therefore we may assume that M hasfinite ϕ-dimension, which implies that m0 is finite.

For the proof we now introduce the following notion: let a1 and a2 realisep0. Choose a model N of finite ϕ–dimension containing a1 and a2, which existsby Lemma 25.5, and put

diff(a1, a2) = dimϕ(x,a1)(N)− dimϕ(x,a2)(N).

This definition does not depend on the model N : if N ′ ≺ N is prime overa1, a2, then by Theorem 25.1 we have for i = 1, 2

dimϕ(x,ai)(N) = dimϕ(x,ai)(N′) + dim(N/N ′),

sodiff(a1, a2) = dimϕ(x,a1)(N

′)− dimϕ(x,a2)(N′).

Clearly we have

diff(a1, a3) = diff(a1, a2) + diff(a2, a3).

This implies that diff(a1, a2) only depends on tp(a1, a2). We will show thatdiff(a1, a2) = 0 for all realisations of p0. This implies that dimϕ(x,a1)(M) =dimϕ(x,a2)(M) for all models M which contain a1 and a2.

For the proof choose an infinite sequence a1, a2, . . . with

tp(ai, ai+1) = tp(a1, a2)

for all i.

Now we use the fact that ℵ1-categorical theories are ω-stable, so the type p0has Morley rank and an extension q0 to {a1, a2, . . .} of the same rank. Let b be


a realisation of q0. Then, by Corollary 24.12, there are at most MD(p0) manydifferent types of the form tp(bai). So let i < j be such that tp(bai) = tp(baj).

Then diff(ai, b) = −diff(b, aj) and

(j − i) diff(a1, a2) = diff(ai, aj) = diff(ai, b) + diff(b, aj) = 0,

implying diff(a1, a2) = 0.

Thanks to the previous lemma we obtain a complete account of the models ofan uncountably categorical theory.

Theorem 25.7 (Baldwin–Lachlan). If T is uncountably categorical, then forany cardinal m ≥ m0 there is a unique model M of T with dimϕ(x,a)(M) = m.These models are pairwise non-isomorphic.

Proof. If m = m0 + β, choose M prime over M0 ∪ {bi : i < β} where thebi ∈ ϕ(C, a0) are independent over M0. Uniqueness follows from Corollary 22.3and non-isomorphy from Lemma 25.6.

Exercise 25.1.All models of an ℵ1–categorical theory are ω-homogeneous.

Exercise 25.2.Let T be strongly minimal and m0 be the dimension of the prime model. Showthat m0 is the smallest number n such Sn+1(T ) is infinite.


26 Computation of Morley Rank

In this section we show that the Morley rank agrees with the dimension of thepregeometry on strongly minimal minimal sets and give some examples of howto compute it in ω-stable fields.

Lemma 26.1. If b is algebraic over aA, we have MR(b/A) ≤ MR(a/A).

Proof. Let MR(a/A) = α. We prove MR(b/A) ≤ α by induction over α.Let d = MD(b/Aa). Choose an L(A)–formula ϕ(x, y) in tp(ab/A) such thatMR(∃yϕ(x, y)) = α and |ϕ(a′,C)| ≤ d for all a′.

We show that the Morley rank of χ(y) = ∃xϕ(x, y) is bounded by α. Forthis consider an infinite family χi(C) of disjoint subclasses of χ(C) defined oversome extension A′ of A. Put ψi(x) = ∃y(ϕ(x, y) ∧ χi(y)). Since any d + 1 ofthe ψi have empty intersection, some ψi(x) has Morley rank β < α. Let b′ beany realisation of χi(y). Choose a′ such that |= ϕ(a′, b′). Then b′ is algebraicover a′A and since a′ realises ψi(x), we have MR(a′/A′) ≤ β. So by inductionwe conclude MR(b′/A′) ≤ β, which shows MRχi ≤ β. So χ does not containan infinite family of disjoint formulas of Morley rank greater or equal to α. SoMRχ ≤ α.

Theorem 26.2. Let ϕ(x) be a strongly minimal formula defined over B anda1, . . . , an a sequence of realisations. Then

MR(a1, . . . , an/B) = dimϕ(a1, . . . , an/B)

.

Proof. By the lemma we may assume that a1, . . . , an are independent over B.Let a1, . . . , an realise the L(B)–formula ψ(x1, . . . , xn). By induction we haveMR(a1, a2, . . . , an/Ba1) = n− 1. So the formula

χa1(x) = ψ(x1, . . . , xn) ∧ x1.= a1

has rank at least n−1. The infinitely many conjugates of χa1 over B are disjointand have rank n− 1 as well. This shows that MRψ ≥ n.

Let B′ ⊃ B be an extension of B. By Corollary 21.7 there is only one typep ∈ Sn(B

′) which is realised by an B′–independent sequence of elements of ϕ(C).So by induction, there is only one n-type of elements of ϕ(C) of rank ≥ n. Thisimplies that ϕ(x1) ∧ . . . ∧ ϕ(xn) has rank n.

Corollary 26.3. Let ϕ(x) be a strongly minimal formula and ψ(x1, . . . , xn) bedefined over B such that ψ implies ϕ(xi) for all i. Then

MRψ = max{dimϕ(a/B) | C |= ψ(a)}.


On strongly minimal, Morley rank is definable:

Corollary 26.4. For any strongly minimal formula ϕ(x) and any formulaψ(x1, . . . , xn, y) which implies ϕ(xi) for all i

{b | MRψ(x1, . . . , xn, b) = k}

is a definable class for every k.

Proof. We show that MRψ(x1, . . . , xn, b) ≥ k is an elementary property of bby induction on n. The case n = 1 follows from the fact that MRψ(x1, b) ≥ 1is equivalent to ∃∞x1ψ(x1, b). This is an elementary property of b since ϕ isstrongly minimal (see the discussion on page 98.) For the induction step weconclude from Corollary 26.3 that MRψ(x1, . . . , xn, b) ≥ k if and only if one ofthe following is true:

• there is an a1 such that MRψ(a1, x2 . . . , xn, b) ≥ k,

• there is an a1 which is is not algebraic over b such thatMRψ(a1, x2 . . . , xn, b) ≥ k − 1.

The first part an elementary property of b by induction. For the second partnote that by induction MRψ(a1, x2 . . . , xn, b) ≥ k − 1 can be expressed by aformula θ(a1, b). The second condition is then equivalent to ∃∞x1θ(x1, b).

For algebraically closed fields, these considerations translate into the follow-ing statement.

Corollary 26.5. Let K be a subfield of a model of TACFp and a a tuple ofelements. Then the Morley rank of a over K equals the transcendence degreeof K(a) over K.

Note that by quantifier elimination definable sets in algebraically closed fieldsare exactly the constructible sets in algebraic geometry. The previous corollaryexpresses the important fact that for a definable set in an algebraically closedfield the Morley rank equals the dimension of its Zariski closure in the sense ofalgebraic geometry (see e.g. [45]).

We now turn to the theory of differentially closed fields of characteristic 0,TDCF0 .

Let K ⊂ F be an extension of differential fields and a an element of F .The dimension of a over K is defined as the transcendence degree of K{a} overK. There is a unique quantifier free type over K of infinite dimension.

Remark 26.6. If the dimension of a over K equals n, then the type of a overK is determined by the d–minimal polynomial f of a over K: f is irreduciblein K[x0, . . . , xn] and f(a, . . . , dna) = 0, (see Remark B 6.7).


Corollary 26.7. TDCF0 is ω–stable. For a differential field K and elements awe have

MR(a/K) ≤ dim(a/K).

If a has infinite dimension, then the type of a over K has Morley rank ω.

Proof. There are at most |K| many d–minimal polynomials over K, so at most|K| many 1–types. Thus, TDCF0 is ω–stable.

We may assume that K is ℵ1–saturated (otherwise we take a extension oftp(a/K) to an ℵ1–saturated field with the same Morley rank. Then the Mor-ley rank stays the same and the dimension does not increase.) We showMR(a/K) ≤ dim(a/K) by induction on dim(a/K). If dim(a/K) = 0, thena is algebraic over K and the Morley rank is 0. Let dim(a/K) = n and let fbe the minimal polynomial of a over K. Apart from tp(a/K), all other typesover K containing f(x, . . . , dnx)

.= 0 have dimension and hence Morley rank

less than n. Since K is sufficiently saturated, this implies that the Morley rankof tp(a/K) is at most n.

By the next remark there are types of rank n for every n. This implies thatthere must be 1–types of rank ≥ ω. Since there is only one type p∞ of infinitedimension, it follows3 that p∞ has rank ω.

Remark 26.8. If a, . . . , dn−1a are algebraically independent over K and dna ∈K, then MR(a/K) = n.

Proof. We prove this by induction on n: Consider the formula ϕ(x) = (dn(x).=

dn(a)). If the claim is true for n − 1, all formulas ϕb(x) = (dn−1(x).= b) have

rank n−1. For all constants c the ϕdn−1(a)+c(x) are contained in ϕ(x). So ϕ(x)has rank n. All a′ which realise ϕ(x) have either at most dimension n − 1 orhave the same type as a over K. This shows that a has Morley rank n overK.

Dimension in pregeometries is additive, i.e. we have dim(ab/B) = dim(a/B)+dim(b/aB). This translates into additivity of Morley rank for elements in thealgebraic closure of strongly minimal sets:

Remark 26.9. Let ϕ be a strongly minimal formula defined over B and a, balgebraic over ϕ(C) ∪B. Then MR(ab/B) = MR(a/B) +MR(b/aB).

Proof. Assume B = ∅ for notational simplicity. Then a is algebraic over somefinite set of elements of ϕ(C). We can split this set into a sequence f of elementswhich are independent over a and a tuple a which is algebraic over fa. By takingnon–forking extensions if necessary, we may assume that f is independent fromab. Now a and a are interalgebraic over f . In the same way we find tuples g

3This is immediate if K is ω–saturated, see Exercise 24.3.


and b in ϕ(C) in which b is algebraic with g independent from abf and so thatb is interalgebraic with b over F = f g. The claim now follows from

MR(ab) = MR(ab/F )

= MR(ab/F ) = MR(a/F ) +MR(b/aF )

= MR(a/F ) +MR(b/aF ) = MR(a) +MR(b/a)

In fact, Exercise 26.5 shows that if F is any infinite B–independent subsetF of ϕ(C) then every element of acl(ϕ(C)∪B) is interalgebraic over FB with atuple in ϕ(C).

Exercise 26.1.Let T be strongly minimal. Show that a finite tuple a is geometrically indepen-dent from B over C if and only if MR(a/BC) = MR(a/C). (See page 235 andExercise C 8.1.)

Exercise 26.2.Let be ψ a formula without parameters. Assume that ψ is almost stronglyminimal, i.e. that there is a strongly minimal formula ϕ defined over some setB such that all elements of ψ(C) are algebraic over ϕ(C) ∪ B. Then for all a, bin ψ(C) and any set C we have MR(ab/C) = MR(a/C) +MR(b/aC).

The following exercise shows that for arbitrary totally transcendental theo-ries the Morley rank need not be additive:

Exercise 26.3.Consider the following theory in a two-sorted language having sorts A and Band a function f : B → A. Assume that sort A is split into infinitely manyinfinite predicates A1, A2, . . . such that any a ∈ An has exactly n preimagesunder f . Let a be a generic element of A, i.e., an element such that MR(a) ismaximal, and choose f(b) = a. Show that MR(ab) = MR(a) = 2, MR(b/a) = 1.

However, this is the worst that can happen: Shelah (see [47], Thm. V 7.8)and Erimbetov (see [14]) proved independently that the Morley rank of B isbounded by α(β + 1) if α ≥ 1 and MR(A) ≤ β and MR(f−1(a)) ≤ α for alla ∈ A. It can be shown that this bound is optimal.

Exercise 26.4.Find an example of a theory without the definable multiplicity property (DMP).

Exercise 26.5.Let ϕ be a strongly minimal formula without parameters and F an infiniteindependent subset of ϕ(C). Then every element of aclϕ(C) is interalgebraicover F with a tuple in ϕ(C).

Chapter 7

Simple theories

So far, we mainly studied totally transcendental theories, a small subclass ofthe class of stable theories, the totally transcendental theories being the moststable ones. Before we turn to stable theories in general, we consider simple (butpossibly unstable) theories, a generalisation which after their first introductionby Shelah [49] gained new attention after the fundamental work of Kim andPillay [30]. Interest in simple theories increased with Hrushovski’s results onpseudo-finite fields, see [25]. Much of the exposition is taken from Casanovas’article [12].

27 Dividing and Forking

We will characterise simple theories by the existence of a well-behaved notionof independence, a relation on types satisfying certain properties. To this endwe here define forking and dividing. In the context of totally transcendentaltheories, these concepts correspond to type extensions of smaller Morley rank.

We begin with a reformulation of the Standard Lemma 15.3 on indiscernibles.

Lemma 27.1 (The Standard Lemma). Let A be a set of parameters, I aninfinite sequence of tuples and J a linear order. Then there is a sequence ofindiscernibles over A of order type J realizing EM(I/A).

Definition 27.2. We say ϕ(x, b) divides over A (with respect to k) if thereis a sequence (bi)i<ω of realisations of tp(b/A) such that (ϕ(x, bi))i<ω is k-inconsistent.1 A set of formulas π(x) divides over A if π(x) implies some ϕ(x, b)which divides over A.

1A family (ϕi(x))x∈I is k–inconsistent if for every k–element subset K of I the set {ϕi |i ∈ K} is inconsistent.

125

CHAPTER 7. SIMPLE THEORIES 126

If ϕ(x, a) implies ψ(x, a′) and ψ(x, a′) divides over A, then ϕ(x, a) dividesover A. Thus ϕ divides over A if and only if {ϕ} divides over A. Also aset π divides over A if and only if a conjunction of formulas from π dividesover A. Note that it makes sense to say that π(x) divides over A for x aninfinite sequence of variables as we may use blind variables without changingthe meaning of dividing.

The following is easy to see:

Remark 27.3. 1. If a 6∈ acl(A), then tp(a/Aa) divides over A.

2. If π(x) is consistent and defined over acl(A), then π(x) does not divideover A.

Lemma 27.4. π(x, b) divides over A if and only if there is a sequence (bi)i<ω ofindiscernibles over A with tp(b0/A) = tp(b/A) and

⋃i<ω π(x, bi) inconsistent.

We may replace ω by any infinite linear order. Note also that b may be atuple of infinite length.

Proof. If (bi)i<ω is a sequence of indiscernibles over A with tp(b0/A) = tp(b/A)and

⋃i<ω π(x, bi) inconsistent there is a conjunction ϕ(x, b) of formulas from

π(x, b) for which Σ(x) = {ϕ(x, bi) | i < ω} is inconsistent. Σ contains somek–element inconsistent subset. This implies that (ϕ(x, bi))i<ω is k–inconsistent.

Assume conversely that π(x, b) divides over A. Then some finite conjunctionϕ(x, b) of formulas from π(x, b) divides. Let (bi)i<ω be a sequence of realisa-tions of tp(b/A) such that (ϕ(x, bi) | i < ω) is k-inconsistent. We may assumeby Lemma 27.1 that (bi)i<ω is indiscernible over A. Clearly,

⋃i<ω π(x, bi) is

inconsistent.

Corollary 27.5. The following are equivalent:

1. tp(a/Ab) does not divide over A.

2. For any infinite sequence of A-indiscernibles I containing b, there existssome a′ with tp(a′/Ab) = tp(a/Ab) and such that I is indiscernible overAa′.

3. For any infinite sequence of A-indiscernibles I containing b, there existsI ′ with tp(I ′/Ab) = tp(I/Ab) and such that I ′ is indiscernible over Aa.

Proof. 2)↔ 3): this is clear by considering appropriate automorphisms. It isalso easy to see that the conclusion of 2) and 3) is equivalent to:

(∗) There exist a′ and I ′ with tp(a′/Ab) = tp(a/Ab) and tp(I ′/Ab) =tp(I/Ab) such that I ′ is indiscernible over Aa′.


1)→ (∗): Let I = (bi)i∈I be an infinite sequence of indiscernibles with bi0 = b.Let p(x, y) = tp(ab/A). Then

⋃i∈I p(x, bi) is consistent by Lemma 27.4; let

a′ be a realisation. By Lemma 27.1, there is I ′′ = (b′′i )i∈I indiscernible overAa′ and realising EM(I/Aa′). Since |= p(a′, b′′i0), there is an automorphismα ∈ Aut(C/Aa′) taking b′′i0 to b. Put I ′ = α(I ′′).

2)→1) Let p(x, y) = tp(ab/A) and let (bi)i<ω be a sequence of indiscernibles overA with tp(b0/A) = tp(b/A). We have to show that

⋃i<ω p(x, bi) is consistent.

By assumption there is a′ with tp(a′/Ab) = tp(a/Ab) such that I is indiscernibleover Aa′. As |= p(a′, b), a′ is a realisation of

⋃i<ω p(x, bi).

The next proposition states a transitivity property of dividing. See Corol-lary 28.17 and its proof.

Proposition 27.6. If tp(a/B) does not divide over A ⊂ B and tp(c/Ba) doesnot divide over Aa, then tp(ac/B) does not divide over A.

Proof. Let b ∈ B be a finite tuple and I an infinite sequence of A–indiscerniblescontaining b. If tp(a/B) does not divide overA, there is some I ′ with tp(I ′/Ab) =tp(I, Ab) and indiscernible over Aa. If tp(c/Ba) does not divide over Aa, thereis I ′′ with tp(I ′′/Aab) = tp(I ′/Aab) and indiscernible over Aac proving theclaim.

Definition 27.7. π(x) forks over A if π(x) implies a disjunction∨l<d ϕl(x) of

formulas ϕl(x) each dividing over A.

Thus, if π(x) divides over A, it forks over A. By definition (and compact-ness), we immediately see the following:

Remark 27.8 (Non-forking is closed). If p ∈ S(B) forks over A, there is someϕ(x) ∈ p such that any type in S(B) containing ϕ(x) forks over A.

Corollary 27.9 (Finite character). If p ∈ S(B) forks over A, there is a finitesubset B0 ⊂ B such that p � AB0 forks over A.

Lemma 27.10. If π is finitely satisfiable in A, then π does not fork over A.

Proof. If π(x) implies the disjunction∨l<d ϕl(x, b), then some ϕl has a realisa-

tion a in A. If the bi, i < ω, realise tp(b/A), then {ϕl(x, bi) | i < ω} is realisedby a. So ϕl does not divide over A.

Lemma 27.11. Let A ⊂ B and let π be a partial type over B. If π does notfork over A, it can be extended to some p ∈ S(B) which does not fork over A.

Proof. Let p(x) be a maximal set of L(B)–formulas containing π(x) which doesnot fork over A. Clearly, p is consistent. Let ϕ(x) ∈ L(B). If neither ϕ nor ¬ϕbelongs to p, then both p ∪ {ϕ} and p ∪ {¬ϕ} fork over A, and hence p forksover A. Thus p is complete.


Exercise 27.1.

1. Let ϕ(x) be a formula over A with Morley rank and let ψ(x) define asubclass of ϕ(C). If ψ forks over A, it has smaller Morley rank than ϕ.

2. Let p be a type with Morley rank and q an extension of p. If q forks overA, it has smaller Morley rank than p.

We will see in Exercise 36.4 that in both statements the converse is also true.

Exercise 27.2.Let p be a type over the model M and A ⊂ M . Assume that M is |A|+–saturated. Show that p forks over A if and only if p divides over A.

Exercise 27.3.A global type which is A–invariant, i.e. invariant under all α ∈ Aut(C/A), doesnot fork over A.

Exercise 27.4.Let M be a κ–saturated and strongly κ–homogeneous model. If p ∈ S(M) forksover a subset A of cardinality smaller than κ, p has κ many conjugates underAut(M/A).

Exercise 27.5.Define the cyclical order on Q by

cyc(a, b, c) ⇔ (a < b < c) ∨ (b < c < a) ∨ (c < a < b).

Show:

1. (Q, cyc) has quantifier elimination.

2. cyc(a, x, b) divides over ∅ if a 6= b.

3. The unique p ∈ S1(∅) forks over ∅.

Exercise 27.6.If tp(a/Ab) does not divide over A and ϕ(x, b) divides over A with respect to k,then ϕ(x, b) divides over Aa with respect to k.


28 Simplicity

In this section, we define simple theories and the notion of forking independencewhose properties characterise such theories. By the absence of binary trees ofconsistent formulas, totally transcendental theories are simple. Using the factthat in simple theories types do not divide over (suitable) small sets, we will seelater that in fact all stable theories are simple.

Definition 28.1. 1. A formula ϕ(x, y) has the tree property (with respectto k) if there is a tree of parameters (as | ∅ 6= s ∈ <ωω) such that:

a) For all s ∈ <ωω, (ϕ(x, asi) | i < ω) is k–inconsistent.

b) For all σ ∈ ωω, {ϕ(x, as) | ∅ 6= s ⊂ σ} is consistent.

2. A theory T is simple if there is no formula ϕ(x, y) with the tree property.

Clearly, for simplicity it suffices to consider formulas ϕ(x, y) without parameters.

Remark 28.2. It is not hard to see that in totally transcendental theories noformula has the tree property. This is immediate for k = 2. In the general casethis follows from Exercise 24.1.

Definition 28.3. Let ∆ be a finite set of formulas ϕ(x, y) without parameters.A ∆–k–dividing sequence over A is a sequence (ϕi(x, ai) | i < δ) such that

1. ϕi(x, y) ∈ ∆

2. ϕi(x, ai) divides over A ∪ {aj | j < i} with respect to k.

3. {ϕi(x, ai) | i < δ} is consistent.

Lemma 28.4. 1. If ϕ has the tree property with respect to k, then for everyA and µ there exists a ϕ–k–dividing sequence over A of length µ.

2. If no ϕ ∈ ∆ has the tree property with respect to k, there is no infinite∆–k–dividing sequence over ∅.

Proof. 1): Note first that we may assume that µ is a limit ordinal. A compact-ness argument shows that for every µ and κ there is a tree (as | ∅ 6= s ∈ <µκ)such that all families (ϕ(x, asi) | i < κ) are k–inconsistent and for all σ ∈ µκ,{ϕ(x, as) | ∅ 6= s ⊂ σ} is consistent. If κ is bigger than 2max(|T |,µ), we finda path σ such that for all s ∈ σ, infinitely many asi have the same type over{at | t ≤ s}. Now (ϕi(x, ai+1) | i < µ) is a ϕ–k–dividing sequence.

2): Suppose there is an infinite ∆–k–dividing sequence over ∅. If ϕ appearsinfinitely many times in this sequence, there is an infinite ϕ–k–dividing sequence(ϕ(x, ai) | i < ω). For each i we choose a sequence (ani | n < ω) with tp(ani /{aj |j < i}) = tp(ai/{aj | j < i}) such that (ϕ(x, ani ) | n < ω) is k–inconsistent.


Then we find parameters bs showing that ϕ has the tree property with respectto k as follows: assume s ∈ i+1ω and ~b = (bs�1, . . . , bs�i) have been defined suchthat tp(a1, . . . , ai−1) = tp(~b). Choose α ∈ Aut(C) with α(a1, . . . , ai−1) = ~b andput bs = α(a

s(i)i ).

It is easy to see that in simple theories for every finite set ∆ and all k thereexists a finite bound on the possible length of ∆–k–dividing sequences.

Proposition 28.5. Let T be a complete theory. The following are equivalent.

a) T is simple

b) For all p ∈ S(B) there is some A ⊂ B with |A| ≤ |T | such that p does notdivide over A.

c) (Local Character) There is some κ such that for all models M and p ∈ S(M)there is some A ⊂M with |A| ≤ κ such that p does not divide over A.

Proof. a)→b): If b) does not hold, there is a sequence (ϕi(x, bi) | i < |T |+) offormulas from p(x) such that every ϕi(x, bi) divides over {bj | j < i} with re-spect to ki. There is an infinite subsequence for which all ϕi(x, y) equal ϕ(x, y)and all ki = k yielding a ϕ–k–dividing sequence.

b)→c): Clear.

c)→a): If ϕ has the tree property, there are ϕ–k–dividing sequences (ϕ(x, bi) |i < κ+). It is easy to construct an ascending sequence of models Mi, (i < κ+)such that bj ∈ Mi for j < i and ϕ(x, bi) divides over Mi. Extend the set ofϕ(x, bi) to some type p(x) ∈ S(M) where M =

⋃i<κ+ Mi. Then p divides over

each Mi.

Corollary 28.6. Let T be simple and p ∈ S(A). Then p does not fork over A.

Proof. Suppose p forks over A, so p implies some disjunction∨l<d ϕl(x, b) of

formulas all of which divide over A with respect to k. Put ∆ = {ϕl(x, y) | l < d}.

We will show by induction that for all n there is a ∆–k–dividing sequenceover A of length n. This contradicts the remark after Lemma 28.4. We willassume also that the dividing sequence is consistent with p(x).

Suppose that (ψi(x, ai) | i < n) is ∆–k–dividing sequence over A, consis-tent with p(x). By Exercise 28.4 we can replace b with a conjugate b′ over Asuch that (ψi(x, ai) | i < n) is a dividing sequence over Ab′. Now one of theformulas ϕl(x, b′), say ϕ0(x, b

′), is consistent with p(x) ∪ {ψi(x, ai) | i < n}. Soϕ0(x, b

′), ψ0(x, a0), . . . , ψn−1(x, an−1) is a ∆–k–dividing sequence over A andconsistent with p(x).


Let p be a type over A and q an extension of p. We call p a forking extensionif q forks over A.

Corollary 28.7 (Existence). If T is simple, every type over A has a non–forkingextension to any B containing A.

Proof. This follows from Corollary 28.6 and Lemma 27.11.

Definition 28.8. A is independent from B over C,

A |C

B,

if for every finite tuple a from A, the type tp(a/BC) does not fork over C.

This definition makes sense since forking of tp(a/BC) does not depend on theenumeration of a and since tp(a/BC) forks over C if the type of a subsequenceof a forks over C. So this is the same as saying that tp(A/BC) does not forkover C.

Definition 28.9. Let I be a linear order. A sequence (ai)i∈I is called

1. independent over A if ai |A{aj | j < i} for all i,

2. Morley sequence over A if it is independent and indiscernible over A,

3. Morley sequence in p(x) over A if it is a Morley sequence over A consistingof realisations of p.

Example 28.10. Let q be a global type invariant over A. Then any sequence(bi)i∈I where each bi realises q � A ∪ {bj | j < i} is a Morley sequence over A.

Proof. Let us call such sequences good. Clearly a subsequence of a good se-quence is good again. So for indiscernibility it suffices to show that all finitegood sequences b0 . . . bn and b′0 . . . b′n have the same type over A. Indeed, usinginduction, we may assume that b0 . . . bn−1 and b′0 . . . b

′n−1 have the same type

and so α(b0 . . . bn−1) = b′0 . . . b′n−1 for some α ∈ Aut(C/A). Then

α(tp(bn/Ab0 . . . bn−1)) = α(q � Ab0 . . . bn−1) = q � Ab′0 . . . b′n−1

= tp(b′n/Ab′0 . . . b

′n−1),

which proves our claim.

We call such a sequence (bi)i∈I a Morley sequence of q over A. Note thatour proof shows that the type of a Morley sequence of q over A is uniquelydetermined by its order type.

Lemma 28.11. If (ai)i∈I is independent over A and J < K are subsets of I,then tp((ak)k∈K/A{aj | j ∈ J}) does not divide over A.


Proof. We may assume that K is finite. The claim now follows from Proposi-tion 27.6 by induction on |K|.

Lemma 28.12 (Shelah). For all A there is some λ such that for any linear orderI of cardinality λ and any family (ai)i∈I there exists an A–indiscernible sequence(bj)j<ω such that for all j1 < . . . < jn < ω there is a sequence i1 < . . . < in inI with tp(ai1 . . . ain/A) = tp(bj1 . . . bjn/A).

Proof. We only need that λ satisfies the following: let τ = supn<ω |Sn(A)|.

1. cf(λ) > τ

2. For all κ < λ and all n < ω there is some κ′ < λ with κ′ → (κ)nτ (seeAppendix C, Definition 10.1).

By Erdös-Rado (see Appendix C, Theorem 10.2) we may take λ = iτ+ .

We now construct a sequence of types p1(x1) ⊂ p2(x1, x2) ⊂ . . . with pn ∈Sn(A) such that for all κ < λ there is some I ′ ⊂ I with |I ′| = κ such thattp(ai1 . . . ain) = pn for all i1 < . . . < in from I ′.

Then we can choose the (bi)i<ω as a realisation of⋃i<ω pi.

If pn−1 has been constructed and we are given κ < λ, we choose κ′ < λwith κ′ → (κ)nτ and some I ′ ⊂ I with |I ′| = κ′ such that tp(ai1 . . . ain−1

/A) =pn−1 for all i1 < . . . < in−1 from I ′. Thus there are I ′′ ⊂ I ′ and pκn withtp(ai1 . . . ain) = pκn for all i1 < . . . < in from I ′′. Since cf(λ) > τ , there is somepn with pκn = pn for cofinally many κ.

The existence of a Ramsey cardinal κ > τ (see p. 239) would directly implythat any sequence of order type κ contains a countable indiscernible subsequence(in fact even an indiscernible subsequence of size κ).

Lemma 28.13. If p ∈ S(B) does not fork over A, there is an infinite Morleysequence in p over A which is indiscernible over B. In particular, if T is simple,for every p ∈ S(A), there is an infinite Morley sequence in p over A.

Proof. Let a0 be a realisation of p. By Lemma 27.11 there is a non–forkingextension p′ of p to Ba0. Let a1 be a realisation of p′. Continuing in thisway we obtain a sequence (ai)i<λ with ai |

AB(aj)j<i for arbitrary λ. By

Lemma 28.12 we obtain a sequence of length ω with the same property andindiscernible over B. The last sentence is immediate by Corollary 28.6.

Proposition 28.14. Let T be simple and π(x, y) be a partial type over A.Let (bi)i<ω be an infinite Morley sequence over A and

⋃i<ω π(x, bi) consistent.

Then π(x, b0) does not divide over A.


Proof. By Lemma 27.1, for every linear order I there is a Morley sequence (bi)i∈Iin tp(b0/A) over A such that Σ(x) =

⋃i∈I π(x, bi) is consistent. Choose I having

the inverse order type of |T |+. Let c be a realisation of Σ. By Proposition 28.5.bthere is some i0 such that tp(c/A ∪ {bi | i ∈ I}) does not divide over A ∪ {bi |i > i0}. This implies that tp(c/A ∪ {bi | i ≥ i0}) does not divide over A ∪ {bi |i > i0}. By Lemma 28.11, tp((bi | i > i0)/Abi0) does not divide over A. Hencetp(c (bi | i > i0)/Abi0) does not divide over A by Proposition 27.6. This impliesthat π(x, bi0) does not divide over A.

Proposition 28.15. Let T be simple. Then π(x, b) divides over A if and onlyif it forks over A.

Proof. By definition, if π(x, b) divides over A, it forks over A. For the converseassume π(x, b) does not divide over A. So if ψ(x, b) =

∨l<d ϕl(x, b) is implied by

π(x, b), it does not divide over A. Let (bi)i<ω be a Morley sequence in tp(b/A)over A, which exists since T is simple. So {ψ(x, bi) | i ∈ ω} is consistent. By thepigeon hole principle there must be some l and some infinite I ⊂ ω such that{ϕl(x, bi) | i ∈ I} is consistent. By Proposition 28.14, ϕl(x, b) does not divideover A. Hence π(x, b) does not fork over A.

Proposition 28.16 (Symmetry). In simple theories, independence is symmet-ric.

Proof. Assume A |CB and consider finite tuples a ∈ A and b ∈ B. Since

a |Cb, Lemma 28.13 gives an infinite Morley sequence (ai)i<ω in tp(a/Cb)

over C, indiscernible over Cb. Let p(x, y) = tp(ab/C). Then⋃i<ω p(ai, y) is

consistent because it is realised by b. Thus, by Proposition 28.14, p(a, y) doesnot divide over C. This proves b |

Ca. Since this holds for all a ∈ A, b ∈ B, it

follows B |CA by Finite character.

Corollary 28.17 (Monotony and Transitivity). Let T be simple, B ⊂ C ⊂ D.Then we have A |

BD if and only if A |

BC and A |

CD.

Proof. One direction of this equivalence, Monotony, holds for arbitrary theoriesand follows easily from the definition. For Transitivity, the other direction, notethat by Proposition 28.15 we may read Proposition 27.6 after replacing finitetuples by infinite ones as

A′ |A

B and C |AA′

B ⇒ CA′ |A

B.

Swapping the left and the right hand side, this is exactly the transitivity. Hencethe claim follows from Proposition 28.16.

Corollary 28.18. (ai)i∈I being independent over A does not depend on theordering of I.


Proof. Let i be an element of I and J,K two subsets such that J < i < K. WriteaJ = {aj | j ∈ J} and aK = {ak | k ∈ K}. We have to show that i |

AaJaK .

Now by Lemma 28.11 we have aK |AAJ i. Monotony yields aK |

AAJi and

by Symmetry we have i |AAJ

AK . The claim follows now from i |AAJ and

Transitivity.

So we can define a family (ai | i ∈ I) to be independent over A if it isindependent for some ordering of I. Clearly (ai)i∈I is independent over A ifand only if ai |

A{aj | j 6= i} for all i. One calls a set B independent over B if

b |A(B \ {b}) for all b ∈ B.

The following lemma is a generalisation of Proposition 28.14.

Lemma 28.19. Let T be simple and I be an infinite Morley sequence over A.If I is indiscernible over Ac, then c |

AI.

Proof. We may assume that I = (ai)i<ω. Consider any ϕ(x, a0, . . . , an−1) ∈tp(c/AI). Put bi = (ani, . . . , ani+n−1). Then by Lemma 28.11 (bi)i<ω is againa Morley sequence over A and {ϕ(x, bi) | i ∈ ω} is consistent since realised byc. We see from Proposition 28.14 that ϕ(x, a0, . . . , an−1) does not fork overA.

Exercise 28.1.If T is simple, there does not exist an ascending chain (pα)α∈|T |+ of forkingextensions.

Exercise 28.2 (Diamond Lemma).Assume T to be simple and p ∈ S(A). Let q be a non–forking extension of p andr any extension of p. Then there is an A–conjugate r′ of r with a non–forkingextension s which also extends q.

p

q r′

s

��

SSS

��

SSS

nf

nf

r′ can be chosen in such a way that the domains of r′ and q are independentover A.

Exercise 28.3.If T is simple and (ai)i∈I is an A-independent sequence, then

aX |AaX∩Y

aY

for all X,Y ⊂ I where aX = {ai | i ∈ X}.


Exercise 28.4.If ϕ(x, b) divides over A and A ⊆ B, there is some A-conjugate B′ of B suchthat ϕ(x, b) divides over B′.

Exercise 28.5.If T is simple, then

ab |A

B ⇐⇒ a |A

B and b |Aa

B.

Exercise 28.6.Assume that T simple and b1 . . . bn |

AC. Then the sequence b1, . . . , bn is

independent over A if and only if it is independent over AC.

Exercise 28.7.If T is simple and Aa |

BC, then a ∈ acl(ABC) implies a ∈ acl(AB).


29 The Independence Theorem

The core of this section is the characterisation of simple theories in terms ofa suitable notion of independence. This is due to Kim and Pillay and willbe applied to pseudo-finite fields in Section 31. We will later specialise thischaracterisation to stable theories. Unless explicitly stated otherwise, weassume throughout this section that T is a simple theory.

Definition 29.1. For any set A we write ncA(a, b) if a and b start an infinitesequence of indiscernibles over A.

A formula θ(x, y) is called thick if there are no infinite antichains, i.e. se-quences (ci)i<ω where ¬θ(ci, cj) for all i < j < ω. By compactness this saysthat there is a bound k < ω on the length of finite antichains. See Exercise 29.2for an explanation of the terminology.

Lemma 29.2. (T arbitrary.) For any set A and n–tuples a, b the following areequivalent:

a) ncA(a, b)

b) |= θ(a, b) for all thick θ(x, y) defined over A.

In particular, ncA is type–definable.

Proof. Let p(x, y) = tp(ab/A). By 15.3, a and b start an infinite sequence ofindiscernibles if and only if there is a sequence (ci)i<ω with |= p(ci, cj) for i < jif and only if for all ϕ ∈ p the complement of ϕ(C) contains arbitrarily longantichains, and so

6|= ψ(a, b) ⇒ ψ is not thick.

Corollary 29.3. (T arbitrary.) If a and b have the same type over a modelM , there is some c such that ncM (a, c) and ncM (c, b).

Proof. We have to show |= ∃z(ϕ(a, z)∧ϕ(z, b)) for every thick formula ϕ(x, y) ∈L(M). We may assume that ϕ is symmetric.2 Since M is a model, there is amaximal antichain a0, . . . , ak−1 of ϕ in M . Thus, for some i we have |= ϕ(a, ai)and hence |= ϕ(b, ai).

In Exercise 32.1 below we give a different proof of the corollary, independentfrom Lemma 29.2.

Lemma 29.4. (T arbitrary.) Let (bi)i<ω be indiscernible over A and (bi)1≤i<ωindiscernible over Aa0b0. Then there is some a1 such that ncA(a0b0, a1b1).

2I.e. |= ∀x, y (θ(x, y) → θ(y, x)).


Proof. Choose ai with tp(aibibi+1 . . . /A) = tp(a0b0b1 . . . /A). Let a′0b0, a′1b1, . . .be a sequence of indiscernibles realizing the EM-type of a0b0, a1b1, . . . over A.Since (bi)1≤i<ω is indiscernible over Aa0b0, we have

tp(a′i1 , bi1 , bi2 , . . . , bin/A) = tp(a0, b0, b1, . . . , bn/A)

for all i1 < . . . < in. So tp(a′0, b0, b1, . . . /A) = tp(a0, b0, b1, . . . /A) and we mayassume that a0b0, a1b1, . . . is indiscernible over A.

