Noether’s normalisation lemma

Petra Izeboud

July 18, 2014

Bachelorproject

Supervisor: Arie Peterson

Korteweg-de Vries Instituut voor Wiskunde

Faculteit der Natuurwetenschappen, Wiskunde en Informatica

Universiteit van Amsterdam

Abstract

Noether’s normalisation lemma proves that any finitely generated k-algebra A can beexpressed as a finite extension ring of a polynomial ring. We can link polynomial ringsto geometric objects using the one-to-one correspondence between prime ideals of thecoordinate ring A(Y ) of an affine variety Y and the closed irreducible subsets of Y . Thesame holds for maximal ideals of A(Y ) and points of Y . Noether’s normalisation lemmathen gets a geometric meaning: every affine variety is a finite cover of some affine space.Moreover for Z an algebraic set in An, we can obtain that the projection of Z to thefirst d coordinates is a finite morphims. To obtain this finite morphism, we only need alinear change of coordinates.

Titel: Noether’s normalisation lemmaAutor: Petra Izeboud, petra.izeboud@live.nl, 10200770Supervisor: Arie PetersonSecond grader: Prof. dr. Gerard van der GeerEnd date: July 18, 2014

Korteweg-de Vries Instituut voor WiskundeUniversiteit van AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.science.uva.nl/math

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Contents

Introduction 5

1 An introduction to commutative algebra 71.1 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Prime ideals, maximal ideal and Spectrum . . . . . . . . . . . . . 81.1.2 Zorn’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.3 Existence of prime ideals and maximal ideals . . . . . . . . . . . . 11

1.2 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.1 The homomorphism and isomorphism theorems . . . . . . . . . . 131.2.2 Generators of a module . . . . . . . . . . . . . . . . . . . . . . . . 141.2.3 Cayley-Hamilton theorem . . . . . . . . . . . . . . . . . . . . . . 151.2.4 The determinant trick . . . . . . . . . . . . . . . . . . . . . . . . 181.2.5 Nakayama’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.3 Noetherian rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.3.1 The ascending chain condition . . . . . . . . . . . . . . . . . . . . 201.3.2 The Noetherian condition . . . . . . . . . . . . . . . . . . . . . . 20

1.4 Finite ring extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.4.1 Finite and integral A-algebras . . . . . . . . . . . . . . . . . . . . 231.4.2 Normalisation of rings . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Noether’s normalisation lemma 262.1 Noether’s normalisation lemma . . . . . . . . . . . . . . . . . . . . . . . 262.2 Proof of Noether’s normalisation lemma . . . . . . . . . . . . . . . . . . 26

2.2.1 Main claim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2.2 Proof of Noether’s normalisation lemma . . . . . . . . . . . . . . 28

3 Further interpretations 303.1 Some basic algebraic geometry . . . . . . . . . . . . . . . . . . . . . . . . 30

3.1.1 Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.1.2 Hilbert’s Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . . 333.1.3 Affine varieties and more bijections . . . . . . . . . . . . . . . . . 353.1.4 The geometric interpretation of example 1.1.1 . . . . . . . . . . . 37

3.2 Maps between algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . 383.2.1 Regular functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.2.2 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.3 Geometric interpretation of Noether’s normalisation lemma . . . . . . . . 43

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Reflection 47

Populaire samenvatting 48

Bibliography 51

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Introduction

Amalie Emmy Noether (1882− 1935) was an influential German mathematician knownfor her groundbreaking contributions to abstract algebra and theoretical physics. DavidHilbert and Albert Einstein described her as the most important woman in the historyof mathematics. This is impressive, but considering that her first seven years work-ing for the university of Erlangen were unpaid and that the four years she worked forthe University of Gottingen she lectured under Hilbert’s name, this is an even biggerachievement.

Noether’s theorem, which explains the fundamental connection between symmetry andconservation laws, played a pivotal role in the development of modern physics. Anotherfield of mathematics in which her research was very important is commutative algebra.In her paper Idealtheorie in Ringbereichen (Theory of Ideals in Ring Domains, 1921)she helped to develop is the theory of ideals in commutative rings which now play amajor role in every algebra course.

Algebraic geometry is the branch of mathematics that classically studies zeroes ofpolynomial equations. Modern algebraic geometry is based on more abstract techniques,mostly from commutative algebra, but the underlying idea is the same as for classic al-gebra. The correspondence between algebra and geometry was already clear in the sev-enteenth century when Descartes introduced the Cartesian coordinate system. Throughthe centuries, algebra and geometry became more and more intertwined: algebraic the-orems have geometrical interpretations and vice versa. Moreover, the tools needed forstudying these fields can hardly be separated anymore.

Besides their intrinsic value and beauty, recent developments have revealed new rea-sons for studying these fields of mathematics as unexpected applications were found inin string-theory and coding theory.

In this thesis we are primarily interested in one theorem; Noether’s normalisationlemma. This is a result in the commutative algebra. Furthermore, we want to build oneof the bridges between commutative algebra and algebraic geometry. The basic idea ofthe bridge is that it is often possible to view a ring A as a certain ring of functions ona space X, and to recover such geometric objects as the set of maximal or prime idealsof A. With this two-way traffic between the different worlds Noether’s normalisationlemma can be translated into a result in algebraic geometry.

In this paper first we discuss underlying results and definitions of commutative algebraand algebraic geometry, before linking the two fields. In chapter 2 we shall define and

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expand on notions from commutative algebra such as modules, Noetherian rings andnormalisations. In chapter 3 we shall see a detailed proof of Noether’s normalisationlemma.

The geometric notions we will develop in chapter 4, are the correspondence betweenideals and (subsets of) affine varieties, their (Zariski) topology, regular functions, mor-phisms and last but not least the geometric interpretation of Noether’s lemma.

About the literature; a lot has been written on these subjects. Of main use for thisthesis were the texts Undergraduate commutative algebra from Miles Reid, Introductionto Algebraic Geometry by Ben Moonen and Syllabus Algebraısche meetkunde by Gerardvan der Geer. For further reading on the subjects, I would refer to Introduction to Com-mutative Algebra by Atiyah and MacDonald, which gives a comprehensive descriptionof most of the notions discussed in this thesis. For further development of the notions aspresented in chapter 4, Algebraic Geometry by Hartshorne is supposed to be the best.

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1 An introduction to commutativealgebra

The aim of this chapter, is to develop concepts needed for a full understanding ofNoether’s normalisation lemma and its proof. To gain an understanding of these con-cepts we will first recall and develop the concepts of ideals, then talk about modulesafter which we get to Noetherian rings and finite ring extensions. The reason to dis-cuss Noetherian rings is that their properties have deep impacts on the structure ofideals and their inclusions. Their most well-known application would probably be theKrull-dimension [3, p.56].

1.1 Ideals

Ideals were first used by Richard Dedekind in the third edition of his book Vorlesungenuber Zahlentheorie in 1876. Later, the concept was greatly expanded on by David Hilbertand Emmy Noether.

We assume that the reader already knows the defintion of a ring and a ring homo-morphism. Throughout this thesis, a ring A is assumed to be commutative and to havean identity element. Most of the definitions and proofs in this chapter come from [?]. Ifnot, this will be clearly stated.

We start with the definition of an ideal;

Definition 1.1 (Ideal). An ideal of a ring R is a subset I ⊂ R such that

i) I is a subgroup of R,

ii) for all x ∈ I and all r ∈ R we have xr ∈ I.

From now on (f1, . . . , fn) and Af1 + . . .+Afn will denote the ideal generated by theelements f1, . . . , fn ∈ A. Recall that an ideal is generated by the elements f1, . . . , fn ifit is the smallest ideal containing these elements. The motivation for the definition ofan ideal comes from the following result and the proof is assumed known.

Theorem 1.2. i) The kernel of a ring homomorphism ϕ : A→ B is an ideal.

ii) If I ⊂ A is an ideal then there exists a ring A/I and a surjective homomorphismϕ : A → A/I such that ker ϕ = I. The map ϕ on A/I is uniquely defined up toisomorphism and is called the quotient homomorphism.

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iii) The map ϕ as defined above induces a map{ideals of A/I

}→{

ideals of A containing I},

I0 → ϕ−1(I0)

which gives a one-to-one correspondence.

Before we continue to more special types of ideals, we recall some more general notionsof rings. For a homomorphism f : K → L where K,L are fields, the image f(K) is asubfield of L. Via the isomorphism f : K → f(K) we view L as a field extension ofK. Such a field extension is called algebraic if for all elements α ∈ L there exists apolynomial f ∈ K[x] such that f(α) = 0 [5, p.47].

For the reader’s convienience we recall here these two definitions.

Definition 1.3 (Unique factorization domain). An integral domain A is called a uniquefactorization domain (UFD) if every k 6= 0 can be written as a product of irreducibleelements, uniquely up to order and units.

Definition 1.4 (Principal ideal domain). An integral domain A is called a principalideal domain (PID) if every ideal is principal, i.e., can is by a single element of A.

1.1.1 Prime ideals, maximal ideal and Spectrum

We call a set S ⊂ A a multiplicative set if 1 ∈ S and

f, g ∈ S =⇒ fg ∈ S.

With this notion of a multiplicative set, we define prime ideals.

Definition 1.5 (Prime ideals). An ideal P ⊂ A is prime if its complement A\P is amultiplicative set.

So for P prime we have 1 ∈ A \P which implies P 6= A. Furthermore we have thatfg ∈ P implies f ∈ P or g ∈ P since if f, g /∈ P , then f, g ∈ A/P which is multiplicativeset so fg ∈ A/P . If (0) is a prime ideal, A is called an integral domain or simply adomain.

Definition 1.6 (Maximal ideal). An ideal m ⊂ A is maximal if m 6= A, and if there isno ideal I strictly between m and A. In other words, for I an ideal and m a maximalideal we find

m ⊂ I ⊂ A =⇒ m = I or I = A.

Note that every maximal ideal is a prime ideal. Furthermore, we call a ring local if ithas a unique maximal ideal m.

We obtain the following results:

Proposition 1.7. i) P is prime ⇐⇒ A/P is an integral domain;

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ii) m is maximal ⇐⇒ A/m is a field.

In chapter 4 we will find that there is a one-to-one correspondence between irreduciblesubsets of affine spaces and prime ideals of k[x1, . . . , xn], and even better, a bijectionbetween the spectrum of a coordinate ring A(Y ) and the closed irreducible subset ofY . The important thing for now is that the the set of all prime ideals gives the firststep to built a bridge between commutative algebra and algebraic geometry! The nextdefinition is therefore not suprising.

Definition 1.8 (Prime spectrum). The set of prime ideals of A is called the spectrumof A, that is,

SpecA ={P∣∣ P ⊂ A is a prime ideal

}.

The maximal spectrum m− SpecA is the set of maximal ideals of A.

Examples

i) The prime ideals of Z are as expected: SpecZ = (0) ∪{

(p)∣∣ p is prime

}.

ii) If k is a field and A = k[X] then a prime ideal of A is (0) or (f) for f an irreduciblepolynomial.

iii) We claim that the prime ideals of Z[Y ] are as follows:

Spec(Z) = 0 ∪ {(f) | for irreducible f ∈ Z[Y ]} ∪ {m |m a maximal ideal}.

Moreover, each maximal ideal can be written of the form m = (p, q) where p is aprime number and q ∈ Z[Y ] is a polynomial that is an irreducible element q ∈ Fp[Y ]mod p.

Proof. Recall that B = Z is a principal ideal domain, that K = Q is its field offractions and let A = B[Y ]. If a prime ideal P ⊂ B[Y ] is 0 or principal there isnothing to prove, so assume that P contains two elements f1, f2 with no commonfactor in B[Y ].

First we claim that f1, f2 also have no common factor in K[Y ]. The proof usesthe lemma of Gauss, which says that for a UFD, the product of two primitivepolynomials is again primitive. This means that the greatest common divisor of itscoefficients is 1.

Suppose that f1 = hg1 and that f2 = hg2 with h, g1, g2 ∈ K[Y ] and deg h ≥ 1.Now since K is a field, it is possible to reduce the expressions for h, g1, g2 to h =ah0, g1 = b1γ1, g2 = b2γ2, where a, b1, b2 ∈ K and h0, γ1, γ2 ∈ B such that theyhave no common primefactors. This is the same as saying that h0, γ1, γ2 ∈ B areprimitive elements of B[Y ], so that with the lemma of Gauss h0γ1 and h0γ2 areagain primitive. This gives us that f1 = hg1 = (ab1)(h0γ1) ∈ B[Y ] which implies

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that ab1 ∈ B, equivalently ab2 ∈ B. Therefore h0|f1 and h|f2, which gives thecontradiction.

Second, we claim that the ideal of A generated by f1 and f2 has a nonzero intersec-tion with B, that is, (f1, f2)∩B 6= 0. To see this, first note that K[Y ] is a principalideal domain, since every polynomial ring in one variable over a field is a PID. Thismeans that the greatest common divisor of f1, f2 is gcd(f1, f2) = 1, and thereforethere exist a, b ∈ K[Y ] such that af1 + bf2 = 1. If c is a common denominator ofall the coefficients of a and b, then c = (ca)f1 + (cb)f2 ∈ B and therefore indeed weget (f1, f2)∩B 6= 0. Note that so far we have only used the fact that B is a uniquefactorization domain.

If P is a prime ideal of A = B[Y ], B ∩ P is a prime ideal of B. To see this, notethat 1 ∈ S ′ = B − (B ∩ P ) and that for a, b ∈ S ′, also a, b ∈ S = A− P , and hencein S. This gives us that S is a multiplicative set so that B ∩ P is indeed prime inB. We have seen in the first two steps that if P is not principal then P ∩ B 6= 0.Now B is a PID, so that every nonzero prime ideal is a maximal ideal.

For P = (f1, f2) to be prime, it is clear we can rewrite this to P = (f1, f2) where f1

is a prime number and f2 ∈ Z[Y ] is a polynomial who modulo f1 is an irreducibleelement f2 ∈ Fp[Y ], so the statement follows.

iv) If we substitute B = k[X], K = k(X) and A = B[Y ] in the above proof we find that

Spec k[X, Y ] ={

0; (f) for irreducible f(X, Y ) ∈ k[X, Y ];m for maximal ideals m},

where we can write m = (p, q) with p = p(X) ∈ k[X] an irreducible polynomialin X and g ∈ k[X, Y ] whose reduction modulo p is an irreducible element g ∈(k[X]/(p)

)[Y ].

We can read through the above proof thinking of B ⊂ K as a general principal idealdomain in its ring of fractions. This approach is easier, more general and hence morebeautiful. There is also a geometric interpretation of example 1.1.1 iii) which we willdiscuss in 3.1.4.

1.1.2 Zorn’s lemma

Our overall aim is to discuss a ring A as a ring of functions on SpecA, so we must atleast show that there exist lots of prime ideals. To prove these results we use a kind ofinduction, called transfinite induction. This type of induction uses the axiom of choice.We will use a particular form of this axiom, Zorn’s lemma, which is very popular withalgebraists. First we give the definition of a partially ordered set.

Definition 1.9 (Partially ordered set). A set Σ is said to be partially ordered if it hasa relation ‘�’which is reflexive, antisymmetric and transitive [9], i.e. for all a, b, c ∈ Σthe following holds:

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i) a � a

ii) if a � b and b � a =⇒ a = b.

iii) if a � b and b � c =⇒ a � c.

Notice that the set of ideals of a ring A are a partially ordered set, where � is givenby ‘⊂’set inclusions. Given a subset S ⊂ Σ, an upper bound of S is an element u ∈ Σsuch that s ≺ u for all s ∈ S. A maximal element of Σ is an element m ∈ Σ such thatm ≺ s does not hold for any s ∈ Σ. A set S is a totally ordered set if for every pairs1, s2 ∈ S either s1 � s2 or s2 � s1.

Axiom 1.10 (Zorn’s lemma). Suppose that Σ is a nonempty partially ordered set andthat any totally ordered subset S ⊂ Σ has an upper bound in Σ. Then Σ has a maximalelement.

This axiom is needed for transfinite induction. The idea of this induction is that ifs ∈ Σ is not maximal, then we can choose s such that s � s and so on, so that we get

s � s � . . . � s(λ) � . . . .

This is a totally ordered subset of Σ, so it has an upper bound in Σ. Now if this upperbound is still not maximal, we can extend the totally ordered set some more. Of course,this cannot prove the axiom, because we need to make infinitely many new choices ateach stage. In set theory it is proved that Zorn’s lemma is equivalent to the axiom ofchoice.

1.1.3 Existence of prime ideals and maximal ideals

The following theorem gives an easy illustration of how Zorn’s lemma is used.

Theorem 1.11. Let A be a ring and I 6= A an ideal; then there exists a maximal idealm of A containing I.

Proof. Define the partially ordered set Σ (ordered by inclusion) as

Σ ={J ( A

∣∣ I ⊂ J}.

Since I ∈ Σ, we get Σ 6= ∅. Now take any totally ordered subset S = (Jλ)λ∈Λ ⊂ Σ. ThenJ = ∪λ∈ΛJλ ∈ Σ and for all Jλ ∈ S we have that Jλ ⊂ J , i.e. all totally ordered subsetsof Σ have an upper bound in Σ. Applying Zorn’s lemma gives us that Σ has a maximalelement, so there exists a maximal ideal containing I.

Corollary 1.12. A = A× t⋃m, where

⋃m denotes the union of maximal ideals and

A× the set of units.

Proof. Suppose a ∈ m for a maximal ideal m. Clearly a is not a unit, since then m = Awhich is a contradiction. Conversely, if a is not a unit then (a) 6= A and so by theprevious theorem a is contained in a maximal ideal.

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The previous theorem can be improved upon to give the existence of prime ideals asfollows:

Theorem 1.13. Let A a ring, S a multiplicative set and I an ideal of A with I ∩S = ∅.Then there exists a prime ideal P of A containing I with P ∩ S = ∅.

Proof. Suppose we can find an ideal P ⊃ I that is maximal with respect to the conditionthat P ∩ S = ∅. To show that this P is also prime let f, g /∈ P . Now P + Af andP + Ag are strictly bigger than P and have a non-empty intersection with S. Letp+ af ∈ S ∩ (P + Af), q + bg ∈ S ∩ (P + Ag). Since S is multiplicative we find

n = (p+ af)(q + bg) = pq + pbg + qaf + abfg = p′ + abfg, with p′ ∈ P,

so that n ∈ S. Since P ∩ S = ∅ it follows that fg 6= P , so P is prime.Therefore it is enough to find an ideal P ⊃ I with P ∩ S = ∅ and maximal to this

condition. We can use Zorn’s lemma as in 1.11 to prove this, where we set Σ = {J ⊂A | I ⊂ J, J ∩ S = ∅, J an ideal}.

1.2 Modules

A module over a ring is a generalization of the notion of vector space over a field.Dedekind was the first to use modules in his work on number theory in 1870 [2, p.15],however the first ‘modern’usage originates from Noether and Schmeider in 1920.

The difference between a vector space and a module is that for in a module thescalars only have to form a ring, which gives a significant generalisation. Easy examplesof modules are ideals and quotient rings. Now results concerning these can be combinedinto a single argument about modules.

A consequence of this generalisation is that modules can become quite a bit morecomplicated to work with than vector spaces. A reason for this is that properties thatdo hold for vector spaces, like the existence of a basis and that this basis has a uniquerank, do not have to hold for modules.

Definition 1.14. For A a ring, a left A-module is an abelian group M with a multipli-cation map

A×M →M, (f,m) 7→ fm

satisfying

i) f(m± n) = fm± fn;

ii) (f + g)m = fm+ gm;

iii) (fg)m = f(gm);

iv) 1Am = m

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for all f, g ∈ A and m,n ∈M .

The definition of a right module is rather similar but with multiplication map M×A→M , written (m, f) → mf with a logical adaptation of the properties mentioned above.From now on module means a left module. A subset N ⊂ M is called a submodule iffm + gn ∈ N for all f, g ∈ A and m,n ∈ N . A homomorphism between A-modulesM,N is a map ϕ : M → N that is A-linear.