Lemma 29.5. Let I be indiscernible over A and J an infinite initial segmentwithout last element.Then I \ J is a Morley sequence over AJ .

Proof. Let I = (ai)i∈I and J = (ai)i∈J . It suffices to show ai |AJ aX for all

i ∈ I \ J and all finite X ⊂ I with X < i. But this follows from Lemma 27.10as tp(aX/AJ ai) is finitely satisfiable in AJ .

Proposition 29.6. If ϕ(x, a) does not fork over A and ncA(a, b), then ϕ(x, a)∧ϕ(x, b) does not fork over A.

Proof. Let I be an infinite sequence of indiscernibles over A containing a andb. We extend I by an infinite initial segment J without last element. Let c bea realisation of ϕ(x, a) independent from J a over A. By Corollary 27.5 we mayassume I to be indiscernible over AJ c.

It follows from Lemma 29.5 that I is a Morley sequence over AJ . So byLemma 28.19 we have c |

AJ I. Transitivity now implies c |AJ I and hence

the claim.

Lemma 29.7. Let ncA(b, b′) and a |

Abb′. Then there is some a′ with

ncA(ab, a′b′).

Proof. Let (bi)i<ω indiscernible over A, b = b0 and b′ = b1. By Corollary 27.5we may assume (bi)1≤i<ω to be indiscernible over Aab. The claim now followsfrom Lemma 29.4.

Proposition 29.8. If ϕ(x, a) ∧ ψ(x, b) does not fork over A, ncA(b, b′) and

a |Abb′, then also ϕ(x, a) ∧ ψ(x, b′) does not fork over A.

Proof. By Lemma 29.7 there is some a′ such that ncA(ab, a′b′). Proposition 29.6implies that ϕ(x, a) ∧ ψ(x, b) ∧ ϕ(x, a′) ∧ ψ(x, b′) does not fork over A.

Corollary 29.9. Assume that ϕ(x, a) ∧ ψ(x, b) does not fork over A, a |Ab′b

and that b and b′ have the same type over some model containing A. Thenϕ(x, a) ∧ ψ(x, b′) does not fork over A.


Proof. By Corollary 29.3 there is some c such that ncA(b, c) and ncA(c, b′) By re-

placing a, if necessary, by a realisation of a non–forking extension of tp(a/Abb′)to Abb′c, which exists by Corollary 28.7, we may assume that a |

Abb′c. Propo-

sition 29.8 yields now first that ϕ(x, a) ∧ ψ(x, c) does not fork over A and thenthat ϕ(x, a) ∧ ψ(x, b′) does not fork over A.

Corollary 29.10. Let a |Mb, a′ |

Ma, b′ |

Mb, |= ϕ(a′, a) ∧ ψ(b′, b) and

assume that a′ and b′ have the same type over M . Then ϕ(x, a) ∧ ψ(x, b) doesnot fork over M .

Proof. Choose a′′ such that tp(a′′a′/M) = tp(bb′/M) and a′′ |Ma′

abb′. Thenby Transitivity we have

a′′ |M

aa′bb′.

It follows that the sequences a, a′, a′′ and a, b, a′′ are both independent over A.This implies

(1) a′ |M

aa′′

and

(2) a |M

a′′b

Since |= ψ(a′, a′′), (1) implies that ϕ(x, a) ∧ ψ(x, a′′) does not fork over M .So ϕ(x, a) ∧ ψ(x, b) does not fork over M by (2) and Corollary 29.9.

Theorem 29.11 (Independence Theorem). Suppose that b and c have the sametype over the model M and suppose that

B |M

C, b |M

B and c |M

C.

Then there exists some d such that d |MBC, tp(d/B) = tp(b/B) and tp(d/C) =

tp(c/C).

Proof. By Corollary 29.10, tp(b/B)∪ tp(c/C) does not fork over M . So we finda d such that d |

MBC, tp(d/B) = tp(b/B) and tp(d/C) = tp(c/C).

Corollary 29.12. Let Bi, i ∈ I, be independent over M and bi such thatbi |

MBi all bi having the same type over M . Then there is some d such that

d |M{Bi | i ∈ I} and tp(d/Bi) = tp(bi/Bi) for all i.

Proof. Well–order I and show the existence of d by recursively constructingpi = tp(d/{Bi | i ∈ I}). The details are left as an exercise.


Theorem 29.13 (Kim-Pillay [30]). Let T be a complete theory and a |0AB a

relation between finite tuples a and sets A and B invariant under automorphismsand having the following properties:

a) (Monotony and Transitivity) a |0ABC if and only if a |0

AB and

a |0AB

C.

b) (Symmetry) a |0Ab if and only if b |0

Aa.

c) (Finite Character) a |0AB if a |0

Ab for all finite tuples b ∈ B.

d) (Local Character) There is a cardinal κ, such that for all a and B thereexists B0 ⊂ B of cardinality less than κ such that a |0

B0B

e) (Existence) For all a, A and C there is a′ such that tp(a′/A) = tp(a/A)and a′ |0

AC.

f) (Independence over models) Let M be a model, a |0Mb, a′ |0

Ma,

b′ |0Mb and tp(a′/M) = tp(b′/M). Then there is some c such that c |0

Mab,

tp(c/Ma) = tp(a′/Ma) and tp(c/Mb) = tp(b′/Mb).

Then T is simple and |0 = | .

We have seen that in simple theories the relation | satisfies the propertiesof the previous theorem (for κ = |T |+).

Proof. We may assume κ to be a regular cardinal, otherwise just replace κ by κ+.

Assume now a |0Ab. We will use Lemma 27.4 to show that tp(a/Ab) does not

divide over A. So, let (bi)i<ω be a sequence of A–indiscernibles starting withb = b0.

Claim: We can find a model M containing A such that (bi)i<ω is indiscernibleover M and bi |0

M{bj | j < i} for all i.

Proof: By Lemma 27.1 we can extend the sequence to (bi)i≤κ. Furthermore,it is easy to construct an ascending sequence of models A ⊂ M0 ⊂ M1 . . .such that for all i < κ all bj , (j < i) is contained in Mi and (bj)i≤j≤κ isindiscernible over Mi. By Local Character there is some i0 such thatbκ |0

Mi0

{bj | i0 ≤ j < κ}. From the Indiscernibility it now follows that

bi |0Mi0

{bj | i0 ≤ j < i} for all i. We can take M = Mi0 and the sequencebi0 , bi0+1, . . ..

Claim: We may assume a |0Mb.


Proof: By Existence we may replace a by a′ with tp(a′/Ab) = tp(a/Ab) anda′ |0

AbM and then apply Monotony and Transitivity.

We now find elements a = a0, a1, . . . such that ai |0M{bj | j ≤ i}, qi(x) =

tp(ai+1/M{bj | j ≤ i}) = tp(ai/M{bj | j ≤ i}) and tp(aibi/M) = tp(ab/M):given a0 . . . , ai, choose a′ with tp(a′bi+1/M) = tp(ab/M). Now apply Inde-pendence over Models to

{bj | j ≤ i} |0M

bi+1, ai |0M

{bj | j ≤ i} and a′ |0M

bi+1

to find ai+1.

This implies that⋃i<ω qi(x) is consistent and contains all p(x, bi) where

p(x, y) = tp(ab/M). By Lemma 27.4, tp(a/Ab) does not divide over A. Sim-plicity of T now follows from Local Character and Proposition 28.5.

It is left to show a |0Ab if tp(a/Ab) does not divide over A. Using Ex-

istence we construct for any λ a sequence (bi)i<λ which is |0A-independent

and for which tp(bi/A) = tp(b/A). If this sequence is sufficiently long, the sameargument as for Lemma 28.13 (but now using Monotony and Finite Char-acter) yields an A–indiscernible sequence (b′i)i<κ which is |0

A-independent as

well and satisfies tp(b′i/A) = tp(b/A) and b = b′0. We now apply Corollary 27.5and obtain a′ such that tp(a′/Ab) = tp(a/Ab) and so that (b′i)i<κ is indiscernibleover Aa′. Local Character and Monotony yield the existence of i0 with

a′ |0A{b′i|i<i0}

b′i0 .

Sinceb′i0 |0

A

{b′i | i < i0}

we geta′ |0

A

b′i0 and hence a |0A

b

from Symmetry and Transitivity using that tp(a′b′i0/A) = tp(a′b′0/A) =tp(ab/A).

Corollary 29.14. The theory of the random graph is simple.

Proof. Define A |0BC by A ∩ C ⊂ B and apply Theorem 29.13,

Exercise 29.1.Let T be simple. Assume that the partial type π(x, b) does not fork over A andthat I is an infinite sequence of indiscernibles over A containing b. Show thatthere is a realisation c of π(x, b) such that c |

AI and I is indiscernible over

Ac.


Exercise 29.2.A symmetric formula is thick if and only if there is no infinite anti-clique, i.e.a sequence (ci)i<ω, where |= ¬θ(x, y) for all i 6= j. This explains the notationncA. Prove the following, without using Lemma 29.2:

1. The conjunction of two thick formulas is thick.

2. If θ(x, y) is thick, then θ∼(x, y) = θ(y, x) is thick.

3. A formula is thick if and only if it is implied by a symmetric thick formula.

Exercise 29.3.Let T be a simple theory and A a set of parameters. Assume that there is anelement b which is algebraic over A but not definable over A. Then the Inde-pendence Theorem does not hold if in its formulation the model M is replacedby A.

Exercise 29.4.Fill in the details of the proof of Corollary 29.12.

Exercise 29.5.Prove directly from the axioms in Theorem 29.13 that

a |0A

B ⇔ a |0A

AB.

Exercise 29.6.Let T be simple and p be a type over a model. Then p has either exactly onenon–forking extension to C (p is stationary) or arbitrarily many.


30 Lascar strong types

In this section we will prove a version of the Independence Theorem 29.11 overarbitrary parameter sets A. For this we have to strengthen the assumption thatb and c have the same type over A to having the same Lascar strong type overA. In what follows T is an arbitrary complete theory.

Definition 30.1. Let A be any set of parameters. The group Autf (C/A) of Las-car strong automorphisms of C over A is the group generated by all Aut(C/M)where the M are models containing A. Two tuples a and b have the same Lascarstrong type over A if α(a) = b for some α ∈ Autf (C/A). We denote this byLstp(a/A) = Lstp(b/A).

It is easy to see that two elements a and b have the same Lascar strong typeover A if and only if there is a sequence a = b0, b1, . . . , bn = b such that for allfor all i < n, bi and bi+1 have the same type over some model containing A.

Lemma 30.2. Assume that T is simple. If ϕ(x, a)∧ ψ(x, b) does not fork overA, Lstp(b/A) = Lstp(b′/A) and a |

Ab′b, then ϕ(x, a) ∧ ψ(x, b′) does not fork

over A.

Proof. Choose a sequence b = b0, b1, . . . , bn = b′ such that for each i < n, bi andbi+1 have the same type over some model containing A. By the properties offorking we may assume that a |

Ab0b1 . . . bn. We thus always have a |

Abibi+1

and the claim follows by induction from Corollary 29.9.

As in the previous proof we will repeatedly use Existence (Corollary 28.7)to assume that we have realisations of types which are independent from othersets. This is also crucial in the following lemma:

Lemma 30.3. Assume T to be simple. For all a,A and B there is some a′such that Lstp(a′/A) = Lstp(a/A) and a′ |

AB.

Proof. By Existence and Symmetry, we can choose a model M ⊃ A witha |

AM . By Existence again we can find a′ such that tp(a′/M) = tp(a/M)

and a′ |MB, so the claim now follows from Corollary 29.3 and Transitivity.

A stronger statement will be proved in Exercise 30.3.

Corollary 30.4. Let Lstp(a/A) = Lstp(b/A). For all a′, B there exists b′ suchthat Lstp(aa′/A) = Lstp(bb′/A) and b′ |

AbB.

Proof. Choose bb′ such that Lstp(aa′/A) = Lstp(bb′′/A) and b′ by the Lemmawith b′ |

AbB and Lstp(b′′/Ab) = Lstp(b′/Ab). It is easy to see that this implies

Lstp(bb′′/A) = Lstp(bb′/A).


Corollary 30.5. Let a |Ab, a′ |

Aa, b′ |

Ab, Lstp(a′/A) = Lstp(b′/A) and

|= ϕ(a′, a) ∧ ψ(b′, b). Then ϕ(x, a) ∧ ψ(x, b) does not fork over A.

Proof. Choose a′′ by Corollary 30.4 such that Lstp(a′′a′/A) = Lstp(bb′/A)and a′′ |

Aa′abb′. Then proceed as in the proof of Corollary 30.5, but use

Lemma 30.2 instead of Corollary 29.9.

Theorem 30.6 (Independence Theorem). Let T be simple and supposeLstp(b/A) = Lstp(c/A),

B |A

C, b |A

B and c |A

C.

Then there exists some d such that d |ABC, Lstp(d/B) = Lstp(b/B) and

Lstp(d/C) = Lstp(c/C).

Proof. By Corollary 30.5, tp(b/B) ∪ tp(c/C) does not fork over A. So we findsome d such that d |

ABC, tp(d/B) = tp(b/B) and tp(d/C) = tp(c/C). The

stronger claim about Lascar strong types is left as Exercise 30.2.

Corollary 30.7. Assume T to be simple and let Bi, i ∈ I, be independent overA and bi such that bi |

ABi all bi having the same Lascar strong type over A.

Then there is some d such that d |A{Bi | i ∈ I} and Lstp(d/Bi) = Lstp(bi/Bi)

for all i.

Proof. Assume the Bi are models containing A. The proof goes then as theproof of Corollary 29.12.

Lemma 30.8. Let T be simple, a |Ab and Lstp(a/A) = Lstp(b/A), there is

an infinite Morley sequence over A containing a and b.

Proof. Consider p(x, y) = tp(ab/A). Starting from a0 = a and a1 = b werecursively construct a long independent sequence (ai) of elements all havingthe same Lascar type over A. If (ai | i < α) is given, the p(ai, y) are realised byelements bi with bi |

Aai and Lstp(bi/A) = Lstp(a/A). By Corollary 30.7 there

is some aα with aα |A{ai | i < α}, |= p(ai, aα) for all i < α and Lstp(aα/A) =

Lstp(a/A). If the sequence is sufficiently long, by Lemma 28.12 there is an A–indiscernible sequence (a′i)i<ω such that |= p(a′i, a

′j) for all i < j and furthermore

the sequence is independent over A because all types (a′j/A{a′i | i < j}) appearin (ai). Since tp(a′1a

′0/A) = tp(ba/A), we may assume a′0 = a and a′1 = b.

Corollary 30.9. Lstp(a/A) = Lstp(b/A) if and only if nc2A(a, b). In particular,the relation EL

A(a, b) ⇔ Lstp(a/A) = Lstp(b/A) is type–definable.


Proof. Choose c with c |Aab and Lstp(c/A) = Lstp(a/A) = Lstp(b/A). By

Lemma 30.8, we have ncA(c, a) and ncA(c, b). Hence

ELA(x, y) ⇔ ∃z (ncA(x, z) ∧ ncA(z, y))

and this is type–definable by Lemma 29.2.

It is an open problem whether in simple theories Lascar strong types are thesame as strong types (see Exercise 35.9).

Exercise 30.1.Show that ncA(a, b) implies ncacl(A)(a, b).

Exercise 30.2.Deduce Theorem 30.6 from the weaker version which claims only the equalitiestp(d/B) = tp(b/B) and tp(d/C) = tp(c/C).

Exercise 30.3.Show that in arbitrary theories ncA(x, a) does not divide over A. Use this toprove a stronger version of Lemma 30.3: Assume that T is simple. For alla,A,B there is some a′ such that ncA(a, a′) and a′ |

AB.

Exercise 30.4.If ncA(a, b), there is some model M containing A such that tp(a/M) = tp(b/M).Conclude that EL

A is the transitive closure of ncA.

Exercise 30.5.A relation R ⊂ Cn × Cn is called bounded if there are no arbitrarily long an-tichains, i.e. sequences (cα | α < κ) with ¬R(cα, cβ) for all α < β < κ. Showthat the intersection of a family Ri, (i ∈ I), of bounded relations is againbounded.

Exercise 30.6.We call a relation A–invariant if it is invariant under all automorphisms inAut(C/A). Show that ncA is the smallest bounded A–invariant relation.

Exercise 30.7.Show that EL

A is the smallest bounded A–invariant equivalence relation.


31 Example: Pseudo-finite fields

We now turn to an important example of simple theories, namely those ofpseudo-finite fields. A perfect field K is called pseudo-finite if it is pseudo-algebraically closed, i.e. if every absolutely irreducible affine variety defined overK has a K-rational point and if its absolute Galois group is Z, i.e., K has aunique extension of degree n for each n ≥ 1. For background on pseudo-finitefields and profinite groups see Appendix B 7.

Proposition 31.1. Let L1 and L2 be regular procyclic extensions of a field Kand let L2 be pseudo-finite. Then L1 can be regularly embedded over K into anelementary extension of L2.

Proof. By Lemma B 7.16, we may assume that N is a common regular procyclicextension of the Li. As a regular procyclic extension of L2, N is 1–free (byAppendix B 7.14). Since L2 is existentially closed in N , N is embeddable overL2 into an elementary extension L′

2 of L2. Let N ′ denote the image of thisembedding. By Appendix B 7.14, L′

2/N′ is regular.

Theorem 31.2. Let L1 and L2 be regular pseudo-finite extensions of K. ThenL1 and L2 are elementarily equivalent over K.

Proof. By Proposition 31.1 we obtain an alternating elementary chain. Its unionis an elementary extension of both L1 and L2.

Corollary 31.3. Let L be pseudo-finite and Abs(L) the relative algebraic closureof the prime field in L, the absolute part of L. The elementary theory of L isdetermined by the isomorphism type of Abs(L). A field K algebraic over itsprime field is the absolute part of some pseudo-finite field if and only if it isprocyclic (this is always true in finite characteristic).

We now fix the complete theory of a pseudo-finite field and work in itsmonster model C.

Corollary 31.4. Let K be a subfield of C, a and b tuples of elements C. Thena and b have the same type over K if and only if the relative algebraic closuresof K(a) and K(b) in C are isomorphic over K via some isomorphism taking ato b.

Proof. Let A and B be the relative algebraic closures of K(a) and K(b), respec-tively. If a and b have the same type over K, then A and B are isomorphic in therequired way by Lemma B 6.13. Conversely, if A and B are isomorphic over Kby such an isomorphism, the claim follows immediately from Theorem 31.2.

Theorem 31.5. In pseudo-finite fields, algebraic independence has all the prop-erties of forking listed in Theorem 29.13.


Proof. We keep working in C. All properties are clear except (Existence) and(Independence over models).

For (Existence): Let K be a subfield of C and L and H two extensionsof K. We may assume that all three fields are relatively algebraically closed inC. By Appendix B 7.16, there is a procyclic extension C of H (not necessarilycontained in C) containing a copy L′ of L/K independent from H over K andsuch that C/L′ is regular. By Proposition 31.1, C can be regularly over Hembedded into C. Let L′′ denote the image of L′ in C. Then L′′ and H areindependent over K and L′′ and L have the same type over K (see p. 32).

For (Independence over models): Let M be an elementary submodelof C and let K and L be field extensions independent over M . Assume furtherthat we are given extensions K ′ and L′ so that K and K ′ as well as L and L′

are independent over M and such that K ′ and L′ have the same type over M .

We may assume that all these fields are relatively algebraically closed inC. Then, if σ is a generator of G(C), the relative algebraic closures of if KK ′

and LL′ in C are the fixed fields of κ′ = σ � (KK ′)alg in (KK ′)alg and ofλ′ = σ � (LL′)alg, respectively.

We now take another field extension H/M (possibly outside C) isomorphicto K ′/M and L′/M and independent of KL over M . Then KK ′ and KH, andLL′ and LH, respectively, are isomorphic over M and the isomorphisms arecompatible with the given isomorphism between K ′ and L′. We transport κ′and λ′ to (KH)alg and (LH)alg via these isomorphism and call the transportedautomorphisms κ and λ. Clearly κ, λ and µ = σ � (KL)alg agree on Kalg andLalg. They also agree on Halg, since κ � Halg and λ � Halg are both the uniqueextension of σ �Malg to G(H).

Since (KH)alg and (LH)alg are independent over Halg, κ and λ extend toan automorphism µ′ of (KH)alg(LH)alg which agrees with µ on KalgLalg. Wewill see in Corollary 32.8 that(

(KH)alg(LH)alg)∩ (KL)alg = KL.

So µ′ and µ have a common extension to some automorphism of(KH)alg(LH)alg(KL)alg which again can be extended to some automorphismτ of (KLH)alg. Let C be the fixed field of τ , so C is procyclic. The relativealgebraic closures of KL, KH and LH in C are isomorphic to the relativealgebraic closures of KL, KK ′ and LL′ in C. Let N be relative algebraicclosure of KL in C. C is a regular extension of N . So, by Proposition 31.1 wefind a regular embedding of C into C over N . The image of H has the requiredproperties.

Note that we did not make use of the fact that M is a model, but only thatM is 1-free and C/M is regular.

We have now proved:


Corollary 31.6. Pseudo-finite fields are simple. Forking independence agreeswith algebraic independence.

Chapter 8

Stable Theories

Recall from Section 16, that a theory is κ-stable if |S(A)| ≤ κ for all sets A ofsize at most κ. We call a theory stable if it is κ-stable for some κ. Anotherequivalent definition will be given in Section 33: a theory is stable if no formulahas the order property. In order to apply the results of the previous chapter tostable theories, we will eventually show that stable theories are simple and thenspecialise the characterisation given in Theorem 29.13 to stable theories. Butbefore that we will introduce some of the classical notions of stability theory,all essentially describing forking in stable theories.

32 Heirs and Coheirs

In this section we fix an arbitrary complete theory T . For types overmodels we here define some special extensions to supersets, viz. heirs, coheirs,and definable type extensions. All these extensions have in common that theydo not add too much new information to the given type. For stable theories,we will see in Section 34 that these extensions coincide with the non–forkingextension (and this is in fact unique).

Definition 32.1. Let p be a type over a model M of T and q ∈ S(B) anextension of p to B ⊃M .

1. We call q an heir of p if for every L(M)-formula ϕ(x, y) such that ϕ(x, b) ∈q for some b ∈ B there is some m ∈ m with ϕ(x,m) ∈ p.

2. We call q a coheir of p if q is finitely satisfiable in M .

It is easy to see that tp(a/Mb) is an heir of tp(a/M) if and only if tp(b/Ma)is a coheir of tp(b/M).

148

CHAPTER 8. STABLE THEORIES 149

The following observation is trivial, but used frequently:

Remark 32.2. Suppose q is an heir of p ∈ S(M). If ϕ(x, b) ∈ q and |= σ(b),then there is some m ∈M with |= σ(m) and ϕ(x,m) ∈ p.

Lemma 32.3. Let q ∈ S(B) be a (co)heir of p ∈ S(M) and C an extension ofB. Then q can be extended to a type r ∈ S(C) which is again a (co)heir of p.

Proof. Suppose q is an heir of p. We have to show that

s(x) = q(x) ∪ {ϕ(x, c) | c ∈ C, ϕ(x, y) ∈ L(M), ϕ(x,m) ∈ p for all m ∈M}

is consistent: if there are formulas ϕ(x, b), ϕ1(x, c1), . . . , ϕn(x, cn) ∈ s(x) withϕ(x, b) ∈ q(x) whose conjunction is inconsistent, then as M is a model and qis an heir of p there would be m,m1, . . . ,mn ∈ M with ϕ(x, m) ∈ p and itsconjunction with ϕ1(x,m1), . . . , ϕn(x,mn) inconsistent. Since ϕi(x,mi) ∈ p,this is impossible. Any type r(x) ∈ S(C) containing s(x) is then an heir of p(x).

If q is a coheir of p, let r be a maximal set of L(C)–formulas containing qwhich is finitely satisfiable in M . Clearly, r is consistent. Let ϕ(x) ∈ L(C). Ifneither ϕ nor ¬ϕ belongs to r, then both r ∪ {ϕ} and r ∪ {¬ϕ} are not finitelysatisfied in M and so neither is r (cf. the proof of Lemma 5.2).

Definition 32.4. A type p(x) ∈ Sn(B) is definable over C if the followingholds: for any L–formula ϕ(x, y) there is an L(C)– formula ψ(y) such that forall b ∈ B

ϕ(x, b) ∈ p if and only if |= ψ(b).

p is definable if it is definable over its domain B.

We write ψ(y) as dp xϕ(x, y) to indicate the dependence on p, ϕ(x, y) and thechoice of the variable tuple x. (So dp has the syntax of a generalised quantifier,cf. [52].) Thus, we have

ϕ(x, b) ∈ p if and only if |= dp xϕ(x, b).

Note that dp xϕ(x, y) is also meaningful for formulas ϕ with parameters in B.

Example:In strongly minimal theories all types p ∈ S(A) are definable. To see this fixϕ0 ∈ p of minimal Morley rank k and minimal degree and consider a formulaψ(x, y) without parameters. The discussion on page 114 shows that ψ(x, a) ∈ pif and only if MR(ϕ0(x)∧¬ψ(x, a)) < k. By Corollary 26.4 this is an A–definableproperty of a.

Lemma 32.5. A definable type p ∈ S(M) has a unique extension q ∈ S(B)definable over M for any set B ⊃M , namely

{ϕ(x, b) | ϕ(x, y) ∈ L, b ∈ B, C |= dp xϕ(x, b)}

and q is the only heir of p.


Proof. The fact that the dp xϕ(x, y) define a type is a first order property ex-pressible in M and is hence true in any elementary extension of M . This provesexistence. On the other hand, if q is a definable extension of p, dq xϕ(x, y)and dp xϕ(x, y) agree on M and hence in all elementary extensions provinguniqueness. Clearly, q is an heir of p. If q′ ∈ S(B) is different from q,then for some ϕ(x, b) ∈ q′ we have 6|= dpxϕ(x, b). But there is no m withϕ(x,m) ∧ ¬dpxϕ(x,m) ∈ p, so q′ is not an heir of p.

Lemma 32.6. A global type which is a coheir of its restriction to a model Mis invariant over M .

Proof. Let q ∈ S(C) be finitely satifiable in M and α ∈ Aut(C/M). Considera formula ϕ(x, c). Since c and α(c) have the same type over M , ϕ(x, c) ∧¬ϕ(x, α(c)) is not satisfiable in M . So ϕ(x, c) ∈ q implies ϕ(x, α(c)) ∈ q.

We conclude by exhibiting coheirs in strongly minimal theories:

Proposition 32.7. Let T be strongly minimal, M a model and B an exten-sion of M . Then tp(a/B) is an heir of tp(a/M) if and only if MR(a/B) =MR(a/M).

Note that in strongly minimal theories MR(a/B) = MR(a/M) is equivalent toa and B being geometrically independent over M (see Exercise 26.1). This isa symmetric notion, which implies that in strongly minimal theories heirs andcoheirs coincide. We will later see in Corollary 34.7 (see also Corollary 36.11)that this is actually true for all stable theories. Note also that this implies thatin strongly minimal theories types over models have a unique extension of thesame Morley rank, i.e. they have Morley degree 1. This is true in all totallytranscendental theories (Corollary 36.12, see also Corollary 36.4.)

Proof. Let k be the Morley rank of p = tp(a/M). Choose a formula ϕ0 ∈ p ofsame rank and degree as p. We saw in the Example on page 32 that the uniqueheir q of p on B is defined by

{ψ(x)L(B)–formula | MR(ϕ0(x) ∧ ¬ψ(x, a)) < k}.

On the other hand this set of formulas must be contained in all extensions of pto B having rank k. So q is also the unique extension of p of rank k.

Corollary 32.8 (Hrushovski-Chatzidakis). Let K,L,H be algebraically closedextensions of an algebraically closed field M . If H algebraically independentfrom KL over M , then(

(KH)alg(LH)alg)∩ (KL)alg = KL.


Proof. We work in the monster model C. Let c be an element of the left handside. So there are tuples a ∈ (KH)alg and b ∈ (LH)alg such that c ∈ dcl(a, b),witnessed by, say, |= ϕ(a, b, c). Furthermore, there are tuples a′ ∈ K and h1 ∈ Hsuch that a is algebraic over a′h1 witnessed by, say, |= ϕ1(a, a

′, h1). Similarly,we find |= ϕ2(b, b

′, h2) for b. By independence and the heir property there areh′1, h

′2 ∈M such that

|= ∃x, y ϕ(a, x, y) ∧ ϕ1(x, a′, h′1) ∧ ϕ2(y, b

′, h′2).

Since K and L are algebraically closed, this implies c ∈ KL.

Exercise 32.1.Use Example 28.10 and Lemma 32.6 to give an alternative proof of Corol-lary 29.3.

Exercise 32.2.Let T be a complete theory, M an ω–saturated model.

1. Let ϕ(x) be a formula over M with Morley rank, and ψ a formula witharbitrary parameters which implies ϕ and has the same Morley rank. Thenψ is realized in M .

2. Let B an extension of M and MR(a/B) = MR(a/M) < ∞. Show thattp(a/B) is a coheir of tp(a/M).

It will follow from Corollaries 34.7 and 36.11 that in totally transcendentaltheories this is true for arbitrary M . In fact this holds for arbitrary theories,see Exercise 36.4.

Exercise 32.3 (Hrushovski-Pillay).Let p(x) and q(y) be global types, and suppose that p(x) is A-invariant. Wedefine a global type p(x) ⊗ q(y) by setting (p ⊗ q) � B = tp(ab/B) for anyB ⊃ A where b realizes q(y) � B and a realizes p � Bb. Show that p(x)⊗ q(y) iswell-defined, and A–invariant if both p(x) and q(y) are.


33 Stability

In analogy with simple theories we here define stable theories via (several equiv-alent) properties of their formulas and note that this definition fits well withthe definition of κ-stability given in Section 16.

In this section, let T be a complete (possibly uncountable) theory.Let Sϕ(B) denote the set of all ϕ–types over B; i.e. maximal consistent sets offormulas of the form ϕ(x, b) or ¬ϕ(x, b) where b ∈ B.

Definition 33.1. Let ϕ(x, y) be a formula in the language of T .

1. The formula ϕ is stable if there is an infinite cardinal λ such that |Sϕ(B)| ≤λ whenever |B| ≤ λ.

The theory T is stable if all its formulas are stable.

2. ϕ has the order property if there are elements a0, a1, . . . and b0, b1, . . . suchthat for all i, j ∈ ω

|= ϕ(ai, bj) if and only if i < j.

3. ϕ(x, y) has the binary tree property if there is a binary tree (bs | s ∈<ω2) of parameters such that for all σ ∈ ω2, {ϕσ(n)(x, bσ�n) | n < ω} isconsistent. (We use the notation ϕ0 = ¬ϕ and ϕ1 = ϕ.)

It is important to note that T is stable if and only if it is κ-stable for someκ, see Exercise 33.7.

Remark 33.2. The notion of ϕ(x, y) having the order property is symmetricalin x and y. This means that if ϕ(x, y) has the order property, then there areelements a0, a1, . . . and b0, b1, . . . such that |= ϕ(ai, bj) if and only if j < i.

Proof. Apply Lemma 27.1 to I = (aibi)i<ω and J = (ω,>).

Theorem 33.3. For a formula ϕ(x, y) the following are equivalent:

a) ϕ is stable.

b) |Sϕ(B)| ≤ |B| for any infinite set B.

c) ϕ does not have the order property.

d) ϕ does not have the binary tree property.

Proof. a)→ d): Let µ be minimal such that 2µ > λ. Then the tree I = <µ2 hascardinality at most λ. If ϕ(x, y) has the binary tree property, by compactnessthere are parameters bs, (s ∈ I), such that for all σ ∈ µ2, qσ = {ϕσ(α)(x, bσ�α) |


α < µ} is consistent. Complete every qσ to a ϕ–type pσ over B = {bs | s ∈ I}.Since the pσ are pairwise different, we have |B| ≤ λ < 2µ ≤ | Sϕ(B)|.

d)→c): Choose a linear ordering of I = ≤ω2 such that for all σ ∈ ω2 andn < ω

σ < σ � n ⇔ σ(n) = 1

If ϕ(x, y) has the order property, then by Lemma 27.1 one can find ai and biindexed by I such that

|= ϕ(ai, bj) if and only if i < j

Now the tree ϕ(x, bs), s ∈ <ω2, shows that ϕ has the binary tree property.

c)→b): Let B be an infinite set of parameters and |Sϕ(B)| > |B|. For any athe ϕ–type of a over B is given by Sa = {b ∈ Bn | |= ϕ(a, b)}. Since |B| = |Bn|we may assume for simplicity that n = 1 and so Sa ⊂ B. Applying the Erdös-Makkai Theorem (see Appendix C Theorem 9.1) to B and S = {Sa | a ∈ C},we obtain a sequence (bi | i < ω) of elements of B and a sequence (ai)i<ω suchthat either bi ∈ Saj ⇔ j < i or bi ∈ Saj ⇔ i < j for all i, j. In the first case ϕhas the order property by definition. In the second case ϕ(x, y) has the orderproperty by Remark 33.2.

b)→ a): Clear.

If ϕ has the binary tree property witnessed, say, by (bs | s ∈ <ω2), thenthe family ϕs =

∧n<|s| ϕ

s(n)(x, bs�n), s ∈ <ω2, is a binary tree of consistentformulas. This shows that totally transcendental theories are stable. We willsee below (Corollary 34.6, also Exercise 33.11)) that stable theories are simple.

Remark 33.4. By Example 37.6 and Exercise 33.7 the theory of any R-moduleis stable (but not necessarily totally transcendental) providing a rich class ofexamples for stable theories. Note that the theory of the random graph issimple by Corollary 29.14 but not stable (see Exercise 33.3).

Exercise 33.1.T is unstable if and only if there is an L–formula ψ(x, y) and elements a0, a1, . . . ,ordered by ψ; i.e., such that

|= ψ(ai, aj) ⇐⇒ i < j.

ψ may contain parameters.

Exercise 33.2.A formula ϕ(x, y) is said to have the independence property (IP) if there areai, i ∈ ω, such that for each A ⊆ ω the set {ϕ(x, ai) : i ∈ A} ∪ {¬ϕ(x, ai) : i 6∈A} is consistent. Show that T is unstable if it contains a formula with theindependence property.

Exercise 33.3.Show that the theory of the random graph is not stable.


Exercise 33.4.A formula ϕ(x, y) is said to have the strict order property (SOP) if there is asequence (ai)i<ω such that

|= ∀y(ϕ(ai, y) → ϕ(aj , y)) ⇔ i ≤ j.

T has the strict order property if there is a formula in T with the strict orderproperty. Show: T has the SOP if and only if there is a partial ordering withinfinite chains definable in T eq. (For the definition of T eq see p. 160.)

Exercise 33.5.Show that a theory with SOP is not simple.

Exercise 33.6.(Shelah) If T is unstable, either there is a formula having the IP or a formulahaving the SOP.

Exercise 33.7.The following are equivalent:

a) T is stable.

b) T is λ–stable for all λ such that λ|T | = λ

c) T is λ–stable for some λ.

Exercise 33.8.Show that for any infinite λ there is a linear order of cardinality greater than λwith a dense subset of size λ.

Exercise 33.9.Show that the set of stable formulas is closed under Boolean combinations, i.e.conjunction, disjunction, and negation. Use this to show that the theory TTree,defined on page 74, is stable.

Exercise 33.10.Fix an L-formula ϕ(x, y). Let Φ denote the class of Boolean combinations offormulas of the form ϕ(x, b). Define the ϕ–rank Rϕ as the smallest functionfrom formulas ψ(x) to {−∞} ∪On ∪ {∞} such that

Rϕ ψ ≥ 0 if ψ is consistent;

Rϕ ψ ≥ β + 1 if there are infinitely many δi ∈ Φ which are pairwiseinconsistent and such that Rϕ(ψ ∧ δi) ≥ β for all i.

Prove:1. Rϕ ψ <∞ if and only if ψ(x) ∧ ϕ(x, y) is stable.

2. If Rϕ ψ <∞, then Rϕ ψ < ω.

Exercise 33.11.If ϕ has the tree property, it is unstable.

This shows that stable theories are simple. We will give a different proof inCorollary 34.6.


34 Definable types

Definability of types turns out to be a crucial feature of stable theories. Weshow here that in stable theories the extensions of a type over a model given byits definition agree with the non–forking extensions (and with heirs and coheirs).

Theorem 34.1. ϕ(x, y) is stable if and only if all ϕ–types are definable.

Proof. Let A be a set of parameters of size ≥ |T |. If all ϕ–types over A aredefinable, there exists no more ϕ–types over A than there are defining formulas,i.e. at most |A| many. So ϕ is stable.

For the converse assume that ϕ(x, y) is stable. Define for any formula θ(x)the degree Dϕ(θ) to be the largest n for which there is a finite tree (bs | s ∈ <n2)of parameters such that for every σ ∈ n2 the set {θ(x)}∪{ϕσ(i)(x, bσ�i) | i < n}is consistent. This is well–defined since ϕ does not have the binary tree property.Now, let p be a ϕ–type over B. Let θ be a conjunction of formulas in p with n =Dϕ(θ) minimal. Then ϕ(x, b) belongs to p if and only if n = Dϕ(θ(x)∧ϕ(x, b)).This shows that p is definable.

Corollary 34.2. T is stable if and only if all types are definable.

Observe that the proof of Theorem 34.1 applies also to a proper class ofparameters. From this we obtain the following important corollary:

Corollary 34.3 (Separation of variables). Let T be stable and let F be a 0–definable class. Then any definable subclass of Fn is definable using parametersfrom F.