Examples of modules

i) Le A = k be a field, a module is just a k-vectorspace.

ii) A ring A is a module over itself, and a submodule I ⊂ A viewed as module isprecisely an ideal.

iii) An abelian group A is a Z-module, with (z, a)→ za.

iv) If A is a subring of a ring B, then multiplication in B makes B an A-module. Sim-ilarly we can view a B-module M as A-module, by restricting scalar multiplicationto elements of A.

The most important modules in commutative algebra are those closely related to thering itself, like ideals and extension rings (which we will talk about later) [7, p. 38].

1.2.1 The homomorphism and isomorphism theorems

We can extend the homo- and isomorphism theorems for vector spaces to modules. Theproofs are rather similar and we will not discuss them here.

Theorem 1.15. i) Let ϕ : M → N an A-module homomorphism, then ker ϕ ⊂ Mand image im ϕ ⊂ N are submodules.

ii) If N ⊂ M is a submodule, there exists a quotient module M/N and a surjectivequotient map ϕ : M → M/N such that ker ϕ = N . The elements of M/N can beconstructed as cosets m+N of N in M .

Theorem 1.16 (Isomorphism theorems). i) If L ⊂M ⊂ N are submodules then

N/M =(N/L

)/(M/L

).

ii) If N is a module and L,M ⊂ N are submodules then(M + L

)/L = M/

(M ∩ L

).

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1.2.2 Generators of a module

We are interested in generators of a module, when we call a module finite and when ithas a basis.

Let M be an A-module. For m1, . . . ,mr ∈M , the linear combinations given by

(m1, . . .mr) =r∑i=1

Ami ={ r∑

i=1

fimi ∈M∣∣∣ fi ∈ A} ⊂M

form a submodule. We can discuss whether this forms the whole of M or not. In thecase of a field, this comes down to the theory of linear independence and bases of vectorspaces. For modules, as mentioned before, this is again a little more complicated andwe will first need some more tools.

Choose Λ a (possible infinite) set. If (Mλ)λ∈Λ is a family of A-modules, their directsum is defined by∑

λ∈Λ

Mλ ={

(mλ)λ∈Λ

∣∣ only finitely many mλ 6= 0}.

Note that although the indexing set of the sum can be infinite, each element involvesonly finitely many indexes. In particular, we can take the sum of A card Λ times:

Acard Λ :=∑λ∈Λ

A ={

(aλ)λ∈Λ

∣∣ only finitely many aλ 6= 0}.

In other words, this is the set of finite sequences given by λ ∈ Λ. It is called a freemodule since it has a basis eλ = (0, . . . , 1, . . .), with 1 in the λth place and zero elsewhere.

In particular, for (mλ)λ∈Λ a set of elements of a fixed module M , we can define ahomomorphism from the direct sum to M ,

ϕ : A cardΛ →M, (fλ)λ∈Λ →∑

fλmλ.

This is well-defined since every element (fλ) ∈ A cardΛ has only finitely many fλ 6= 0which implies that the sum is finite. Therefore the image of ϕ is the set of all finitelinear combinations

∑fλmλ of the given elements mλ.

With these tools in mind we can create the notions of generators, a basis and finitenessof a module.

Definition 1.17 (Family of generators). The (possible infinite) set (mλ)λ∈Λ is a familyof generators of M if ϕ : A cardΛ →M is surjective.

Definition 1.18 (Finite A-module). If M is an A-module and there exists a finite setof generators of M , M is called finite, or finite as an A-module.

Note that finite is used througout to mean finitely generated as A-module.

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Definition 1.19 (Basis of a module). The set (mλ)λ∈Λ is a basis of the module M ifϕ : A cardΛ → M is an isomorphism. If there exists a basis for M , it is called a freemodule.

One important property should be pointed out immediately: any quotient of a finitemodule is also finite. Indeed, if m1, . . . ,mr generate M then their quotients m1, . . . ,mr

in M/N generate M/N .Simple examples of finite A-modules are A itself and A⊕A. In Reid p. 41 it says “A

free module is a lucky accident”. The following examples illustrate this.

Examples

i) Let I be a nontrivial ideal to zero. Then the A-module A/I is generated by oneelement, namely the image of 1A, but it is not free. To see this let r be an elementof R/I and let i ∈ I. Then by definition, ir = 0, so r. This implies that no subsetof R/I is linearly independent. For I = (0) this argument clearly does not hold.

ii) For R a ring and {Fi}i∈N a countable family of free R-modules, their direct sum⊕i∈NFi is again a free R-module. Since every Fi was free, for all i ∈ N we have abasis for Fi, which we denote by βi. Then ∪βi gives a basis for ⊕i∈NFi, hence it isa free R-module.

iii) If A is an integral domain and 0 6= f ∈ A then A[1/f ] is usually not finitelygenerated as an A-module. Since A[1/f ] = A + Af−1 + Af−2 + . . ., a family ofgenerators can be given by {1, f−1, f−2, . . .}, which is an infinite set.

iv) The notion of being finitely generated as A-module must not be confused with thenotion of being finitely generated as ring. A finitely generated ring over R meansthat there exists a set of elements G = {x1, . . . , xn} of A such that the smallestsubring of A containing G and R is A itself. For example, the polynomial ring A[x]is finitely generated as A-algebra by {x}, but not as A-module! To see this, notethat for a finite number of generators G = (mλ)

kλ=1, there is an n ∈ N such that

xn has a higher power then all the elements in G. Then xn is not in the image ofϕ : Ak →M , so ϕ is not surjective.

v) Note that R is a Q-module, but that it is clearly not finitely generated.

1.2.3 Cayley-Hamilton theorem

Let N : kn → kn be an endomorphism, where N is represented as a matrix. From linearalgebra we know that for ‘fairly ’general matrices, N can be diagonalised. This meansthat N can be written as N = diag(λ1, . . . , λn) by choosing a basis of kn made up ofeigenvectors ei. Then the equation (N − λi · In)ei = 0 holds, where In is the identitymatrix. In words; this product kills the basis vector ei. If we then look at the product

(N − λ1 · In) · · · (N − λn · In),

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we see that all the basis vectors are cancelled out, so that it equals zero.Now define pN(x) =

∏ni=1(x−λi). This is the characteristic polynomial of N , that is,

pN = det(x · In −N). By the equation above, pN(N) = 0.This is a result from linear algebra and does not automatically hold for rings. However

the Cayley-Hamilton Theorem shows that the same thing holds for any n × n matrixwith entries in a ring A.

To get some feel for this theorem we will first write out the case when n = 2.

Example 1.20. (Cayley-Hamilton Theorem for n = 2) For a 2× 2 matrix

N =

(a bc d

),

the characteristic polynomials pN(x) = det(x · In − N) = x2 − (a + d)x + (ad − bc).Let M denote the space of 2 × 2-matrices with entries in a ring A. Define p : M → Mwhere p(S) = S2− (a+ d)S+ (ad− bc). Then the Cayley-Hamilton theorem states thatp(N) = 0. Indeed we find

p(N) = N2 − (a+ d)N + (ad− bc)I2

=

(a2 + bc ab+ dbac+ dc bc+ d2

)− (a+ d)

(a bc d

)+ (ad− bc)

(1 00 1

)= 0.

Theorem 1.21 (Cayley-Hamilton Theorem). Let A be a commutative ring, N = (aij)an n× n matrix with entries in A and write pN(x) =det(x · In − n). Then

pN(N) = 0 (as an n× n matrix).

Proof. Let ϕ : An → An be the homomorphism of the free module An where ϕ is themap taking

ek = (0, . . . , 1, . . . , 0)→ (0, . . . , 1, . . . , 0)N = (ak1, . . . , akn), (1.1)

the k-th row of N . We introduce the notation µf : An → An for the map m→ fm. Wewrite A′[ϕ] for the subring generated by µf for f ∈ A, and ϕ. This is a commutative ring,since ϕ is A-linear by definition of the multiplication map. With this formal constructionAn can be seen as a module over A′[ϕ] where the map ϕ becomes a multiplication by aring element. It follows that A′[ϕ] consists of the n×n matrices generated by the scalarmultiples of In and N .

Applying this to the elements ek we find

ϕek =n∑i=1

akiei for k = 1, . . . , n, (1.2)

which we can rewrite ton∑i=1

(ϕδki − aki)ei = 0, (1.3)

16

where δki is the Kronecker symbol. Now write ∆ = (ϕδki− aki), which denotes an n× nmatrix with coefficients in A′[ϕ].

We claim that det∆ = 0. To see how this will imply the theorem, consider pN(x) =det(x · In −N) ∈ A[x]. Define the map ϑ as

ϑ : A[x]→ A′[ϕ],m∑i=0

aixi →

m∑i=0

aiNi. (1.4)

This is the map that sends a polynomial in variable x ∈ A to a matrix with coeffi-cients in A. Since ϑ is a homomorphism, it commutes with the algebraic operationsinvolved in taking the determinant. This means that pN(x) = det(x · In −N) maps topN(N) = det ∆. So if det∆ = 0, we have pN(N) = 0.

Claim 1.22 (det∆ = 0). To prove the claim, let adj ∆ be the adjoint matrix of ∆. Thisis the transpose of the cofactor matrix C of A. Here Cij = (−1)i+jAij and so the adjointmatrix is given by (adj A)ij = Cji. The adjoint matrix is defined this way to make theproduct of A with its adjoint matrix yields a diagonal matrix, whose diagonal entriesare detA. In our case this yields

(adj ∆) ·∆ = ∆ · (adj ∆) = det∆ · In. (1.5)

Let bkj denote the entries in adj ∆. Now since we have det∆ ∈ A′[ϕ] ⊂ End An, it isenough to prove that (det∆) · ej = 0 for j = 1, . . . , n. To see this we multiply (2.3) withbkj and sum over k. We get:

n∑k=1

bkj ·n∑i=1

(ϕδki − aki)eI = 0. (1.6)

Note that this equivalent to saying that

adj ∆ ·∆ · ei = In · det ∆ · ei = 0, (1.7)

where the second equality holds because of (2.5). In other words, we find (det ∆)ej = 0for each j and therefore det∆ = 0.

This implies the theorem.

The next theorem we will need for several theorems in Chapter 4 and is called the’going up theorem‘. The last part of the proof is similar to that of the Cayley-Hamiltontheorem.