Proof. Let ψ(a,C) be a definable subclass of Fn. The type q = tp(a/F) isdefinable over a subset of F by Corollary 34.2. Thus,

ψ(a,C) = {f ∈ Fn | |= dq xψ(x, f)}.

If the property in the conclusion of Corollary 34.3 holds for a 0–definableclass F (in a not necessarily stable theory T ), then F is called stably embedded.

At first glance, the next lemma looks mysterious. In essence it states thatin stable theories heirs and coheirs coincide. We need it in the proof of Corol-lary 36.3.

Lemma 34.4 (Harrington). Let T be stable and let p(x) and q(y) be globaltypes. Then for every formula ϕ(x, y) with parameters

dp xϕ(x, y) ∈ q(y) ⇔ dq yϕ(x, y) ∈ p(x).


Proof. Let p, q and ϕ be definable over A. We recursively define sequences aiand bi, i ∈ ω: if a0, . . . , an−1 and b0, . . . , bn−1 have been defined, let bn be arealisation of q � Aa0, . . . , an−1 and an a realisation of p � Ab0, . . . , bn. Then wehave for i < j

|= ϕ(ai, bj) ⇔ |= dq yϕ(ai, y) ⇔ dq yϕ(x, y) ∈ p(x)

and for j ≤ i

|= ϕ(ai, bj) ⇔ |= dp xϕ(x, bj) ⇔ dp xϕ(x, y) ∈ q(y).

Since ϕ does not have the order property, the claim follows.

Lemma 34.5. Let p ∈ S(C) be a global type.

1. If p is definable over A, then p does not divide over A.

2. If T is stable and p does not divide over the model M , p is definable overM .

Note that for global types dividing and forking coincide (Exercise 27.2).

Proof. 1): Consider a formula ϕ(x,m) ∈ p and an infinite sequence of indis-cernibles m = m0,m1, . . . over A. If p is definable over A, all ϕ(x,mi) belongto p. So ϕ(x,m) does not divide over A by Lemma 27.4.

2): Now assume that T is stable and p does not divide over the model M .We will show that p is an heir of p � M . By Corollary 34.2 and Lemma 32.5this implies that p is definable over M . So assume that ϕ(x, b) ∈ p, we want toshow that ϕ(x, b′) ∈ p for some b′ ∈M .

Let I = (bi)i<ω be a Morley sequence of a global coheir extension of tp(b/M)over M starting with b0 = b (see Example 28.10 and Lemma 32.6). Sincetp(a/Mb) does not divide over M , Lemma 27.5 implies that we may assume thatI is indiscernible over Ma. So we have |= ϕ(a, bi) for all i. By Corollary 34.2,the type q = tp(a/M{bi | i < ω}) is definable. Assume that the parameters ofdq xϕ(x, y) are in M{b0, . . . , bn−1}. Since tp(bn/M{b0, . . . , bn−1}) is a coheir oftp(b/M), and since |= dq xϕ(x, bn), there is a b′ ∈ M with |= dq xϕ(x, b

′). Thisimplies |= ϕ(a, b′) and so ϕ(x, b′) ∈ tp(a/M) = p �M .

Corollary 34.6. Stable theories are simple.

Proof. Let p be a type over a model M . Then p is definable over some A ⊂Mof cardinality ≤ |T |. Let p′ be the global extension of p given by the defini-tion over A. By Lemma 34.5.1, p′ and hence also p does not divide over A.Proposition 28.5 implies that T is simple.


This implies in particular that forking and dividing coincides in stable the-ories (see Proposition 28.15).

Corollary 34.7. Let T be a stable theory, p a type over a model M and A anextension of M . Then p has a unique extension q ∈ S(A) with the followingequivalent properties.

a) q does not fork over M .

b) q is definable over M .

c) q is an heir of p.

d) q is a coheir of p.

Proof. By Lemma 34.5, q does not fork over M if and only if it is definableover M . Since p is definable, we know by Lemma 32.5 that there is a uniqueextension q which is definable over M , and which is also the unique heir of p.

To prove the equivalence with d) we may assume that A = M ∪ {a} fora finite tuple a. Fix a realisation b of q. Then q = tp(b/Ma) is a coheir ofp = tp(b/M) if and only if tp(a/Mb) is an heir and hence, by the first part ofthe proof, a non–forking extension of tp(a/M). Now forking symmetry and thefirst part of the proof imply the desired.

Exercise 34.1.Find a theory T and a type p over the empty set such that no definition of pdefines a global type. (A definition which defines a global type is called a gooddefinition of p, see Theorem 36.1).

Exercise 34.2.Let T be an arbitrary complete theory andM be a model. Consider the followingfour properties of a global type p :

(D) p is definable over M .

(C) p is a coheir of p �M .

(I) p is M–invariant.

(H) p is an heir of p �M .

Use the example T = TDLO and M = Q to show that D→I, C→I, D→H arethe only logical relations between these notions.

Exercise 34.3.Let p(x) ∈ S(M) be definable over B. Then, for any n, the map

r(y1, . . . , yn) 7→ {ϕ(x, y1, . . . , yn) | dp xϕ ∈ r}


defines a continuous section πn : Sn(B) → Sn+1(B). Show that this defines abijection between all types definable over B and all “coherent” families (πn) ofcontinuous sections Sn(B) → Sn+1(B).

Exercise 34.4.Let ϕ(x) be a formula without parameters and let M be a model of T . Showthat ϕ(M) is stably embedded in M (i.e. every M -definable relation of ϕ(M) isdefinable over ϕ(M)) if and only if for all n, every p(x1, . . . , xn) ∈ Sn(ϕ(M))which contains ϕ(x1) ∧ . . . ∧ ϕ(xn) has a unique extension p′ ∈ Sn(M). If ϕ is(absolutely) stably embedded and p is definable, show that p′ is definable overϕ(M).

Exercise 34.5.Call a formula ϕ0(x) stable if ϕ0(x) ∧ ϕ(x, y) is stable for all ϕ(x, y). We call atype stable if it contains a stable formula. Prove:

1. Types with Morley rank are stable.

2. Stable types are definable.

Exercise 34.6.Let T be stable, and p ∈ S(A). Show that p is definable over C if p is finitelysatisfiable in C. Furthermore for every ϕ(x, y), dp xϕ(x, y) is a positive Booleancombination of formulas ϕ(c, y), c ∈ C.

Exercise 34.7.We call q a weak heir of p ∈ S(M) if the heir property holds for all ϕ(x, y)without parameters. Show that in stable theories, weak heirs are in fact heirs.

Exercise 34.8.In Corollary 34.7 prove the equivalence of c) and d) directly from Lemma 34.4.


35 Elimination of imaginaries and T eq

This section is an excursion outside the realm of stable theories: for a model Mof an arbitrary theory T and any 0–definable equivalence relation E(x, y) on n–tuples, we now consider the equivalence classes of Mn/E as elements of a newsort of so–called imaginary elements. Adding these imaginaries makes manyarguments more convenient. For certain theories, these imaginaries are alreadycoded in the original structure. However, if this is not already the case, thenadding imaginaries leads to a new theory T eq which does have this property sothat we do not run into an infinite regression.

Let D be a definable class in Cn. A finite tuple d ⊂ C is called a canonicalparameter if d is fixed by the same automorphisms of C which leave D invariant.Lemma 23.10 implies that D is definable over d, and by Corollary 23.12 (1) dis determined by D up to interdefinability. We write d = pDq, or d = pϕ(x)qif D = ϕ(C). Note that the empty tuple is a canonical parameter for every0–definable class.

Definition 35.1. T eliminates imaginaries if any class e/E of a 0–definableequivalence relation E on Cn has a canonical parameter d ⊂ C.

Theorem 35.2. If T eliminates imaginaries, then the following holds:

1. Every definable class D ⊂ Cn has a canonical parameter c.

2. Every definable type p ∈ S(C) has a canonical base. This is a set B whichis pointwise fixed by the same automorphisms which leave p invariant.

Proof. Write D = ϕ(C, e). Define the equivalence relation E by

y1Ey2 ⇐⇒ ∀x ϕ(x, y1) ↔ ϕ(x, y2)

and let d be a canonical parameter of e/E. Then d is a canonical parameter ofD.

If dp is a definition of p, the set B = {pdp xϕ(x, y)q | ϕ(x, y) L–formula} isa canonical base of p.

Lemma 35.3. Assume that T eliminates imaginaries. Let A be a set of para-meters and D a definable class. Then the following are equivalent:

a) D is acl(A)–definable.

b) D has only finitely many conjugates over A.

c) D is the union of equivalence classes of an A–definable equivalence relationwith finitely many classes (a finite equivalence relation).


Proof. Let d be a canonical parameter of D. Then D is definable over acl(A) ifand only if d belongs to acl(A). On the other hand D has as many conjugatesover A as d. So a) and b) are equivalent.

For the equivalence of b) and c), first notice that any class of an A–definablefinite equivalence relation has only finitely many conjugates over A, which yieldsc) → b). For the converse, let D = D1, . . .Dn be the conjugates of D over A.Consider the finite equivalence relation E(c, c′) defined by

c ∈ Di if and only if c′ ∈ Di for all i.

Clearly D is a union of E–classes. E is definable and since it is invariant underall A-automorphisms of C, it is in fact definable over A.

Note that elimination of imaginaries was only used for b) → a).

The previous results show why it is convenient to work in a theory elim-inating imaginaries. It is easy to see that a theory eliminates imaginaries ifevery 0-definable equivalence relation arises from fibers of a 0-definable func-tion. While not all theories have this property (e.g. the theory of an equivalencerelation with infinitely many infinite classes), we now show how to extend anycomplete theory T to a theory T eq (in a corresponding language Leq) whichdoes:

Let Ei(x1, x2), (i ∈ I), be a list of all 0–definable equivalence relations onni–tuples. For any model M of T we consider the many–sorted structure

M eq = (M,Mni/Ei)i∈I ,

which carries the home sort M and for every i the natural projection

πi :Mni →Mni/Ei

The elements of the sorts Si = Mni/Ei are called imaginary elements, theelements of the home sort are real elements.

The M eq form an elementary class axiomatised by the (complete) theoryT eq which, in in the appropriate many-sorted language Leq, is axiomatized bythe axioms of T and for each i ∈ I by

∀y ∃x πi(x).= y (y a variable of sort Si)

and∀x1, x2 (πi(x1)

.= πi(x2) ↔ Ei(x1, x2))

The algebraic (definable, respectively) closure of in M eq is denoted by acleq

(dcleq, respectively).

The first two statements of the following proposition explain why we considerT eq as an inessential expansion of T .


Proposition 35.4. 1. Elements of Ceq are definable over C in a uniformway.

2. The 0–definable relations on the home sort of Ceq are exactly the same asthose in C.

3. T eq eliminates imaginaries,

Proof. 1: Every element of sort Si has the form πi(a) for an ni–tuple a from C.

2: We show that every Leq–formula ϕ(x) with free variables from the home–sortis equivalent to an L–formula ϕ∗(x) by induction on the complexity of ϕ. If ϕ isatomic, it is either an L–formula or of the form πi(x1)

.= πi(x2), in which case

we set ϕ∗ = Ei(x1, x2). We let ∗ commute with negations, conjunctions andquantification over home–sort variables. Finally, if y is a variable of sort Si, weset (

∃yψ(x, y))∗

= ∃x′ψ(x, πi(x′))∗.

3: We observe first that Ceq is the monster model of T eq, i.e. that every typep(y) over a set A is realized in Ceq: By part 1 we may assume that A ⊂ C.If y is of sort Si, the set Σ(x) = p(πi(x)) is finitely satisiable. By part 2 Σ isequivalent to a set Σ∗ of L–formulas. Σ∗ has a realisation b, which gives us arealisation πi(b) of p.

It is now clear that πi(e) is a canonical parameter of the class e/Ei. By theproof of Theorem 35.2 this implies that every relation in Ceq which is definablewith parameters from C has a canonical parameter in Ceq. On the other hand,by part 1, every definable relation in Ceq is definable in C

Corollary 35.5. T eliminates imaginaries if and only if in T eq every imaginaryis interdefinable with a real tuple.

Proof. Since every automorphism of C extends (uniquely) to an automorphismof Ceq, a real tuple d is a canonical parameter of e/Ei in the sense of T if andonly if it is a canonical paramter in the sense of T eq. But this is equivalent tod being interdefinable with πi(e).

The proof of the following criterion for elimination of imaginaries shows howT eq can be used:

Lemma 35.6. The following are equivalent:

a) T eliminates imaginaries and has at least two 0–definable elements.

b) Every 0–definable equivalence relation on Cn is the fibration of a 0–definablefunction f : Cn → Cm.


Proof. b) → a): If E is the fibration of a 0–definable function f : Cn → Cm, wehave pe/Eq = f(e). To see that there are at least two 0–definable elements lookat the following equivalence relation on C2:

x1x2E y1y2 ⇔ (x1 = x2 ↔ y1 = y2).

This has two classes, which are both 0–definable. If E is the fibration of a0–definable π : C2 → Cm the two images of π are two different 0–definable m–tuples.

a) → b): Let E be a 0–definable equivalence relation on Cn. Every e/E isinterdefinable with an element of some power Cme . So by Exercise 23.11, e/Ebelongs to a 0–definable D ⊂ Cn/E with a 0–definable injection f : D → Cme .A compactness argument shows that we can cover Cn/E by finitely many 0–definable classes D1, . . . ,Dk with 0–definable injections fi : Di → Cmi . We mayassume that the Di are pairwise disjoint, otherwise we replace Di by Di \ (D0 ∪. . . ∪ Di−1). Now, using the two 0–definable elements, we can find, for somebig m, 0–definable injections gi : Cmi → Cm with pairwise disjoint images. Theunion of the gifi is a 0–definable injection from Cn/E into Cm.

Using 1. and 2. of the previous proposition, one can see that in general allproperties of T which concern us here are preserved when going from T to T eq.Here are some examples:

Lemma 35.7. 1. T is ℵ1–categorical if and only if T eq is ℵ1–categorical.

2. T is λ–stable if and only if T eq is λ–stable.

3. T is stable if and only if T eq is stable.

Proof. 1. is clear.

For 2: Let A be a set of parameters in T eq of cardinality λ. A is containedin the definable closure of some set B of cardinality λ of the home sort. Forany p ∈ S(B) we may first take the unique extension of p to dcleq(B) and thenits restriction to A. This defines a surjection S(B) → S(A). Notice that wenow have to specify not only the number of variables but also the sorts for thevariables in the types.

If S′(A) consists of types of elements of the sort Cn/E, Sn(A) denotes then–types of the home sort and π the projection Cn → Cn/E, then tp(b/A) 7→tp(π(b)/A) defines a surjection Sn(A) → S′(A). This shows that T eq is stable ifT is.

Of course 3. follows from 2. and Exercise 33.7. We still give a direct proof asan example of how to translate between T and T eq. Let ϕ(y1, y2) be a formula


in T eq with the order property. If y1 and y2 belong to Cn1/E1 and Cn2/E2,respectively, there are tuples ai of the home sort such that

|= ϕ(π1(a1), π2(a2)) if and only if i < j.

By Proposition 35.4 the formula ϕ(π1(x1), π2(x2)) is equivalent to some L–formula ψ(x1, x2), which has the order property in T .

For applications the following special cases are often useful:

Definition 35.8.

1. T eliminates finite imaginaries if every finite set of n–tuples has a canon-ical parameter.

2. T has weak elimination of imaginaries if for every imaginary e there is areal tuple c such that e ∈ dcleq(c) and c ∈ acl(e).

Lemma 35.9. T eliminates imaginaries if and only if it has weak eliminationof imaginaries and eliminates finite imaginaries

Proof. This follows from the observation that T has weak elimination of imag-inaries if and only if every imaginary e is interdefinable with the canonicalparameter of a finite set of real n–tuples: Indeed, if e ∈ dcleq(c) and c ∈ acl(e),and {c1, . . . , cm} are the conjugates of c over e, then e is interdefinable withp{c1, . . . , cm}q. If conversely e is interdefinable with p{c1, . . . , cm}q, then e ∈dcleq(c1 . . . cm) and c1 . . . cmx ∈ acl(e).

Lemma 35.10 (Lascar-Pillay). Let T be strongly minimal and acl(∅) infinite.Then T has weak elimination of imaginaries.

Proof. Let e = c/E be an imaginary. It suffices to show that c/E containsan element algebraic over e or, more general, that every non-empty definableX ⊂ Cn contains an element of A = acl(pXq). We proceed by induction on n.For n = 1 there are two cases: if X is finite, it is a subset of A. If X is infinite,almost all elements of acl(∅) belong to X. If n > 1, consider the projectionY of X to the first coordinate. Y contains an element a of acl(pY q), which isa subset of A. By induction the fibre Xa contains an element b of acl(pXaq),which is also a subset of A. So (a, b) is in X ∩A.

Corollary 35.11. The theory TACFp of algebraically closed fields of character-istic p eliminates imaginaries.

Proof. By the preceding lemmas it suffices to show that every theory of fieldseliminates finite imaginaries. Let S = {c0, . . . , ck−1} be a set of n–tuples ci =(ci,j)j<n. Consider the polynomial

p(X,Y0, . . . , Yn−1) =∏i<k

(X −∑j<n

ci,jYj).


An automorphism leaves p fixed if and only if it permutes S. So the coefficientsof p serve as a canonical parameter of S.

Lemma 35.12. A totally transcendental theory in which every global type hasa canonical base in C has weak elimination of imaginaries.

Proof. Let e = c/E be an imaginary and α the Morley rank of the class c/E.Let p be a global type of Morley rank α which contains E(x, c). By assumptionp has a canonical base d ⊂ C. Since there are only finitely many such p, d isalgebraic over e. Also e is definable from d since for an automorphism α fixingp, c/E and αc/E cannot be disjoint, so they must be equal. Clearly we mayassume that d is a finite tuple (see also Exercises 35.1 and 35.7).

Corollary 35.13. TDCF0 eliminates imaginaries.

Proof. By quantifier elimination every global type p(x) is axiomatised by itsquantifier free part pqf (x), which is equivalent to a union of qi(x, dx, . . . , di),i = 0, 1, . . ., where the qi(x0, . . . , xi) are quantifier free pure field-theoretic types.If Ci is the canonical base of qi in the sense of TACF0 , C0∪C1∪ . . . is a canonicalbase of p.

Exercise 35.1.Let D be a definable class. Assume that there is a set D which is fixed by thesame automorphisms which leave D invariant. Show that D contains a canonicalparameter of D.

Exercise 35.2.A theory T has weak elimination of imaginaries if and only if for every definableclass D there is a smallest algebraically closed set over which D is definable.

Exercise 35.3.Use Exercise 35.2 to prove that the theories T∞ and TDLO (not easy) have weakelimination of imaginaries. Show also that TDLO has elimination of imaginaries,but T∞ does not.

Exercise 35.4.Show that all extensions of p ∈ S(A) to acl(A) are conjugate over A. Moregenerally this is true for every normal extension B of A. These are sets whichare invariant under all α ∈ Aut(C/A). Note that normal extensions must besubsets of acl(A).

Exercise 35.5.An algebraic type over A has a good definition (see Exercise 34.1 or p. 166) overB ⊂ A if and only if it is realised in dcl(B).

Exercise 35.6.Let d be a canonical parameter of D. Then d is 0–definable in the L ∪ {P}–structure (C,D).


Exercise 35.7.Let T be totally transcendental and p a global type.

1. Show that p has a finite canonical base in Ceq.

2. If p has a canonical base D ⊂ C, then it has a finite base d ⊂ C.

Exercise 35.8.Show that Lemma 35.12 is true for stable theories. (Hint: In the proof of 35.12replace p by a suitable E(x, y)–type.)

Exercise 35.9.Define the strong type of a over A as stp(a/A) = tp(a/ acleq(A)). Show thatstp(a/A) is axiomatised by

{E(x, a) : E(x, y) A–definable finite equivalence relation.}


36 Properties of forking in stable theories

Except in Theorem 36.10 we assume throughout this section that Tis stable. We now collect the crucial properties of forking in stable theories.As with simple theories, we will see in Theorem 36.10 that these propertiescharacterise stable theories and forking. Since we already know that stabletheories are simple, some of these properties are immediate. For completenessand reference we restate them in the context of stable theories.

Let p ∈ S(B) be defined by L(B)–formulas dp xϕ. We call the definitiongood if it defines a global type (or, equivalently, if it defines a type over somemodel containing B).

Theorem 36.1. Let T be stable. A type p ∈ S(B) does not fork over A ⊂ B ifand only if p has a good definition over acleq(A).

Proof. If p does not fork over A, p has a global extension p′ which does notfork over A. Let M be any model which contains A. By Lemma 34.5, p′ isdefinable over M , so the canonical base of p′ belongs to M eq. By Exercise 23.2the canonical base belongs to acleq(A).

If conversely p has a good definition over acleq(A), p does not fork overacleq(A) and therefore does not fork over A.

Definition 36.2. A type is stationary if and only if it has a unique non–forkingextension to any superset.

Corollary 36.3 (Uniqueness). If T is stable and eliminates imaginaries, thenany type over an algebraically closed set is stationary.

Proof. Let A = acleq(A). Let p′ and p′′ two global non–forking extensions ofp ∈ S(A). Consider any formula ϕ(x, b), and let q(y) be a global non–forkingextension of tp(b/A). By Theorem 36.1, p′, p′′ and q are definable over A. Nowwe apply Harrington’s Lemma 34.4:

ϕ(x, b) ∈ p′ ⇔ dp′ xϕ(x, y) ∈ q ⇔ dq yϕ(y, x) ∈ p

⇔ dp′′ xϕ(x, y) ∈ q ⇔ ϕ(x, b) ∈ p′′

Corollary 36.4. In a stable theory, types over models are stationary.

Proof. This is immediate by the above proof since we can replace acleq(A) every-where by M . It follows also formally from Corollary 36.3 since M eq = dcleq(M)is an elementary substructure of Ceq and so algebraically closed.


For the remainder of this section we also assume that T eliminatesimaginaries. In view of T eq (cp. Section 35) this is a harmless assumption.

As stable theories are simple we may first collect some of the properties offorking established in Section 28.

We keep usinga |A

B

to express that tp(a/AB) does not fork over A.

Theorem 36.5. If T is stable, forking independence has the following proper-ties:

1. (Monotony and Transitivity) Let A ⊂ B ⊂ C and q ∈ S(C). Then q doesnot fork over A if and only if q does not fork over B and q � B does notfork over A.

2. (Symmetry)

a |A

b =⇒ b |A

a

3. (Finite character) If p ∈ S(B) forks over A, there is a finite subset B0 ⊂ Bsuch that p � AB0 forks over A.

4. (Local character) For p ∈ S(A) there is some A0 ⊂ A of cardinality atmost |T | such that p does not fork over A0

5. (Existence) Every type p ∈ S(A) has a non–forking extension to any setcontaining A.

6. (Algebraic Closure)

(a) p ∈ S(acl(A)) does not fork over A.

(b) If tp(a/Aa) does not fork over A, then a is algebraic over A.

Proof. This is contained in 28.17, 28.16, 27.9, 28.5, 28.7 and 27.3.

The following properties do not hold in arbitrary simple theories.

Theorem 36.6. Assume T is stable.

1. (Conjugacy) If A ⊂ M and M is strongly κ–homogeneous for some κ >max{|T |, |A|}, then all non–forking extensions of p ∈ S(A) to M are con-jugate over A.

2. (Boundedness) Any p ∈ S(A) has at most 2|T | non–forking extensions forevery B ⊃ A.


Proof. For 1. let q1 and q2 be non–forking extensions of p to M . Any A–automorphism of M which takes q1 � acl(A) to q2 � acl(A) (see Exercise 35.4)takes q1 to q2. Since types over algebraically closed sets are stationary byCorollary 36.3, the claim now follows.

To prove 2. let A0 be a subset of A of cardinality at most |T | such that pdoes not fork over A0. Then p has at most as many non–forking extensions asp � A0 has extensions to acl(A0).

Corollary 36.7. Let T be stable and p ∈ S(A). Then p is stationary if andonly if it has a good definition over A.

Proof. Assume first that p is stationary and let q be the global non–forkingextension. q is definable and invariant under all automorphisms over A, so q isdefinable over A by Lemma 23.10. This shows that p has a good definition overA. For the converse assume that p has a good definition over A. So p has anon–forking global extension p′, definable over A by 36.1. Since all global non–forking extensions of p are conjugate over A, and p′ is fixed by all automorphismsover A, p′ is the only global non–forking extension of p.

Let p ∈ S(A) be a stationary type. We call p based on B if p is parallelto some stationary type q defined over B, i.e. if p and q have the same globalnon–forking extension (see Exercise 38.4). The canonical base Cb(p) of p is thecanonical base of the non–forking global extension of p (see Theorem 35.2).

Lemma 36.8. A stationary type p ∈ S(A) is based on B if and only ifCb(p) ⊂ dcl(B). So p does not fork over B ⊂ A if and only if Cb(p) ⊂ acl(B).

Proof. Let r be the global non–forking extension of p and q = r � B. Assumethat p is based on B. Then q is stationary and r the unique non–forking exten-sion of q. By Corollary 36.7 q has a good definition over B, which also definesr. So r is definable over B, which means Cb(p) ⊂ dcl(B).

If, conversely, r is definable over B, we know by Theorem 36.1 that r doesnot fork over B and that q is stationary by Corollary 36.7.

The last statement follows from the easy fact that p does not fork over B ifand only if p is based on acl(B).

For A ⊂ B let N(B/A) be the set of all types over B that do not fork overA. By Remark 27.8, N(B/A) is closed in S(B). For future reference we recordthe following useful fact:

Theorem 36.9. (Open mapping) The restriction map π : N(B/A) −→ S(A) isopen.


Proof. It is easy to see that we may replace B by C. If π(q) = π(q′) for someq, q′ ∈ N(C/A), then q and q′ are conjugate. So if O is a (relative) open subsetof N(C/A), then

O′ = π−1(π(O)) =⋃

{α(O) | α ∈ Aut(C/A)}

is again open. SoS(A) \ π(O) = π(N(C/A) \O′)

is closed since it is the image of a closed set.

Theorem 36.10 (Characterisation of Forking). Let T be a complete theory andn > 0. T is stable if and only if there is a special class of extensions of n–types,which we denote by p @ q, with the following properties.

a) (Local character) There is a cardinal κ such that for q ∈ Sn(C) there isC0 ⊂ C of cardinality at most κ such that q � C0 @ q.

b) (Weak Boundedness) For all p ∈ Sn(A) there is a cardinal µ such that phas, for any B ⊃ A, at most µ extensions q ∈ Sn(B) with p @ q.

If @ satisfies in addition

c) (Invariance) @ is invariant under Aut(C),

d) (Existence) For all p ∈ Sn(A) and A ⊂ B, there is q ∈ Sn(B) such thatp @ q,

e) (Transitivity) p @ q @ r implies p @ r,

f) (Weak Monotony) p @ r and p ⊂ q ⊂ r implies p @ q,

then @ coincides with the non–forking relation.

Proof. Assume properties a) and b). First, choose µ′ big enough so that for allA0 of cardinality at most κ all n–types over A0 have at most µ′ @–extensionsto any supserset.

Let A be a set of parameters. Then the number of n–types over A is boundedby the product of the number of subsets A0 of A of cardinality at most κ, timesa bound for the number of types p over A0, times a bound for the number of@–extensions of p ∈ Sn(A0) to A. So we have

|Sn(A)| ≤ |A|κ · 2max(κ,|T |) · µ′

and it follows that T is λ–stable if λκ = λ and λ ≥ max(|T |, µ′), hence stableby Exercise 33.7.


Now let @ have the properties a) to f). Consider a type p ∈ Sn(A) with anextension q ∈ Sn(B).

Assume first that p @ q. Let µ be the cardinal given Boundedness applied top. By Exercise 27.4 there is an extension M of B such that every r ∈ Sn(M)which forks over A has more than µ conjugates over A. By Existence andTransitivity q has an extension r to M such that p @ r. By Invariance wehave p @ r′ for all conjugates r′. So r has no more than µ conjugates, whichimplies that r does not fork over A and that q is a non–forking extension of p.

Now assume that q is a non–forking extension of p. Choose an extension M of Bwhich is sufficiently saturated in the sense of Theorem 36.6.1. Let r ∈ Sn(M) bea non–forking extension of q and r′ ∈ Sn(M) such that p @ r′. By the above r′is a non–forking extension of p. So r and r′ are conjugate over A. This impliesp @ r and p @ q by Weak Monotony.

Corollary 36.11. Let T be totally transcendental, p ∈ S(A) and q an extensionof p to some superset of A. Then q is a non–forking extension if and only ifMR(p) = MR(q). Hence p is stationary if and only if it has Morley degree 1.

Proof. In a totally transcendental theory, extensions having the same Morleyrank satisfy the conditions of Theorem 36.10 (see Section 24).

The same is true for types with Morley rank in stable theories (see Exercise36.4). It follows in particular that in totally transcendental theories for any typep ∈ S(A) there is a finite set A0 ⊆ A such that p does not fork over A0. Stabletheories with this property are called superstable, see section 37.

Corollary 36.12. In a totally transcendental theory, types over models haveMorley degree 1.

Proof. This follows from Corollaries 36.4 and 36.11.

Corollary 36.13. If T is strongly minimal, we have A |BC if and only if

A and C are algebraically independent over B in the pregeometry sense, i.e. ifdim(a/B) = dim(a/BC) for all finite tuples a ∈ A.

Proof. This follows from Theorem 26.2 and Corollary 36.11.

Corollary 36.14. Let K ⊂ F1, F2 be differential fields contained in a model ofTDCF0 . Then F1 |

KF2 if and only if F1 and F2 are algebraically independent

over K.

Proof. By Exercises 9.2 and 28.7, F1 |KF2 implies the algebraic independence.

For the converse we may assume that K is algebraically closed. So let F1 andF2 be algebraically independent over K. By existence of non–forking extensionschoose a copy F ′ of F1 satisfying the same type over K and forking independent


of F2 over K. Then F ′ is algebraically independent of F2 over K. Since K isalgebraically closed, F ′ and F1 satisfy the same type over F2 in the sense offield theory. Since F ′, F1 and F2 are d-closed and F ′ and F1 are isomorphicas d-fields, we conclude that F ′ and F1 have the same type over F2. Thus,F1 |

KF2.

Exercise 36.1 (Finite equivalence relation theorem).Let A ⊂ B and let tp(a/B) 6= tp(b/B) be types which do not fork over A. Thenthere is anA-definable finite equivalence relation E with q1(x)∪q2(y) ` ¬E(x, y).

Exercise 36.2.If a and b are independent realisations of the same type over acl(A), thentp(a/Ab) is stationary.

Exercise 36.3.Prove:

1. Let p ∈ S(A) be stable and q an extension of p. Then q does not fork overA if and only if q has a good definition over acleq(A).

2. Stable types over sets which are algebraically closed in Ceq are stationary.

Exercise 36.4.

1. Let T be stable, p ∈ S(A) a type with Morley rank and q an extension ofp to some superset of A. Then q is a non–forking extension if and only ifMR(p) = MR(q). It follows that a type with Morley rank is stationary ifand only if it has Morley degree 1.

2. Show that the same is true for an arbitrary theory T .

Exercise 36.5.Assume that T is stable. For any p ∈ S(A) there is some A0 ⊂ A of cardinalityat most |T | such that p is the unique non–forking extension of p � A0 to A. If phas Morley rank, A0 can be chosen as a finite set.

Exercise 36.6 (Forking multiplicity).(T stable) Define the multiplicity of a type p as the number mult(p) of its globalnon–forking extensions. Show:

1. If p is algebraic, mult(p) is the number of realisations of p. (This was ourprevious definition of the multiplicity of algebraic types. See page 94 andRemark 20.3.)

2. If T is countable, then mult(p) is either finite or 2ℵ0 .

3. If p has Morley rank, show that mult(p) = MD(p).


Exercise 36.7.Let G be a totally transcendental group. Show:

1. If a and b are independent elements, then MR(a · b) ≥ MR(a). Equalityholds if and only if a · b and b are independent.

2. Assume that G is ω–saturated. Then an element a is generic, i.e. MR(a) =MR(G), if and only if for all b ∈ G

a | b ⇒ a · b | b.

It can be shown that all ω–saturated stable groups contain generic elements,i.e., elements satisfying property 2), see Poizat [41], or Wagner [54].

Exercise 36.8 (Group configuration).Let G be a totally transcendental group, and let a1, a2, a3 ∈ G be independentgeneric elements, i.e. elements of maximal rank in Th(G). Put b1 = a1 · a2,b2 = a1 · a2 · a3 and b3 = a2 · a3. We consider these six elements as the pointsof a geometry with “lines” A0 = {a1, b1, a2}, A1 = {a2, b3, a3}, A2 = {a1, b2, b3}and A3 = {b1, b2, a3}. It is easy to see that every permutation of the four linesgives rise to an automorphism of this geometry.

a1 a3

b1

b2

b3

a2

AAA �

��

AAA �

��

QQ

QQ ��

QQ��

Show:

1. Each point an a line is algebraic over the other two points on the line.

2. Any three non-collinear points are independent.

Any family of points a1, a2, a, a3, b1, b2, b2, b3 with these properties1 is calleda group configuration. Hrushovski proved that whenever a totally transcenden-tal structure contains a group configuration, there is a group definable in thisstructure whose Morley rank equals the Morley rank of any of the points. Formore details see Bouscaren [10], Wagner [54] or Pillay [39].

Exercise 36.9.Let T be an arbitrary complete theory, not necessarily stable. For any set ofparameters A the map S(acl(A)) −→ S(A) is open. (For stable theories, this isjust the Open Mapping Theorem.)

1It is easy to see that 2) can equivalently be replaced by: 2a) Any two points on a line areindependent and 2b) Any two lines are independent over their intersection.


37 SU-rank and the stability spectrum

We saw that in totally transcendental theories forking is governed by the Mor-ley rank. The SU-rank, which we define here, generalizes this to superstabletheories. We use it to show that the stability spectrum of countable theories israther restrictive: there are only four possibilities for the class of cardinals inwhich a countable theory is stable.

Definition 37.1. Let T be a simple theory. We define SU(p) ≥ α for a type pby recursion on α:

SU(p) ≥ 0 for all types p;

SU(p) ≥ β + 1 if p has a forking extension q with SU(q) ≥ β;

SU(p) ≥ λ (for a limit ordinal λ) if SU(p) ≥ β for all β < λ.

and the SU-rank SU(p) of p as the maximal α such that SU(p) ≥ α. If there isno maximum, we set SU(p) = ∞.

Lemma 37.2. Assume T to be simple. Let p have ordinal valued SU-rank andlet q be an extension of p. Then q is a non–forking extension of p if and only ifq has the same SU-rank as p. If p has SU-rank ∞, then so does any non–forkingextension.

Proof. It is clear that the SU-rank of an extension cannot increase. So it isenough to show for all α that SU(p) ≥ α implies SU(q) ≥ α whenever q is anon–forking extension of p ∈ S(A). The interesting case is where α = β + 1is a successor ordinal. Then p has a forking extension r with SU(r) ≥ β. Bythe Diamond Lemma (Exercise 28.2) there is an A–conjugate r′ of r with anon–forking extension s which also extends q. By induction SU(s) ≥ β. But sis a forking extension of q, so SU(q) ≥ β.

Since every type does not fork over a set of cardinality at most |T |, thereare at most 2|T | different SU-ranks. Since they form an initial segment of theordinals, all ordinal ranks are smaller than (2|T |)+. (Actually one can provethat they are smaller than |T |+.) It follows that every type of SU-rank ∞ hasa forking extension of SU-rank ∞.

Definition 37.3. A simple theory is supersimple if every type does not fork overa finite subset of its domain. A stable, supersimple theory is called superstable.

Note that totally transcendental theories are superstable.

Lemma 37.4. T is supersimple if and only if every type has SU-rank <∞.

Proof. If SU(p) = ∞, there is an infinite sequence p = p0 ⊂ p1 ⊂ . . . of forkingextensions of SU–rank ∞. The union of the pi forks over every finite subset of


its domain. If p ∈ S(A) forks over every finite subset of A, there is an infinitesequence A0 ⊂ A1 ⊂ . . . of finite subsets of A such that p � Ai+1 forks over Ai.This shows that p � ∅ has SU-rank ∞.

Let T be a complete theory. The stability spectrum Spec(T ) of T is the classof all infinite cardinals in which T is stable.

Theorem 37.5. Let T be a countable complete theory. There are four cases:

1. T is totally transcendental. Then Spec(T ) = {κ | κ ≥ ℵ0}.

2. T is superstable but not totally transcendental. Then Spec(T ) = {κ | κ ≥2ℵ0}.

3. T is stable but not superstable. Then Spec(T ) = {κ | κℵ0 = κ}.

4. T is unstable. Then Spec(T ) is empty.

Proof. 1: This follows from Theorem 16.5.

2: Let T be superstable and |A| = κ. Since every type over A does not forkover a finite subset of A, an upper bound for the size of S(A) can be computedas the product of

• the number of finite subsets E of A,

• the number of types p ∈ S(E),

• the number of non–forking extensions of p to A.

So we have |S(A)| ≤ κ·2ℵ0 ·2ℵ0 = max(2ℵ0 , κ). If T is not totally transcendental,the proof of Theorem 16.5 shows that T cannot be stable in cardinals smallerthan 2ℵ0 .

3: If T is stable, then T is κ–stable whenever κℵ0 = κ by Exercise 33.7. If T isnot superstable, the proof of Lemma 37.4 shows that there is a type p over theempty set of infinite SU-rank with a forking extension p′ of infinite SU-rank.Let q be a non–forking global extension of p′ and let κ ≥ ℵ0. By Exercise 27.4 qhas κ many different conjugates qα, (α < κ). Choose A0 of size κ such that allpα = qα � A0 are different. By Lemma 37.2 the pα have again infinite SU-rank.Continuing in this manner we get a sequence A0 ⊂ A1 ⊂ . . . of parameter setsand a tree of types pα0,...,αn ∈ S(An+1), (n < ω,αi < κ). We may assumethat all Ai have cardinality κ. Each path through this tree defines a type overA =

⋃n<ω An. So we have |S(A)| ≥ κℵ0 .

4: This follows from Theorem 33.3.

The spectrum of uncountable theories is more difficult to describe, see [47,Chapter III].


Example 37.6 (Modules). For any R-module M , the LM(R)-theory of M isκ-stable if κ = κ|R|+ℵ0 .

See [42] or [56] for more on the model theory of modules.

Proof. Let B be a subset of some model N of Th(M), |B| ≤ κ. Let SN (B)denote the set of all complete types over B which are realized in N . Every typetp(a/B) is axiomatised by

tp±(a/B) = tp+(a/B) ∪ tp−(a/B),

wheretp+(a/B) = {ϕ(x, b) | ϕ pp–formula, b ∈ B, N |= ϕ(a, b)}

andtp−(a/B) = {¬ϕ(x, b) | ϕ pp–formula, b ∈ B, N |= ¬ϕ(a, b)}.

Clearly, tp− is determined by tp+.

By Corollary 9.8, tp+(a/B) contains - up to equivalence - at most one for-mula ϕ(x, b) for any pp–formula ϕ(x, y). Hence tp+(a/B) is determined by apartial map f from the set of pp–formulas to the set of finite tuples in B in thesense that it is axiomatised by {ϕ(x, f(ϕ)) : ϕ pp–formula}. Hence we have

|SN (B)| ≤ (|B|+ ℵ0)|R|+ℵ0 .

Thus T is κ-stable in every κ with κ = κ|R|+ℵ0 .

Example 37.7 (Separably closed fields). The theory SCFp,e of separably closedfields is stable for all κ with κ = κℵ0 -

Proof. Let L be a model of SCFp,e. Fix a p–basis b1, . . . , be and consider thecorresponding λ-functions λν . Now let K be a subfield of cardinality κ. ByTheorem 9.20 every type tp(a/K) is axiomatized by Boolean combinations ofequations t(x) .= 0 where the t(x) are L(c1, . . . , ce, λν)ν∈pe–terms with parame-ters from K. To compute an upper bound for the number of types over K wemay assume that K contains the p–basis and is closed under the λ-functions. Itis now easy to see that every t(x) is equivalent to a term p(λν1(x), . . . , λνk(x))where p(X1, . . . , Xk) ∈ K[X1, . . . , Xk] and the λνi are iterated λ-functions:

λν = λν1...νm = λν1 ◦ . . . ◦ λνm

So, if λν0 , λν1 , . . . is a list of all iterated λ-functions, the type of a over Kis determined by the sequence of the quantifier free L–types of the tuples(λν0(a), . . . , λνn(a)) over K. By Example 16.2 for each n there are only κ manysuch types. So we we can bound the number of types over K by κℵ0 .

Exercise 37.1.Show that the types of SU-rank 0 are exactly the algebraic types. Show alsothat a type is minimal if and only if it is stationary and has SU–rank 1.


Exercise 37.2.Let T be an arbitrary theory. Define the U-rank (or Lascar rank) U(p) of atype p ∈ S(A) as its SU–rank, except for the clause

U(p) ≥ β + 1 if for any κ there is a set B ⊃ A to which p has atleast κ many extensions q with U(q) ≥ β.

Show that in stable theories U–rank and SU–rank coincide.

Exercise 37.3.Show that in simple theories SU(p) ≤ MR(p).

Exercise 37.4.Let ϕ(x, y) be a formula without parameters and k natural number. Define therank D(p, ϕ, k) by

D(p, ϕ, k) ≥ β + 1 if p has an extension q with D(q, ϕ, k) ≥ β and whichcontains a formula ϕ(x, b) which divides over the do-main of p with respect to k.

Show for simple T :

1. D(p, ϕ, k) is bounded by a natural number which depends only on ϕ andk.

2. Let q be an extension of p. Then p is a non–forking extension of p if andonly if D(p, ϕ, k) = D(q, ϕ, k) for all ϕ and k.

Exercise 37.5.A countable theory is ω–stable if and only if it is superstable, small and if everytype (over a finite set) has finite multiplicity.

Exercise 37.6 (Lachlan).Show that in an ℵ0–categorical theory there are only finitely many strong 1-types over a finite set. Conclude that an ℵ0–categorical superstable theory isω-stable.

Note that there are stable ℵ0–categorical theories which are not ω-stable(see [22]).

Exercise 37.7.Let T be a stable theory. Assume that there is a sequence of definable equiva-lence relations E0 ⊂ E1 ⊂ . . . such that every Ei–class contains infinitely manyEi+1–classes. Show that T is not superstable.

Exercise 37.8.Use Exercise 37.7 to show that a superstable group has no infinite descendingsequence of definable subgroups G = G0 ≥ G1 ≥ G2 ≥ G3 ≥ . . . each of infiniteindex in the previous one. Conclude from this that SCFp,e is not superstable ife > 0.


Exercise 37.9.Prove that a module M is totally transcendental if and only if it has the dcc onpp–definable subgroups of M . M is superstable if and only if there is no infinitedescending sequence of pp–definable subgroups each of which is of infinite indexin its predecessor.

Chapter 9

Prime extensions

In this chapter we return to questions around the uniqueness of prime extensions.We will now prove their uniqueness for totally transcendental theories and forcountable stable theories having prime extensions.

38 Indiscernibles in stable theories

We assume throughout this section that T is complete, stable andeliminates imaginaries. Indiscernibles in stable theories are in fact indis-cernible for every ordering of the underlying set. More importantly, we showthat they form a Morley sequence in some appropriate stationary type.

A family I = (ai)i∈I of tuples is totally indiscernible over A, if

C |= ϕ(ai1 , . . . aik) ↔ ϕ(aj1 , . . . ajk)

for all L(A)–formulas ϕ and sequences i1 . . . ik, j1 . . . jk of pairwise distinctindices.

Lemma 38.1. If T is stable, indiscernibles are totally indiscernible.

Proof. Assume that I = (ai)i∈I is indiscernible over A, but not totally in-discernible over A. By Lemma 27.1 we may assume (I,<) = (Q, <). Be-cause any permutation on {1, . . . n} is a product of transpositions of neigh-bouring elements, there are some L(A)–formula ϕ(x1, . . . , xn), rational numbersr1 < . . . < rn and some j ∈ Q such that

|= ϕ(ar1 , . . . , arj , arj+1 , . . . , arn)

and|= ¬ϕ(ar1 , . . . , arj+1 , arj , . . . , arn).

178

CHAPTER 9. PRIME EXTENSIONS 179

The formula ψ(x, y) = ϕ(ar1 , . . . , x, y, . . . , arn) orders the elements (ar), (rj <r < rj+1). By Exercise 33.1, this contradicts stability.

Let p ∈ S(A) be a stationary type. Recall that a Morley sequence in p is aset of realisations of p independent over A. Morley sequences (aα)α<λ are easyto construct as follows: choose aα realising the unique non–forking extensionof p to A ∪ {aβ | β < α}. By Corollary 28.18, {aα | α < λ} is independentover A. Since any Morley sequence arises in this way (independently of theenumeration), the Morley sequences of p are uniquely determined by λ up toisomorphism over A. If p is not an algebraic type, the aα are pairwise distinctand hence indiscernible over A.

Theorem 38.2. If T is stable, then any infinite sequence of indiscernibles overA is a Morley sequence for some stationary type defined over some extension ofA.

Proof. Let I = (ai)i∈I be indiscernible over A. Notice that for all formulasϕ(x, b) the set

Jϕ = {i ∈ I| |= ϕ(ai, b)}is finite or cofinite in I: otherwise for all J ⊂ I the set of formulas

{ϕ(ai, y) | i ∈ J} ∪ {¬ϕ(ai, y) | i 6∈ J}

would be consistent. So there would be 2|I| many ϕ–types over I, contradictingstability of T .

This shows that for every ϕ either Jϕ or I \Jϕ is bounded by some kϕ (whichdepends only on ϕ).

The average type of I is a global type defined by

Av(I) = {ϕ(x, b) | b ∈ C, |= ϕ(ai, b) for all but finitely many i ∈ I}.

By the preceding remarks, this is a complete type. Let I0 be an infinitesubset of I. Since ϕ(x, b) ∈ Av(I) if and only if {i ∈ I0 | |= ϕ(ai, b)} containsmore than kϕ many (and hence infinitely many) elements, Av(I) is definableover I0. Hence Av(I) does not fork over I0 and its restriction to I0 is stationary(cf. Theorem 36.1 and Corollary 36.7.)

It is easy to see that all ai, i ∈ I \ I0, realise the type

p = Av(I) � AI0.

As this is also true for all I ′0 ⊃ I0, we see that ai, i ∈ I \ I0, forms a Morleysequence for p.

At the beginning of the proof we can now replace I by an infinite set ofindiscernibles I ′ containing I as a coinfinite subset which shows I to be aMorley sequence for p′ = Av(I ′) � A(I ′ \ I).


Exercise 38.1.If p is stationary and q a non–forking extension of p, then any Morley sequenceof q is also a Morley sequence for p.

Exercise 38.2.Let p ∈ S(A) be stationary and I a Morley sequence of p.

a) Let B ⊃ A and I0 ⊂ I such that B |AI0

I. Then I\I0 is a Morley sequenceof the non–forking extension of p to B.

b) Av(I) is the non–forking global extension of p.

Exercise 38.3.We call indiscernibles I0 and I1 parallel if there is some infinite set J such thatI0J and I1J are indiscernible. Show that I0 and I1 are parallel if and only ifthey have the same average type.

Exercise 38.4.Stationary types are called parallel if they have the same global non–forkingextension. Show that two types are parallel if and only if two (or all) of theirinfinite Morley sequences are parallel.

Exercise 38.5.Show the converse of Lemma 38.1: if all indiscernibles sequences are totallyindiscernible, then T is stable.


39 Totally transcendental theories

Let T be a totally transcendental theory. In this section we will prove thefollowing theorem:

Theorem 39.1 (Shelah [46]). Let T be totally transcendental.

1. A model M is a prime extension of A if and only if M is atomic over Aand does not contain an uncountable set of indiscernibles over A.

2. Prime extensions are unique.

We first aim to show that a constructible set does not contain an uncountableset of indiscernibles.

Lemma 39.2. Let I be indiscernible over A and B a countable set. Then Icontains a countable subset I0 such that I \ I0 is indiscernible over ABI0.

Proof. By Theorem 38.2, I is a Morley sequence of some stationary types oversome extension A′ of A. Since T is totally transcendental, we only need to extendA by finitely many elements (this follows using Exercise 36.5 and Exercise 38.1).So we may assume A′ = A. For every finite tuple b in B there is some finite setI0 such that b |

AI0I. In this way we find a countable set I0 with

BI0 |AI0

I.

It now follows from Exercise 38.2 that I\I0 is a Morley sequence over ABI0.

We need a bit of set theory: a club D ⊂ ω1 is a closed and unbounded subsetwhere closed means that sup(α ∩D) ∈ D for all α ∈ ω1.

Theorem 39.3 (Fodor). Let D ⊂ ω1 be a club and f : D → ω1 a regressivefunction, i.e., f(α) < α for all α ∈ D. Then f is constant on an unboundedsubset of ω1.

Proof. Suppose not, then for each α ∈ ω1, the fibre Dα = {x ∈ D : f(x) = α}is bounded. Construct a sequence α0 < α1 < . . . of elements of D as follows.Let α0 be arbitrary. If αn is constructed, choose for αn+1 an upper bound of⋃β<αn

Dβ . So we have for all γ ∈ D

γ ≥ αn+1 ⇒ f(γ) ≥ αn.

For δ = supn<ω αn, this implies f(δ) ≥ δ. A contradiction.


Recall from Section 17 that a set B = {bα}α<µ is a construction over A if alltp(bα/ABα) are isolated where as above Bα = {bβ | β < α}). A subset C ⊂ Bis called construction closed if for all bα ∈ C the type tp(bα/ABα) is isolated bysome formula over A ∪ (Bα ∩ C).

The following lemma holds for arbitrary T .

Lemma 39.4. Let B = {bα}α<µ be a construction over A.

1. Any union of construction closed sets is construction closed.

2. Any b ∈ B is contained in a finite construction closed subset of B.

3. If C ⊂ B is construction closed, then B is constructible over AC.

Proof. The first part is clear. For the second part, let b = bα ∈ B. Since thetype of b = b0 is isolated over A, we can do induction on α. As {bβ}β<µ isa construction over A, tp(bα/ABα) is isolated by some formula ϕ(x, c) wherec = bβ1 . . . bβn with βi < α. By induction, each bβi is contained in a finiteconstruction closed set Ci. Thus C1 ∪ . . . ∪Cn ∪ {bα} is finite and constructionclosed.

For 3. we assume A = ∅ to simplify notation. We will show that the typetp(bα/CBα) is isolated for all α. This is clear if bα ∈ C. So assume bα 6∈ C.From the assumption it is easy to see that C is isolated over Bα+1 where theisolating formulas only contain parameters from Bα+1 ∩ C ⊂ Bα.

We thus havetp(C/Bα) ` tp(C/Bαbα).

If ϕ(x) isolates the type tp(bα/Bα), then ϕ(x) also isolates the type tp(bα/CBα).

Lemma 39.5. If B is constructible over A, then B does not contain an un-countable set of indiscernibles over A.

Proof. We assume A = ∅. Let I = {cα}α<ω1 be indiscernible. By Lemma 39.4we can build a continuous sequence Cα of countable construction closed subsetsof B such that cα ∈ Cα+1. By Lemma 39.2 there is a club D consisting of limitordinals such that for all δ ∈ D the set {cα | α ≥ δ} is indiscernible over Cδ.Each cδ is isolated over Cδ by a formula with parameters from Cδ′ with δ′ < δ.By Fodor’s Theorem there is some δ0 such that for some cofinal set of δ’s fromD the parameters can be chosen in Cδ0 . Assume that δ1 < δ2 are such elements.Then cδ1 and cδ2 have the same type over Cδ0 . But this is impossible since

tp(cδ2/Cδ0) ` tp(cδ2/Cδ0cδ1).


Let M be a model and A ⊂M . We call a subset B of M normal in M overA if for every element b ∈ B all realisations of tp(b/A) in M are contained in B.

Lemma 39.6. Let T be a (not necessarily totally transcendental) theory whichhas prime extensions. If M is atomic over A and B is a normal subset of Mcontaining A, then M is atomic over B.

Proof. Let c be a tuple from M , so tp(c/A) is isolated. Since the isolated typesare dense over B by Theorem 14.7, there is some d ∈ M atomic over B andrealizing the type tp(c/A). Let tp(d/B) be isolated by ψ(x, d0) for some tupled0 ∈ B. Since c and d satisfy the same type over A there is some c0 ∈ M suchthat tp(cc0/A) = tp(dd0/A). Then c satisfies ψ(x, c0) and as B is normal, wehave c0 ∈ B. It follows easily that ψ(x, c0) is complete over B as well.

Proof of 39.1(1). If M is a prime extension of A, then M is atomic over A byCorollary 17.7 and since M can be embedded over A into some constructibleprime extension, M does not contain an uncountable set of indiscernibles overA by Lemma 39.5.

For the converse assume again A = ∅, i.e. supposeM is atomic over ∅ withoutuncountable set of indiscernibles. In order to prove that M can be embeddedinto any model N we enumerate all types over ∅ realised in M as (pµ)µ<ν andrecursively extend the empty map to the normal sets Cµ =

⋃α<µ pα(M). That

this is possible follows from the following

Claim: Let M be atomic over B, p ∈ S(B) and B ⊂ C ⊂ M normal over B.Then any elementary map C −→ N can be extended to C ∪ p(M).

We proof the claim by induction on the Morley rank of p and note that theclaim is clear if p is algebraic.

Assume inductively that the claim is proved for all types of Morley rank lessthan α (over arbitrary sets B). Then any given elementary map f : C −→ Nwith B ⊂ C ⊂ M and C normal over B can be extended to C ∪ {a ∈ M |MR(a/B) < α}.

Let now p ∈ S(B) with MR(p) = α. Let {ci}i<µ be a maximal set ofrealisations of p in M independent over B. By Exercise 39.1, {ci}i<µ splits intoa finite number of indiscernible sequences, which implies that µ is countable,and we can assume that µ ≤ ω. Let Bi = B ∪ {c0, . . . , ci−1} and Ci = C ∪ {a ∈M | MR(a/Bi) < α}, so B = B0. By maximality we have p(M) ⊂

⋃i<µ Ci. As

M is atomic over B, M is also atomic over Bi and since Ci is normal over Bieven atomic over Ci. If f has been extended to Ci, we may extend f to CiBi+1

by the atomicity. By induction hypothesis applied to Bi+1 and CiBi+1 there isan extension to Ci+1.

The proof of 39.1 (1) can easily be symmetrised to yield 39.1 (2).


Example:[48] Let L contain a binary relation symbol Eα for every ordinal α < ω1 and letT be the theory stating that each Eα is an equivalence relation such that E0

consists of only one class and for any α < β < ω1 each Eα–equivalence class isthe union of infinitely many Eβ–equivalence classes. T is complete, stable (butnot totally transcendental) and admits quantifier elimination. Every consistentL(A)–formula ϕ(x) can be completed over A. Therefore there exists a model Mwhich is constructible over the empty set.

In M any chain (Kα)α<ω1 of Eα–equivalence classes has non-empty inter-section: otherwise there would be a countable subset A of M and some limitordinal η < ω1 such that

(i) A is construction closed (with respect to a fixed construction);

(ii) A ∩⋂α<ηKα = ∅;

(iii) A ∩Kα 6= ∅ for all α < η.

M/A is atomic. Let c ∈ Kη; then tp(c/A) is isolated by some formula ϕ(x; a);by (ii) there exists some α < η such that a ∩ Kα = ∅. By (iii) there is somed ∈ A ∩Kα; since a ∩Kα = ∅, it would follow |= ϕ(d; a); but this is impossibleas ϕ isolates tp(c/A). Therefore we have ∩(Kα)α<ω1 6= ∅.

Now let a ∈M and let N be the set of all b from M for which there is someordinal α < ω1 with |= ¬aEαb; N is also prime model, but M and N are notisomorphic.

Exercise 39.1.Let T be totally transcendental, and I be an independent set of realisationsof p ∈ S(A). Then I can be decomposed into a finite number of I1,. . . ,In ofindiscernible sets over A.


40 Countable stable theories

We assume throughout this section that T is countable and stable.For such T we will show that prime extensions, if they exist, are unique. Themain point is Shelah’s result that in this situation subsets of constructible setsare again constructible.

The proof of Theorem 39.1(1) showed that atomic extensions of A with-out uncountable sets of indiscernibles are constructible and hence prime. Theuniqueness of such prime extensions then also follows directly from the followingtheorem which holds for arbitrary theories.

Theorem 40.1 (Ressayre). Constructible prime extensions are unique.

Proof. It suffices to prove the theorem for constructible prime extensions M andM ′ over the empty set. Let f0 : E0 → E′

0 be a maximal elementary map betweenconstruction closed subsets E0 ⊂ M and E′

0 ⊂ M ′. If E0 6= M , there is someproper construction closed finite extension E1 of E0. Since E1 is atomic overE0 by Lemma 39.4.3 and Corollary 17.6 there is an extension f1 : E1 → E′

1 off0. Then E′

1 need not be construction closed, but there is a construction closedfinite extension E′

2 of E′1. Similarly there exists a (not necessarily construction

closed) set E2 ⊂ M and some extension f1 to an isomorphism f2 : E2 → E′2.

Continuing in this way we obtain an ascending chain of elementary isomorphismsfi : Ei → E′

i. Then E∞ := ∪Ei and E′∞ := ∪E′

i are construction closed andf∞ := ∪fi is an elementary isomorphism from E∞ to E′

∞, contradicting themaximality of f0.

Theorem 40.2 (Shelah [48]). If T is stable and countable, then any subset ofa set constructible over A is again constructible over A.

We immediately obtain the following corollary:

Corollary 40.3. If T is a countable stable theory with prime extensions (seeExercise 17.1), then all prime extensions are unique.

For the proof of the theorem we need the following lemma:

Lemma 40.4. Let T be countable and stable; if A and B are independent overC and B′ is countable, then there is a countable subset C ′ of A with A and BB′

independent over CC ′.

Proof. Using the properties of forking, we find a countable subset C ′ ⊆ A withABC |

BCC′ B′; then A |

BCC′ B′, and since A |

CC′ B we have A |CC′ BB

′.

Proof of 40.2. Let B be constructible over A and D a subset of B. We mayassume that B is infinite since finite sets are constructible. If E is an arbitrary


construction closed subset of B and E′ ⊂ B a countable extension of E (i.e.E′ \E is countable), then there is some countable construction closed extensionE′′ of E′. Similarly, for any E withD |

A(D∩E)E and every countable extension

E′ of E there is a countable extension E′′ with D |A(D∩E′′)

E′′ (40.4).

Applying these closure procedures alternatingly countably many times, onesees that for any construction closed subset E of B with

D |A(D∩E)

E

and for any countable extension E′ of E there exists a construction closed count-able extension E′′ of E′ such that

D |A(D∩E′′)

E′′.

In this way we obtain a continuous chain (Cα)α<ξ of construction closed setswith C0 = ∅,

⋃α<ξ Cα = B, countable differences Cα+1 \ Cα and

D |A(D∩Cα)

Cα.

For each α we can choose an ω–enumeration of D ∩ (Cα+1 \ Cα). Theseenumerations can be composed to an enumeration of D. In order to show thatthis enumeration is a construction of D it suffices to show that every initialsegment d of D ∩ (Cα+1 \Cα) is atomic over A(D ∩Cα). This follows from theOpen Mapping Theorem (36.9) as d is atomic over ACα.

Chapter 10

The fine structure ofℵ1–categorical theories

41 Internal types

By the results in Sections 22 and 25 we know that models of ℵ1–categoricaltheories are (minimal) prime extensions of strongly minimal sets. We will seein the next section that in this case the prime extensions are obtained in a par-ticularly simple way. Unless stated otherwise, we assume in this sectionthat T is totally transcendental.

We need the concept of an internal type. We fix a 0–definable infinitesubclass F = ϕ0(C) of C.

Definition 41.1. A partial type π(x) is called F–internal if for some set B, theclass π(C) is contained in dcl(FB).

Note that by compactness π is F-internal if and only if there is some finiteconjunction ϕ of formulas in π which is F-internal.

Example:Let G be a group. Let

M = (G,A)

be a two–sorted structure where A is a copy of G without the group structure.Instead, the structure M contains the map

π : G×A→ A

defined as π(g, a) = ga. Then M is the prime extension of G and clearly thetype p of an element in A is G–internal, in fact, p(M) ⊆ dcl(G, a) for any a ∈ A.We will see in Corollary 41.6 that this is the typical picture for internal types.

187

CHAPTER 10. STRUCTURE OF ℵ1–CATEGORICAL THEORIES 188

Lemma 41.2. A type p ∈ S(A) is F–internal if and only if there is some set ofparameters B and some realisation e of p such that e ∈ dcl(FB) and e |

AB.

Proof. If p(C) ⊂ dcl(FB), we just choose e as a realisation of p independent ofB over A.

Conversely, given a realisation e of p with e |Ab and e ∈ dcl(Fb), we choose

a Morley sequence b0, b1, . . . of tp(b/ acleq(A)) of length |T |+. For e′ ∈ p(C) thereis some i such that e′ |

Abi as otherwise the pi = tp(e′/ acleq(A)b1 . . . bi−1)

form a chain of forking extensions of length |T |+ contradicting Exercise 28.1.If p is stationary, then eb and e′bi realise the same type over A and hencee′ ∈ dcl(Fbi). So p(C) ⊂ dcl(FB) for B = {b0, b1 . . .}. If p is not stationary,then by the previous argument all extensions of p to acleq(A) are F–internal.By taking unions we obtain a set B such that the realisations of all extensionsof p to acleq(A) are contained in dcl(FB).

Lemma 41.3. A consistent formula ϕ is F-internal if and only if there is adefinable surjection from some Fn onto ϕ(C).

Proof. If there is a B–definable surjection Fn → ϕ(C), ϕ(C) is contained indcl (FB). For the converse assume that ϕ(C) ⊂ dcl (FB). Then, by Exer-cise 23.11, for every e ∈ ϕ(C) there is a B–definable class De ⊂ Fne and aB–definable map fe : De → C such that e is in the image of fe. A compactness ar-gument shows that there is a finite number of definable classes D1, . . . ,Dm ⊂ Fnand definable maps fi : Di → C such that ϕ(C) is contained in the union of thefi(Di). Fix a sequence of distinct elements a1, . . . , am ∈ F and an element b ofϕ(C). Define f : Fn+1 → ϕ(C) by setting f(x, y) = fi(x) if y = ai and x ∈ Diand f(x, y) = b otherwise. Then f is a surjection from Fn+1 onto ϕ(C).

Lemma 41.4. Let T be an arbitrary theory and F a stably embedded 0–definableclass. If a and b have the same type over F, they are conjugate under an elementof Aut(C/F).

Proof. We construct the automorphism as the union of a long ascending se-quence of elementary maps α : A ∪ F → B ∪ F which are the identity on F.Assume that α is constructed and consider an element c ∈ C. Since F is stablyembedded, the type of cA over F is definable over some subset C of F. Choosesome d ∈ C with tp(dB/C) = tp(cA/C). We can then extend α to an elemen-tary map α′ : {c} ∪A ∪ F → {d} ∪B ∪ F. To see this assume that |= ϕ(c, a, f),where a ∈ A and f ∈ F. Then ϕ(x, y, f) belongs to the type of cA over F, soϕ(x, y, f) belongs also to the type of dB over F which shows |= ϕ(d, α(a), f).

A groupoid is a category where all morphisms are isomorphism. A groupoidis connected if there are morphisms between any two objects. A definablegroupoid G is a groupoid whose objects are given by a definable family (Oi)i∈I


of classes and whose morphisms by a definable family (Mi,j)i,j∈I of bijectionsOi → Oj . We use the notation HomG(Oi, Oj) and AutG(Oi) for Mi,j and Mi,i.

We denote by Feq ⊂ Ceq the collection of those 0–definable equivalenceclasses having representatives in F.

Theorem 41.5 (Hrushovski’s Binding Groupoid). Let T be totally transcen-dental, E and F be 0–definable and assume that E is F–internal and non-empty.Then E is an object of a 0–definable connected groupoid G with the followingproperties

a) E is not the only object of G. The objects other than E are subsets of Feq.

b) AutG(E) = Aut(E/F).

AutG(E) as a definable group is Poizat’s Groupe de Liaison ([41]).

Proof. By Lemma 41.3 there is a definable surjection from some power Fn to E.This induces a bijection f : O → E for some definable class O of Feq. Let f = fcbe defined from a parameter c and O = Od from a parameter d. By Lemma 34.3we find d in F. Let M0 be an atomic model of T , so we may assume d ∈ F∩M0.Since, by Lemma 17.4, the isolated types are dense over any parameter set, wemay assume that the type of c over F∩M0 is isolated, say by a formula ϕ(x, a).It is easy to see that ϕ(x, a) isolates a type over F. By extending d if necessarywe may assume that d = a. Now let ψ(y) isolate the type of d.

The objects of G are E and the Oe where e realises ψ(y). HomG(Oe,E) isthe set of all fc′ where c′ realises ϕ(x, a). We claim that HomG(Oe,E) is a rightcoset of Aut(E/F). We can then set

HomG(E, Oe) = {f−1 | f ∈ HomG(Oe,E)}HomG(E,E) = {f ◦ g−1 | f, g ∈ HomG(Oe,E)}

HomG(Oe, Od) = {f−1 ◦ g | f ∈ HomG(Od′ ,E), g ∈ HomG(Oe,E)}.

To see that HomG(Oe,E) is a right coset of Aut(E/F) we have to show that forsome c′ we have

HomG(Oe,E) = Aut(E/F) ◦ fc′ .

This follows easily from the fact that, by Lemma 41.4, the elements of HomG(Oe,E)have the form fα(c) for automorphisms α ∈ Aut(C/F) and from the formula

fα(c′) = α ◦ fc′ .

Recall that a group G acts regularly on a set A if for all a, b ∈ A there existsa unique g ∈ G with ga = b.


Corollary 41.6 (Binding group). Let T be a totally transcendental theory, Eand F be 0–definable and assume that E is F–internal. Then the following holds.

1. There is a definable group G ⊂ Feq, the binding group, and a definableclass A on which G acts regularly and such that E ⊂ dcl(Fa) for all a ∈ A.G, A, the group operation and the action of G on A are definable withparameters from F.

2. Aut(E/F) is a 0–definable permutation group, Aut(E/F) acts regularly onA and is definably isomorphic to G.

Proof. Let G be the groupoid of Theorem 41.5. Fix any object Oi differentfrom E. Set G = AutG(Oi) and A = HomG(Oi,E). Now replace the definablebijections in G and A by their canonical parameters in order to obtain elementsof Feq and Ceq respectively.

Exercise 41.1.Prove that every element of dcleq(F) is interdefinable with an element of Feq.

Exercise 41.2.Let T be arbitrary, F 0–definable and C a subset of F. Show that tp(a/F) isdefinable over C if and only if tp(a/C) ` tp(a/F).

Exercise 41.3 (Chatzidakis-Hrushovski, [13]).Let T be arbitrary and F 0–definable. Show that the following are equivalent:

a) F is stably embedded.

b) Every type tp(a/F) is definable over a subset C of F.

c) For every a there is a subset C of F such that tp(a/C) ` tp(a/F).

d) Every automorphism of F extends to an automorphism of C.


42 Analysable types

Throughout this section we assume that T is a stable theory elimi-nating imaginaries.

Definition 42.1. Let F be a 0–definable class. A type p ∈ S(∅) is called F–analysable if for every realisation a of p there is a sequence of tuples a0, . . . , an =a in dcl(a) such that tp(ai/a0 . . . ai−1) is F–internal for i = 0, . . . n.

Theorem 42.2 (Hrushovski [23]). Let T be ℵ1–categorical and F a 0–definablestrongly minimal set. Then every type p ∈ S(∅) is F–analysable.

We need some preparation in order to prove this theorem:

Theorem 42.3. Let p ∈ S(A) be a stationary type and I an infinite Morleysequence for p. Then Cb(p) ⊂ dcl(I).

Proof. By Exercise 38.2 b) the average type Av(I) is the non–forking extensionof p to the monster model. The proof of Theorem 38.2 implies that Av(I) isbased on I.

Definition 42.4. 1. We call types p, q ∈ S(A) almost orthogonal if anyrealisation of p is independent over A from any realisation of q. The typesare orthogonal if for any set B ⊇ A all non–forking extensions of p to Bare almost orthogonal to all non–forking extensions of q to B. Note thatalgebraic types are orthogonal to all types.

2. A theory T is called unidimensional if all stationary non-algebraic typesare pairwise non-orthogonal.

Let F be a strongly minimal set. We call a type p ∈ S(A) orthogonal to F iffor every realisation b of p, any c ∈ F and any extension B of A over which F isdefined we have

b |A

B =⇒ b |B

c.

If q is any type of Morley rank 1 containing F(x), this is equivalent to p beingorthogonal to q.

Lemma 42.5. Let T be ℵ1–categorical and F a strongly minimal set. Then nonon-algebraic type is orthogonal to F.

Proof. Let p ∈ S(A) be orthogonal to F. We choose a model M containingA over which F is defined, a realisation b of p independent from M over A,and a model N containing M and b. It follows from the assumptions that b isindependent from F(N) over M .


On the other hand N is the (minimal) prime extension of M ∪F(N). Henceb is atomic over Mc for some tuple c ∈ F(N) and isolated by, say, ψ(x, c). ByCorollary 34.7 the type tp(c/Mb) is an heir of tp(c/M). Therefore there mustbe some m ∈ M realizing ψ(x, c). But this m can only be b itself. So we seethat b |

Ab, hence b ∈ acl(A) by Theorem 36.5.6. and p is algebraic.

Proof of 42.2. By induction on α = MR(p). If α = 0, p is algebraic and hencetrivially internal (in any definable class). If α > 0, we apply Lemma 42.5 andfind a realisation b of p, some c ∈ F and some set of parameters B such thatb | B and b 6 |

Bc. By finite character of forking we may assume that B is

finite.

Let D be the canonical base of tp(cB/ acl(b)). As cB |Db, but cB 6 | b, we

have b 6 | D.Let (ciBi) be an infinite Morley sequence for tp(cB/D). By Theorem 42.3, wehave

D ⊂ dcl(c0B0c1B1 . . .).

It follows from b | B (and D ⊂ acl(b)) that we must have D | B andhence D | B0B1 . . .. This shows that the type of any tuple d ∈ D is F–internalby Lemma 41.2. We choose a finite tuple d ∈ D with b 6 | d. Then we haveMR(b/d) < α. We may absorb the parameter d into the language and applythe induction hypothesis to T (d) = T ∪ {ϕ(d) : ϕ ∈ tp(d)} to find a sequenceb1, . . . , bn = b such that bi ∈ dcl(db) and the types tp(bi/db1 . . . bi−1) are F–internal. Setting b0 = d, we would be done if we knew that d ∈ dcl(b). For thiswe replace d by the canonical parameter d′ of the finite set {d1, . . . , dk} of conju-gates of d over b. We have thus achieved d′ ∈ dcl(b). Because d′ ∈ dcl(d1 . . . dk)the type of d′ is F–internal and since d ∈ acl(d′), we have MR(b/d′) ≤ MR(b/d) <α. We can use this d′ to finish the proof.

Theorem 42.6 (Baldwin [2]). ℵ1–categorical theories have finite Morley rank.

To prove Theorem 42.6 we need the following definition which allows us toextend additivity of Morley rank beyond strongly minimal sets (see also Remark26.9 and Exercise 26.3):

Definition 42.7. Let f : B → A be a definable surjection. We say that thefibers of f have definable Morley rank if there is a finite bound for the Morleyrank of the fibers f−1(a) and if for every definable B′ ⊂ B and every k′ the class{a ∈ A | MR(f−1(a) ∩ B′) = k′} is definable.

Remark 42.8. If B is a power of a strongly minimal set, the fibers of f havedefinable Morley rank by Corollary 26.4.

For the next statement remember that the Morley rank of the empty set isdefined as −∞.


Lemma 42.9. If the fibers of f : B → A have definable Morley rank and MR(A)is finite, we have

MR(B) = maxk<ω

(MR(Ak) + k),

where Ak = {a ∈ A | MR(f−1(a)) = k}.

Proof. We leave it as an exercise (Exercise 42.1) to show that MRB ≥ MRAk+kfor all k. For the converse we may assume that all fibers have Morley rank kand that A has Morley degree 1 and Morley rank β. We show MRB ≤ β + kby induction on β.

Let Bi be a family of disjoint definable subsets of B. We want to show thatone of the Bi has smaller Morley rank than β+k. Consider any a ∈ A. Then forone i the fiber f−1(a) ∩ Bi must have rank smaller than k. So the intersectionof all Aik = {a | MR(f−1(a) ∩ Bi) = k} is empty, which implies that one of theAik has smaller rank that β. Induction yields MRBi < β + k.

Proof. (of Theorem 42.6) It is enough to prove that every element a has finiteMorley rank over ∅. Each a has an analysing sequence a0, . . . , an = a whereall types tp(ai/a0 . . . ai−1) are F–internal. We prove by induction on n that thetuple a0 . . . an has finite Morley rank. By Lemma 26.1 this implies that an hasfinite Morley rank.

By induction hypothesis, a0 . . . an−1 is contained in a 0–definable set A offinite Morley rank. Since tp(an/a0 . . . an−1) is F–internal, an is contained in an(a0 . . . an−1)–definable set which is an image of some power of F by a definablemap. So we may assume that a0 . . . an belongs to a 0–definable set B whichprojects onto A by the restriction map π : B → A and such that the fibersπ−1(a) are definable images of some power of F. By Corollary 42.8 the fibersof π have definable Morley rank. If the rank of the fibers is bounded by k,Lemma 42.9 bounds the rank of B by MRA+ k.

We end this section with a different characterisation of ℵ1-categorical theo-ries due to Erimbetov [14].

Theorem 42.10. A countable theory T is ℵ1–categorical if and only if it isω–stable and unidimensional.

Proof. Assume first that T is ℵ1-categorical. Let F be strongly minimal, definedover A, p and q be two stationary types over A. By Lemma 42.5 there is anextension A ⊂ B, realisations a, b, c1, c2 of p, q and F such that a |

AB,

b |AB, a 6 |

Bc1,b 6 | B c2. That means that c1 ∈ acl(aB)\acl(B), c2 ∈ acl(bB)\

acl(B). So c1 and c2 have the same type over B and we may assume that c1 = c2.But then c1 6 |

Bc1 implies a 6 |

Bb and p and q are not orthogonal.

For the converse assume that T is ω–stable and unidimensional. The proofof the Baldwin–Lachlan Theorem shows that it is enough to prove that there


are no Vaughtian pairs M � N for strongly minimal formulas ϕ(y) defined overM . Let a be any element in N \M and p(x) = tp(a/M). By assumption thereis an extension M ′ of M and an element c ∈ ϕ(C) \M ′ such that a |

MM ′ and

c ∈ acl(aM ′). Let δ(a,m′, y) be a formula which isolates the type of c over aM ′.Then the following sentences are true in M ′

dp x∃yδ(x,m′, y)

dp x∀y(δ(x,m′, y) → ϕ(y))

∀y dp x¬δ(x,m′, y)

So we find an m ∈ M for which the corresponding sentences are true in M .This implies that there is an b ∈ B such that N |= δ(a,m, b) and that all suchb lie in ϕ(N) \M . So M � N is not a Vaughtian pair for ϕ.