Theorem 1.23 (The going up theorem). Let B be finite over a subring A and ma amaximal ideal in A. Then there exists a maximal ideal mb in B such that mb ∩A = ma.

17

Proof. Suppose that B 6= maB. Then maB is contained in a maximal ideal mb andma ⊂ mb∩A ⊂ A. From the maximality of ma and the fact that 1 /∈ mb we can concludethat ma = mb ∩ A.

To show that B 6= maB, consider B =∑m

i=1Abi with bI ∈ B. If B = maB then

bi =∑

xijbj

with xij ∈ ma. We can view this as a matrix equation (I−X)b = 0, where X = (xij) andb is the column vector (b1, . . . , bm)t. Multiplying by adj (I −X) we find, the same wayas in the Cayley-Hamilton theorem that det(I −X) = 0, but det(I −X) = 1 mod ma

which gives the contradiction. Hence B 6= maB and we are done.

1.2.4 The determinant trick

The next theorem we discuss has many applications in commutative algebra and isneeded to prove a relation between finite algebras and integral elements.

Theorem 1.24 (The determinant trick). Let M be a finite A-module generated by nelements and let ϕ : M → M be a homomorphism. If I is an ideal of A such thatϕ(M) ⊂ IM , then ϕ satisfies a relation of the form

ϕn + an−1ϕn−1 + . . .+ a1ϕ+ a0 = 0, (1.8)

where ai ∈ I i for i = 0, 2, . . . n− 1.

Note that (2.8) gives a relation between endomorphisms of the module M , that is arelation in the ring A′[ϕ] ⊂ End M as introduced in the proof of the Cayley-HamiltonTheorem. The result is proved in a similar manner as the Cayley-Hamilton Theorem.

Proof. Let m1, . . . ,mn be the set of generators of M . By assumption we have thatϕ(mi) ∈ IM , so we can write

ϕ(mi) =n∑i=1

aijmj with aij ∈ I. (1.9)

As in the proof of the Cayley-Hamilton Theory, 2.9 can be written rewritten as

n∑j=1

(δijϕ− aij)mj = 0, (1.10)

with δijϕ− aij ∈ A′[ϕ]. Write ∆ = (δijϕ− aij). As in the proof of the Cayley-HamiltonTheorem, Let bij denote the entries of the adjoint matrix of ∆. If we then multiply(2.10) by bki and sum over i, we deduce in the same way as in the Cayley-Hamiltonproof that det ∆ = 0 ∈ A′[ϕ]. Expanding out the determinant gives the relation in thestatement.

18

1.2.5 Nakayama’s lemma

One application of the determinant trick is Nakayama’s lemma, named after its author.The lemma is also known as the Krull-Azumaya Theorem, Krull was the first whodiscovered a special case of the theorem after which the general case was discoveredby Azumaya. The lemma is a consequence of the Cayley-Hamilton theorem and it isimportant because it allows us to make many deductions about properties of M as anA-module from the properties of M/mM as vector space over the residue field k = A/m.

To prove Nakayama’s lemma the following result is essential:

Theorem 1.25. If M is a finite A-module, I is an ideal of A and M = IM holds, thenthere exists an element x ∈ A such that x ≡ 1 mod I and that xM = 0.

Proof. Since M = IM with I an ideal, we have for ϕ = IdM that ϕ(M) ⊂ IM , so we canapply the determinant trick to ϕ. This gives us a relation of the form ϕn + an−1ϕ

n−1 +. . .+ a1ϕ+ a0 = 0 for some ai ∈ m. Since (IdM)i = IdM , this gives us the relation

(1 + b)IdM = 0, where b = an−1 + . . .+ a0 ∈ I.

Let x = 1 + b Then x ≡ 1 mod I and we find that xM = (1 + b)IdMM = 0.

Theorem 1.26 (Nakayama’s lemma). Let (A,m) be a local ring (as defined in 1.1), andM a finite A-module; then M = mM implies that M = 0. More generally, for I anideal such that (1 + I) ⊂ A× and M = IM , the same holds. Here A× denotes the unitsof A.

Proof. Since m is the unique maximal ideal of A and since M = mM by assumption,we know from the previous theorem that there exists an element x ∈ A such thatx ≡ 1 mod m and xM = 0. Since x ≡ 1 mod m where (1 + m) ⊂ A× by the definitionof a local ring, it follows that x is an invertible element of A. This implies that M =x−1 · xM = 0.

In exactly the same way it follows that M = IM also implies M = 0. Note that thecondition (1 + I) ⊂ A× is needed to guarantee that x is a unit in A.

One of the consequences of Nakayama’s lemma is the result below.

Corollary 1.27. If M is a finitely generated A-module and f : M →M is a surjectiveendomorphism, then f is an isomorphism.

1.3 Noetherian rings

Now that we have discussed the fundamental notions of ring theory, we will focus on amore specific type: Noetherian rings. These rings play an important role in commutativeas well as in non-commutative algebra. The reason they are so important has to do withthe fact that the Noetherian property gives a ring-theoretic version of finiteness.

19

For example, all polynomial rings in finite variables over a field are Noetherian. Thismakes it possible to prove that any infinite set of polynomial equations can be viewedas a finite set with the same solutions [2, p. 108].

Hilbert was the first to study the finiteness of the polynomial ring k[X1, . . . , Xn]. Heused their finiteness to prove that rings of invariants are finitely generated. His abstractway of proving the existence of a finite set of generators was uncommon in his time anddiscarded as “theology, not mathematics” [7, p. 49]. It was Emmy Noether who in 1920started working on a condition called the ‘ascending chain condition’. She used it as afiniteness condition on rings and this is one of the equivalent conditons for a ring to beNoetherian. Hence it will be discussed first.

1.3.1 The ascending chain condition

For a partially ordered set Noether defined the ascending chain condition:

Definition 1.28 (The ascending chain condition). A partially ordered set Σ has theascending chain condition (a.c.c.) if for every chain

s1 � s2 � . . . � sk � . . .

there exists a k ∈ N such that sn = sn+1 = . . ., for all n ≥ k.

In logic, this definition gives a finiteness condition which allows arguments by induc-tion, even if the partially ordered set is infinite. Clearly, a set Σ has the a.c.c. if andonly if every non-empty S ⊂ Σ has a maximal element. For example, vector subspacesof a finite dimensional vector space satisfy the a.c.c.

1.3.2 The Noetherian condition

There are three equivalent ways to define Noetherian rings.

Definition 1.29 (Noetherian rings). A ring A is Noetherian if one of the followingequivalent conditions hold:

i) The set Σ of ideals of A has the a.c.c.; in other words, every increasing chain ofideals

I1 ⊆ I2 ⊆ . . . ⊆ Ik ⊆ . . .

eventually stops, that is Ik = Ik+1 = . . . for some k.

ii) Every nonempty set S of ideals has a maximal element.

iii) Every ideal I ⊂ A is finitely generated.

A proof of the equivalence can be found in [7, Chap. 3].

20

At first sight it might seem easier to work with conditions ii) or iii), but as a rule,the Noetherian condition i) is both more general and practical to work with [7, p. 51].

Some examples of Noetherian rings:

i) Any field k is a Noetherian ring, since a field has only two ideals; (0) and k. Clearlythe a.c.c. holds.

ii) Any principal ideal domain A is Noetherian, since every ideal is generated by oneelement. To make an ascending chain I1 ⊆ I2 ⊆ . . ., let Im = (a1), Im+1 = (a2), forIm ( Im+1, am had to lose a prime factor.

Since it can only have finitely many prime factors, this chain will eventually stop.

Some examples of non-Noetherian rings.

Noetherian rings are a very broad class and most of the rings usually encountered incommutative algebra are Noetherian [2, p. 108]. However there are lots of easy examplesof non-Noetherian rings to be found:

i) The polynomial ring A = k[X1, . . . , Xn, . . .] in an infinite number of variables isnon-Noetherian. To see this, we can define the ascending chain (Ik) by

(X1) ⊂ (X1, X2) ⊂ (X1, X2, X3) ⊂ . . . .

Clearly there does not exist a k such that In = In+1 for all n ≥ k. So A is notNoetherian.

ii) Consider the ring of polynomials in x, y of the form f(x, y) = a + xg(x, y), wherea ∈ k with k a field and g ∈ k[x, y]. So f has no pure power yi for i > 0. The ringA can be written as

A ={f(x, y) =

∑aijx

iyj∣∣∣i, j ≥ 0 and i > 0ifj 6= 0

}= k[x, xy, xy2, . . . , xyn, . . .].

Define the chain (xy) ⊆ (xy, xy2) ⊆ (xy, xy2, xy3) ⊆ . . .. Since y /∈ A,xyn+1 /∈ (xy, xy2, . . . xyn) and thus (xyn) ( (xyn+1) and the chain does not breakoff. It follows that A is not Noetherian.

It is possible for a Noetherian ring to have a non-Noetherian subring. This followsfrom the examples above since the non-Noetherian ring k[x, xy, xy2, . . . , xyn, . . .] is asubring of k[x, y] which was Noetherian as discussed above. The fact that k[x, y] isNoetherian also follows from the following very important theorem, proven by Hilbertin 1890:

Theorem 1.30 (Hilbert’s Basis Theorem). If A is a Noetherian ring then the polynomialring A[X] is Noetherian as well.

21

Proof. We have to prove that any ideal I ⊂ A[X] is finitely generated. For this, definethe sets Jn ⊂ A by

Jn = {a ∈ A | there exists a f ∈ I such that f = aXn + bn−1Xn−1 + . . .+ b0}.

Note that this gives an ideal in A since I is an ideal. Furthermore, since for f ∈ I alsoXF ∈ I, we have that Jn ⊂ Jn+1 and thus an increasing chain of ideals

J1 ⊂ J2 ⊂ . . . ⊂ Jk ⊂ . . . ,

which must eventually break off since A is Noetherian by assumption. Let k ∈ N suchthat Jk = Jk+i for all i ∈ N.

For each m ≤ k the ideal Jm ⊂ A is finitely generated, so that we can write Jm =(am,1, . . . , am,rm). Then by definition of Jm, for all am,j with 1 ≤ j ≤ rm there exists apolynomial fm,j ∈ I of degree m having the leading coefficient am,j. Now write down

F ={fm,j

}m≤k,1≤j≤rm

⊂ I.