Exercise 42.1.Let f : B → A definable and all fibers of Morley rank ≥ α. Then MRB ≥α+MRA.


43 Locally modular strongly minimal sets

In this section, we let T denote a complete stable theory and ϕ(x) a stronglyminimal formula without parameters.

We call ϕ modular if its pregeometry is modular in the sense of Definition 8.9in Appendix C, i.e. if for all relatively algebraically closed A,B in ϕ(C)

(1) dim(A ∪B) + dim(A ∩B) = dim(A) + dim(B).

T is locally modular if (1) holds whenever A ∩ B contains an element not inacl(∅).

It is easy to see that ϕ is modular if and only if any two relatively alge-braically closed subsets A and B of ϕ(C) are independent over their intersection:

A |A∩B

B.

(cf. Appendix C, Lemma 8.10). In fact this holds for arbitrary sets B, notnecessarily contained in ϕ(C):

Lemma 43.1. If ϕ is modular, then

A |A∩B

B

for algebraically closed B and any A which is relatively algebraically closed inϕ(C).

Proof. Let C be the intersection of B and ϕ(C). It is enough to show that Ais independent from B over C. For this we may assume that B is the algebraicclosure of a finite set and the elements of A are algebraically independent overC. We have to show that the elements of A remain algebraically independentover B. Choose a B-independent sequence A0, A1, . . . of sets realising the sametype as A over B. For any i the intersection of acl(A0 . . . Ai) and acl(Ai+1) iscontained in B. So by local modularity A0 . . . Ai and Ai+1 are independent overC. This implies that the elements of A0 ∪ A1 . . . are algebraically independentover C. By Exercise 38.2 for some i the elements of Ai∪Ai+1 . . . are algebraicallyindependent over B. Hence also the elements of A are algebraically independentover B.

Definition 43.2. A formula ψ(x) without parameters is 1-based if

A |acleq(A)∩ acleq(B)

B

for all B and all subsets A of ψ(C).


For a set B and a tuple a, the strong type stp(a/B) = tp(a/ acleq(B)) isstationary (cf. Exercise 35.9). We denote by Cb(a/B) the canonical basis ofstp(a/B). Note that Cb(a/B) is a subset of Ceq.

Lemma 43.3. ψ is 1–based if and only if

Cb(a/B) ⊂ acleq(a)

for all sets B and finite tuples a in ψ(C).

Proof. Let C be the intersection acleq(a)∩ acleq(B). If T is 1–based, the strongtype of a over B does not fork over C. So by Lemma 36.8 Cb(a/B) is con-tained in C ⊂ acleq(a). If conversely Cb(a/B) is contained in acleq(a), it is alsocontained in C and a |

Cb(a/B)B implies a |

CB.

Corollary 43.4. 1. 1–basedness is preserved under adding and removing pa-rameters, i.e., ψ is 1–based if and only if ψ is 1–based in CA for any setA of parameters.

2. If ψ is 1–based and if every element of ψ′(C) is algebraic over ψ(C), thenψ′(C) is 1–based.

Proof. 1. If ψ is 1–based, then CbA(a/B) = Cb(a/AB) ⊂ acleq(a) ⊂ acleqA(a).If conversely ψ is 1–based in CA and a ∈ ψ(C) and B are given, we may as-sume that a,B is independent from A. We have than Cb(a/B) = Cb(a/AB) =CbA(a/B) ⊂ acleqA(a). Since Cb(a/B) is also contained in acleq(B), we con-clude Cb(a/B) ⊂ acleq(a).

2. First note that if c is algebraic over a, then c and B are independent overCb(a/B), so we have Cb(c/B) ⊂ acleq Cb(a/B). Now let a′ be a finite tuplefrom ψ′(C) and B any set. Choose a tuple a1 from ϕ(C) over which a′ is alge-braic. Then choose a2 which realises the type of a1 over a′ and is independentfrom a1 over a′. We have then Cb(a′/B) ⊂ acleq Cb(a1/B) ∩ acleq Cb(a2/B) ⊂acleq(a1) ∩ acleq(a2) ⊂ acleq(a′).

Theorem 43.5. Let T be totally transcendental and ϕ a strongly minimal for-mula without parameters. Then the following are equivalent.

a) ϕ is locally modular.

b) ϕ is 1–based.

c) Every family of plane curves in ϕ has dimension at most 1. This means thatfor all B and elements a, b of ϕ(C), if tp(ab/B) has Morley rank 1, thenCb(ab/B) has Morley rank at most 1 over the empty set.


Proof. a→b: If ϕ is locally modular, ϕ becomes modular if we add a name forany element x ∈ ϕ(C) \ acleq(∅) to the language. If B and a ∈ ϕ(C) are given,we choose x independent of a,B. It follows from Lemma 43.1 that a and Bxare independent over acleq(ax)∩acleq(Bx). This implies Cb(a/Bx) ⊂ acleq(ax).On the other hand, we have Cb(a/Bx) = Cb(a/B) ⊂ acleq(B). Since x and Bare independent over a, we have acleq(ax) ∩ acleq(B) ⊂ acleq(a). This impliesCb(a/B) ⊂ acleq(a).

b→c: Write d = Cb(ab/B). Then MR(ab/d) = 1. If the Morley rank of d isnot zero, ab and d are dependend over the empty set. By our assumption d ∈acleq(ab), so we have MR(abd) ≤ 2. Since MR(ab/d) = 1, we have MR(d) ≤ 1using Remark 26.9.

c→a: Let x be a non-algebraic element of ϕ(C). By Lemma 8.11 of Appendix Cwe have to show the following: For all elements a, b and sets B in ϕ(C), ifMR(ab/x) = 2, MR(ab/Bx) = 1, there is a c ∈ acl(Bx) such that MR(ab/cx) =1. We may assume that a 6∈ acl(Bx). Consider the imaginary element d =Cb(ab/Bx). Since d is contained in acleq(Bx), a is not algebraic over d. Also,since d is algebraic over ab by assumption, x is not algebraic over d. So a andx have the same type over d and we can find an element c such that xc and abhave the same type over d. Now b ∈ acl(ad) implies c ∈ acl(xd) ⊂ acl(Bx). Wehave also d ∈ acleq(xc), which implies MR(ab/cx) = 1.

Proposition 43.6. A totally transcendental theory T which contains an infinitedefinable field is not one-based.

This is true for stable theories, see [38].

Proof. Let K be an infinite definable set with a definable field structure. Wemay assume that everything is definable over the empty set. Let α be the Morleyrank of K. We call an element x of K generic if MR(x) = α. We note first thatif p = (x, y) is an element of the line ga,b = {(x, y) | ax + b = y} which is notalgebraic over a, b, then a, b ∈ dcl Cb(p/a, b). This follows from the fact, thattwo lines intersect in at most one point. So it is enough to find such p and ga,bwith (a, b) not algebraic over p.

For this we choose four independent generic elements a, b, a′, b′ and let p =(x, y) be the intersection of ga,b and ga′,b′ . Since x and b′ are interdefinableover a,b,a’, we can conclude that x, a, b, a′ are generic and independent. So p isnot algebraic over a, b. Since y and b are interdefinable over x, a we have thatx, y, a, a′ are generic and independent. This implies that a, b is not algebraicover p.

The converse of the previous proposition for strongly minimal theories wasknown as Zilber’s conjecture. This conjecture, namely that for any non-locallymodular strongly minimal theory T an infinite field is definable in T eq, wasrefuted by Hrushovski in [24]. A variant of his construction of a new strongly


minimal set will be given in the next section. However, Hrushovski and Zilberproved in their fundamental work [26] that the conjecture holds for so-calledZariski structures.


44 Hrushovski’s Examples

To end the book, we present a modification of Hrushovski’s ab initio example ofa new strongly minimal set [24]. This counterexample to Zilber’s conjecture hasbeen the starting point of a whole new industry constructing new uncountablycategorical groups [6], fields [5], [7], and geometries [4], [51]. The dimensionfunction defined below also reappears in Zilber’s work around Shanuel’s Con-jecture [57].

Following Baldwin [4] (see also [51]), we construct an almost strongly mini-mal projective plane as a modified Fraïssé limit: instead of considering structureswith all their substructures we restrict the amalgamation to so-called strong sub-structures and embeddings. This will allow us to keep control over the algebraicclosure of sets so that the resulting structure is uncountably categorical.

Recall that a projective plane is a point-line geometry such that any twolines meet in a unique point, any two points determine a unique line throughthem and there are four points no three of which are collinear. For convenience,we consider a projective plane as a bipartite graph where the bipartition dividesthe vertices into points and lines, and incidence between a point and a line isrepresented by an edge. In these terms a projective plane is a bipartite graphwith the property that the distance between any two vertices is at most 3,the smallest cycles in the graph have length 6, and any element has at least 3neighbours.

We fix a language L = {P,E} for bipartite graphs where P is a predicatedenoting the partition and E denotes the edges. For a finite graph A, let e(A)denote the number of edges in A and |A| the number of vertices. We defineδ(A) = 2|A| − e(A) and more generally for finite subgraphs A,B ⊆ M we putδ(A/B) = δ(AB) − δ(B). We call a finite set B ⊆ M strong in M , B ≤ M , ifδ(A/B) ≥ 0 for all finite A ⊆M containing B.

Note that if A and B are disjoint subsets of a graph M , then

δ(AB) = δ(A) + δ(B)− e(A,B)

where e(A,B) denotes the number of edges connecting a vertex in A to one inB.

Let K be the class of graphs A, bipartite with respect to P , not containingany 4-cycles, and such that δ(A′) ≥ 4 for any finite subgraph A′ of A with|A′| ≥ 3. Note that this implies δ(A) ≥ 0 for all A ∈ K.

From now on all finite graphs considered are in K.

Definition 44.1. We call a proper strong extension B over A minimal if itcannot be split into two proper strong extensions A ≤ C ≤ B. For A,B finitesubgraphs of a graph M , we say that B is i-simple over A if A and B aredisjoint, δ(AB/A) = i and for every proper nonempty subset C of B we have


δ(AC/C) > i. A graph B is minimally i-simple over A if it is i-simple over Abut not i-simple over any proper nonempty subset of A.

Remark 44.2. It is easy to see that a 0-simple extension B of A is minimally0-simple over the set of elements of A which are incident with an element of B.Note also that if A ≤ B and C ⊂ B, then A ∩ C ≤ C (see Exercise 44.1).

A minimal extension B over A is i-simple for i = 0, 1 or 2. More precisely,either B is 0-simple over A, or it is of the form B = A∪ {b} where b is incidentto at most one element of A. If b is incident with exactly two vertices a1, a2of A, then b is 0-simple over A and minimally 0-simple over {a1, a2}. Clearlyδ(b/A) equals 1 if b is incident to an element of A and 2 otherwise. Any 1-simpleextension is of this form: if B is 1-simple over A with |B| > 1, then for anyb ∈ B we would have δ(bA/A) = 2, i.e. no element of B is connected to anyelement of A. Then δ(AB) = δ(A) + δ(B) > δ(A) + 1, a contradiction.

We next fix a function µ from pairs (A,B) of finite graphs in K with Bminimally 0-simple over A into the natural numbers satisfying the followingproperties:

1. The function µ depends only on the isomorphism type of (A,B).

2. If A = {a1, a2}, B = {b} and b is connected to a1 and a2, then µ(A,B) = 1;otherwise µ(A,B) ≥ max{δ(A), 3}.

For any graph N and any minimally 0-simple pair (A,B) with A ⊆ N wedefine χN (A,B) to be the number of pairwise disjoint graphs B′ ⊆ N such thatB and B′ are isomorphic over A.

Let now Kµ be the subclass of K consisting of those N ∈ K for which forevery minimally 0-simple pair (A,B) with A ⊆ N we have χN (A,B) ≤ µ(A,B).

We need the following lemma:

Lemma 44.3. Let C0, C1, C2 ∈ Kµ, C0 ⊆ C1, C2 with C1 = C1 \ C0 minimally0-simple over A0 ⊆ C0. If D = C1 ⊗C0 C2 /∈ Kµ, one of the following happens:

1. χC2(A0, C1) = µ(A0, C1) and there is an isomorphic copy of C1 over C0

inside C2, or

2. there is some minimally 0-simple pair (A,B) with

a) χD(A,B) > µ(A,B),b) A ⊆ A0 ∪ C1, andc) at least one copy of B is properly contained in C1.

Proof. Without loss of generality we may assume that C0 = A0. Suppose thatD = C1 ⊗C0 C2 /∈ Kµ. If D contains a 4-cycle (a0, a1, . . . , a4 = a0), then since


this cannot be entirely contained in either C1 or C2, without loss of generalitywe have a1 ∈ C1, a3 ∈ C2, a0, a2 ∈ C0. Since C1 is 0-simple over C0, it followsthat C1 = {a1} and a3 is the isomorphic copy of a1 over C0.

Now suppose that there is some minimally 0-simple pair (A,B) with χD(A,B) >µ(A,B). By the previous paragraph, we may assume that |C1| > 1 and henceχD(A,B) > µ(A,B) ≥ 3. Then for every copy B′ of B over A, every element ofA is connected to some element of B′ because B is minimally 0-simple over A.

First consider the case A ⊆ C2. Since C2 ∈ Kµ there is some copy B′ of Bover A with B′∩C1 6= ∅. If B′ 6⊆ C1, then δ(B′∩C2/A) > 0 and thus δ(B′/C2) <0, contradicting C2 ≤ D. So B′ ⊆ C1 and since δ(B′/A) = δ(B′/C2) = 0, by0-simplicity of C1 over A0 we see that B′ = C1 and A = A0.

Next consider the case A 6⊆ C2, i.e. A ∩ C1 6= ∅. Then since B is minimally0-algebraic over A, no copy of B is contained in C2 = C2\C0. Now suppose thatthere are k copies B1, . . . , Bk of B over A contained in C2 and that the copiesBk+1, . . . , Bk+l intersect both B2 and C1. Since each Bi contains vertices whichare connected to the vertices of A ∩ C1, it follows immediately that δ(A/C2) ≤δ(A/C2 ∩A)− k ≤ δ(A)− k.

Since the Bi are 0-simple over A, we have for each i = k + 1, . . . , k + l:

δ(Bi/C2 ∪A ∪Bk+1 ∪ . . . ∪Bi−1) < 0

This implies

δ(k+l⋃i=k+1

Bi/C2 ∪A) ≤ −l.

Hence

0 ≤ δ(k+l⋃i=k+1

Bi ∪A/C2) ≤ δ(A/C2)− l ≤ δ(A)− (k + l).

Thus at most δ(A) many copies of B over A are not contained in C1, leavingat least one copy of B over A inside C1. Since each element of A is connected tosome element of this copy, we see that A must be contained in A0∪ C1, finishingthe proof.

As in Section 13 we say that M ∈ Kµ is Kµ-saturated if for all finite B ≤Mand strong extensions C of B with C ∈ Kµ there is a strong embedding of Cinto M fixing B elementwise.

Theorem 44.4. The class Kµ is closed under substructures, and has the jointembedding and the amalgamation property with respect to strong embeddings.There exists a countable Kµ-saturated structure Mµ which is unique up to iso-morphism. In particular, in Mµ any partial isomorphism f : A → A′ withA,A′ ≤Mµ extends to an automorphism of Mµ.


Proof. Clearly, Kµ is closed under substructures. We will show that Kµ hasthe joint embedding and the amalgamation property with respect to strongembeddings. Then exactly as in the proof of Theorem 13.4 we obtain a count-able Kµ-saturated structure Mµ which is unique up to isomorphism. In par-ticular, in Mµ any partial isomorphism f : A → A′ with A,A′ ≤ Mµ extendsto an automorphism of Mµ. Since the empty graph is in Kµ and strong inany nonempty A ∈ Kµ, it suffices to prove the amalgamation property. LetC0, C1, C2 ∈ Kµ, C0 ≤ C1, C2, Ci = Ci \C0. We have to find some D ∈ Kµ suchthat the natural embeddings C0 ≤ C1, C2 ≤ D are all strong. We prove theamalgamation property by induction on |C1|.

Given graphs A ⊆ B,C we denote by B⊗AC the trivial amalgamation of Band C over A, obtained as the graph whose set of vertices is the disjoint unionB \A)∪ (C \A)∪A with incidence and predicate P induced by B and C. Notethat B,C ≤ B⊗AC. It is easy to verify that if δ(F ) ≥ 4 for all finite subgraphsF of B,C with |F | ≥ 3, then the same is true for D.

Case 1: If there is some proper subset X of C1 with δ(XC0) = δ(C0), thenC0 ≤ C0X ≤ C1 and by induction hypothesis we may amalgamate C0X withC2 over C0 to get D′, and then amalgamate D′ with C1 over XC0 to obtain D.

Case 2: There is no proper subset X of C1 with δ(XC0) = δ(C0).

Case 2.i): If δ(C1/C0) = 0, then C1 is 0-simple over C0. If D = C1 ⊗C0

C2 6∈ Kµ and C2 does not contain an isomorphic copy of C1 over C0, then bythe Lemma 44.3 there is some minimally 0-simple pair (A,B) with B properlycontained in C1 and A ⊆ C1, but A 6⊆ C0. Then 0 = δ(B/A) > δ(B/A ∩ C0).But since C1 is a strong extension of C0, this is not possible.

Case 2.ii): Suppose δ(C1/C0) = 1. If there is a proper subset X of C1 withδ(X/C0) = 1, then we have C0 ≤ XC0 ≤ C1 by the failure of Case 1 and we canuse induction. Otherwise, C1 = {b} is 1-simple over C0. Then D = C1 ⊗C0 C2

contains no k-cycle for k < 6. To see that D ∈ Kµ note first that there is no0-simple extension (A,B) inside D with b ∈ B. If C ⊆ D is minimally 0-simpleover F ⊆ D with b ∈ F , then 1 = χD(C,F ) ≤ µ(C,F ) as b is connected to aunique element of C0. So we have D ∈ Kµ.

Case 2.iii): If δ(C1/C0) > 1, then by the failure of Case 1 we must haveδ(b/C0) ≥ 1 for all b ∈ C1. First suppose that there is some b ∈ C1 withδ(bC0/C0) = 1. Then we can use Case 2.ii) to amalgamate b and C2 over C0

into some D′. Again by the failure of Case 1 it follows that bC0 is strong in C1

and we can use the induction hypothesis to amalgamate C1 with D′ over bC0.

Otherwise we have δ(bC0/C0) = 2 for all b ∈ C1, and so no element of C1 isconnected to any element of C0. Then D = C1⊗C0 C2 is in Kµ since D containsno k-cycle for k < 6 and any minimally 0-simple pair (A,B) is either entirelycontained in C1 or in C2.


Remark 44.5. It is easy to see that the Kµ-saturated structure Mµ is a projec-tive plane: since any two vertices of the same colour form a strong substructureof Mµ, for any two pairs of such vertices with the same colouring there is anautomorphism taking one pair to the other. Since there are pairs of verticesof the same partition at distance 2 in the graph, the same is true for any suchpair. Thus any two points lie on a common line and any two lines intersect ina point. Uniqueness is immediate since there are no 4-cycles. The existenceof four points in Mµ no three of which are collinear follows from the fact thatthe corresponding graph, an 8-cycle of pairwise distinct elements is containedin Kµ.

We now turn to the model theoretic properties of Mµ.

Theorem 44.6. Let Tµ be the theory (in the language of bipartite graphs)axiomatising the class of models M such that:

1. M is a projective plane;

2. M ∈ Kµ;

3. M ∪ C 6∈ Kµ for each C which is 0-simple over M .

Then Tµ = Th(Mµ).

Proof. Note first that this forms an elementary class whose theory Tµ is con-tained in Th(Mµ): clearly, 1. is a first-order property, which holds in Mµ bythe previous remark. For each minimally 0-simple pair (A,B) we can expressthat χM (A,B) ≤ µ(A,B), so 2. is first-order expressible and holds in Mµ byconstruction. For 3. notice that if C is 0-simple over M , then by Lemma 44.3the fact that M ∪C 6∈ Kµ is witnessed by some minimally 0-simple pair (A,B)with B ⊆ C and A ⊆ C∪C0 where C is minimally 0-simple over C0. Since thereare only finitely many isomorphism types of such (A,B), this can be expressedin a first order way. It again holds in Mµ by construction.

Since Mµ is a Kµ-saturated model of this theory, for the proof it now sufficesto show the following:

Claim: M is an ω-saturated model of Tµ if and only if it is Kµ-saturated.

Let M |= Tµ be ω-saturated. To show that M is Kµ-saturated, let A ≤ Mand A ≤ B ∈ Kµ finite. By induction we may assume that B is minimal over A.

If B is a 0-simple extension of A minimally 0-simple over, say, A0 ⊆ M ,then since M ⊗A0 B 6∈ Kµ, by Lemma 44.3 there is an isomorphic copy of Bover A0 inside M and since there are no connections between A \ A0 and B,this isomorphism is in fact over A. Otherwise B = A∪{b} where b is connectedto at most one element of A. By w-saturation there is another vertex b′ ∈ M


either not connected to any element of A or connected to the same element ofA as b.

Conversely, suppose M is Kµ-saturated, so M is in Kµ. Suppose that forsome C which is 0-simple overM we haveM∪C ∈ Kµ. If A0 ≤M is such that Cis minimally 0-simple over A0, then by Kµ-saturation we may embed C over A0

into M . Iterating this ad libitum, we contradict the fact that M ∈ Kµ. HenceM is a model of Tµ, which is ω-saturated because M is partially isomorphic toMµ.

Since every saturated model of Tµ is partially isomorphic to Mµ, it nowfollows that Tµ is the complete theory of Mµ

Definition 44.7 (Coordinatisation). Let Π = (P,L, I) be a projective plane,let ` ∈ L be a line and let x1, x2 ∈ P be points not on `. Let D` denote the setof points on `. Then every point of Π is in the definable closure of D`∪{x1, x2}:for y ∈ Π \ D` ∪ {x1, x2} the lines `1 = (y, x1), `2 = (y, x2) meet ` in twodistinct elements y1, y2 ∈ D`. Then y ∈ dcl(x1, x2, y1, y2). This process is calledcoordinatisation.

Let cl(B) = clM (B) be the smallest strong subgraph of M containing B.Note that cl(B) is finite if B is. We also define

d(A) = min{d(B) | A ⊆ B ⊆M} = δ(cl(A)).

Similarly, we put d(A/B) = d(AB)− d(B).

We will show that for any vertex x ∈ Mµ the set Dx of neighbours of x isstrongly minimal with dimension function given by d. To this end we start withthe following easy lemma:

Lemma 44.8. Let M and M ′ be models of Tµ. Then tuples a ∈M and a′ ∈M ′

have the same type if and only if the map a 7→ a′ extends to an isomorphism ofcl(a) to cl(a′). In particular, d(a) depends only on the type of a.

Lemma 44.9. Suppose M ∈ Kµ, A strong in M , and δ(B/A) = 0. Then thereexist only finitely many copies of B over A in M , so cl(A) ⊆ acl(A).

Proof. By decomposing into minimal extensions, we may assume that B is 0-simple over A. We claim that any two copies of B over A in M are disjoint, andhence there are only finitely many of them.

Let B1, B2 ⊆M be distinct copies of B over A. Since A is strong in M andB1 is 0-simple over A, we have B1A ≤M . So by the last part of Remark 44.2,B2A ∩ B1A is strong in B2A. By 0-simplicity we must have B2A ∩ B1A = Aproving the claim.

Proposition 44.10. For any model M of Tµ and any a ∈ M , the set Da isstrongly minimal (with dimension function given by d).


Proof. We work over the parameter a. We may assume that M = Mµ. Sincefor x1 6= x2 ∈ Da, the path (x1, a, x2) is strong in M , there is a unique type ofpairs of distinct elements. Let X ⊆ Da and x ∈ Da. Note that if |X| ≥ 2, thena ∈ cl(X), so d(x/X) ≤ 1.

• If d(x/X) = 0, then by the previous lemma, x is algebraic over cl(X) andhence over X.

• If d(x/X) = 1, then cl(X) ∪ {x} is strong in M , so any other elementx′ ∈ Da with d(x′/X) = 1 is conjugate to x over X, showing thatthere is a unique non-algebraic type over X. The claim now follows fromLemma 21.3.

Together with coordinatisation, the strong minimality of Da now impliesthat Tµ is almost strongly minimal : a complete theory is called almost stronglyminimal if there is a strongly minimal formula ϕ (possibly containing parame-ters) such that every model M is in the algebraic closure of ϕ(M) ∪ a for somefinite tuple a ⊆M .

Theorem 44.11. The theory Tµ is almost strongly minimal, not 1–based andof Morley rank 2.

Proof. By Proposition 44.10 and coordinatisation, Tµ is almost strongly mini-mal. By Theorem 26.2 the Morley rank is given by the dimension function ofthe strongly minimal set Da, say, which again coincides with d. Hence for anyx ∈M we have MR(x) = d(x) = 2.

To see that Tµ is not 1–based, let ` be a line of M and p be a point on`, so clearly ` ∈ Cb(p/`). Then MR(p/`) = 1 and MR(`) = MR(p) = 2. ByRemark 26.9 (and Exercise 26.2), we have

3 = MR(p`) = MR(`/p) +MR(p),

so ` 6∈ acl(p).

To show that this yields indeed a counterexample to Zilber’s Conjecture, wefinish by showing that

Proposition 44.12. There is no infinite group definable in Tµeq.

Proof. Again it suffices to work in M = Mµ. We work over the parameters`, p1, p2, p3, p4 ∈ M where ` is a line, p1, p2 are points not on ` and p3, p4are points on `. By Exercise 44.3, acl(`, p1, . . . , p4) is infinite. Hence, byLemma 35.10 the theory Th(M, `, p1, . . . p4) has weak elimination of imaginaries.


Suppose now that G is an infinite group definable in M eq. Then G is totallytranscendental (even uncountably categorical) and so every type has Morleyrank. Let a1 ∈ G be an element realizing a generic type of G, i.e. a type ofmaximal Morley rank. Then by weak elimination of imaginaries and coordi-natisation 44.7, there are x1, . . . xm ∈ D` such that a1 ∈ dcl(x1, . . . xm) andx ∈ acl(a). By Theorem 26.2, the Morley rank of G, which is equal to theMorley rank of a1, is at most m, say MR(a) = n. Let a2 be another realisationof tp(a1) independent from a1, and put a3 = a1 · a2. Let b1 realise tp(a1) andindependent from a1, a2. Put b2 = a1 · b1, b3 = a−1

3 · b2 = a−12 · b1. Now put

A0 = {a1, a2, a3}, A1 = {a3, b2, b3}, A2 = {a1, b1, b2} and A3 = {a2, b1, b3}. LetAij = Ai ∩Aj for i, j = 0, . . . 3 (see Exercise 36.8). Then

Ai |Aij

Aj and hence also acl(Ai) |Aij

acl(Aj).

We move back to working in D`: by coordinatisation and weak eliminationof imaginaries, we replace each of ai, bj , i, j = 1, 2, 3 by a tuple xi, yj ⊂ D`

with ai ∈ dcl(xi), bj ∈ dcl(yj) and xi ∈ acl(ai), yj ∈ acl(bj) to obtain finite setsE0 = {x1, x2, x3}, . . . , E3 = {x2, y1, y3} ⊂ D`.

Since cl(A) ⊆ acl(A), we have cl(Ei) |Eij

cl(Ej) for i, j = 0, . . . 3. Let

E =⋃3i=0 cl(Ei). Since x1, x2, y1 are algebraically independent, Theorem 26.2

implies

δ(E) ≥ d(E) ≥ RM(E) = 3n.

On the other hand, note that by Proposition 44.10, for the strong subsetscl(Ei), Morley rank agrees with the d- and the δ-value. So cl(Ei) |

cl(Eij)cl(Ej)

implies that e(cl(Ei) \ cl(Eij), cl(Ej) \ cl(Eij)) = 0.

Noting that δ(cl(Ei)) = 2n, i = 0, . . . 3, and δ(cl(Eij)) = n for the six choicesof i 6= j ≤ 3, we can now use Exercise 44.2 to calculate δ(E) as

δ(E) = 4 · 2n− 6 · n = 2n.

Thus n = 0 and G is finite, a contradiction.

We now have the promised counterexample to Zilber’s Conjecture:

Corollary 44.13. Let Ta be the induced theory of Tµ on the strongly minimalset Da (after adding the parameter a to the language). Then Ta is stronglyminimal, not locally modular and does not interpret an infinite field.

Proof. By Corollary 43.4 since Tµ is almost strongly minimal over Da and not1–based, the strongly minimal set Da itself cannot be locally modular. Hence


the induced theory Ta is strongly minimal and not 1-based. If Ta would interpretan infinite field, then so would Tµ, which it doesn’t.

Exercise 44.1.Prove that if A ≤ B and C ⊂ B, then A ∩ C ≤ C.

Exercise 44.2.If A,B are subgraphs ofM , then δ(AB) = δ(A)+δ(B)−δ(A∩B)−e(A\B,B\A).

Exercise 44.3.Let M =Mµ, and `, p1, . . . , p4 ∈M where ` is a line, p1, p2 are points not on `,and p3, p4 are points on `. Show that acl(`, p1, . . . , p4) is infinite. (Hint: For anyk ≥ 3, a 2k-cycle (x0, x1, . . . , x2k−1, x2k = x0) is a 0-simple extension of a graph A ifevery xi has a unique neighbour in A.)

Exercise 44.4.Show that for any a ∈ Aut(Mµ) acts 2-transitively on Da: for any two pairsof elements x1 6= x2 and y1 6= y2 in Da, there is some g ∈ Aut(Bµ) such thatg(x1) = y1 and g(x2) = y2.

Appendix A

Set Theory

In this appendix we collect some facts from set theory presented from the naivepoint of view and refer the reader to [27] for more details. In order to talk aboutclasses as well as sets we begin with a brief axiomatic treatment.

1 Sets and classes

In modern mathematics, the underlying axioms are mostly taken to be ZFC,i.e. the Zermelo-Fraenkel axioms (ZF) including the axiom of choice (AC) (seee.g. [27], p.3). In order to talk about the monster model we may work in Bernays-Gödel set theory (BG) which is formulated in a two–sorted language, one typeof objects being sets and the other type of objects being classes, the element-relation being defined between sets and sets and between sets and classes only.Since the axioms are less commonly known, we give them here following [27],p.70. We use lower case letters as variables for sets and capital letters forreferring to classes. BG has the following axioms:

1. (a) Extensionality: ∀u(u ∈ x↔ u ∈ y) → x = y.(b) Pairing: For any sets x and y, there is a set {x, y}.(c) Union: For every set x, the set

⋃x exists.

(d) Power Set: For every set x, the power set P (x) exists.(e) Infinity: There is an infinite set.

2. (a) Class extensionality: ∀u(u ∈ X ↔ u ∈ Y ) → X = Y .(b) Comprehension: ∀Y1 . . .∀Yn∃X X = {x | ϕ(x, Y1, . . . Yn)} where ϕ is

a formula in which only set-variables are quantified.(c) Replacement: If a class F is a function and x is a set, then

{F (z) : z ∈ x} is a set.

208

APPENDIX A. SET THEORY 209

3. Regularity: Every nonempty set has an ∈-minimal element.

For BGC we add:

5. Global Choice: There is a function F such that F (x) ∈ x for every non-empty set x.

A model M of ZF becomes a model of BG by taking the definable subsetsof M as classes. This shows that BG is a conservative extension of ZF: any set-theoretical statement provable in BG is also provable in ZF. Similarly, BGC is aconservative extension of ZFC, see [16]. For a historical discussion see also [8].


2 Cardinals

Two sets are said to have the same cardinality if there is a bijection betweenthem. To each set x we can associate a cardinal |x| (its cardinality) in sucha way that two sets have the same cardinal if and only if they have the samecardinality.1

The cardinality of the empty set is denoted by 0, a one-element set hascardinality 1 etc.

Definition 2.1. For cardinals κ and λ we define

κ ≤ λ,

if λ is the cardinality of x and κ the cardinality of a subset of x.

It is easy to see that ≤ is reflexive and transitive. The Cantor–BernsteinTheorem (see [27] 3.2) states that two sets which can be injectively mappedto each other have the same cardinality. Thus, ≤ is anti–symmetric and hencedefines a partial ordering on the class of cardinals.

For any two sets x, y, by Zorn’s lemma (see [27] 5.4) there is a maximalbijection f between subsets x0 of x and a subset y0 of y. If both x0 and y0 wereproper subsets, we could extend f . So either f is a bijection between x0 and yor between x and y0, showing that ≤ is in fact a linear ordering.

Definition 2.2. A well–ordering of X is a linear ordering such that any non–empty subset of X contains a smallest element or, equivalently, such thatX doesnot contain infinite properly descending chains. If X is a proper class ratherthan a set, we also ask that for all x ∈ X the set {y | y < x} of predecessors isa set.

Lemma 2.3. The class of cardinals is well–ordered by ≤.

Proof. By the well–ordering theorem (which, like Zorn’s Lemma is equivalentto the Axiom of Choice, see [27] 5.1) any set can be well–ordered. Using Zorn’slemma one can see as above that for any two well–orderings one them of isisomorphic to an initial segment2 of the other one.

If x is a well–ordering with cardinality κ, then every smaller cardinal is thecardinality of an initial segment of x. Since the initial segments of a well–ordering are well–ordered by inclusion, the claim follows.

1It is tempting to define |x| as the class of all sets having the same cardinality as x, butthen a cardinal would not be a set.

2 An initial segment of a partial order X is a subset which contains for each of its elementsall smaller elements of X.


ℵ0 denotes the cardinality of the natural numbers (which of course are well–ordered). Sets of cardinality ℵ0 are called countable. All proper initial segmentsare finite, so ℵ0 is the smallest infinite cardinal.

Any family (κi)i∈I of cardinals has an upper bound (e.g. the cardinality of⋃i∈I xi, where κi = |xi|). Thus there is a least upper bound supi∈I κi.

Sums, products, and powers of cardinals are defined by disjoint union, carte-sian power and sets of maps, respectively. Thus

|x|+ |y| = |x ∪ y|(1)|x| · |y| = |x× y|(2)|x||y| = |yx|(3)

where we assume in (1) that x and y are disjoint. In (3), the set yx denotes theset of all functions from y to x.

It is easy to see that these operations satisfy the same rules as the corre-sponding operations on the natural numbers. E.g.,

(κλ)µ

= κλ·µ.

The following is known as Cantor’s Theorem, see [27, 3.1, 3.5]) for a proof.The first part also follow from Theorem A 3.2.

Theorem 2.4.

1. If κ is infinite, then κ · κ = κ.

2. 2κ > κ.

Note that 2.4.2. implies that there is no largest cardinal. Since every set ofcardinals has an upper bound, the class of cardinals is not a set.

Corollary 2.5.

1. If λ is infinite, then κ+ λ = max(κ, λ).

2. If κ > 0 and λ are infinite, then κ · λ = max(κ, λ).

3. If κ is infinite, then κκ = 2κ.

Proof. Let µ = max(κ, λ). Then µ ≤ κ + λ ≤ µ + µ ≤ 2 · µ ≤ µ · µ = µ and ifκ > 0, then µ ≤ κ · λ ≤ µ · µ = µ.Finally,

2κ ≤ κκ ≤ (2κ)κ= 2κ·κ = 2κ.

Corollary 2.6. The set of all finite sequences of elements of a non–empty setx has cardinality max(|x|,ℵ0).


Proof. Let κ be the cardinality of all finite sequences in x. Clearly, |x| ≤ κ andℵ0 ≤ κ. On the other hand

κ =∑n∈N

|x|n ≤(supn∈N

|x|n)· ℵ0 = max(|x|,ℵ0),

because

supn∈N

|x|n =

1, if |x| = 1

ℵ0, if 2 ≤ |x| ≤ ℵ0

|x|, if ℵ0 ≤ |x|.

For every cardinal κ, κ+ denotes the smallest cardinal greater than κ, thesuccessor cardinal of κ. By 2.4.2, we have κ+ ≤ 2κ. The Generalised ContinuumHypothesis (GCH) states that

κ+ = 2κ for all infinite κ.

GCH is independent of ZFC (assuming these axioms are consistent, see e.g. [27]14.32).


3 Ordinals

The class On of ordinals is characterised up to isomorphism by properties sharedby the class of cardinals:

a) On is well–ordered;

b) On is a proper class.

The well–ordering of On yields the following principle of transfinite induction:

Let E be a property of ordinals. Assume that whenever all ordinalsless than α have property E, then α itself has this property. Thenall ordinals have property E.

Let (X,<) be a well–ordering and f a maximal isomorphism between an initialsegment of X and an initial segment of On. It is easy to see that f is defined onall of X. If X is a proper class, then f is an isomorphism between X and On.If X is a set, then f is an isomorphism between X and a proper initial segment,which is of the form

{β | β < α}.

It is easy to see that f , and hence α are uniquely determined by (X,<). (Ifg : X → On is another isomorphism to an initial segment, just consider thesmallest x such that f(x) 6= g(x).) This α is called the order type of (X,<)

α = otp(X,<).

To simplify notation one often identifies an ordinal with its set of predecessors3

α = {β | β < α}.

In this way, α is its own order type.

By the well–ordering theorem any cardinal is the cardinality of an ordinal.In axiomatic set theory, a cardinal is identified with the smallest ordinal havingthis cardinality. In this way the class of cardinals is a subclass of On: an ordinalκ is a cardinal if and only if all ordinals α < κ have smaller cardinality.

For any ordinal α its successor is defined as α ∪ {α}; it is the smallestordinal greater than α. Starting from the smallest ordinal 0 = ∅, its successoris 1 = {0}, then 2 = {0, 1} and so on, yielding the natural numbers. The ordertype of the natural numbers is denoted by ω = {0, 1, . . .}, the next ordinal isω + 1 = {0, 1, . . . . , ω}, et cetera. Notice that the natural numbers and ω = ℵ0

are cardinals.3 In axiomatic set theory ordinals are defined in this way.


By definition, a successor ordinal β contains a maximal element α (so β isthe successor of α) and we write β = α+ 1. For natural numbers n, we put

α+ n = α+ 1 + . . .+ 1︸︷︷︸n− times

.

Ordinals > 0 which are not successor ordinals are called limit ordinals. When-ever {αi | i ∈ I} is a non-empty set without biggest element, supi∈I αi is a limitordinal. Any ordinal can be uniquely written as

λ+ n,

with λ = 0 or a limit ordinal.

Theorem 3.1. Let G be an assignment such that for every function f definedon an ordinal, G(f) is a set. Then there is a unique function F defined on allof On satisfying the recursion formula

F (α) = G(F � α).

Proof. It is easy to see by induction that for all β there is a unique functionFβ defined on β and satisfying the recursion formula for α < β. Put F =⋃β∈On Fβ .

Example:The aleph-function assigns ordinals to all infinite cardinals. It is defined as

ℵα = the smallest infinite cardinal greater than all ℵβ (β < α).

We may thus talk about successor and limit cardinals. Similarly, we write ωαto denote the ordinal corresponding to ℵα.

Example:For every cardinal µ the beth function is defined as

iα(µ) =

µ, if α = 0,

2iβ(µ), if α = β + 1,

supβ<λ iβ(µ), if α is a limit ordinalλ.

A finite sequence of length n of elements from A is a function s : n→ A.

<ωA =⋃n<ω

nA

denotes the set of all finite sequences of elements from A.

Lemma 3.2 (The Gödel well–ordering, see [27, 3.5]). There is a bijection On →On×On, which induces a bijection κ→ κ× κ for all infinite cardinals κ. Thisis a well–ordering of On×On, which has the property that for each pair (α, β)the class of predecessors is actually a set. This implies that there is a (unique)


Proof. Define

(α, β) < (α′, β′) ⇔(max(α, β), α, β

)<lex

(max(α′, β′), α′, β′)

where <lex is the lexicographical ordering on triples. This is a well–orderingwith the additional property that for all α, β the class Xα,β of predecessors of(α, β) is a set. This implies that there is a (unique) order-preserving bijectionγ : On × On → On. We show by induction that γ maps κ × κ to κ for everyinfinite cardinal κ, which in turn implies κ · κ = κ (see A 2.4.1). Since theimage of κ × κ is an initial segment, it suffices to show that Xα,β has smallercardinality than κ for every α, β < κ. We note first that Xα,β is contained inδ × δ with δ = max(α, β) + 1. Since κ is infinite, we have that the cardinalityof δ is smaller that κ. Hence by induction |Xα,β | ≤ |δ| · |δ| < κ.

For any linear order (X,<) we can easily construct a well-ordered cofinalsubset, i.e. a subset Y such that for any x ∈ X there is some y ∈ Y with x ≤ y.

Definition 3.3. The cofinality cf(X) is the smallest order type of a well orderedcofinal subset of X.

It is easy to see that cf(X) is a regular cardinal where an infinite cardinalκ is regular if cf(κ) = κ. Successor cardinals and ω are regular. The existenceof weakly inaccessible cardinals, i.e., uncountable regular limit cardinals cannotbe proven in ZFC.

Appendix B

Fields

4 Ordered fields

Let R be an integral domain. A linear < ordering on R is compatible with thering structure if for all x, y, z ∈ R

x < y → x+ z < y + z

x < y ∧ 0 < z → xz < yz

A field (K,<) together with a compatible ordering is an ordered field .

Lemma 4.1. Let R be an integral domain and < a compatible ordering of R.Then the ordering < can be uniquely extended to an ordering of the quotientfield of R.

Proof. Puta

b> 0 ⇔ ab > 0.

It is easy to see that in an ordered field sums of squares can never be negative.In particular, 1, 2, . . . are always positive and so the characteristic of an orderedfield is 0. A field K in which −1 is not a sum of squares is called formally real.

Lemma 4.2. A field has an ordering if and only if it is formally real.

Proof. A field with an ordering is formally real by the previous remark. Forthe converse first notice that Σ�, the set of all sums of squares in K, is a

216

APPENDIX B. FIELDS 217

semi-positive cone, i.e., a set P such that

Σ� ⊂ P(1)P + P ⊂ P(2)P · P ⊂ P(3)

−1 6∈ P(4)

The first and third condition easily imply

x ∈ P \ 0 ⇒ 1

x∈ P.

Therefore, condition (4) is equivalent to P ∩(−P ) = 0. It is also easy to see thatfor all b, the set P + bP has all the properties of a semi-positive cone, exceptpossibly (4). Condition (4) holds if and only if b = 0 or −b 6∈ P . We now chooseP as a maximal semi-positive cone. Then

x ≤ y ⇔ y − x ∈ P

and we obtain a compatible ordering of K.

Corollary 4.3 (of the proof). Let K be formally real1 and let a be an elementof K. There is an ordering of K making a negative if and only if a is not thesum of squares.

Proof. If a 6∈ Σ�, then Σ�− a · Σ� is a semi-positive cone.

Definition 4.4. An ordered field (R,<) is real closed if

a) every positive element is a square,

b) every polynomial of odd degree has a zero.

(R,<) is called a real closure of the subfield (K,<) if R is real closed andalgebraic over K.

The field of real numbers is real closed. Similarly, the field of real algebraicnumbers is the real closure of Q. More generally, any field which is relativelyclosed in a real closed field is itself real closed.

Theorem 4.5. Every ordered field (K,<) has a real closure, and this is uniquelydetermined up to isomorphism over K.

Proof (Sketch). Existence: Let K≥0 be the positive cone of (K,<) and let L be a fieldextension of K. The ordering of K can be extended to L if and only if

P = {x1y21 + · · ·+ xny2n | xi ∈ K≥0, yi ∈ L}

1It suffices to assume that the characteristic of K is different from 2.


is a semi-positive cone of L, i.e. if −1 6∈ P . Therefore we may apply Zorn’s Lemma toobtain a maximal algebraic extension R of K with an ordering extending the orderingof K. We claim that R is real closed.Let r be a positive element of R and assume that r is not a square. Since the orderingcannot be extended to L = R(

√r), there are ri ∈ R≥0 and si, ti ∈ R such that

−1 =∑

ri(si√r + ti)

2.

Then −1 =∑ri(s

2i r + t2i ), contradicting the fact that the right hand side is positive.

Thus every element of R is a square (and the ordering of R is unique).Let f ∈ R[X] be a polynomial of minimal odd degree n without zero (in R). Clearly,f is irreducible. Let α be a zero of f (in the algebraic closure of R) and put L = R(α).Since L cannot be ordered, −1 is a sum of squares in L. So there are polynomialsgi ∈ R[X] of degree less than n such that f divides h = 1 +

∑g2i . The leading

coefficients of the g2i are squares in R and so cannot cancel out. Hence the degree ofh is even and less than 2n. But then the polynomial hf−1 has odd degree less than nand no zero in R since h does not have a zero. A contradiction.

Uniqueness: Let R and S be real closures of (K,<). It suffices to show that R andS are isomorphic over K as fields. By (the easy characteristic 0 case of) Lemma 6.13below it is enough to show that an irreducible polynomial f ∈ K[X] has a zero in Rif and only if it has one in S. As pointed out by Knebusch this follows from the nextlemma which we prove at the end of the section.

Lemma 4.6 (J.J. Sylvester). Let f be irreducible2 in K[X] and (R,<) a real closedextension of (K,<). The number of zeros of f in R equals the signature of the traceform of the K-algebra K[X]/(f).

Notice that the uniqueness depends on the given ordering. For a formally realfield there may be different orderings leading to non-isomorphic real closures.

The Fundamental Theorem of Algebra holds for arbitrary real closed fields3:

Theorem 4.7. Let R be real closed. Then C = R(√−1) is algebraically closed.

Proof. Notice that all elements of C are squares: One square root of a+b√−1 is given

by √√a2 + b2 + a

2±

√√a2 + b2 − a

2

√−1,

where we choose ± according to the sign of b.

Let F be a finite extension of C. We claim that F = C. We may assume thatF is a Galois extension of R. Let G be a 2–Sylow subgroup of Aut(F/R) and Lthe fixed field of G. Then the degree L/R is odd. The minimal polynomial of agenerating element of this extension has the same degree and is irreducible. But since

2It suffices that f is non–constant and all zeros in acl(K) have multiplicity 1.3Since in real closed fields the ordering is uniquely determined by the field structure, it

makes sense to say that a field is real closed without specifying its ordering.


all irreducible polynomials over R of odd degree are linear, we have L = R. ThereforeG = Aut(F/R) and hence also H = Aut(F/C) are 2–groups. 2-groups are soluble(even nilpotent). So if H is non–trivial, it has a subgroup of index 2 and then C hasa field extension of degree 2; but this is impossible since every element is a square.Hence H = 1 and F = C.

Corollary 4.8. If R is real closed, the only monic irreducible polynomials are:

• Linear polynomialsX − a,

(a ∈ R)

• Quadratic polynomials(X − b)2 + c,

(b, c ∈ R, c > 0)

Proof. Since all non–constant polynomials f ∈ R[X] have a zero in R(√−1), all irre-

ducible polynomials must be linear or quadratic. Any monic polynomial of degree 2 isof the form (X − b)2 + c. It is reducible if and only if it has a zero x in R if and onlyif c ≤ 0 (namely x = b±

√−c).

Finally we prove Sylvester’s Lemma (4.6). Let K be an ordered field, R a realclosure of K and f ∈ K[X] irreducible. Consider the finite dimensional K–algebraA = K[X]/(f). The trace TrK(a) of a ∈ A is the trace of left multiplication by a,considered as a vector space endomorphism. The trace form is a symmetric bilinearform given by

(a, b)K = TrK(ab).

Let a1, . . . , an be a basis of A diagonalising ( , )K , i.e., (ai, aj)K = λiδij . The signatureof such a form is defined as the number of positive λi minus the number of negativeλi. Sylvester’s Theorem (from linear algebra) states that the signature is independentof the diagonalising basis. By tensoring A with R, we obtain the R–algebra

AR = A⊗R ∼= R[X]/(f).

The basis a1, . . . , an is also a diagonalising R–basis for the trace form of AR with thesame λi and hence the same signature as the trace form of A. We now split f in R[X]into irreducible polynomials g1, . . . , gm. Since f does not have multiple zeros, the giare pairwise distinct. By the Chinese Remainder Theorem we have

R[X]/(f) ∼= R[X]/(g1)× · · · ×R[X]/(gm).

This shows that the trace form of R[X]/(f) is the direct sum of the trace forms ofthe R[X]/(gi), and hence its signature is the sum of the corresponding signatures.Sylvester’s Lemma follows once we show that the signature of the trace form is equalto 1 for a linear polynomial, while the signature is 0 for an irreducible polynomial ofdegree 2.

If g is linear, then R[X]/(g) = R. The trace form (x, y)R = xy has signature 1. Ifg is irreducible of degree 2, then R[X]/(g) = R(

√−1) and TrR(x+ y

√−1) = 2x. The

trace form is diagonalised by the basis 1,√−1. With respect to this basis we have

λ1 = 2 and λ2 = −2 and so its signature is zero.


5 Differential Fields

In this section, all rings considered have characteristic 0. Let R be acommutative ring and S an R-module. (We mainly consider the case that S isa ring containing R.) An additive map d : R→ S is called a derivation if

d(rs) = (dr)s+ r(ds)

For any ring S the ring of dual numbers is defined as

S[ε] = {a+ bε | a, b ∈ S},

where ε2 = 0. The following is an easy observation:

Lemma 5.1. Let S be a ring containing R and let π : S[ε] → S; a + bε 7→ a.Any derivation d : R→ S defines a homomorphism

td : R→ S[ε]; r 7→ r + (dr)ε

inverting π. Conversely, any such homomorphism arises from a derivation.

Lemma 5.2. Let S be a ring containing the polynomial ring R′ = R[x1, . . . , xn]and let d : R→ S be a derivation. For any sequence s1, . . . , sn of elements of Sthere is a unique extension of d to a derivation d′ : R′ → S taking xi to si fori = 1, . . . n.

Proof. Extend td : R→ S[ε] via td′(xi) = xi+siε to a homomorphism td′ : R′ →

S[ε].

Lemma 5.3. Let R be a subring of the field S and let R′ be an intermediate fieldalgebraic over R. Then every derivation d : R → S can be uniquely extended toR′.

Proof. It suffices to consider the following two cases:

1. R′ is the quotient field of R: if a is a unit, then a+ bε is a unit in S[e] (withinverse a−1 − ba−2ε.) Therefore, td takes elements 6= 0 of R to units in S[ε] andthus can be uniquely extended to R′.

2. R is a field and R′ = R[a] a simple algebraic extension: Let f(x) be theminimal polynomial of a over R. Then td can be extended to R′ if and only if(tdf)(x) has a zero of the form a+ bε in S[ε]. But (tdf)(x) = f(x) + (fdε)(x),hence

(5) (tdf)(a+ bε) = f(a+ bε) + fd(a)ε =(∂f∂x

(a)b+ fd(a))ε.

Since char(S) = 0, f is separable, so ∂f∂x (a) 6= 0 and d can be extended to R′ by

d′(a) = a− fd(a)(∂f∂x

(a))−1

.


Let C = {c ∈ K | dc = 0} denote the set of constants of K. Clearly, C is asubring containing 1. The previous lemma implies that C is a subfield which isrelatively algebraically closed in K.

Corollary 5.4. Let R be a subring of the field K. Any derivation d : R → Kcan be extended to a derivation of K.

Remark 5.5. Let (K, d) be a differential field and F a field extension of K,a, b ∈ Fn. Assume that the ideal of all f in K[x] with f(a) = 0 is generated byI0. Then the following are equivalent:

a) There is an extension of d to F with da = b.

b) For all f ∈ I0 we have∂f

∂x(a)b+ fd(a) = 0.

Proof. It is clear that a) implies b) because

d(f(a)) =∂f

∂x(a)b+ fd(a).

In order to show the converse we have to extend the homomorphism td : K →F [ε] to K[a] in such a way that a is mapped to a+bε. This is possible if and onlyif (tdf)(a+ bε) = 0 for all f ∈ I0. By (5) this implies ∂f

∂x (a)b+ fd(a) = 0.

Let (F, d) be a saturated model of TDCF0 and V ⊂ Fn an irreducible affinevariety (see [45, Chapter 1] for basic algebraic geometry). The torsor T (V ) ⊂F 2n is defined by the equations

f(x) = 0 and∂f

∂x(x)y + fd(x) = 0

for all f ∈ I(V )4. Clearly (a, da) ∈ T (V ) for all a ∈ V .

Remark 5.5 states that for any small subfield K over which V is defined,and any (a0, b0) ∈ T (V ) such that a0 ∈ V is generic over K there is a pair(a, da) ∈ T (V ) satisfying the same field–type over K as (a, b). This alreadyproves one direction of the following equivalence:

Remark 5.6 (Pillay). An algebraically closed differential field (K, d) is a modelof TDCF0 if and only if for every irreducible affine variety V defined over K andevery regular section s : V → T (V ) of the projection T (V ) → V there is somea ∈ V (K) with da = s(a).

Proof. We show that an algebraically closed differential field (K, d) with thisproperty is a model of the axioms of TDCF0 .

4I(V ) is called the vanishing ideal of V .


Let f and g be given with f irreducible. As in the proof of Theorem 9.22we find a field extension F = K(α, . . . , dnα) in which α, . . . , dn−1α are alge-braically independent over K and f(α, . . . , dnα) = 0. Let β be the inverse ofg(α, . . . , dn−1α) ∂f∂xn (α, . . . , d

nα). Note that K[α, . . . , dnα, β] is closed under d.Putting c = (α, . . . , dnα, β), let (c, dc) = s(c) for some tuple s of polynomialsin K[x0, . . . , xn+1]. Then s defines a section V → T (V ) where V ⊂ Kn+1 isthe variety defined by f = 0. By assumption there is some (a, b) ∈ V (K) with(a, b, da, db) = s(a, b). Hence f(a, . . . , dna) = 0 and g(a, . . . , dn−1a) 6= 0.

Remark 5.7 (Linear differential equations). Let K be a model of TDCF0 andA an n× n–matrix with coefficients from K. The solution set of the system ofdifferential equations

(6) dy = Ay

is an n–dimensional C–vector space.

Proof. Choose an n×n–matrix Y over an extension of K whose coefficients arealgebraically independent over K. Lemma 5.2 shows that d can be extended toK(Y ) by dY = AY . Since K is existentially closed, there is a regular matrixover K, which we again denote by Y , such that dY = AY . The columns of Yare n linearly independent solutions of (6). Such a matrix is called fundamentalsystem of the differential equation. We show that any solution y is a C-linearcombination of the columns of Y : let z be a column over K with y = Y z. Then

AY z = d(Y z) = (dY )z + Y dz = AY z + Y dz.

Hence Y dz = 0, so dz = 0, i.e. the elements of z are constants.

Remark 5.8. Let K be a differential field and K ′ an extension of K withfields of constants C and C ′, respectively. Then K and C ′ are linearly disjointover C, i.e. any set of C-linearly independent elements of K remains linearlyindependent over C ′ (see Section B.6).

Proof. Let a0, . . . , an be in K and linearly dependent over C ′. Then the columnsof

B =

a0 . . . anda0 . . . dan...

...dna0 . . . dnan

are linearly dependent. Let m < n be maximal with

Y =

a0 . . . amda0 . . . dam...

...dma0 . . . dmam


regular. We want to conclude that a0, . . . , am+1 is linearly dependent over C.So we may assume n = m+1. Then the last row of B is a K–linear combinationof the first m+ 1 rows. Thus for some m×m matrix A all columns of Y and

z =

an...

dman

are solutions of the differential equation dy = Ay. The proof of Remark 5.7shows that z is a C–linear combination of the columns of Y .


6 Separable and regular field extensions

Definition 6.1. Two rings R,S contained in a common field extension aresaid to be linearly disjoint over a common subfield k if any set of k-linearlyindependent elements of R remains linearly independent over S. Note that thenR is also linearly disjoint over K from the quotient field of S.

Remark 6.2. It is easy to see that this is equivalent to saying that the ringgenerated by R and S is canonically isomorphic to the tensor product R ⊗k S.So linear disjointness is a symmetric notion.

If S is a normal algebraic extension of K, so invariant under automorphimsof Kalg/K, then the linear disjointness of R and S over K does not depend onthe choice of a common extension of R and S (see Exercise 35.4 for a similarphenomenon).

Recall that an algebraic field extension L over K is separable if every elementof L is a zero of a separable polynomial f ∈ K[X] and galois if L is normal andseparable over K.

Lemma 6.3. If F and L are field extensions of K with L galois over K, thenF and L are linearly disjoint over K if and only if F ∩ L = K.

Proof. One direction is clear (see also the remark belowy). For the other direc-tion assume F ∩ L = K. We may assume that L/K is finitely generated. SinceL/K is separable, it is finitely generated, say L = K(a). Let f be the minimalpolynomial of a over F . Since alle roots of f belong to L, f is in L[X] andtherefore in K[X]. It follows that [K(a) : K] = [F (a) : F ] and F and L arelinearly disjoint over K.

That h1, . . . , hn are algebraically independent over L means that the mono-mials in h1, . . . , hn are linearly independent over L. This observation impliesthat L and H are algebraically independent over K if they are linearly disjointover K. The converse holds if L/K is regular :

Definition 6.4. A field extension L over K is regular if L and Kalg are linearlydisjoint over K in some common extension.

Lemma 6.5. If L is a regular extension of K and H/K is algebraically inde-pendent from L/K, then L and H are linearly disjoint over K.

The reader may note that in the special case where H = Kalg this is justthe definition of regularity.

Proof. Let li ∈ L, hi ∈ H,∑i<n lihi = 0, but not all hi = 0. Since L and H

are independent over Kalg, the type tp(L/KalgH) is an heir of tp(L/Kalg) by


Corollaries 26.5 or 36.13. Thus, there is a non-trivial n–tuple h′ ∈ Kalg suchthat l · h′ = 0. Since L/K is regular, there is a non–trivial h′′ ∈ K such thatl · h′′ = 0. This proves the claim.

Lemma 6.6. Let K be a field. There is a natural bijection between isomorphismtypes of field extensions K(a1, . . . , an) of K with n generators5 and prime idealsP in K[X1, . . . , Xn]. Regular extensions correspond to absolutely prime ideals,i.e. ideals which generate a prime ideal in Kalg[X1, . . . , Xn].

Proof. We associate with L = K(a1, . . . , an) the vanishing ideal P = {f ∈K[X1, . . . , Xn] | f(a1, . . . , an) = 0}. Conversely if P is a prime ideal the quo-tient field of K[X1, . . . , Xn]/P is an extension of K generated by the cosets ofX1, . . . , Xn.

Let I be the ideal generated by P inKalg[X1, . . . , Xn]. ThenK[a1, . . . , an]⊗KKalg and Kalg[X1, . . . , Xn]/I are isomorphic as K–algebras. Now L/K is regu-lar if and only if K[a1, . . . , an]⊗KKalg is an integral domain, which means thatI is prime.

Remark 6.7. It is easy to see that the case tr.deg(a1, . . . , an) = n − 1 cor-responds exactly to the case where P is a principal ideal, generated by anirreducible polynomial f ∈ K[X1, . . . , Xn]. f is uniquely determined up to aconstant factor. ([31, VII, Ex.26])

Definition 6.8. A field extension L over K is separable if, in some commonextension, L and the perfect hull of K are linearly disjoint over K. The perfecthull is the smallest perfect field containing K, so in characteristic 0 equals K,and in characteristic p is the union of all Kp−n = {a ∈ Kalg | apn ∈ K}. Thisextends the definition in the algebraic case.

Again this does not depend on the choice of the common extension L and theperfect hull of K. Note that regular extensions are separable.

Let K be a field of characteristic p > 0. For any subset A we call the fieldKp(A) the p–closure of A. This defines a pregeometry on K (see Section C8)whose associated dimension is the degree of imperfection. A basis of K in thesense of this pregeometry is called a p–basis. If b = (bi | i ∈ I) is a p–basis ofK, then the products bν =

∏i∈I b

νii defined for multi–indices ν = (νi | i ∈ I)

where 0 ≤ νi < p and almost all νi equal to zero, form a linear basis of Kover Kp. If the degree of imperfection of K is a finite number e, it follows that[K : Kp] = pe.

Remark 6.9. L is a separable extension of K if and only if L and Kp−1

arelinearly disjoint over K. It follows that L/K is separable if and only if a p–basisof K stays p–independent in L.

5By quantifier elimination this is just Sn(K) in the theory of Kalg.


Proof. Consider a sequence (ai) of elements of L which is linearly independentover K. Then (ai) is independent over Kp−1

by assumption which implies that(api ) is independent over K. Then (api ) is independent over Kp−1

implying that(ai) is independent over Kp−2

. In this way one can show inductively that (ai)

is independent over Kp−n for all n.

Lemma 6.10. If b1, . . . , bn are p–independent in K, they are algebraically in-dependent over Fp.

Proof. b1, . . . , bn−1 form a p–basis of F = Fp(b1, . . . , bn−1). So K/F is separableby Remark 6.9. If an element c ∈ K is algebraic over F , it is separably algebraicover F . So c is also separably algebraic over KpF , which is only possible if cbelongs to KpF . Since bn does not belong to KpF , it follows that bn is notalgebraic over F .

Lemma 6.11. 1. Let R be an integral domain of characteristic p and let b bea p–basis of R in the sense that every element of R is a unique Rp–linearcombination of the bν . Then b is also a p–basis of the quotient field of R.

2. Let b be p–basis of K and L is a separable algebraic extension of K, thenb is also a p–basis of L.

Proof. 1): Let K be the quotient field of R. Clearly b is p–independent inK. Since the b are algebraic over Kp, Kp[b] is a subfield of K which containsR = Rp[b]. So K = Kp[b].

2): We may assume that L/K is finite. Since Lp and K are linearly disjoint overKp, we have [LpK : K] = [Lp : Kp] = [L : K]. It follows that LpK = L.

Lemma 6.12. The field L = K(a1, . . . , an) is separable over K if and only ifthere is a transcendence basis a′ ⊆ {a1, . . . , an} of L/K such that L is separablyalgebraic over K(a′).

Proof. This follows from the fact that every irreducible polynomial in K[X] is ofthe form f(Xpk) for some k ≥ 0 and some separable polynomial f ∈ K[X].

Lemma 6.13. Algebraic field extensions of K are isomorphic over K if andonly if the same polynomials in K[X] have a zero in these extensions.

Proof. Consider two algebraic extensions L and L′ of K. A compactness argu-ment shows that L and L′ are isomorphic over K if and only if they contain, upto isomorphy over K, the same finitely generated subextensions. The conditionthat the same polynomials in K[X] have zeros in L and L′ is equivalent to Land L′ having the same simple subextensions. So the lemma is clear if L andL′ are separable over K.


For the general case it suffices to prove the following: let L be a finitealgebraic extension of K and L′ an extension of K such that for every a ∈ Lthere is an a′ ∈ L′ such thatK(a) is isomorphic toK(a′) under an automorphismfixing K and taking a to a′. Then L′ contains a copy of L over K.

By the above we may assume that K has characteristic p and is infinite.The extension L/K contains a separable extension F/K such that L is purelyinseparable over F . Let us first remark that for every a ∈ L, the field F (a) issimply generated over K. To see this choose a generator c of F/K and a powerq of p with aq ∈ F . For all d ∈ K the element a + dc also generates L overF . We claim that for some d ∈ K, (a + dc)q generates F/K. If not, infinitelymany of the (a+ dc)q are contained in the same intermediate field K ⊂ F ′ ( Fas there are only finitely many of these. So there are d1 6= d2 ∈ K such that(a + d1c)

q, (a + d2c)q ∈ F ′. This implies cq ∈ F ′. But since F/K is separable,

cq is also a generator of F/K, a contradiction.

By assumption, for every a ∈ L there is an embedding fa : F (a) → L′

over K. Let f1, . . . , fn be a list of all embeddings of F into L′. We construct asequence a0, a1, . . . as follows. Assume that a0 . . . , an−1 are already constructed.Consider for each i the field Li/F generated by all aj with faj � F = f i. We arefinished once one of the Li equals L. As long this is not the case, by Neumann’sLemma 9.9, we find an an ∈ L, which does not yet belong to any of the Li.Adding an to the appropriate Li this construction stops after finitely manysteps yielding a finite set A of generators of L over F such that all fa for a ∈ Aagree on F . These fa combine to an embedding of L = F (A) into L′.

Exercise 6.1.Show that elementary extensions are regular.


7 Pseudo-finite fields and profinite groups

Definition 7.1. A field K is pseudo algebraically closed (PAC) if every ab-solutely irreducible affine variety defined over K has a K–rational point.

See [45, Chapter 1] for basic algebraic geometry.

Lemma 7.2. Let K be a field. The following are equivalent:

a) K is PAC,

b) Let f(X1, . . . , Xn, T ) ∈ K[X1, . . . , Xn, T ] be absolutely irreducible and ofdegree greater than 1 in T and let g ∈ K[X1, . . . , Xn]. Then there exists(a, b) ∈ Kn ×K such that f(a, b) = 0 and g(a) 6= 0.

c) K is existentially closed in every regular field extension.

Proof. a)→b): Let C be an algebraically closed field containing K. The zeroset of f defines an absolutely irreducible variety X ⊂ Cn × C over K. Then theset V = {(a, b, c) | f(a, b) = 0, g(a)c = 1} is absolutely irreducible as well anddefined over K.

b)→c): Let L/K be a finitely generated regular extension. Since regular ex-tensions are separable, by Lemma 6.12 we find a transcendence basis a of L/Ksuch that L is separable algebraic over K(a). Then there is b separable algebraicover K(a) such that L = K(a1, . . . , an, b). Let now c1, . . . , cm be elements ofL satisfying certain equations over K.6 We need c′i ∈ K satisfying the sameequations. Write ci =

hi(a,b)g(a) for polynomials hi(x, y) and g(x) over K. The

vanishing ideal of (a, b) over K is generated by some f(x, y), which is absolutelyirreducible since L/K is regular. There are a′, b′ ∈ K such that f(a′, b′) = 0

and g(a′, b′) 6= 0. Now the c′i =hi(a

′,b′)g(a′) satisfy all equations satisfied by the ci.

c)→a): Let V be an absolutely irreducible variety over K. If c ∈ V is a pointin C generic over K, then K(c)/K is regular by Lemma 6.6.

Corollary 7.3. “PAC” is an elementary property.

Proof. Being absolutely irreducible is an elementary property of the coefficientsin K because the theory of algebraically closed fields has quantifier elimination.

Definition 7.4. A profinite group is a compact7 topological group with a neigh-bourhood basis for 1 consisting of subgroups.

6As we are working in an infinite field, we don’t need inequalities.7Note that for us compact spaces are Hausdorff.


Profinite groups can be presented as inverse limits of finite groups, see [55]for more details.

Definition 7.5. A profinite group is called procyclic if all finite (continuous)quotients are cyclic.

Lemma 7.6 (see [55]). Let G be a profinite group. The following are equivalent:

a) G is procyclic;

b) G is an inverse limit of finite cyclic groups;

c) G is (topologically) generated by a single element a, i.e. the group abstractlygenerated by a is dense in G.

By a generator for a profinite group we always mean a topological generator.

Definition 7.7. Z is the inverse limit of all Z/nZ, with respect to the canonicalprojections.

Lemma 7.8. A profinite group G is procyclic if and only if for every n thereis at most one closed subgroup of index n. If G has exactly one closed subgroupof index n for each n > 0, then G is isomorphic to Z.

Proof. It suffices to show that a finite group G of order n is cyclic if and only iffor each k dividing n there is a unique subgroup of order k. If a is a generatorfor G and H a subgroup of G of order k, then an/k is a generator for H, provinguniqueness. For the other direction, note that by assumption any two elementshaving the same order generate the same subgroup. Thus, if G has an elementof order k, G contains exactly ϕ(k) such elements. The claim now follows fromthe equality

n =∑k|n

ϕ(k).

Lemma 7.9. Let G → H be an epimorphism of procyclic groups. Then anygenerator of H lifts to a generator of G.

Proof. As it suffices to prove the claim for finite cyclic groups G and H, we mayagain assume that G and H are p–groups. If H = 0, we choose any generatoras the preimage. If H 6= 0, any preimage of a generator of H is a generatorof G.

For a field K we write G(K) for its absolute Galois group Aut(Kalg/K).


Definition 7.10. A perfect field K is called procyclic if G(K) is procyclic, and1–free if G(K) ∼= Z.

By Lemma 7.8 a perfect field is procyclic (1–free, respectively) if and only ifit has at most one (exactly one, respectively) extension of degree n in Kalg forevery n. Thus, being procyclic or 1–free is an elementary property of K.

Definition 7.11. A perfect 1–free PAC–field is called pseudo-finite.

Remark 7.12. Being pseudo-finite is an elementary property. A theorem ofLang and Weil (see [32]) on the number of points on varieties over finite fieldsshows that infinite ultraproducts of finite fields are PAC and hence pseudo-finite.

Proposition 7.13. Let K be perfect and L/K a field extension. Then thefollowing are equivalent:

a) L/K regular,

b) K is relatively algebraically closed in L,

c) the natural map G(L) → G(K) is surjective.

Proof. Using remark 6.2, it is easy to see that a) implies c). Clearly, c) impliesb). To see that b) implies a) assume that K is relatively algebraically closedin L. Since every finite extension between K and Kalg is generated by a singleelement, it suffices to show for a ∈ Kalg that the minimal polynomial f of aover L has coefficients in K.

But this follows as the coefficients of f can be expressed by conjugates ofa ∈ Kalg ∩ L.

If L is perfect and N/L and L/K are regular extensions, then also N/K isregular.

Corollary 7.14. Let L/K be regular, L procyclic and K 1–free. Then L is1–free. If N/L is an extension of L such that N/K is regular, then N/L isregular.

Proof. If G(L) is procyclic and π : G(L) → Z = G(K) is surjective, then π isan isomorphism. Let ρ : G(N) → G(L) be the natural map. If ρπ is surjective,then so is ρ.

Any profinite group G acts faithfully and with finite orbits on the set Sconsisting of all left cosets of finite index subgroups. Since point stabilisersare open subgroups of G, this action is continuous with respect to the discretetopology on S. We use this easy observation to prove the following:

Lemma 7.15. Any procyclic field has a regular pseudo-finite field extension.


Proof. Let K be a procyclic field. Let S be as in the previous remark for G = Zand let L′ = K(Xs)s∈S with the action of G on L′ given by the action of Gon S. Then L′ is a Galois extension of L = Fix(G) with Galois group Z andL′ and L are regular extensions of K. Let σ ∈ Aut(KalgL′/L) be a commonextension of a generator of G(K) and a generator of Gal(L′/L). Extend σ tosome σ′ ∈ Aut(Lalg/L) and consider the fixed field L′′ of σ′′. Then L′′ is a 1–freeregular extension of K. It is now easy to construct, by a long chain, a regularprocyclic extension N of L′′ which is existentially closed in all regular procyclicextensions. N is again 1–free. In order to see that N is PAC, we consider aregular extension N ′ of N . Let σ ∈ G(N ′) be a lift of a generator of G(N) andN ′′ the fixed field of σ in N ′alg. Then N ′′ is a regular procyclic extension of N .By construction, N is existentially closed in N ′′ and hence also in N ′.

Procyclic fields have the amalgamation property for regular extensions:

Lemma 7.16. If L1, L2 are regular procyclic extensions of a field K, there isa common procyclic regular extension H of L1 and L2 in which L1 and L2 arelinearly disjoint over K.

Proof. Let L1 and L2 be regular procyclic extensions of a field K (so K is alsoprocyclic). We may assume that the Li are algebraically independent over K insome common field extension. Let σ be a (topological) generator of G(K). ByLemma 7.9, we can lift σ to generators σi of G(Li). By Lemma 6.5, L1 and L2

are linearly disjoint. As Lalg1 Lalg

2 is the quotient field of the tensor product ofLalg1 and Lalg

2 over Kalg, the σi generate an automorphism of Lalg1 Lalg

2 which canbe extended to an automorphism τ of (L1L2)

alg. The fixed field of τ is procyclicand a regular extension of L1 and L2.

By Lemma 7.15 this also implies that pseudo-finite fields have the amalga-mation property for regular extensions.

Appendix C

Combinatorics

8 Pregeometries

In this section, we collect the necessary facts and notions about pregeometries,existence of bases and hence a well-defined dimension, modularity laws etc. Firstrecall Definition 20.5:

Definition. A pregeometry (X, cl) is a setX with a closure operator cl : P(X) →P(X) such that for all A ⊂ X and a, b ∈ X

a) A ⊂ cl(A)

b) cl(A) is the union of all cl(A′), where the A′ range over all finite subsetsof A.

c) cl(cl(A)) = cl(A)

d) (exchange property) a ∈ cl(Ab) \ cl(A) ⇒ b ∈ cl(Aa)

Remark 8.1. The following structures are pregeometries:

1. A vector space V with the linear closure operator.

2. For a field K with prime field F , the relative algebraic closure cl(A) =F (A)alg

3. The p–closure in a field K of characteristic p > 0, i.e. cl(A) = Kp(A).

Proof. We prove the exchange property for algebraic dependence in a field K asan example.1 We may assume that A is a subfield of K. Let a, b be such that

1For the p–closure see [9, §13].

232

APPENDIX C. COMBINATORICS 233

a ∈ A(a)alg. So there is a non–zero polynomial F ∈ A[X,Y ] with F (a, b) = 0.If we also assume that b /∈ A(b)alg, it follows that F (a, Y ) = 0. This impliesa ∈ Aalg.

A pregeometry in which points and the empty set are closed, i.e., in which

cl′(∅) = ∅ and cl′(x) = {x} for all x ∈ X

is called geometry. For any pregeometry (X, cl), there is an associated geometry(X ′, cl′) obtained by setting X ′ = X•/ ∼, and cl′(A/ ∼) = cl(A)•/ ∼ where ∼is the equivalence relation on X• = X \ cl(∅) defined by cl(x) = cl(y). Startingfrom a vector space V , the geometry obtained in this way is the associatedprojective space P(V ). The important properties of a pregeometry are mostlyin fact properties of the associated geometry.

Definition 8.2. Let (X, cl) be pregeometry. A subset A of X is called

1. independent if a 6∈ cl(A \ {a}) for all a ∈ A;

2. a generating set if X = cl(A);

3. a basis if A is an independent generating set.

Lemma 8.3. Let (X, cl) be a pregeometry with generating set E. Any indepen-dent subset of E can be extended to a basis contained in E. In particular everypregeometry has a basis.

Proof. Let B be an independent set. If x is any element in X \ cl(B), B∪{x} isagain independent. To see this consider an arbitrary b ∈ B. Then b 6∈ cl(B\{b}),whence b 6∈ cl(B \ {b} ∪ {x}) by exchange.

This implies that for a maximal independent subset B of E, we have E ⊂cl(B) and therefore X = cl(B).

Definition 8.4. Let (X, cl) be a pregeometry. Any subset S gives rise to twonew pregeometries, the restriction (S, clS) and the relativisation (X, clS), where

clS(A) = cl(A) ∩ S,clS(A) = cl(A ∪ S).

Remark 8.5. Let A be a basis of (S, clS) and B a basis of (X, clS). Then the(disjoint) union A ∪B is a basis of (X, cl).

Proof. Clearly A ∪ B is a generating set. Since B is independent over S, wehave b 6∈ clS(B \ {b}) = cl(A ∪ B \ {b}) for all b ∈ B. Consider an a ∈ A. Wehave to show that a 6∈ cl(A′ ∪ B), where A′ = A \ {a}. As a 6∈ cl(A′), we letB′ be a maximal subset of B with a 6∈ cl(A′ ∪ B′). B′ 6= B would imply thata ∈ cl(A′ ∪B′ ∪{b}) for any b ∈ B \B′ which again would imply b ∈ cl(A∪B′),a contradiction.