Note that F gives a finite set, so that if we can prove that F generates I, we have thatI is finitely generated, which is what we need to prove. To see that F generates I welook at a polynomial f ∈ I. If f has degree m, then its leading coefficient A is in Jm,hence if m ≥ k, then a ∈ Jm = Jn so that a =

∑bian,i with bi ∈ A and

deg f −∑

biXm− nfm,i < m.

For m ≤ n something similar holds: then a ∈ Jm so that a =∑biami

with bi ∈ A andf −

∑bifm,i has degree smaller than m.

By induction on m, it follows that every f ∈ I can be written as a linear combinationof elements of F . This proves that any ideal of A[X] is finitely generated, so that A[X]is a Noetherian ring.

Corollary 1.31. If A is Noetherian and ϕ : A → B a ring homomorphism, such thatB is a finitely generated extension ring of ϕ(A), then B is Noetherian.

In particular, it follows from this theorem that any finitely generated algebra over Zor a field is Noetherian.

1.4 Finite ring extensions

In this section we will talk about finite ring extensions. It is this finiteness conditionsthat will naturally lead to Noether’s normalisation lemma.

Let L be an extension field of K. An element y ∈ L is algebraic over K if there exists a0 6= f ∈ K[Y ] such that f(y) = 0. If f is also monic, then the relation f(y) = 0 is calledan integral dependence relation for y [7, p. 58]. In a field, algebraic dependence impliesintegral dependence since it costs nothing to divide by the head coefficient. However ina general ring this implication does not hold. Note also that for y1, y2 ∈ L integral overK the same holds for y1 + y2 and y1y2.

22

1.4.1 Finite and integral A-algebras

For A, B rings, a homomorphism ϕ : A → B is an A-algebra. Moreover, B is an A-module with ψ : A× B → B defined by ψ(a, b) = ϕ(a)b. If A ⊂ B, thenB is called anextension ring. Usually this is denoted by ϕ(A) = A′ ⊂ B.

In this section we are mostly interested in finite A-algebras, these are defined as:

Definition 1.32 (Finite and integral A-algebra). Let B be an extension ring of A.

i) An A-algebra B is finite, or is finite over A, if it is finite as an A-module.

ii) An element y ∈ B is integral over A if there exists a monic polynomial f ∈ A′[Y ] ⊂B, with f 6= 0, such that f(y) = 0. The algebra B is integral over A if every b ∈ Bis integral.

The proof does not present any difficulties and can be found in [7, p.60].The next theorem gives us a relations between the above definitions.

Theorem 1.33. Let ϕ : A→ B be an A-algebra and y ∈ B; then the following conditionsare equivalent:

i) The element y is integral over A

ii) The subring A′[y] ⊂ B generated by A′ = ϕ(A) and y is finite over A

iii) There exists an A-subalgebra C ⊂ B such that A′[y] ⊂ C and C is finite over A.

1.4.2 Normalisation of rings

Definition 1.34 (integral closure). The integral closure of A in B is defined as thesubset

A ={y ∈ B

∣∣ y is integral over A}⊂ B.

A ring A is called integrally closed in B if A = A.

A special case worth considering is a ring embedded in its field of fractions.

Definition 1.35 (Normal). An integral domain A is called normal if it is integrallyclosed in its field of fractions, i.e.

A = A ⊂ K = Frac A. (1.11)

The integral closure of A in its field of fractions K is logically called the normalisationof A. We will now see a few examples of the normalisation of some rings.

23

Examples

i) Let A be a UFD with field of fractions K. Then no element of K\A is integral overA and hence A is normal.

Proof. Let x = a/b ∈ K with a ∈ A and b ∈ A\{0} such that a and b do not haveany irreducibles in common. suppose x is integral over A, then we have

xn + cn−1xn−1 + . . .+ c1x+ c0 = 0.

Substituting x = a/b and multiplying with bn ∈ A gives

an + cn−1an−1b+ . . .+ c1ab

n−1 + c0bn = 0.

This shows that b divides an. Suppose b’s factorisation has a irreducible factor p.Then p has to be in an’s factorisation and hence in a’s, which gives a contradiction.So b has no irreducibles in its factorisation, i.e. b is a unit. This means that x ∈ Aand so we get A = A.

ii) Let A = k[X, Y ]/(Y 2 −X2 −X3). The normalisation of A is k[t] where t = Y/X.

Proof. We first want to see if A is an integral domain, i.e. A has no zero divisors.This is the same as checking that the polynomial h(X, Y ) = Y 2 − X2 − X3 isirreducible in k[X, Y ]. To see this consider the polynomial Y 2 −X2 −X3, a monicpolynomial of degree 2 in Y with coefficients in k[X]. Now it is clear that there isno f(X) ∈ k[X] such that f(X)2 = X2 +X3, so h is irreducible.

Now consider the field of fractions K of A. Since the class of X in A is not 0,t = Y/X ∈ K is well defined. Hence k[t] ⊂ K. Furthermore we find

t2 = (Y/X)2 = (X2 +X3)/X2 = X + 1.

Take f(T ) = Y 2 −X2T 2 ∈ A[T ] then

f(t) = Y 2 −X2(X + 1) = Y 2 −X3 −X2 ≡ 0,

so t is integral over A and hence k[t] ⊂ A, where A denotes the integral closure ofA.

On the other hand, x = t2−1 and y = tx = t3− t, so x, y ∈ k[t] and hence A ⊂ k[t].So we see that the field of fractions K of A is also the field of fractions of k[t] andwe have

A ⊂ k[t] ⊂ A ⊂ K.

Now k[t] is normal since it is a UFD so k[t] = k[t] and since A ⊂ k[t] implies that

A ⊂ k[t], we find that k[t] = A. So k[t] gives indeed the normalisation of A.

24

One way to connect the notions of integral and finite is by the tower laws. The lawsprovide an easy way to determine when a ring B is finite over A. This is done byconsidering integral elements.

Theorem 1.36. Let B be an extension ring of A.

i) If A ⊂ B ⊂ C are extension rings such that C is a finite B-algebra, and B a finiteA-algebra. Then C is finite over A.

ii) If A ⊂ B ⊂ C with C integral over B and B integral over A then C is integral overA.

iii) If y1, . . . ym ∈ B are integral over A, then A[y1, . . . , ym] is finite over A; In particu-lar, every f ∈ A[y1, . . . , ym] is integral over A.

iv) The subset A ={y ∈ B

∣∣ y is integral over A}⊂ B is a subring of B; Moreover

˜A = A.

For the first statement their is a similar result for fieldextension, for which the proofis the same, the other statements follows rather quickly from i) and a proof of them canbe found in [7, p.61].

25

2 Noether’s normalisation lemma

We have now discussed everything needed to understand and prove the following re-sult. We will follow the proof as given in [7, p. 63]. The original proof of Noether’snormalisation lemma is due to Nagata in the 1950s.

2.1 Noether’s normalisation lemma

Theorem 2.1 (Noether’s normalisation lemma). Let k be a field, and let A be a finitelygenerated k-algebra. Then there exist elements z1, . . . , zm ∈ A such that

i) z1, . . . , zm are algebraically independent over k;

ii) A is finite over B = k[z1, . . . , zm].

The theorem states that a finitely generated extension k ⊂ A can be written as

k ⊂ B = k[z1, . . . , zm] ⊂ A (2.1)

where k ⊂ B is a polynomial extension, and B ⊂ A is finite.

2.2 Proof of Noether’s normalisation lemma

The proof is divided into three steps. First, the main claim to prove the lemma isdiscussed. For the proof of the main claim, a second lemma, ‘the horrible lemma’isneeded (Don’t worry, it’s not horrible at all). After these results are established, anelegant proof of the Noether normalisation lemma is given.

2.2.1 Main claim

Lemma 2.2 (Main claim). Suppose that A = k[y1, . . . , yn], and that

0 6= F ∈ k[Y1, . . . , Yn]

is such that F (y1, . . . , yn) = 0. Then there exist y∗1, . . . , y∗n−1 ∈ A such that yn is integral

over A∗ = k[y∗1, . . . , y∗n−1] and A = A∗[yn].

Proof. Choose integers r1, . . . , rn−1 (which we will specify later) and set

y∗i = yi − yrii for i = 1, . . . , n− 1

26

Now defineG ∈ A byG(y∗1, . . . , y∗n−1, yn) = F (y∗i +y

rin , yn). By hypothesis F (y1, . . . , yn) =

0 and it follows that

G(y∗1, . . . , y∗n−1, yn) = F (y∗1 + yr11 , . . . , y

∗n−1 + y

rn−1

n−1 , yn)

= F (y1 − yr11 + yr11 , . . . , yn−1 − yrn−1

n−1 + yrn−1

n−1 , yn)

= F (y1, . . . , yn)

= 0

To prove the claim it is sufficient show that for suitably chosen r1, . . . , rn−1, the re-lation for yn over k[y∗1, . . . , y

∗n−1] is an integral depence relation. Recall that G ∈

k[y∗1, . . . , y∗n−1][Y ] gives an integral dependence relation for yn if G 6= 0, G is monic

and if G(yn) = 0. For G defined above, G(yn) = 0, so what remains, is to show thattheir exist ri such that G 6= 0 and that G = amG

′ where am 6= 0 is a constant and G′ ismonic in yn.

Write F as the sum of monomials F =∑

m amym with am ∈ A and unequal to zerofor all m. Write y = (y1, . . . , yn).We can rewrite F as:

F =∑m

am

n∏i=1

ymii .

where m = (m1, . . . ,mn). Since G(y∗1, . . . , y∗n−1, yn) = F (y∗i + yrin , yn), I can write

G =∑m

am

( n−1∏i=1

(y∗i + yrin )mi

)ymnn .

The term am

(∏n−1i=1 (y∗i + yrin )mi

)ymnn splits up into a polynomial of different terms,

with a unique term of highest degree, namely amy∑n−1

i=1 rimin ymn

n = amy∑n

i=1 rimin .

To prove that G 6= 0, it suffices to show that

m 6= m′ =⇒n∑i=1

rimi 6=n∑i=1

rim′i. (2.2)

If this implication holds, then max{∑n

i=1 rimi

∣∣ m such that am 6= 0}

is assumed ex-actly once. This implies that the highest order term of G in yn can not be cancelledout so then G would not be 0. Since ai are elements of a field, G can be written asG = amG

′, where am 6= 0 and G′ is a monic, non-zero polynomial in yn.Then G′ is the monic polynomial which gives the integral dependence relation for yn,

i.e. yn is integral over A∗ = k[y∗1, . . . , y∗n−1] and A = A∗[yn].

It is therefore sufficient to prove that there exist integers r1, . . . , rn such that (3.2)holds. This is proven in the following

27

Lemma 2.3 (The Horrible lemma). Suppose{Y m}

is a finite set of moniomials inY1, . . . , Yn, so any f ∈

{Y m}

is of the form Y m11 Y m2

2 . . . Y mnn . Then there exists a

system of integers w1, . . . , wn−1, wn = 1 such that the weight of a monomial, defined as

w(Y m) :=n∑i=1

wimi,

distinguishes the monomial, that is,

m 6= m′ =⇒ w(Y m) 6= w(Y m′).

Proof. The proof is by induction on n. Assume n = 1. Then w(Y m1) = w1m1. Clearlym1 6= m′1 implies w1m1 6= w1m

′1. Assume the hypothesis holds for i = 1, . . . , n − 1

and consider the set of monomials in n variables. Write each monomial as Y m11 Y m∗

where m∗ = (m2, . . . ,mn) and Y m∗ is a monomial in n− 1 variables. By the inductionhypothesis there exist weights w2, . . . , wn such that they distinguish the monomial, i.e.

m∗ 6= m∗′

=⇒ w∗(Y m∗) =n∑i=2

wimi 6=n∑i=2

w′im′i = w∗(Y m∗

′

)

Now choose w1 > max{w(Y m∗)

∣∣ Y m∗ is a monomial in n−1 terms}

. Then for m 6=m′ at least m1 6= m′1 or m∗ 6= m∗′ and in both cases it follows that w(Y m) 6= w(Y m′).If both inequalities occur, the condition that w1 > maxw(Y m∗) shows that the requiredinequality still holds.

This lemma shows the existence of suitable r1, . . . rn, since we can take the weightsfrom the lemma. Hence the main claim, lemma 2.2, holds.

2.2.2 Proof of Noether’s normalisation lemma

The proof of Noether’s normalisation lemma proceeds by an induction argument, inwhich we use the main claim. For the reader’s convenience, we recall the statement ofthe lemma

Theorem 2.4 (Noether normalisation lemma). Let k be a field, and let A be a finitelygenerated k-algebra. There exist elements z1, . . . , zm ∈ A such that

i) z1, . . . , zm are algebraically independent over k;

ii) A is finite over B = k[z1, . . . , zm].

Proof. Let A be generated as k-algebra by the elements y1, . . . , yn. If n = 0 there isnothing to proof. For n > 0 with y1, . . . , yn algebraically independent over k, takezi = yi and we are done.

Now assume y1, . . . , yn are algebraically dependent over k, so there exists a non-zeropolynomial F ∈ k[Y1, . . . , Yn] such that F (y1, . . . , yn) = 0. By the main claim, there

28

then exist y∗1, . . . , y∗n−1 ∈ A such that yn is integral over A∗ = k[y∗1, . . . , y

∗n−1] and such

that A = A∗[yn].Since A∗ has n − 1 generators it follows from the induction hypothesis that there

exist z1, . . . , zm ∈ A∗ that are algebraically independent over k with A∗ finite overB = k[z1, . . . , zm].

With the tower laws we find, since yn ∈ A integral over A∗, that A∗[yn] = A is finiteover over A∗. So for the inclusions B = k[z1, . . . , zn] ⊂ A∗ ⊂ A we have that B ⊂ A∗,A∗ ⊂ A are finite. Once more by applying the tower laws, it follows that A is finite overB. This completes the proof.

Noether’s normalisation lemma is an important result in the commutative algebrabecause it is used to prove several important theorems, like Hilbert’s weak and strongNullstellensatz. We will them in chapter 4.

29

3 Further interpretations

In order to consider the geometric implications of Noether’s normalisation lemma, it isnecessary to first discuss some basic concepts of algebraic geometry. Roughly, algebraicgeometry is the study of certain geometric objects – called algebraic varieties – that aredefined by polynomial equations [10, p.1].

3.1 Some basic algebraic geometry

From now on k denotes an algebraically closed field. Note here that algebraic closedfiels are infinite. The affine n-space An is defined as An = kn. The reason this specialnotation is introduced, is that it should not be confused with kn as k-vector space. Inparticular, 0 does not play a special role.

A polynomial f ∈ k[x1, . . . , xn] defines a function f : An → k, given by (a1, . . . , an)→f(a1, . . . , an). Furthermore, these k-valued functions form a k-algebra, hence the map ϕdefined as

ϕ: k[x1, . . . , xn]→ {functions An → k}f → f

gives a k-algebra homomorphism. In [11, p.1] it is proven that for k = k, the homomor-phism ϕ is injective. The proof uses the fact that for k to be equal to k, it has to haveinfinitely many elements. This result implies that we can simply identify a polynomialf ∈ k[x1, . . . , xn] with the associated function f : An → k.

From now on, we will write f for f .

3.1.1 Zariski topology

The Zariski topology is a topology chosen for algebraic varieties, it is due to OscarZariski and was introduced in the 1950s. It is an interesting topology because it is verydifferent from metric topologies, like the Euclidean topology on Cn. For example, An isnot a Hausdorff space for n > 0!

Definition 3.1 (Zariski topology). The Zariski topology on An is the topology for whichthe closed sets are the subsets of the form Z (S), where S ⊂ k[x1, . . . , xn]. Here Z (S) ⊂An is called the zero set or the zero locus of S and is defined by

Z (S) ={P = (a1, . . . , an) ∈ An

∣∣ f(P ) = 0 for all f ∈ S}.

30

That this really defines a topology on An follows from the following

Theorem 3.2. i) If S is a subset of k[x1, . . . , xn] and I ⊂ k[x1, . . . , xn] is the idealgenerated by S then Z (S) = Z (I).

ii) We have Z(0)

= An and Z(k[x1, . . . , xn]

)= ∅

iii) If{Sα}α∈A is a collection of subsets of k[x1, . . . , xn] then Z

(∪α∈ASα

)= ∩α∈AZ (Sα).

iv) If I and J are ideals of k[x1, . . . , xn] then Z (IJ) = Z (I) ∪Z (J).

The proof presents no difficulties and we will not discuss it here. The first partof this theorem shows that the zero locus of a subset S ⊂ k[x1, . . . xn] is the sameas the zero locus of the ideal I generated by S. Since k[x1, . . . , xn] is Noetherian, ifit is finitely generated, which implies that there exist a finite number of polynomialsf1, . . . , fr ∈ k[x1, . . . , xn] with I = (f1, . . . , fr). By i) of the previous proposition thezero set of S is then the set of points P ∈ An for which

f1(P ) = . . . = fr(P ) = 0.

It follows that the subsets of the form Z (S) ⊂ An are the zero sets given by a finitenumber of polynomial equations. We define algebraic sets as the set of solutions of asystem of polynomial equations, i.e. closed sets with respect to the Zariski topology.

Examples

i) Consider the Zariski topology on the affine line A1 = k. Every ideal in k[x] isprincipel, so every algebraic set is the zero locus of a single polynomial. Since apolynomial f 6= 0 only has finitely many roots, this implies that the algebraic setsin A1 are just the finite subsets, together with the whole space (corresponding tof = 0). Therefore the Zarisky topology is the co-finite topology. This is the topologywhich has as open sets the empty set and all subsets of X whose complements arefinite.

ii) Every point P = (a1, . . . , an) ∈ An is closed in the Zariski topology since the zerolocus of S =

{x1 − a1, . . . , xn − an

}⊂ k[x1, . . . , xn] is precisely P . In theorem 3.6

we shall see that the map P 7→ (x1−a1, . . . , xn−an) establishes a bijection betweenAn and the set of maximal ideals of k[x1, . . . , xn].

iii) An example of a closed subset in A2 is the union of the circle Z (x2 + y2 − 1)with the line y = 1 and the point (2,−1). This set is the zero locus of S ={(x2 + y2 − 1)(y − 1)(x− 2), (x2 + y2 − 1)(y − 1)(y + 1)}.

For a subset Y ⊂ An we define I (Y ) in a dually fashion as Z (I);

I (Y ) = {f ∈ k[x1, . . . xn] | f(P ) = 0 for all P ∈ Y }.

Note that this forms an ideal of k[x1, . . . , xn] and that this is the ideal containing allpolynomials that vanish on Y .

31

With the definitions of Z and I we have obtained the diagram

and we may wonder whether or not these maps are mutually inverse.

Example 3.3. To show that this is not the case, we look at the ideals (x) and (x2) ink[x] and see that their zero loci are the same:

Z((x))

={P = (a1, . . . , an) ∈ An

∣∣ f(P ) = 0 for all f ∈ (x)}

= {0} = Z((x2)

).

As we will see later on, we do get a bijection if we restrict to radical ideals (as definedin 3.7) and closed subsets of An with respect to the Zariski topology. First we shalldiscuss the following “inclusion-reversing”theorem.

Theorem 3.4. i) For the ideals I1 ⊂ I2 ⊂ k[x1, . . . , xn] it holds that that Z (I1) ⊃Z (I2); the map Z is inclusion-reversing.

ii) The map I is inclusion reversing as well: Y1 ⊂ Y2 ⊂ An implies I (I1) ⊃ I (Y2).

iii) For Y ⊂ An we have that Z (I (Y )) = Y , the closure of Y in the Zariski topologyon An.

Proof. Part i) and ii) follow quickly from the definition so we will only prove iii). Fromthe definitions it is clear that Y ⊆ Z

(I (Y )

)and since the zero locus of a subset is

closed by the definition of the Zariski topology we find that Y ⊆ Z(I (Y )

).