A is called a basis of S and B a basis over or relative to S.

Lemma 8.6. All bases of a pregeometry have the same cardinality.

Proof. Let A be independent and B a generating subset of X. We show that

|A| ≤ |B|.

Assume first that A is infinite. Then we extend A to a basis A′. Choose forevery b ∈ B a finite subset Ab of A′ with b ∈ cl(Ab). Since the union of the Abis a generating set, we have A′ =

⋃b∈B Ab. This implies that B is infinite and

|A| ≤ |A′| ≤ |B|.

Now assume that A is finite. |A| ≤ |B| follows immediately from the fol-lowing exchange principle: For any a ∈ A \ B there is b ∈ B \ A such thatA′ = {b} ∪ A \ {a} is independent. Proof: Since a ∈ cl(B), B cannot be con-tained in cl(A \ {a}). Choose b in B but not in cl(A \ {a}). It follows from theexchange property that A′ is independent.

Definition 8.7. The dimension dim(X) of a pregeometry (X, cl) is the cardi-nality of a basis. For a subset S of X let dim(S) be the dimension of (S, clS)and dim(X/S) the dimension of (X, clS).

By Remark 8.5 we have

Lemma 8.8.dim(X) = dim(S) + dim(X/S).

The dimensions in our three standard examples are:

• The dimension of a vector space.

• The transcendence degree of a field.

• The degree of imperfection of a field of finite characteristic (see [9, §13,Ex.1]).

Let (X, cl) be a pregeometry. For arbitrary subsets A and B one sees easilythat

dim(A ∪B) + dim(A ∩B) ≤ dim(A) + dim(B).

One may hope for equality to hold if A and B are closed:

Definition 8.9. We call a pregeometry (X, cl) modular, if

(1) dim(A ∪B) + dim(A ∩B) = dim(A) + dim(B)

for all cl–closed A and B.


The main examples are

• trivial pregeometries where cl(A ∪B) = cl(A) ∪ cl(B) for all A,B.

• vector spaces with the linear closure operator.

Let K be a field of transcendence degree at least four over its prime field F .The following argument shows that the pregeometry of algebraic dependenceon K is not modular: Choose x, y, x′, y′ ∈ K algebraically independent overF . From these elements we can compute a, b ∈ K such that ax + b = y andax′+b = y′. Since the elements x, x′, a, b generate the same subfield as x, y, x′, y′,they are also algebraically independent. This implies that F (x) and F (x′) areisomorphic over F (a, b)cl, where cl denotes the relative algebraic closure in K.This isomorphism maps y to y′ and therefore we have

F (x, y)cl ∩ F (a, b)cl ⊂ F (x, y)cl ∩ F (x′, y′)cl = F cl.

So K is not modular since

tr.degF (x, y, a, b) + tr.degF = 3+ 0 < 2 + 2 = tr.degF (x, y) + tr.degF (a, b).

Let us call two sets A and B (geometrically) independent over C if allsubsets A0 ⊂ A and B0 ⊂ B which are both independent over C are disjointand their union is again independent over C. The following is then easy to see:

Lemma 8.10. For a pregeometry (X, cl) the following are equivalent:

1. (X, cl) is modular.

2. Any two closed A and B are independent over their intersection.

3. For any two closed sets A and B we have dim(A/B) = dim(A/A ∩B).

Lemma 8.11. A pregeometry (X, cl) is modular if and only if for all a, b, Bwith dim(ab) = 2, dim(ab/B) = 1, there is c ∈ cl(B) such that dim(ab/c) = 1.

Proof. If (X, cl) is modular and a, b, B are as in the lemma, then ab and B aredependent, but independent over the intersection of cl(ab) and cl(B). Let c bean element of the intersection which is not in cl(∅). Then dim(ab/c) = 1.

Assume that the property of the lemma holds. We show that the third condi-tion of Lemma 8.10 is satisfied. For this we may assume that n = dim(A/A∩B)is finite and proceed by induction on n. The cases n = 0, 1 are trivial. So as-sume n ≥ 2. Let a1, . . . an be a basis of A over A ∩ B. We have to show thatdim(a1, . . . , an/B) = n. By induction we know that dim(a1 . . . an−2/B) = n−2.So it is enough to show that dim(an−1an/B

′) = 2 where B′ be the closureof {a1, . . . , an−2} ∪ B. If not, by our assumption there is c ∈ B′ such thatdim(an−1an/c) = 1.


Definition 8.12. (X, cl) is locally modular if (1) holds for all closed sets A,Bwith dim(A ∩B) > 0.

Remark 8.13. Clearly, (X, cl) is locally modular if and only if for all x ∈X \ cl(∅) the relativised pregeometry (X, clx) is modular.

An affine subspace of a vector space V is a coset of a subvector space. Theaffine subspaces of V define a locally modular geometry, which is is not modular,because of the existence of parallel lines. Note that in this example, points havedimension one, lines dimension two, etc.

The arguments on page 235 show also that a field of transcendence degree atleast five is not locally modular. Just replace F by a subfield of transcendencedegree 1.

Exercise 8.1.Consider a pregeometry (X, cl). Let A |cl

CB be the relation of A and B being

independent over C and A |0CB the relation cl(AC) ∩ cl(BC) = cl(C). Show

the following:

1. |cl has the following properties as listed in Theorem 29.13: Monotony,Transitivity, Symmetry, Finite Character and Local Charac-ter.

2. |0 has Weak Monotony (cf. Theorem 36.10), Symmetry, FiniteCharacter and Local Character. Monotony holds only if |0 =

|cl , i.e. if X is modular.

Exercise 8.2.Prove that trivial pregeometries are modular.

Exercise 8.3 (P. Kowalski).Let K be a field of characteristik p > 0 with degree of imperfection at least 4.Prove that K with p–dependence is not locally modular.

Exercise 8.4.(X, cl) is modular if and only if for all c ∈ cl(A ∪ B) there are a ∈ cl(A) andb ∈ cl(B) such that c ∈ cl(a, b).

Exercise 8.5.The set of all closed subsets of a pregeometry forms a lattice, where the infimumis intersection and the supremum of X and Y is X tY = cl(X ∪Y ). Show thata pregeometry is modular if and only if the lattice of closed sets is modular, i.e.if for all closed A,B,C

A ⊂ C ⇒ A t (B ∩ C) = (A tB) ∩ C.

Exercise 8.6.Let (X, cl) be a pregeometry of uncountable dimension. Suppose that for all


closed B of countable dimension the automorphism group Aut(X/B) acts tran-sitively on X \B. Then (X, cl) is locally modular if and only if for all closed Bof countable dimension and all a, b with dim(a, b) = 2 and dim(a, b/B) = 1 andevery σ ∈ Aut(X/B) the two pairs (a, b) and (σ(a), σ(b)) are not independentover ∅. Conclude that for any finite A, X is locally modular if and only if XA

is locally modular.


9 The Erdös–Makkai Theorem

Theorem 9.1 (Erdös–Makkai). Let B be an infinite set and S a set of subsetsof B with |B| < |S|. Then there are sequences (bi | i < ω) of elements of B and(Si | i < ω) of elements of S such that either

bi ∈ Sj ⇔ j < i(2)

or

bi ∈ Sj ⇔ i < j(3)

for all i, j ∈ ω.

Proof. Choose a subset S ′ be a subset of S of the same cardinality as B suchthat any two finite subsets of B which can be separated by an element of S canbe separated by an element of S ′. The hypothesis implies that there must bean element S∗ of S which is not a Boolean2 combination of elements of S ′.

Assume that for some n, three sequences (b′i | i < n) in S∗, (b′′i | i < n) inB \ S∗ and (Si | i < n) in S ′ have already been constructed. Since S∗ is not aBoolean combination of S0, . . . , Sn−1, there are b′n ∈ S∗ and b′′n ∈ B \ S∗ suchthat for all i < n

b′n ∈ Si ⇔ b′′n ∈ Si.

Choose Sn as any set in S ′, which separates {b′0, . . . , b′n} and {b′′0 , . . . , b′′n}.

Now, an application of Ramsey’s theorem shows that we may assume thateither b′n ∈ Si or b′n 6∈ Si for all i < n. In the first case we set bi = b′′i and get(2), in the second case we set bi = b′i+1 and get (3).

2It actually suffices to consider positive Boolean combinations.


10 The Erdös–Rado Theorem

Definition 10.1. For cardinals κ, λ, µ we write κ → (λ)nµ (κ arrows λ) toexpress that for any function f : [κ]n → µ there is some A ⊂ κ with |A| = λ,such that f is constant on [A]n. In other words, every partition of [κ]n into µpieces has a homogeneous set of size λ.

With this notation Ramsey’s Theorem 15.5 states that

ω → (ω)nk for all n, k < ω.

In an analogous manner one can define a cardinal κ to be a Ramsey cardinalif κ → (κ)<ω2 . In other words, if for any n a partition of [κ]n into two classesis given, there is a set of size κ simultaneously homogeneous for all partitions.A Ramsey cardinal κ satisfies κ → (κ)<ωγ for all γ < κ (see [29] 7.14). Moregenerally, an uncountable cardinal κ is called weakly compact if it satisfies κ→(κ)22. Such cardinals are weakly inaccessible and their existence cannot beproven from ZFC.

Theorem 10.2 (Erdös-Rado).

i+n (µ) → (µ+)n+1

µ .

Proof. This follows from µ+ → (µ+)1µ and the following lemma:

Lemma 10.3. If κ+ → (µ+)nµ, then (2κ)+ → (µ+)n+1µ .

Proof. We note first that the hypothesis implies µ ≤ κ. Now let B be a set ofcardinality (2κ)+ and f : [B]n+1 → µ be a colouring. If A is a subset of B, wecall a function p : [A]n → µ a type over A. If b ∈ B \A, the type tp(b/A) is thefunction which maps each n–element subset s of A to f(s ∪ {b}). If |A| ≤ κ,there are at most 2κ many types over A. Thus an argument as in the proof ofLemma 23.2 shows that there is some B0 ⊂ B of cardinality 2κ such that forevery A ⊂ B0 of cardinality at most κ every type over A which is realised in Bis already realised in B0.

Fix an element b ∈ B \B0. We can easily construct a sequence (aα)α<κ+ inB0 such that every aα has the same type over {aβ | β < α} as b. By assumption{aα | α < κ+} contains a subset A of cardinality µ+ such that tp(b/A) isconstant on [A]n. Then f is constant on [A]n+1.

Appendix D

Solutions of Exercises

Exercise 2.3We consider formulas which are built using ¬,∧,∃,∀ and we move the quantifiersoutside using

¬∃xϕ ∼ ∀x¬ϕ¬∀xϕ ∼ ∃x¬ϕ

(ϕ ∧ ∃xψ(x, y)) ∼ ∃z (ϕ ∧ ψ(z, y))(ϕ ∧ ∀xψ(x, y)) ∼ ∀z (ϕ ∧ ψ(z, y))

In the last equivalence we have replaced the bound variable x by a variable z,which does not occur freely in ϕ.

Exercise 3.1The theory TDLO of dense linear orders without endpoints is axiomatised in LO

by

• ∀x ¬x < x

• ∀x, y, z (x < y ∧ y < z → x < z)

• ∀x, y (x < y ∨ x .= y ∨ y < x)

• ∀x, z (x < z → ∃y (x < y ∧ y < z))

• ∀x∃y x < y

• ∀y∃x x < y

The class of all algebraically closed fields can be axiomatised by the theoryTACF :

240

APPENDIX D. SOLUTIONS OF EXERCISES 241

• TF (field axioms)

• For all n > 0:

∀x0, . . . xn−1 ∃y x0 + x1y + . . .+ xn−1yn−1 + yn

.= 0

Exercise 3.3Let A and B be two elementarily equivalent L–structures. It is easy to seethat A and B are isomorphic if L is finite. Let L be arbitrary and assumefor contradiction that A and B are not isomorphic. Then for every bijectionf : A → B there is a Zf ∈ L which is not respected by f . If L0 is the set ofall the Zf , A � L0 and B � L0 are not isomorphic, which contradicts our firstobservation. One should note that the isomorphy follows also from Exercise 23.1and Lemma 12.3.

Exercise 4.2Hint: Let T = {ϕ : M |= ϕ for all M ∈ C}. For M |= T construct an ultraprod-uct of L–structure in C having the same theory.)

Exercise 6.2For every prime p, TACFp has a model which is the union of a chain of finitefields.

Exercise 7.1We imitate the proof of Lemma 7.1. That a) implies b) is clear. For the converseconsider an element y1 of Y1 and Hy1 , the set of all elements of H which containy1. b) implies that the intersection of the sets in Hy1 is disjoint from Y2. So afinite intersection hy1 of elements of Hy1 is disjoint from Y2. The hyi , y1 ∈ Y1,cover Y1. So Y1 is contained in the union H of finitely many of the hyi . Hseparates Y1 from Y2.

b) ⇒ a):For any model A1 of T1 let HA1 be the set of all sentences from H which aretrue in A1. b) implies that HA1 and T2 cannot have a common model. By theCompactness Theorem there is a finite conjunction ϕA1 of sentences from HA1

inconsistent with T2. Clearly,

T1 ∪ {¬ϕA1 | A1 |= T1}

is inconsistent since any model A1 of T1 satisfies ϕA1 . Again by compactness T1implies a disjunction ϕ of finitely many of the ϕA1

. This ϕ is in H and separatesT1 from T2.

Exercise 8.1Hint: For any simple existential formula write down the equivalence to a quan-tifier free formula. Show that this is equivalent to an ∀∃–sentence and that Tis axiomatised by these.


Exercise 9.1Like the proof of 9.2 with order preserving automorphisms replaced by edgepreserving ones.

Exercise 9.2The cases TACF ,TRCF are easy. So we concentrate on TDCF0 . Clearly, thealgebraic closure of the differential field generated by A is contained in themodel theoretic algebraic closure. For the converse, let K0 be an algebraicallyclosed differential field and a0 an element not contained in K0. If dim(a0/K0) isinfinite, then a0 and all its derivatives have the same type over K, so a0 is notnot model–theoretically algebraic over K0. If dim(a0/K0) = n > 0, considerthe minimal polynomial f of a0 over K0. Let K1 be a d-closed extension of K0

containing a0. Then f remains irreducible over K1 and there is some a1 whoseminimal polynomial over K1 is f . Now extend K1 to some field K2 containinga1 etc.. In this way we obtain an infinite sequence of distinct elements havingthe same type over K0 as a0, showing that a0 is not algebraic over K0 in thesense of model theory.

Exercise 9.3Let K be algebraically closed. X = {a | K |= γ(a, b)} a definable subset of Kn

and f : X → X is given by an n–tuple of polynomials f(x, b). We may assumethat γ(x, z) is quantifier free. We want to show that K satisfies

∀y(∀x(ϕ(x, y) → ϕ(f(x, y), y)

)∧

∀x, x′(ϕ(x, y) ∧ ϕ(x′, y) ∧ f(x, y) .= f(x′, y) → x

.= x′

)→

∀x′(ϕ(x′, y) → ∃x(ϕ(x, y) ∧ f(x, y) .= x′)

)).

This is obviously true in finite fields (even in all finite LR–structures) and logi-cally equivalent to an ∀∃–sentence. So the claim follows from Exercise 6.2.

For the second part use Exercise 23.13 and proceed like before.

Exercise 11.1[ϕ] is a singleton if and only if [ϕ] is non–empty and cannot be divided intotwo non-empty clopen subsets [ϕ ∧ ψ] and [ϕ ∧ ¬ψ]. This means that for all ψeither ψ or ¬ψ follows from ϕ modulo T . So [ϕ] is a singleton if and only if ϕgenerates the type

〈ϕ〉 = {ψ(x) | T ` ∀x (ϕ(x) → ψ(x))},

which of course must be the only element of [ϕ].

This shows that [ϕ] = {p} implies that ϕ isolates p. If, conversely, ϕ isolatesp, this means that 〈ϕ〉 is consistent with T and contains p. Since p is a type, itfollows p = 〈ϕ〉.

Exercise 11.2a): The sets [ϕ] are a basis for the closed subsets of Sn(T ). So the closed setsof Sn(T ) are exactly the intersections

⋂ϕ∈Σ[ϕ] = {p ∈ Sn(T ) | Σ ⊂ p}.


b): X is the union of a sequence of countable nowhere dense sets Xi. We mayassume that the Xi are closed, i.e. of the form {p ∈ Sn(T ) | Σi ⊂ p}. ThatXi has no interior means that Σi is not isolated. The claim follows now fromCorollary 10.3.

Exercise 11.3Let X = {tp(a0, a2, . . .) | the ai enumerate a model of T}. Consider for everyformula ϕ(v, y) the set Xϕ = {p ∈ Sω | (∃yϕ(v, y) → ϕ(v, vi)) ∈ p for some i}.The Xϕ are open and dense and X is the intersection of the Xϕ.

Exercise 11.5The homeomorphism from Sm(aB) to the fiber above tp(a/B) is given bytp(c/aB) 7→ tp(ca/B).

Exercise 14.1If T is countable, not small, Sn(T ) uncountable, say. Prove that there is aconsistent formula ϕ such that both [ϕ] and [¬ϕ] are uncountable. Inductivelywe obtain a binary tree of consistent formulas (cf. proof of Theorem 16.5.2).

Exercise 15.2To ease notation we replace the partition by a function γ : [A]n → {1, . . . , k}.Fix a non-principal ultrafilter U on A, i.e. an ultrafilter which does not containany finite set. For each s ∈ [A]n−1 chose a c(s) such that {a ∈ A | γ(s ∪ {a}) =c(s)} belongs to U . Construct a sequence a0, a1, . . . of distinct elements suchthat γ(s ∪ {an}) = c(s) for any s ∈ [{a0, . . . , an−1}]n−1. Apply induction to crestricted to [{a0, a1, . . .}]n−1.

Exercise 16.1Let T define a linear ordering of the universe. By Exercise 33.8 there is a linearordering J of bigger cardinality than κ which has a dense subset of cardinalityκ. A compactness argument shows that T has a model with a subset B whichis order-isomorphic to J . Let A be a dense subset of B of cardinality κ. Thenall elements of B have different types over A and so |S(A)| ≥ |B| > κ.

Exercise 17.1a ⇒ b: Let A a countable subset of the model M. That A has a prime extensionmeans that TA = Th(MA) has a prime model. Now apply Theorem 14.7.

b ⇒ c: Assume that A is contained in the model M and that the isolatedtypes are not dense over A. So there is a consistent L(A)–formula ϕ, whichdoes not contain a complete L(A)–formula. Add a predicate P for the set Aand consider the L ∪ {P}–structure (M, A). Choose a countable elementarysubstructure (M0, A0) which contains the parameters of ϕ. Then ϕ does notcontain a complete L(A0)–formula.

c ⇒ a: Like the proof of 17.3.

Exercise 19.4Clearly, T (q) is complete λ-stable if and only if T is. It is also clear that if T (q)has a Vaughtian pair then so does T . For the converse use a construction as in


Theorem 19.2 to find a Vaughtian pair M ≺ N such that q is realized in N.

Exercise 19.6Hint: Use Exercise 19.5.

Exercise 20.1If p ∈ S(A) is not algebraic, then all n-types

q(x1, . . . xn) = {xi, i = 1, . . . n, satisfies p and the xi are pairwise distinct}

are consistent and hence realised in M.

Exercise 20.2If ci ∈ acl(Acn) for some i < n, we realise tp(c0 . . . cn/A) such that the c0 . . . cn−1

is disjoint from acl(AB). Then cn cannot belong to B. If ci 6∈ acl(Acn) for alli < n, we realise the tp(cn/A) and then tp(c0 . . . cn−1/Acn), both disjoint fromB.

Exercise 23.3Choose special models Ai of Ti of the same cardinality.

Exercise 23.4If θ does not exist, the set T ′ of all L–sentences θ such that ` ϕ1 → θ or` ¬ϕ2 → θ is consistent. Choose a complete L–theory T which contains T ′ andapply Exercise 23.3 to T1 = {ϕ1} ∪ T and T2 = {¬ϕ2} ∪ T ′.

Exercise 23.5Let |A| < κ and p = p(xi)i<κ) be a κ–type over p. Denote by pα the restrictionof p to the variables (xi)i<α. Construct a realisation (ai)i<κ of p inductively: If(ai)i<α realises pα, choose aα as a realisation of pα+1((ai)i<α, xα).

Exercise 23.10If B ⊂ dcl(A), every formula with parameters in A is equivalent to a formulawith parameters in A. So every type over A axiomatises a type over B. Thisproves 1→2. For the converse show that b ∈ dcl(A) if tp(b/A) has a uniqueextension to Ab.

Exercise 23.11If ϕ(a, b) is a formula witnessing b ∈ dcl(a), let D equal the set of elementsx for which there is a unique y with ϕ(x, y) and let f : D → E denote thecorresponding map, which we may assume to be surjective. If also a ∈ dcl(b)witnessed by ψ(y, x) and function g : E1 → D1, then we get a 0-definable bijec-tion {(x, y) ∈ E× D1|f(x) = y and g(y) = x}.

Exercise 23.12Use Exercise 9.2.

Exercise 23.13Use Exercise 23.12 and compactness.

Exercise 23.15By induction on n. We distinguish two cases. First assume that for some i < 0,


Hn ∩Hi has finite index in Hn. We can then cover every coset of Hn by finitelymany cosets of Hi. Since G is not a finite union of cosets of H0, . . . ,Hn−1, weare done. Now assume that all H ′

i = Hn∩Hi have infinite index in Hn. Assumefor contradiction that G is a finite union of cosets of the H0, . . . ,Hn. Since Hn

has infinite index in G, there must be a coset of Hn which is covered by a finitenumber of cosets of H0, . . . , Hn−1. This implies that Hn is a finite union ofcosets of the H ′

i, i < n, which is imposible by induction.

Exercise 23.16To see that Exercise 23.15 implies Exercise 23.14 consider the subgroups Hi :=Gci , i ≤ n which have infinite index in G. The finitely many cosets ajHi withaj(ci) ∈ B, i ≤ n do not cover G, so there is some g ∈ G such that for all i ≤ nwe have g(ci) /∈ B.

For the converse, let H1, . . . Hn ≤ G be subgroups of infinite index, andconsider the action of G on the disjoint union of the G/Hi by left translation.By Exercise 23.14, for any a1, . . . an ∈ G there is some g ∈ G such that for alli ≤ n we have g(1 ·Hi) /∈ aiHi, proving Exercise 23.15.

Exercise 24.1Let I be that set of all i for which ϕ∧ψi has rank α. The hypothesis implies thatall k–element subsets of I contain two indices i, j such that ϕ ∧ ψi 6∼α ϕ ∧ ψj .So |I| ≤ (k − 1)MDϕ.

Exercise 24.2Let T be ω–stable and A countable. Choose elements bi and cij with S1(A) ={tp(bi/A) | i ∈ ω} S1(Abi) = {tp(cij/Abi) | j ∈ ω}. Then S2(A) = {tp(bicij/A) |j ∈ ω} is countable. Similarly, one can show that all the Sn(A) are countable.

If there is a binary tree of formulas in n variables, there is a countable sub-language L0 ⊂ L for which the countable theory T0 = {ϕ | ϕ L0–sentence, T `ϕ} = T � L0 is not ω–stable for n–types. Hence T0 is not ω–stable and T0 andT are not totally transcendental.

Exercise 24.4Let G0 be the intersection of all definable subgroups of finite index. G0 isdefinable by Remark 24.7. If N is a finite subgoup which is normalised by G,the centraliser of N in G is a definable group of finite index in G0.

Exercise 24.5Let M be an ω–saturated model and let p be a type over M . Let ϕ(x,m) ∈ pbe of minimal rank α and degree n. If n > 1, there is a formula ψ(x, b) suchthat ϕ(x,m)∧ψ(x, b) and ϕ(x,m)∧¬ψ(x, b) both have Morley rank α. Choosea ∈ M with tp(a/m) = tp(b/m). Then both formulas ϕ(x,m) ∧ ψ(x, a) andϕ(x,m) ∧ ¬ψ(x, a) have rank α and one of these formulas belongs to p, contra-dicting the maximal choice of ϕ(x,m).

Exercise 24.6Assume that there is a formula ϕ(x, b) of rank ≥ |T |+. Construct a binary


tree of formulas1, ϕs(x, ys), (s ∈ <ω2), below ϕ(x, b) so that for all k and allα < |T |+ there are parameters as such that MRϕs(x, as) ≥ α for all s with|s| = k. Conclude that MRϕ(x, b) = ∞,

Exercise 24.8Let a be in acl(A) and a1, . . . , an the conjugates of a over A. Then ϕ(x, a) andϕ(x, a1) ∧ . . . ∧ ϕ(x, an) have the same Morley rank.

Exercise 26.1In a pregeometry a finite set A is independent from B over C if and only ifdim(A/BC) = dim(A/C). Now use Theorem 26.2.

Exercise 26.2Assume that abC is independent from B and apply Remark 26.9.

Exercise 27.2If p forks over A, there is a ϕ(x,m) ∈ p which implies a conjunction

∨l<d ϕl(x, b)

of formulas which fork over A. Choose a tuple b′ in M which realises the typeof b over Am. The formulas ϕl(x, b′) fork over A and one of them belongs to p.

Exercise 27.3Let q be A–invariant. Use Lemma 27.4 to show that q does not divide over A:If π(x, b) belongs to q, then all the π(x, bi) also belong to q. That q does notfork over A follows from Exercise 27.2.

Exercise 27.4p containes a formula ϕ which divides over A. So there are κ many αi ∈Aut(M/A) such that the system of all αi(ϕ) is k–inconsistent. This impliesthat κ many of the αi(p) must be distinct.

Exercise 27.5p(x) forks over the empty set since it implies the disjunction of cyc(0, x, 3) andcyc(2, x, 1).

Exercise 27.6This is just a variant of Proposition 27.6. Let I = (bi | i < ω) a sequence ofA-indiscernibles containing b such that (ϕ(xbi) | i < ω) is k–inconsistent. Sincetp(a/Ab) does not divide over A, we can assume that I is indiscernible over Aa.

Exercise 28.1Let (pα)α∈|T |+ be a chain of types with pα ∈ S(Aα). Their union

⋃α∈|T |+ pα

does not fork over a subset A′ of⋃α∈|T |+ Aα of cardinality at most |T |. This

follows from Proposition 28.5 (Local Character) and Proposition 28.15. SinceA′ ⊆ Aα for some sufficiently large α ∈ |T |+, from that index on the chain doesnot fork anymore.

1in the sense of Definition 14.8


Note that this property is just a reformulation of Local Character for κ =|T |+.

Exercise 28.2Let q = tp(b/B) and r = tp(c/C). Find an A–automorphism, which maps c tob and C to C ′ such that B |

AbC ′. Then r′ = tp(b/C ′) and s = tp(b/BC ′) are

as required: we have B |Ab, which together with B |

AbC ′ yields B |

AC ′b

from which b |C′ B.

Exercise 28.3By monotony and symmetry it suffices to show that aX and aY \X are inde-pendent over A. So we can assume that X and Y are disjoint and, by finitecharacter, that X and Y are finite. We proceed by induction on |X ∪ Y |. Letz be the maximum of X ∪ Y . By symmetry we may assume that z ∈ Y . Thenaz is independent from aX∪Y \{z}. The claim follows now from the inductionhypothesis and transitivity.

Exercise 28.4This follows from: If (bi | i < ω) is indiscernible over A, there is an A–conjugateB′ of B such that (bi | i < ω) is indiscernible over B′.

Exercise 28.5Up to symmetry this is only a reformulation of Monotony and Transitivity.

Exercise 28.6For ease of notation we restrict to the following special case:If b1b2 is independent from C, then we have

b1 | b2 ⇐⇒ b1 |C

b2.

Proof: By Corollary 28.18 both sides of this equivalence are equivalent to thetriple (b1, b2, C) being independent. (For the direction from right to left itsuffices in fact to assume that b1 and b2 are independent from C individually.)

Exercise 28.7The hypothesis implies a |

ABC. Now the claim follows from Remark 27.3.

Exercise 29.1Let I be the sequence b = b0, b1 . . .. Use Proposition 29.6 and induction on nto show that

⋃{π(x, bi) | i < 2n} does not fork over A. So we find a realisation

of⋃{π(x, bi) | i < ω} which is independent from I over A. That one can find

c in such a way that I is indiscernible over Ac follows from Lemma 15.3 andFinite Character.

Exercise 29.21. Let (ci | i < ω) be an antichain for θ1 ∧ θ2. Then by Ramsey’s theorem thereis an infinite A ⊂ ω such that (ci | i ∈ A) is an antichain for θ1 or for θ2.

2. If θ∼(x, y) is not thick, it has, by compactness, antichains (ci | i ∈ I) indexedby arbitrary linear orders I. If I∼ is the inverse order, (ci | i ∈ I∼) is an


antichain for θ.

3. Consider θ ∧ θ∼.

Exercise 29.3Choose an A–conjugate c of b, different from b, and set B = b and C = c.

Exercise 29.5Show that a |0

AA with Existence and use Monotony and Transitivity.

Exercise 29.6Let p ∈ S(M) with two different non–forking extensions to B ⊃ M . LetB0, B1, . . . be an M–independent family of conjugates of B. Then on eachBi there are two different extensions q0i , qi of p. Now by the independence theo-rem for any function ε : ω → 2 there is a nonforking extension qε of p to ∪i<ωBi,which extends every each qε(i)i .

Exercise 30.1If θ is thick and defined over A, the conjunction of the A–conjugates of θ isthick and defined over A.

Exercise 30.2Extend B and C to models MB and MC such that MB |

AMC , b |

AMB and

c |AMC . Now it suffices to find a d such d |

AMBMC , tp(d/MB) = tp(b/MB)

and tp(d/MC) = tp(c/MC).

Exercise 30.3We have to show that for every thick θ(x, y) the formula θ(x, a) does not divideover A. So let a = a0, a1, . . . be indiscernible over A. Then |= θ(ai, aj) for alli, j. This shows that {θ(x, ai) | i < ω} is finitely satisfiable. If T is simple andB is any set, chose a′′ independent from B over A such that ncA(a′′, a). Finallychoose a′ such that tp(a′/Aa) = tp(a′′/Aa) and a′ |

AaB.

Exercise 30.4If I is an infinite sequence of indiscernibles over A, I is indiscernible over somemodel which contains A.

Exercise 30.5Use the Erdös-Rado-Theorem C10.2.

Exercise 30.6Let R be bounded and A–invariant and a0, a1, . . . indiscernible over A. Showthat R(ai, aj) for all i < j.

Exercise 30.7This follows from Exercise 30.6.c

Exercise 32.1Consider a Morley sequence of a global coheir extension of tp(a/M) = tp(b/M)over M


Exercise 32.2We may assume that ϕ has Morley degree 1. Then the MR(ψ ∧ ¬ψ) = β issmaller than the Morley rank of ϕ. Choose a family of M–definable formulasϕi of rank β which define disjoint subsets of ϕ(C). The for some i the rank ofψi ∧ ¬ψ must be smaller than β. ψ ∧ ψ is realised in M by induction on therank of ϕ.

The second part follows from the first.

Exercise 32.3If b′ is another realisation of q(y) � B and a′ realizes p(x) � Bb′ there is a B–automorphism α taking b′ to b. If p(x) is A–invariant, α(a′) realizes tp(a/Bb)and so p(x) ⊗ q(y) is well-defined. The same proof shows p(x) ⊗ q(y) to beA–invariant if p, q both are A-invariant.

Exercise 33.1If ϕ(x, y) has the order property witnessed by a0, a1, . . . and b0, b1, . . ., then thesequence a0b0, a1b1, . . . is ordered by the formula ϕ′(xy, x′y′) = ϕ(x, y′). Theconverse is obvious.

Exercise 33.3The formula xRy has the binary tree property.

Exercise 33.5If ϕ(x, y) has SOP, the formula ψ(x, y1, y2) = ϕ(y1, x) ∧ ¬ϕ(y2, x) has the treeproperty with respect to k = 2.

Exercise 33.6Hint: if T is unstable, there are a formula ϕ(x, y) and indiscernibles (aibi | i ∈ Q)with |= ϕ(ai, bj) ⇔ i < j. If ϕ does not have the independence property,there are finite disjoint subsets J,K of Q such that ΦJ,K(y) = {ϕ(ai, y) | i ∈J} ∪ {¬ϕ(ai, y) | i ∈ K} is inconsistent. Not all of J can be less than allelements of K. Choose J and K minimising the number of inversions F ={(j, k) ∈ J ×K | k < j}. Choose (j, k) ∈ F so that the interval (k, j) does notcontain any elements of J ∪K. Write J = J0 ∪ {j} and K = K0 ∪ {k}. ThenΦJ0∪{k},K0∪{j}(y) is consistent and the formula (with parameters)∧

ΦJ0,K0(y) ∧ ¬ϕ(x, y)

has the strict order property.

Exercise 33.7a) ⇒b): A type p ∈ S(A) is determined by the family of all pϕ, the ϕ–parts ofp.

Hence|S(A)| ≤

∏ϕ

|Sϕ(A)| ≤∏ϕ

|A| = |A||T |.

So if |A| = λ and λ|T | = λ, then |S(A)| ≤ λ.


b) ⇒ c): Clear.

c ⇒ a): Follows directly from 33.3, a ⇒b.

Exercise 33.8See the proof of 33.3, a)→ d). Let A =

{0, 12 , 1

}. We embed I into µA by

extending sequences from I0 to a sequence of length µ with the constant value 12 .

We order I by the ordering induced from the lexicographic ordering of µA.

Exercise 33.9Clearly, Sϕ(B) = S¬ϕ(B). If ϕ ∨ ψ has the order property witnessed byai, bj , i, j ∈ ω, use Ramsey’s Theorem to find an infinite subset J ≤ ω suchthat ai, bj , i, j ∈ J witness the order property for either ϕ of ψ. The sameargument works for ϕ ∧ ψ.

Exercise 33.101. Rϕ ψ = ∞ implies as in Theorem 24.6 that there is a binary tree of consistentformulas of the form ψ∧δ, δ ∈ Φ. Now we follow the proof of Theorem 16.5 andconclude first that over some countable A there are incountable many ϕ–typeswhich contain ψ. This implies then that ϕ(x, y) has the binary tree property.Again by the proof of 24.6 this implies that Rϕ ψ = ∞.

2. If β < Rϕ ψ there is a formula ϕ(x, a) such that ψ(x) ∧ ϕ(x, a) and ψ(x) ∧ϕ(x, a) have at least rank β. So if ω ≤ Rϕ ψ, there ϕ would have binary treesof arbitrary finite height and ϕ had the binary tree property.

Exercise 33.11Assume that the tree property of ϕ is witnessed by the parameters A = {as |s ∈ <ωω}. If ϕ has the tree property with respect to k = 2, it is easy to seethat ϕ has the binary tree property.

For the general case we make use of Exercise 33.10: if ϕ is stable, all ϕ–ranks are less than ∞. It follows that there is a sequence σ ∈ ωω such thatthe ϕ–ranks of the formulas

∧1≤i<n ϕ(x, aσ�i) are strictly decreasing, which is

impossible.

Exercise 34.1Let T the theory of all equivalence relations with three, infinite classes. Thereis only one 1–type p over the empty set. p has no good definition.

Exercise 34.2Six of the eight possible cases are realised by 1–types. For the cases (¬D, C, I,¬H), (¬D, ¬C, I, H) use 2–types and for the case (¬D, ¬C, I, ¬H) a 3–type.

Exercise 34.4Consider the Boolean algebra of all M–definable subsets of ϕ(M)n and thesubalgebra of ϕ(M)–definable subsets. The two algebras coincide if and only ifthey have the same Stone spaces (see p. 61). For the second part note that – ifϕ(C) has a least two elements – then for every ψ(x, y) there is a formula χ(x, z)


such that every class ψ(x, b) which is a subclass of ϕ(C)n has the form χ(C, c)for some c ∈ ϕ(C).

Exercise 34.51. Prove that formulas with Morley rank are stable. The proof that totallytranscendental theories are stable on page 153 is similar.

2. It is enough to find dp xϕ(x, y) for stable ϕ(x, y). Use the proof of Theo-rem 34.1.

Exercise 34.6The proof follows the pattern of the proof of the Erdös–Makkai–Theorem, 9.1.Assume that B∗ = {b ∈ B | ϕ(x, b) ∈ p} is not a positive Boolean combinationof sets of the form {b ∈ B : |= ϕ(c, b)}, c ∈ C. Construct three sequences(b′i | i < n) in B∗, (b′′i | i < n) in B \ B∗ and (ci | i < n) in C such that for alli < n

|= ϕ(ci, b′n) ⇒ |= ϕ(ci, b

′′n)

and for all i ≤ n|= ϕ(cn, b

′i) and |= ¬ϕ(cn, b′′i ).

Exercise 34.7Hint: If q is a weak heir of p, then Dϕ(q) = Dϕ(p) where Dϕ(p) is defined as theminimum of Dϕ(θ) for θ ∈ p. The argument in Theorem 34.1 now shows that qis definable over M .

Exercise 34.8Let p be the global extension of tp(a/M) which is definable over M . ByLemma 32.5 tp(a/Mb) is an heir of tp(a/M) if and only if tp(a/Mb) ⊂ q,i.e. if and only if ϕ(x, b) ∈ tp(a/Mb) ⇔ |= dp xϕ(x, b). So if q is the global M–definable extension of tp(b/M), we have that tp(a/Mb) is an heir of tp(a/M)if and only if ϕ(x, b) ∈ tp(a/Mb) ⇔ |= dp xϕ(x, y) ∈ q(y). Lemma 34.4 impliesnow that tp(a/Mb) is a heir of tp(a/M) if and only if tp(b/Ma) is an heir oftp(b/M).