32

Conversely, we know that Y = Z (J) for some ideal J ⊂ k[x1, . . . , xn]. From thedefinitions we find that J ⊂ I

(Z (J)

)and since by ii) I (Y ) ⊃ I (Y ) = I

(Z (J)

), we

obtain that J ⊂ I (Y ). By i) this gives that Y ⊃ Z(I (Y )

). So we get Z

(I (Y )

)=

Y .

Note that our aim was to describe a bridge between rings and subsets of An. Thedefinitions of Z and I give us the first step in the right direction. In the next sectionwe elaborate on this.

3.1.2 Hilbert’s Nullstellensatz

The existence of a bijection between An and the maximal ideals of k[x1, . . . xn] is givenby Hilbert’s Weak Nullstellensatz. Together with the ‘strong’Nullstellensatz we obtaina bijection between radical ideals and closed subsets of An. An important result neededfor proving Hilbert’s Nullstellensatz is the next theorem.

Theorem 3.5. Let k be a field and let K be a field extension which is finitely generatedas a k-algebra. Then K is algebraic over k.

We will only give an example of how this can be proven. Suppose K is the nonalgebraic field extension k(x) of k. We claim that if f1, . . . fm ∈ K, that the k-algebraA they generate is always strictly smaller than K, which will give a contradiction. Tosee this, choose c ∈ k such that c is not a pole of one of the rational functions fi. Thisimplies that no element of A can have a pole at c, so that 1/(x − c) is not in A. Since1/(x− c) is in K, A is strictly smaller than K indeed.

A full proof is for example given in a note in [1].With the theorem above we can proof Hilbert’s Weak Nullstellensatz.

Theorem 3.6 (Hilbert’s Weak Nullstellensatz). Let k be an algebraically closed field.The map that sends a point P = (a1, . . . , an) ∈ An to the maximal ideal mP = (x1 −a1, . . . , xn − an) ⊂ k[x1, . . . , xn] defines a bijection

An → {maximal ideals of k[x1, . . . , xn]}.

Proof. The important thing to notice here is that if we can prove that indeed all themaximal ideals are of the form mP as defined above, the bijection follows immediately.To prove this, let m be a maximal ideal of A = k[x1, . . . , xn]. Then as discussed inthe second chapter, A/m is a field which is finitely generated as a k-algebra. By theprevious theorem it is also an algebraic extension of k and therefore equal to k, since kis algebraically closed.

This implies that each xi maps to some ai ∈ k under the natural map A→ A/m = k.A fortiori, (x1− a1, . . . , xn− an) ⊂ m. Since (x1− a1, . . . , xn− an) is maximal, it equalsm. This completes the proof.

Before we get to Hilbert’s Nullstellensatz we first need the notion of a radical ideal.

33

Definition 3.7 (Radical ideal). For a ring A and an ideal I ⊂ A, the radical of I is theideal

√I = {r ∈ A | rm ∈ I for some m ∈ N} ⊃ I.

We call I radical if I =√I.

Example 3.8. For an example we look at some subset Y ⊂ An and the ideal I (Y ) ⊂k[x1, . . . , xn] associated with this. Now for r ∈ k[x1, . . . , xn] such that rm ∈ I (Y ) forsome m ∈ N, we find that rm(P ) = 0 for all P ∈ Y , so that r(P ) = 0 must also hold forall P ∈ Y . Therefore r ∈ I (Y ) and thus I (Y ) ⊂ k[x1, . . . , xn] is a radical ideal.

The lack of symmetry that I (Z (I)) contains I but does not have to equal I, makesthat the maps Z and I do not define a bijection between the ideals of k[x1, . . . , xn]and closed subsets of An. In example 3.3, where we found that Z

((x2)

)= Z

((x)),

we see that (x2) is not radical since√

(x2) = (x) 6= (x2) . In general, it is clear thatif fm ∈ I for some m ≥ 1, that f ∈ I (Z (I)). The Hilbert Nullstellensatz gives thatthe condition on I to be radical is sufficient to find the equality we look for. This thenresults in the bijection between radical ideals of k[x1, . . . , xn] and closed subsets of An.

Theorem 3.9 (Hilbert’s Nullstellensatz). Let k be an algebraically closed field. For anyideal I ⊂ k[x1, . . . , xn],

I (Z (I)) =√I.

Proof. The inclusion√I ⊂ I

(Z (I)

)is clear. It remains to show the reverse inclusion.

This means that we need to prove that if a polynomial g vanishes on the zero locus ofI = (f1, . . . , fn) that some power of g lies in I. Here I = (f1, . . . , fn) lies in k[x1, . . . , xn],which we have identiefied with polynomial functions on kn.

If I is an ideal in k[x1, . . . , xn] such that Z (I) = ∅, then I must be k[x1, . . . , xn], theunit ideal. For if not, I would be in some maximal ideal. But by the weak form ofthe Nullstellensatz, every maximal ideal vanishes at one point. We conclude that thecommon zeroes of I cannot be empty unless I is the unit ideal.

Suppose g ∈ I(Z (I)

). The trick, found by Rabinowitsch, is to consider the ideal

J in k[x1, . . . , xn, xn+1], generated by I and xn+1g − 1. Note that Z (J) = ∅ since gvanishes on the zero locus of I, so xn+1g − 1 does not. As discussed above, this impliesthat J is the unit ideal. We can thus express 1 as

1 = p1f1 + . . .+ pmfm + pm+1 · (xn+1g − 1),

where the p’s are polynomials in x1, . . . , xn+1. If we take the image of this equationunder the homomorphism

k[x1, . . . , xn+1]→ k(x1, . . . , xn)

xn+1 → 1/g

34

we find the equation

1 = p1

(x1, . . . , xn,

1

g

)f1 + . . .+ pm

(x1, . . . xn,

1

g

)fm.

If we multiply this by a power of g to clear the denominators, we find that some powerof g lies in I, whence g lies in

√I. We have proven Hilbert’s Nullstellensatz.

Note that the condition on k to be algebraically closed is necessary. To see this takeR and the ideal (x2 + 1) ⊂ k[x]. We see that

√(x2 + 1) = (x2 + 1). On the other hand,

we find that

Z((x2 + 1)

)= {P ∈ R | f(P ) = 0 for all f ∈ (x2 + 1)} = ∅,

so that

I(Z (x2 + 1)

)= I (∅) = {f ∈ k[x] , | f(P ) = 0 for all P ∈ Y }= k[x] 6= (x2 + 1) =

√(x2 + 1).

Corollary 3.10. From the previous results in this section it follows that the map

{closed subsets of An} → {radical ideals of k[x1, . . . , xn]}Z → I (Z),

defines a bijection.

3.1.3 Affine varieties and more bijections

To order to make the bridge between algebraic geometry and commutative algebrastronger, we first link An to prime ideals of k[x1, . . . , xn]. In order to do this, we needthe notion of irreducible sets.

Definition 3.11 (Irreducible). A topological space X is reducible if it is the union oftwo proper closed subsets, i.e. if there exist Y, Z ( X, closed in X, such that X = Y ∪Z.The space X is irreducible if it is not reducible.

The next theorem gives a link between prime ideals and irreducible subsets of An.

Theorem 3.12. A closed subset Y ⊂ An is irreducible if and only if the correspondingideal I (Y ) ⊂ k[x1, . . . , xn] is a prime ideal.

Proof. Let Y be an irreducible set and f, g ∈ k[x1, . . . , xn] be elements such that fg ∈I (Y ). It follows that Y ⊂ Z (f) ∪ Z (g). By the definition of irreducible sets thismeans that either Y ⊂ Z (f) or that Y ⊂ Z (g). This implies that either f or g is anelement of I (Y ) such that this is a prime ideal.

Conversely suppose that I (Y ) is prime, but that Y = Y1 ∪ Y2 is reducible. SinceYi 6= Y , there exists a fi ∈ I (Yi) −I (Y ). Then f1f2 vanishes on Y1 ∪ Y2 = Y , whichimplies that f1f2 ∈ I (Y ). Since I (Y ) was a prime ideal, this implies that one of f1

and f2 is in I (Y ). This gives a contradiction.

35

Corollary 3.13. With this we find the bijection

I : {Closed irreducible subsets of An} → {Prime ideals of k[x1, . . . , xn]}.

Definition 3.14 (Noetherian). A topological space X is Noetherian if for every de-scending chain of closed subsets

Z1 ⊃ Z2 ⊃ Z3 ⊃ . . . ,

there exists a k ∈ N such that Zk = Zk+i for all i ∈ N.

Notice that for Noetherian rings we considered ascending chains, whereas in the corre-sponding property for topological spaces we consider descending chains. In proposition3.4 we saw that for the descending chain (Zi)i∈N, we get an ascending chain of ideals{I (Zi)}i∈N, which makes the two properties correspond naturally.

An example of a topological Noetherian space is An itself. This follows from thebijection between the closed subsets of An and radical ideals of k[x1, . . . , xn] togehterwith the fact that k[x1, . . . , xn] is a Noetherian ring.

The following property of Noetherian spaces is important in defining affine varietiesand their coordinate rings.

Theorem 3.15. Let Y be a non-empty closed subset of a Noetherian topological space X.Then Y can be written as a finite union of closed irreducible subsets, say Y = Z1 ∪ . . .∪Zm, such that Zi * Zj for i 6= j. The collection of closed irreducible sets {Z1, . . . , Zn}is uniquely determined by Y and the Zi are called the irreducible components of Y .

We shall not prove this theorem because it is rahter long and not really to the pointhere. For a good proof, see for instance [10, p.5].

Definition 3.16 (Affine variety). An affine variety is a closed irreducible subset of An.A quasi-affine variety is a non-empty open part of an affine variety.

Since An is Noetherian, theorem 3.15 gives us that every closed subset Y of An canbe written uniquely as a union of its irreducible subsets.

Classically, an algebraic variety was defined to be the set of solutions of a system ofpolynomial equations. Modern definitions as the one here have generalized this, whileattempting to preserve the geometric meaning. In the definition as given below werequire that an algebraic variety be irreducible; the ones that are not irreducible are thealgebraic sets, as we have used before. With these ingredients we can define a coordinatering of a subset Y ⊂ An.

Definition 3.17 (Coordinate ring). For Y ⊂ An we call

A(Y ) = k[x1, . . . , xn]/I (Y )

the coordinate ring of Y .

36

From proposition 3.12 follows that

A(Y ) is a domain ←→ Y is irreducible.