Exercise 35.1D is definable from some some tuple d ∈ D. Each such d is a canonical parameterof D.

Exercise 35.2Let e be an imaginary and A the smallest algebraically closed set in the homesort over which e is definable. Then e is definable from a finite tuple a ∈ A.Since every automorphism which fixes e leaves A invariant, all elements of Aare algebraic over e.

For the converse let a ∈ acl(e) be a real tuple over which e is definable. ThenA = acl(a) is the smallest algebraically closed set over which e is definable.


Exercise 35.3T∞ and TDLO have the following property: If A, B are finite sets and if thetupels a and b have the same type over A ∩ B, then there is a sequence a =a0, b0, . . . , an, bn = b such that ai and bi have the same type over A and bi andai+1 have the same type over B. This implies that for every definable class Dthere is a smallest set over which D is definable.

T∞ does not eliminate imaginaries since no finite set with at least two ele-ments has a canonical parameter.

Exercise 35.4Let q1 and q2 be extensions of p to B. Choose realisations a1, a2 of q1, q2,respectively. There is an α ∈ Aut(C/A) taking a1 to a2. Since α(B) = B, wehave α(q1) = q2.

Exercise 35.5Let p ∈ S(A) be algebraic. If p is realised by b ∈ dcl(B) dp xϕ(x, y) = ϕ(b, y)is a good definition of p over B. Conversely, if q is an extension of p to M ,then q is realised by some b in M and dq x(x

.= y) defines the set {b}. So if q is

definable over B, then b ∈ dcl(B).

Exercise 35.6Let d ba a canonical parameter of D = ϕ(C, d). If d′ has the same type as d, wehave

ϕ(C, d′) = D ⇒ d′ = d.

By compactness this is true for all d′ which satisy some ψ(x) ∈ tp(d). Considerthe L ∪ {P}–formula θ(x, P ) = ψ(x) ∧ ∀y (ϕ(y, x) ↔ P (y)).

Exercise 35.71. Let ϕ(x, a) ∈ p have the same Morley rank as p and be of degree 1. Thenpdx ϕ(x, y)q is a canonical base of p.

2. Use 1. and Exercise 35.1.

Exercise 35.9If stp(a/A) 6= stp(b/A) there is an acl(A)-definable class D = ϕ(x, a) withϕ(a, a) and 6= ϕ(b, a). By Lemma 35.3, D is the union of equivalence classes ofan A-definable finite equivalence relation Ea, proving the claim.

Exercise 36.1Let P be the set of all strong types over A which are consistent with p, and Qbe the set of all strong types which are consistent with q. P and Q are closedsubsets of S(acleq(A)) and disjoint since strong types are stationary. So theycan be separated by a formula ϕ(x) over acleq(A). By Lemma 35.3 ϕ(C) is aunion of classes of a finite A–definable equivalence relation E(x, y).

Exercise 36.3Use the second part of Exercise 34.5. The first claim can now be proved likeTheorem 36.1.


The second claim is proved like Corollary 36.3, but we must be more carefuland use the ϕ–rank introduced in Exercise 33.10. We call the the minimal ϕ–rank of a formula in a type p the ϕ–rank of p. Let A = acleq(A), and p ∈ S(A)a stable type and p′ and p′′ two non–forking global extensions. p′ and p′′ aredefinable over A. We want to show that ϕ(x, b) ∈ p′ ⇔ ϕ(x, b) ∈ p′′. Wemay assume that ϕ(x, y) is stable (containing parameters from A). Let q(y) bea global extension of tp(b/A) which has the same ϕ∼–rank, where ϕ(x, y)∼ =ϕ(y, x). Since there are only finitely many possibilites for the ϕ–part of q, theϕ–part is definable over A. Now the claim follows from an adapted version of34.4.

We have still to show that p has a non–forking global extension, i.e. an exten-sion which is definable over A. Choose for every stable ϕ(x, y) a global extensionof p whith the same ϕ–rank. By the above the ϕ–part pϕ of this extension isdefinable over A and as such is uniquely determined. It remains to show thatthe union of all pϕ is consistent. Consider a finite sequence ϕ1(x, y), . . . , ϕn(x, y)of stable formulas. Choose a stable formula ϕ(x, y, z) such that every instanceϕi(x, b) has the form ϕ(x, b, c) for some choice of c. Then pϕ contains all pϕi fori = 1, . . . , n.

Exercise 36.41. Argue as in the second part of the proof of Theorem 36.10. Replace p @ qby MR(p) = MR(q).

2. By the first part of Exercise 34.5 p is stable Let q be global extension of p.If MR(p) = MR(q), then q has only finitely many conjugates over A. Since q isdefinable, this implies that q is definable over acleq(A). So q does not fork over Aby Exercise 36.3. Now assume that q does not fork over A. Using Exercise 24.8we see that we can assume that A = acleq(A). Let q′ be an extension of p withthe same Morley rank. Then q′ does not fork over A. So be Exercise 36.3 q = q′.

Exercise 36.5Choose A1 ⊂ A of cardinality at most |T | over which p does not fork. Let (pi)be the non–forking extensions of p � A1 to A. For each L–formula ϕ(y, y) thereare only finitely many different piϕ. Hence there is a finite subset Aϕ of A suchthat for all i

(pi � Aϕ)ϕ = (p � Aϕ)ϕ =⇒ piϕ = pϕ.

Now put A0 = A1 ∪⋃ϕAϕ.

If p has Morley rank, choose ϕ ∈ p having the same Morley rank and degreeas p. Any set A0 containing the parameters of ϕ does the job.

Exercise 36.61: Easy.

2: Exercise 36.5 shows that it is enough to consider types p over a countable setA. The multiplicity is the number of extensions of p to acl(A). These extensions


form a separable compact space. By Exercise 35.4, either all or none of them areisolated. In the first case the space is finite, in the other case it has cardinality2ℵ0 .

3: Use Exercise 36.4.

Exercise 36.71) a and a · b are interalgebraic over b. This implies that MR(a) = MR(a/b) =MR(a · b/b) ≤ MR(a · b). If MR(a) = MR(a · b), we have MR(a · b/b) = MR(a · b)and a · b and b are independent.

2) Let b be independent from a. If a is generic, MR(a · b) cannot be bigger thanMR(a), so a · b and b are independent by part 1). For the converse we chooseb generic. Part 1) (with sides reversed) implies that a · b is also generic. If a · band b are independent, it follows that MR(a) = MR(a · b) and a is generic.

Exercise 36.8For 1) notice that each line Ai consists of two elements and their product.

For 2) note that by Exercise 36.7, if a, b, c are independent generics, then a, b, b·care again independent generics. If one applies this rule repeatedly starting witha1, a2, a3, one obtains every non–collinear triple of our diagram.

Exercise 37.1It follows from Remark 27.3 and Symmetry that a type is algebraic if and only ifit has no forking extensions. A type has SU–rank 1 if and only if if the algebraicand the forking extensions coincide. So a type is minimial if and only if it hasSU–rank 1 and has only one non–forking extension to every set of parameters.

Exercise 37.2Use Exercise 27.4.

Exercise 37.3This follows from Exercise 27.1.

Exercise 37.4The first claim follows from Lemma 28.4(2). and the remark thereafter. Thesecond claim is easily proved using the Diamond Lemma (Exercise 28.2) andExercise 27.6.

Exercise 37.5Totally transcendental theories are superstable by Corollary 36.11. It followsalso that the multiplicity of a type over arbitrary sets is finite, namely equal toits Morley degree. If T is superstable, one can compute an upper bound for thenumber of types over a set A of cardinality κ as in the proof of Theorem 37.5 (2).If T is small, there are only countably many types over a finite set E. If we knowalso that all p ∈ S(E) have finite multiplicity, we have |S(A)| ≤ κ · ℵ0 · ℵ0 = κ.

Exercise 37.7


We can assume that all Ei are 0–definable. Choose a sequence a0, a1, . . . suchthat |= Ei(ai, ai+1) and pai/Eiq is not algebraic over a0 . . . ai−1. Let b be anelement in the intersection of all ai/Ei, A = {a0, a1, . . .} and p = tp(b/A). Thenfor all i we have b 6 |

a0,...,ai−1pai/Eiq by Remark 27.3. This shows that p forks

over each finite subset of A.

Exercise 37.8Define Ei(x, y) as xy−1 ∈ Gi. For any imperfect field K of finite characteristikp set Gi = Kpi . x− y ∈ Kpi+1

.

Exercise 37.9Half of the claim follows from Remark 24.7 and Exercise 37.8. Assume that Mhas the dcc on pp-definable subgroups. Then for every element a and every set Bof parameters the positive type tp+(a/B) contains a smallest element ϕ0(x, b).So there are at most max(|T |, |A|) many types over A. This shows that M � R0

is ω–stable for every countable subring, so M is totally transcendental (see Ex-ercise 16.4). Now assume that there is no infinite sequence of pp-subgroupswith infinite index in each other. Then tp+(a/B) contains a formula ϕ0(x, b0)such that tp+(a/B) is axiomatised by formulas ϕ(x, b) where ϕ(M, 0) is a sub-group of finite index in ϕ0(M, 0). There are max(|T |, |A|) many possibilities forϕ0(x, b0) and for each ϕ(x, y) finitely many possibilites. So the number of typesover A is bounded by max(2|T |, |A|) and M must be superstable. Indeed, theproof of Theorem 37.5.3 shows that otherwise for every κ there would be a setA of cardinality κ with |S(A)| ≥ κℵ0 .

Exercise 38.1Let p ∈ S(A) and q a non–forking extension to B. Let I be a Morley sequenceof q so I is independent over B. Since every element of I is independent fromB over A, I is independent over A as well (cp. Exercise 28.6), hence a Morleysequence of p.

Exercise 38.2a): I \ I0 is independent from B over A, hence independent over B. Theelements of I realise the non–forking extension of p to B.

b): Let B ⊇ A and q the non–forking extension of p. We extend I to a verylong sequence I ′ indiscernible over A. Then I ′ is still a Morley sequence of p.So if we choose I0 ⊂ I ′ with |I0| ≤ |T |+ |B| and B |

AI0I ′, then I ′ \ I0 is an

infinite Morley sequence of q having the same average type as I, so q ⊂ Av(I).

Exercise 38.3If I0 and I1 are parallel, hence I0J and I1J indiscernible, then

Av(I0) = Av(I0J ) = Av(I1J ) = Av(I1).

If conversely p = Av(I0) = Av(I1), note that by the proof of Theorem 38.2there are sets B0 and B1 over which p does not fork and such that I0 and I1are Morley sequences of the stationary types p � B0 and p � B1. Let J be a


Morley sequence of p � B0I0B1I1. Then I0J and I1J are Morley sequences ofp � B0 and p � B1, respectively.

Exercise 38.4Let p and q be stationary types with infinite Morley sequences I and J . Thenthe average types Av(I) and Av(J ) are the global non–forking extensions of pand q, respectively. Now p and q are parallel if and only if Av I = AvJ , i.e., ifand only if I and J are parallel.

Exercise 39.1Let p1,. . . ,pn be the extensions of p to acleq(A) and let Ii be the elements of Iwhich realise pi.

Exercise 41.1If d is in dcleq(F) the set {d } is definable over F. The proof of Theorem 35.2shows that {d} has a canonical parameter in Feq.

Exercise 41.3a→ b was implicitly used in 41.4.

b→ a: this is the proof of 34.3.

b↔ c: this is Exercise 41.2.

b→ d: same as the proof of 41.4

d→ a: Assume that ϕ(a,F) is not definable with parameters from F. Let ai bean enumeration of C. Construct a sequence ϕi of partial automorphisms of F sothat the domain of ϕi contains some f with |= ϕ(a, f) ⇔|= ¬ϕ(ai, ϕi(f)).

Exercise 42.1Induction on MRA.

Exercise 44.4This follows because any path (x1, a, x2) is strong in Mµ.

AppendixExercise 6.1Let L be an elementary extension ofK. Show first that if S is an integral domainwhich contains K, then L ⊗K S is again an integral domain. So L ⊗K Kalg isan integral domain.

Exercise 8.11. Symmetry is clear from the definition. For the other properties show firstthat a finite set A is independent from B over C iff dim(A/BC) = dim(A/C).This implies Monotony and Transitivity. Finite Character and LocalCharacter follow from the fact that for finite A and any D there is a finiteD0 ⊂ D such that dim(A/D) = dim(A/D0).


2. Symmetry, Finite Character and Weak Monotony are clear. If Ais finite and D is an set, let D0 be a basis of cl(A) ∩ D. D0 is finite andA |0

D0D. This shows Local Character. A |cl

CB always implies A |0

CB.

If |0 satisfies Monotony, the converse is true: We may assume that B =

{b1, . . . , bn} is finite. Then A |0CB implies A |0

Cb1...bibi+1 for all i. But this

is the same as A |clCb1...bi

bi+1, from which follows that A |clCB.

Exercise 8.3Choose a, b, x, c ∈ K p–independent. Set F0 = Kp(c), F1 = Kp(c, a, b) andF2 = Kp(c, x, ax+b). Then F0 has p–dimension 1, F1 and F2 have p–dimension 3and F1F2 has p–dimension 4. To show that F0 = F1∩F2, prove that dimF0 F1 =dimF0 F2 = p2 and dimF0(F1 + F2) = 2p2 − 1.

Exercise 8.5First show that in any pregeometry for any closed A ⊂ B: If dim(A/B) is finite,then it is the longest length n of a proper chain B = C0 ⊂ C1 ⊂ . . . ⊂ Cn = Bof closed sets Ci.

Bibliography

[1] J. T. Baldwin and A. H. Lachlan. On strongly minimal sets. J. SymbolicLogic, 36:79–96, 1971.

[2] John T. Baldwin. αT is finite for ℵ1–categorical T . Trans. Amer. Math.Soc., 181:37–51, 1973.

[3] John T. Baldwin. Fundamentals of Stability Theory. Perspectives in Math-ematical Logic. Springer Verlag; Berlin, Heidelberg, New York, London,Paris, Tokyo, 1988.

[4] John T. Baldwin. An almost strongly minimal non-Desarguesian projectiveplane. Trans. Amer. Math. Soc., 342(2):695–711, 1994.

[5] A. Baudisch, A. Martin-Pizarro, and M. Ziegler. Red fields. J. SymbolicLogic, 72(1):207–225, 2007.

[6] Andreas Baudisch. A new uncountably categorical group. Trans. Amer.Math. Soc., 348(10):3889–3940, 1996.

[7] Andreas Baudisch, Martin Hils, Amador Martin Pizarro, and Frank O.Wagner. Die böse farbe. Journal de l’Institut de Mathèmatiques de Jussieu.

[8] Paul Bernays. Axiomatic set theory. Dover Publications Inc., New York,1991. With a historical introduction by Abraham A. Fraenkel, Reprint ofthe 1968 edition.

[9] N. Bourbaki XI Algèbre, chapitre 5. Corps commutatifs. Hermann, Paris,1959.

[10] Elisabeth Bouscaren. The group configuration—after E. Hrushovski. InThe model theory of groups (Notre Dame, IN, 1985–1987), volume 11 ofNotre Dame Math. Lectures, pages 199–209. Univ. Notre Dame Press, NotreDame, IN, 1989.

[11] Steven Buechler. Essential stability theory. Perspectives in MathematicalLogic. Springer-Verlag, Berlin, 1996.

[12] Enrique Casanovas. Simplicity.

258

BIBLIOGRAPHY 259

[13] Zoé Chatzidakis and Ehud Hrushovski. Model theory of difference fields.Trans. Amer. Math. Soc., 351(8):2997–3071, 1999.

[14] M. M. Erimbetov. Complete theories with 1-cardinal formulas. Algebra iLogika, 14(3):245–257, 368, 1975.

[15] Ju. L. Eršov. Fields with a solvable theory. Dokl. Akad. Nauk SSSR,174:19–20, 1967. English transl., Soviet Math. Dokl., 8:575-576, 1967.

[16] Ulrich Felgner. Comparison of the axioms of local and universal choice.Fund. Math., 71(1):43–62. (errata insert), 1971.

[17] Steven Givant and Paul Halmos. Introduction to Boolean algebras. Under-graduate Texts in Mathematics. Springer, New York, 2009.

[18] Victor Harnik. On the existence of saturated models of stable theories.Proc. Amer. Math. Soc., 52:361–367, 1975.

[19] Wilfrid Hodges. Model theory, volume 42 of Encyclopedia of Mathematicsand its Applications. Cambridge University Press, Cambridge, 1993.

[20] Wilfried Hodges. Model Theory. Encyclopedia of Mathematics and itsApplications. Cambridge University Press, 1993.

[21] Wilfried Hodges. A Shorter Model Theory. Cambridge University Press,1997.

[22] Ehud Hrushovski. A stable ℵ0-categorical pseudoplane. Preprint, 1988.

[23] Ehud Hrushovski. Unidimensional theories are superstable. Ann. PureAppl. Logic, 50:117–138, 1990.

[24] Ehud Hrushovski. A new strongly minimal set. Ann. Pure Appl. Logic,62(2):147–166, 1993. Stability in model theory, III (Trento, 1991).

[25] Ehud Hrushovski. A non-PAC field whose maximal purely inseparableextension is PAC. Israel J. Math., 85(1-3):199–202, 1994.

[26] Ehud Hrushovski and Boris Zilber. Zariski geometries. J. Amer. Math.Soc., 9(1):1–56, 1996.

[27] Thomas Jech. Set theory. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. The third millennium edition, revised and expanded.

[28] Klaus Kaiser. Über eine Verallgemeinerung der Robinsonschen Modellver-vollständigung. Z. Math. Logik Grundlag. Math., 15, 1969.

[29] Akihiro Kanamori. The higher infinite. Springer Monographs in Mathe-matics. Springer-Verlag, Berlin, second edition, 2003. Large cardinals inset theory from their beginnings.

BIBLIOGRAPHY 260

[30] Byunghan Kim and Anand Pillay. From stability to simplicity. Bull. Sym-bolic Logic, 4(1):17–36, 1998.

[31] Serge Lang. Algebra. Addison–Wesley Publishing Company, second edition,1984.

[32] Serge Lang and André Weil. Number of points of varieties in finite fields.Amer. J. Math., 76:819–827, 1954.

[33] Daniel Lascar. Stability in Model Theory. Longman, New York, 1987.

[34] Angus Macintyre. On ω1-categorical theories of fields. Fund. Math.,71(1):1–25. (errata insert), 1971.

[35] David Marker. Model theory, volume 217 of Graduate Texts in Mathematics.Springer-Verlag, New York, 2002. An introduction.

[36] M. Morley. Categoricity in power. Trans. Amer. Math. Soc., 114:514–538,1965.

[37] Anand Pillay. An introduction to stability theory, volume 8 of Oxford LogicGuides. The Clarendon Press Oxford University Press, New York, 1983.

[38] Anand Pillay. The geometry of forking and groups of finite Morley rank.J. Symbolic Logic, 1995.

[39] Anand Pillay. Geometric stability theory, volume 32 of Oxford Logic Guides.The Clarendon Press Oxford University Press, New York, 1996. OxfordScience Publications.

[40] Bruno Poizat. Cours de Théorie des Modèles. Nur Al-Mantiq Wal-Ma’rifah,Villeurbanne, 1985.

[41] Bruno Poizat. Groupes Stables. Nur Al-Mantiq Wal-Mari’fah, Villeurbanne,1987.

[42] Mike Prest. Model theory and modules, volume 130 of London MathematicalSociety Lecture Note Series. Cambridge University Press, Cambridge, 1988.

[43] V. A. Puninskaya. Vaught’s conjecture. J. Math. Sci. (New York),109(3):1649–1668, 2002. Algebra, 16.

[44] Gerald E. Sacks. Saturated model theory. W. A. Benjamin, Inc., Reading,Mass., 1972. Mathematics Lecture Note Series.

[45] Igor R. Shafarevich. Basic algebraic geometry. 1. Springer-Verlag, Berlin,second edition, 1994. Varieties in projective space, Translated from the1988 Russian edition and with notes by Miles Reid.

[46] Saharon Shelah. Uniqueness and characterization of prime models over setsfor totally transcendental first-order-theories. J. Symbolic Logic, 37:107–113, 1972.

BIBLIOGRAPHY 261

[47] Saharon Shelah. Classification Theory. North Holland, Amsterdam, 1978.

[48] Saharon Shelah. On uniqueness of prime models. J. Symbolic Logic, 43:215–220, 1979.

[49] Saharon Shelah. Simple unstable theories. Ann. Math. Logic, 19(3):177–203, 1980.

[50] Joseph R. Shoenfield. Mathematical logic. Association for Symbolic Logic,Urbana, IL, 2001. Reprint of the 1973 second printing.

[51] K. Tent. Very homogeneous generalized n-gons of finite Morley rank. J.London Math. Soc. (2), 62(1):1–15, 2000.

[52] Jouko Väänänen. Barwise: abstract model theory and generalized quanti-fiers. Bull. Symbolic Logic, 10(1):37–53, 2004.

[53] Frank Wagner. Simple Theories. Kluwer Adacemic Publishers, Dordrecht,2000.

[54] Frank O. Wagner. Stable groups, volume 240 of London Mathematical So-ciety Lecture Note Series. Cambridge University Press, Cambridge, 1997.

[55] John S. Wilson. Profinite groups, volume 19 of London Mathematical So-ciety Monographs. New Series. The Clarendon Press Oxford UniversityPress, New York, 1998.

[56] Martin Ziegler. Model theory of modules. Ann. Pure Appl. Logic,26(2):149–213, 1984.

[57] Boris Zilber. Analytic and pseudo-analytic structures. In Logic Colloquium2000, volume 19 of Lect. Notes Log., pages 392–408. Assoc. Symbol. Logic,Urbana, IL, 2005.

Index

0, 2130–definable set, 161, 2132, 213α–minimal formula, 114α–strongly minimal formula, 112Av(I), 179∧i<m, 17

∼α, 112∀, 14∧, 14(A, R), 9(A, a1, . . . , an), 9A ⇒∆ B, 37A ⇒∃ B, 38A ∼= B, 7A ≺ B, 26A � K, 9A ⊂ B, 8AB , 10∀∃–formula, 39|A|, 7a |

BC, 167

A ≺1 B, 45Abs(L), 145A |

CB, 131

acl(A), 94acleq(A), 160A ≡ B, 23ℵ0, 211ℵα, 214α+ 1, 214α+ n, 214A |= ϕ[~b], 15[A]n, 77Aut(A), 8Aut(A/B), 10

~bax , 15B–definable set, 16iα, 214Cb(a/B), 196Cb(p), 168cf, 215clS , 233clS , 233Diag(A), 18dim(N/M), 118dimϕ(M), 100pDq, 159pϕq, 159dcl(A), 108dcleq(A), 160deg(p), 94dimϕ(N/M), 117dimS(X/S), 234dim(X), 234dp xϕ(x, y), 149D(p, ϕ, k), 176.=, 14∃, 14↔, 14∃∀–formula, 39∃>n, 92EM(I/A), 77〈∅〉B, 8∃∞, 92ϕ(π), 19ϕ(t1, . . . , tn), 17ϕ(x1, . . . , xn), 16f : A →∆ B, 37[ϕ], 60ϕ(A), 16Feq, 189, 190ϕ–dimension, 100

262

INDEX 263

over parameters, 100f(p), 61ϕ–type, 152G(K), 229G(k), 229h : A

≺−→ B, 25h : A

∼→ B, 7h : A → B, 7→, 14κ–categorical theory, 35κ ≤ λ, 210κ–homogeneous structure, 105κ+, 212κ–saturated, 82κ–stable, 81, 152, 154κ–universal structure, 105k–inconsistent Familie, 125K–saturated, 69LAG, 6Lalg, 95λν , 55L(B), 10L(C), 10Leq, 160LG, 6Lsets, 6LM(R), 49LN, 6L∅, 6LO, 6LOR, 6LR, 6Lsep, 55LSk, 79Lstp(a/A), 142M eq, 160MD(a/A), 114MD(p), 114MDϕ, 113MR(T ), 111MR(a/A), 114MR(p), 114MRϕ, 111MB , see ABMDα ϕ, 113

M eq, 160|=, 107mult(p), 171N(B/A), 168¬, 14ncA, 136, 141n–type, 31∨i<m, 17

∨, 14otp(X,<), 2130–dimensional space, 60ω, 213<ωA, 214ω–homogeneous, 65ω–stable, 81On, 213ω–saturated, 64p–basis, 225, 226p–closure, 225pϕ, 249ϕ–rank, 154, 253p � A, 61P(V ), 233Rϕ, 154〈S〉B, 8supi∈I αi, 214supi∈I κi, 211SA(B), 31SAn (B), 31S(B), 31Sn(B), 31Sϕ(B), 152S(T ), 60Sn(T ), 60stp(a/A), 165SU(p), 173T eq, 160t(t1, . . . , tn), 13t(x1, . . . , xn), 13tA[~b], 12tA[a1, . . . , an], 13|T |, 22T∀, 38TACF , 23, 51, 240TACF0 , 52

INDEX 264

TACFp , 52T∀∃, 40TAG, 21TDLO, 23, 240T -e.c, 45T eq, 160T ≡ S, 22Th(A), 22TKH, 46TF, 21TM(R), 49tp(a), 73tp(a), 31tp(a/B), 31tpA(a/B), 31T ` ϕ, 22tpqf(a), 72T ` S, 22TR, 21TRCF, 52TTree, 74TRG, 57⊥, 17>, 17SCFp,e, 54SCFp(c1, · · · , ce), 54SCFp,∞, 54TSk(L), 79T∞, 48⋃i∈I Ai, 9

U(p), 176v0, v1, . . ., 12yx, 211Z, 229non–forking extension

existence, 131, 139, 16717. Hilbert’s Problem, 53

abelian group, 21absolute Galois group, 229absolute part of a pseudo-finite field,

145absolutely prime, 225additivity of Morley rank, 123, 192affine subspace, 236Aleph function, 214

algebraicclosure, 94element, 94formula, 94type, 94

algebraically closed field, 35, 51, 52, 94,97, 150

almost orthogonal type, 191almost strongly minimal, 124, 205Amalgamation Property, 69analysable type, 191antichain, 136assignment, 12atom, 62atomic

diagram, 18extension, 87formula, 14structure, 73tuple, 73

automorphism, 8automorphism group, 8average type, 179axiom, 21

1–based formula, 195basic formula, 17basis, 233, 234Bernays-Gödel Set theory, 208Beth function, 214Beth’s Interpolation Theorem, 108BG, 111, 208binary tree, 75, 82binary tree property, 152binding group, 190binding strength, 15Boolean algebra, 61bounded relation, 144

canonicalbase, 159, 168, 196parameter, 159

Cantor-Bendixson rank, 116cardinal, 210, 213

arithmetic, 211Ramsey, 239

INDEX 265

regular, 215successor, 212weakly compact, 239weakly inaccessible, 215, 239

cardinality, 210cardinality of a structure, 7categoricity, 35chain

continuous, 32elementary, 27Lemma, 27of structures, 9

Charakterisierungssatz, 139class

elementary, 23, 47closed

sets of types, 62closedness of the set of non–forking ex-

tensions, 127closure

definable, 108club, 181cofinal, 215cofinality, 215commutative ring, 21Compactness Theorem, 28complete

formula, 62theory, 22

components of a formula, 112conjugacy, 167conjugate over, 107conjunction, 14conjunctive normal form, 20conservative extension, 62consistent

set of formulas, 21, 107theory, 21

constant, 6constant term, 14construction, 182continuous chain, 32Continuum Hypothesis, 212contructible set, 85countable, 211

theory, 58

cyclical order, 128

definablebijection, 97class, 107closure, 108set, 16type, 149

definable bijection, 97degree

of imperfection, 54, 234degree of a type, 94degree of imperfection, 225density, 74derivation, 220descending chain condition (dcc), 113diagram

atomic, 18elementary, 25

Diamond-lemma, 134differential field, 220differentially

closed field, 56differentially closed field, 122, 170dimension

ϕ–dimension, 100over parameters, 100

of a pregeometry, 234of extensions of differential fields,

122directed

family, 8elementary, 27

partial order, 8disjoint formulas, 75disjunction, 14disjunctive normal form, 20dividing

formula, 125sequence, 129set of formulas, 125

domain, 7of a type, 31

dual numbers, 220

Ehrenfeucht-Mostowski-type, 77

INDEX 266

elementaryclass, 23, 47, 160diagram, 25directed family, 27embedding, 25equivalence, 23extension, 26map, 25, 61property, 98substructure, 25

eliminationof ∃∞, 92of finite imaginaries, 163of imaginaries, 159, 161

weak, 163of quantifiers, 42

embedding, 7elementary, 25

equality symbol, 14equivalence, 14equivalent

formulas, 16, 38, 107modulo T , 38theories, 22

exchange property, 95existence of non–forking extensions, 131,

139, 167existential formula, 18

primitive, 42simple, 42

existential quantifier, 14existentially closed

structure, 45substructure, 45

expansion, 9extension, see field extension

atomic, 87conservative, 62elementary, 26forking, 131minimal, 87of a structure, 8of types, 61prime, 85

field, 21

algebraically closed, 35, 51, 52, 94,97, 150

differential, 220differentially closed, 56, 122, 170formally real, 2161–free, 230ordered, 216procyclic, 230pseudo algebraically closed, 228pseudo-finite, 230real closed, 217, 218separably closed, 54, 176

field extensiongalois, 224normal, 224regular, 224separable, 224, 225

filter, 20finite character of forking, 127, 139,

167finite equivalence relation, 159Finite equivalence relation theorem, 171finitely axiomatisable, 32, 33finitely generated substructure, 8finitely satisfiable

set of formulas, 31theory, 28

forkingextension, 131independence, 131multiplicity, 171set of formulas, 127symmetry, 133, 139, 167

formally real field, 216formula

1–based, 195α–strongly minimal, 112α–minimal over A, 114∀∃, 39∃∀, 39algebraic, 94atomic, 14basic, 17complete, 62dividing, 125existential, 18

INDEX 267

primitive, 42simple, 42

has Morley rank, 111isolating, 58locally modular, 195positive primitive, 49quantifier free, 17stable, 152, 158stably embedded, 155, 190strongly minimal, 97symmetric, 136thick, 136, 141universal, 18with the binary tree property, 152with the order property, 152with the tree property, 129

formulasdisjoint, 75equivalent, 16, 38, 107

Fraïssé limit, 701–free field, 230free occurrence, 15function symbol, 6Fundamental Theorem of Algebra, 218

galois field extension, 224GCH, 212generated

substructure, 8type, 242

generating set, 233generic, 124, 221

type, 206generic element, 172generic point, 228generic type, 206geometry, 233global type, 107good definition of a type, 166graph, 32

connected, 32random, 57, 66, 72, 140, 153

group configuration, 172Groupe de Liaison, 189groupoid, 188

connected, 188

definable, 188

heir, 148, 149Henkin–constants, 28Henkin–theory, 28Hereditary Property, 69Hilbert Basis Theorem, 81Hilbert’s 17. Problem, 53Hilbert’s Nullstellensatz, 52home sort, 160homomorphism, 7

imaginary elements, 160implication, 14independence

forking, 131geometric, 235

independence property, 153Independence Theorem, 138, 143independent

family, 134sequence, 131set, 134, 233

indiscernibles, 76parallel, 180total, 178

induced theory, 98induction, 213inductive theory, 39infinitesimal, 33initial segment, 210interdefinable, 108internal type, 187Interpolation Theorem, 108interpretation, 7invariant

class, 108, 109, 144relation, 67type, 128

isolatedset of formulas, 58

isolating formula, 58isomorphic over a set, 87isomorphic structures, 7isomorphism, 7

Joint Consistency, 108

INDEX 268

Joint Embedding Property, 69

Kaiser hull, 46

λ-functions, 55language, 6Lascar rank, 176Lascar strong type, 142limit ordinal, 214linearly disjoint, 222, 224local character of forking, 130, 139, 167locally finite, 67locally modular

pregeometry, 236formula, 195

logical symbols, 14

many–sorted structure, 10map

elementary, 25, 61matroid, 95meager set, 62minimal

prime extension, 87set, 97type, 99, 175

model, 17, 107consisting of constants, 29prime, 73

model companion, 45model complete theory, 44modular

lattice, 236pregeometry, 234theory, 195

module, 49monotony, 133, 139, 167

weak, 169monster model, 106–110Morley degree

of a formula, 113of a type, 114

Morley rankof T , 111of a formula, 111of a type, 114

Morley sequence, 131, 179in p, 131of q over A, 131

Morleyisation, 42multiplicity, 94, 171

negation normal form, 18negation symbol, 14normal field extension, 224normal form

conjunctive, 20disjunctive, 20negation, 18prenex, 20

normal subset, 164, 183nowhere dense set, 62Nullstellensatz, 52

omitting a set of formulas, 58Omitting Types, 58order property, 152order type, 213ordered field, 216ordinal, 213

limit, 214orthogonal type, 191

PAC field, 228parallel

indiscernibles, 180type, 168, 180

parameter set, 107parentheses, 14, 15partial type, 58perfect hull, 110, 225plane

projective, 199Poizat, Bruno, 189, 268positive primitive formula, 49pp–definable subgroup, 50pp–formula, 49predicate, 6pregeometry, 95

modular, 234prenex normal form, 20preservation theorems, 36

INDEX 269

primeabsolutely, 225extension, 85

minimal, 87model, 73structure, 42

primitive existential formula, 42procyclic

field, 230group, 229

profinite group, 228projective plane, 199projective space, 233property

independence, 153order, 152

pseudo algebraicallyclosed field, 228

pseudo-finite field, 230

quantifierexistential, 14universal, 14

quantifier elimination, 42quantifier free formula, 17

Ramsey cardinal, 239Ramsey’s Theorem, 77random graph, 57, 66, 72, 140, 153rank

SU, 173U, 176phi–rank, 154, 253Cantor-Bendixson, 116Lascar, 176Morley, 111

real closed field, 217, 218real closure, 217real elements, 160realisation of a set of formulas, 31realisation set, 16recursion theorem, 214regular action, 189regular cardinal, 215regular field extension, 224relation

bounded, 144relation symbol, 6relativisation of a pregeometry, 233restriction

of a pregeometry, 233of a structure, 9of a type, 61, 62

ring, 21Robinson’s

Joint Consistency Lemma, 108Test, 44

satisfaction, 15saturated structure, 64, 82, 104–106sentence, 17separable field extension, 224, 225separably closed field, 54, 176separating sentence, 36Separation Lemma, 36set

minimal, 97set of formulas

dividing, 125Shelah’s Lemma about indiscernibles,

132simple existential formula, 42simple theory, 129skeleton, 69Skolem function, 79Skolem theory, 79small theory, 66SOP, 154sort, 160

home, 160special structure, 104stability spectrum, 174stable

κ–stable, 81, 152, 154ω–stable, 81formula, 152, 158theory, 152type, 158

stably embedded formula, 155, 158, 190Standard Lemma on indiscernibles, 77,

125stationary type, 141, 166

INDEX 270

Stone duality, 61Stone space, 61strict order property, 154strong type, 165

Lascar strong type, 142strongly κ–homogeneous structure, 105strongly minimal

almost, 124, 205formula, 97theory, 97type, 97

structure, 7κ–homogeneous, 105κ–saturated, 82κ–universal, 105ω–saturated, 64atomic, 73existentially closed, 45many–sorted, 10, 160saturated, 82special, 104strongly κ–homogeneous, 105

sublanguage, 9substitution, 13

lemma, 13, 17substructure, 8

elementary, 25existentially closed, 45finitely generated, 8generated, 8

substructure complete theory, 44successor cardinal, 212supersimple theory, 173superstable theory, 173SU-rank, 173Sylvester, J.J., 218symmetric formula, 136symmetry of forking, 133, 139, 167

Tarski’sChain Lemma, 27Test, 26

term, 12constant, 14

TheoremBeth’s Interpolation, 108

Cantor’s, 211Cantor–Bernstein, 210of Cantor, 35of Löwenheim–Skolem, 34of Lachlan, 88of Morley, 103

downwards, 89of Ryll–Nardzewski, 64of Vaught, 66Ramsey’s, 77Ressayre’s, 185Vaught’s Two-cardinal Theorem, 90

theoremFinite equivalence relation, 171

theory, 21κ–stable, 81, 152, 154ω–stable, 81complete, 22consistent, 21countable, 58equivalent, 22finitely satisfiable, 28induced, 98inductive, 39κ–categorical, 35model complete, 44modular, 195of abelian groups, 21of commutative rings, 21of fields, 21simple, 129small, 66stable, 152substructure complete, 44supersimple, 173superstable, 173totally transcendental, 82unidimensional, 191universal, 38with prime extensions, 87

thick formula, 136, 141totally transcendental theory, 82trace, 219transcendence degree, 234transitivity, 133, 139, 167tree, 75

INDEX 271

tree property, 129trivial pregeometry, 235tuple

atomic, 73type, 31, 60

algebraic, 94almost orthogonal, 191analysable, 191based on, 168definable over C, 149generated, 242global, 107internal, 187invariant, 128Lascar strong type, 142minimal, 99, 175of a set, 32of an element, 31orthogonal, 191parallel, 168, 180partial, 58quantifier free, 72stable, 158stationary, 141, 166strong, 165strongly minimal, 97

type-definable class, 109

ultra-homogeneous, 69ultrafilter, 20unidimensional theory, 191unique decomposition, 12, 15uniqueness of non–forking extensions,

166universal

formula, 18quantifier, 14theory, 38

universe, 7U-rank, 176

valid sentence, 22vanishing ideal, 225variable, 12Vaught’s conjecture, 67Vaught’s Test, 34

Vaught’s Two-cardinal Theorem, 90Vaughtian pair, 90vector space, 49

weak elimination of imaginaries, 163weak monotony, 169weakly compact cardinal, 239weakly inaccessible cardinal, 215, 239well–ordering, 210

theorem, 210