This gives us that for an affine variety Y the coordinate ring is a finitely generatedk-algebra.

Example 3.18. Consider V ⊂ A2, where V = {(x, y) ∈ A2 | y2 = x} is the parabolagiven by y2 = x. Its coordinate ring is

A(V ) = k[x, y]/I (Y ) = k[x, y](y2 − x) ∼= k[y].

Corollary 3.19. Let Y be an affine variety. Then the map Z → I (Z)A(Y ) from

{closed irreducible subsets of Y } → {prime ideals of A(Y )}

gives a bijection. This map restricts to a bijection

Y → {maximal ideals of A(Y )}

where P = (a1, . . . , an) is mapped to mPA(Y ), a maximal ideal of A(Y ).

Let Y ⊂ An be an affine variety with coordinate ring A(Y ), and let ϕ : k[x1, . . . , xn]→A(Y ) be the canonical homomorphism. For an ideal IA(Y ) of the coordinate ring we can

write IA(Y ) = (f 1, . . . , fn) where fi ∈ k[x1, . . . , xn] and fi denotes the class (fi mod I (Y ) ∈A(Y ). Then ϕ−1(IA(Y )) = I (Y ) + (f1, . . . , fn), where the part on the right is indepen-dent of the chosen representatives fi.

For an ideal IA(Y ) ⊂ A(Y ) we need a new definition of the zero locus. We define thisas

Z (IA(Y )) = Z(ϕ−1(IA(Y ))

)⊂ Y.

Informally speaking, Z (IA(Y )) is obtained by adjoining the equations f1 = . . . = fn = 0to the equations for Y .

3.1.4 The geometric interpretation of example 1.1.1

We now make a little side step to give the geometric interpretation of example 1.1.1,where we found that

SpecZ[X] ={

(0)}∪{

(f)∣∣ for irreducible f ∈ Z[Y ]

}∪{m∣∣m is a maximal ideal in Z[X]

},

and that each maximal ideal is of the form m = (p, q) where p is a prime number andq ∈ Z[Y ] is a polynomial that modulo p is an irreducible element q ∈ Fp[Y ].

37

If k is an algebraically closed field, then the maximal ideals of k[X, Y ] are of the form(X − a, Y − b). This corresponds to the point (a, b) ∈ k2.

In a more general case where k is not necessarily algebraically closed, if we want ageometric interpretation of this maximal ideal as a point of the (X, Y ) plane, we mustallow X, Y to take values in an extension field F . This will help us in the same waythat pairs of complex conjugate roots help to understand roots of real polynomials.Therefore we look at the algebraic extension field k[X, Y ]/m = F of k. The quotienthomomorphism h : k[X, Y ]→ F maps X to a and Y to b and we find

m = k[X, Y ] ∩ (X − a, Y − b) ⊂ F [X, Y ],

where (X − a, Y − b) is the maximal ideal of F [X, Y ].

3.2 Maps between algebraic sets

Now it is time to talk about maps between algebraic sets. Recall that algebraic setsover (possibly) non-reducible algebraic varieties. We would like to single out specialmaps, the morphisms. We start with the definition of a regular function, these are easymorphisms; they map to the ground field k = A1. Regular functions should be thought ofas the analogue of continuous functions in topology, or holomorphic functions in complexanalysis. Of course, in algebraic geometry we want them to be algebraic functions, i.e.(quotients of) polynomial functions.

3.2.1 Regular functions

In this paragraph two definitions of regular functions will be given. The first is intuitivelyclearer, the second is needed for a fuller understanding of the concept. We start withthe definition of the field of rational (not to be confused with regular) functions.

Let X be a quasi-affine variety, let U1, U2 be non-empty open subsets of Y , let fi be arational function on Ui for i = 1, 2.. Two pairs (U1, f1) and (U2, f2) are called equivalentif f1 and f2 restricted to an open subset of U12 = U1 ∩ U2, are the same function. Theequivalence classes form a field K(X), with addition and multiplication defined by

[U1, f1] + [U2, f2] = [U12, f1 + f2], [U1, f1] · [U2, f2] = [U12, f1f2],

where we restrict f1 and f2 to functions on U12.

Definition 3.20 (Field of rational functions). Let X ⊂ An be an affine variety. Thequotient field K(X) as we just defined it is called the field of rational functions on X.

Formally, the quotient field K of an integral domain R is defined as the set of pairs(f, g) with f, g ∈ R and g 6= 0, modulo the equivalence relation

(f, g) ≡ (f ′, g′)←→ fg′ − gf ′ = 0.

Such a formal element (f, g) is usually denoted by fg

and we think of it as the formalquotient of two ring elements.

38

Definition 3.21 (ring of regular functions). Let X ⊂ An be an affine variety and P ∈ Xa point and A(X) the coordinate ring of X. We call

OX,P =

{f

g

∣∣∣∣ f, g ∈ A(X) and g(P ) 6= 0

}⊂ K(X)

the local ring of X at the point P . Elements of OX,P can be seen as the rational functionsthat are regular at P . For U ⊂ X a quasi-affine variety, we define

OX(U) =⋂P∈U

OX,P

as the ring of regular functions on U . This defines a subring of K(X) and an elementof OX(U) is called a regular function.

Example 3.22. In definition 3.21 we defined regular functions on an open subset ofan affine variety X ⊂ An to be rational functions, i.e. elements in the quotient fieldK(X), with certain properties. This means that every such function can be written asthe “quotient”of two elements in the coordinate ring A(X). This does not mean howeverthat we can always write a regular function globally as the quotient of two polynomials ink[x1, . . . , xn]. For example ([4, p.40]) let X ⊂ A4 be the variety defined by the equationx1x4 = x2x3, and let U ⊂ X be the open subset of the points in X for which x2 6= 0 orx4 6= 0. Then x1

x2is defined at all points of X where x2 6= 0 and in a similar way for x3

x4where x4 6= 0. If the functions are both defined on an open subset U , they coincide, sothat [

U,x1

x2

=x3

x4

]∈ K(X).

This gives rise to a regular function on U . However, there is no representation of thisfunction as a quotient of two polynomials in k[x1, x2, x3, x4] that works on all of U so wehave to use different representations at different points.

As we will usually want to write down regular functions as quotients of polynomi-als, we are interested in how they can be patched together from different polynomialrepresentations. The following definition of regular functions is therefore useful andequivalent to 3.21. For a proof of this equivalence, we refer to Gathmann, AlgebraicGeometry, Chapt. 2.

Definition 3.23 (Regular function). Let Y ⊂ An be a quasi-affine variety. Then wehave the following definitions:

i) Let P ∈ Y . Then a function f : Y → k is said to be regular at the point Pif there exists an open subset U ⊂ Y containing P together with polynomialsg, h ∈ k[x1, . . . , xn] where h is nowhere zero on U such that f |U = g/h as functionson U .

ii) The function f is called regular if it is regular at all points of Y .

39

iii) For U ⊂ Y an open subset we write OY (U) for the k-algebra of regular functionson U .

It is easy to see that sums, products and scalar multiples of regular functions areagain regular and that the ring of functions that are regular on all of X is precisely thecoordinate ring A(X).

Example 3.24. In this example ([4, p.41]), we will give an explicit computation ofthe ring OX(U) for U ⊂ X the complement of the zero locus of a single polynomialf : X → k, so

U = {P ∈ X | f(P ) 6= 0}.

We claim that

OX(U) = A(U)f :=

{g

f r

∣∣∣∣ g ∈ A(X) and r ≥ 0

}.

It is obvious that A(U)f ⊂ OX(U), so let us prove the converse. Let ϕ ∈ OX(U) ⊂ K(U)and define the ideal J of the coordinate ring of U as

J = {g ∈ A(U) | gϕ ∈ A(U)} ⊂ A(X).

If we can show that f r ∈ J for some r, we get that ϕ ∈ A(U)f and hence the equalitywe look for.

Let P ∈ U . Since ϕ ∈ OX(U), we can write ϕ = hg

with g nonzero in a neighbourhoodof P . In particular g ∈ J , so J contains an element that does not vanish at P . We findthat

Z (I (U)) + J ⊂ Z(f)

= Z {P ∈ X | f(P ) = 0}.

By theorem 3.4 we know that I(Z (f)

)⊂ I

(Z (I(U) + J)

)so that f ∈ I

(Z (I(U) +

J)), where f is a representative of f in k[x1, . . . , xn]. This implies that f r ∈ I (U) + J

for some r by the Nullstellensatz, and so f r ∈ J . Hence we find that the algebra ofregular functions OX(U) equals A(U)f .

As we discussed at the start of this paragraph, regular functions can be viewed asthe algebraic-gemetric analogue of continous functions in topology. The next importantresult will therefore not suprise you!

Proposition 3.25. For Y a quasi-affine variety and f : Y → k a regular function, f iscontinous for the Zariski topologies on Y and k.

Proof. Since continuity is a local notion and since we can write any regular functionf : Y → k locally as f = g/h for polynomials g 6= 0 and h, it suffices to consider thecase where f = g/h on Y . Here g and h are continous functions on Y so that f iscontinuous too with respect to the Zariski topology.

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3.2.2 Morphisms

With the notion of regular functions in the back of our head, we get to morphismsbetween quasi-affine varieties.

Definition 3.26 (morphism). For X, Y quasi-affine varieties, a morphism from X toY is a continuous map ϕ : X → Y such that for every open U ⊂ Y and every regularfunction f ∈ OY (U), the function

f ◦ ϕ : ϕ−1(U)→ k

is regular on ϕ−1(U) ⊂ X.

Definition 3.27. If ϕ : X → Y is a morphism, we denote by

ϕ∗ : OY (Y )→ OX(X)

the homomorphism of k-algebras given by f → f ◦ ϕ.

Intuitively it should be clear that morphisms are maps that are locally given by rationalmaps.

The simplest example of a morphism is the identity map on X. Another example isthe inclusion map i : X → An, for X a quasi-affine variety. For a morphism to an affinevariety we have a more interesting example which shows that these morphisms can notjust locally, but even globally be given by polynomials.

Example 3.28. We claim that for X ⊂ An, Y ⊂ Am affine varieties, every morphismf : X → Y corresponds uniquely to a k-algebra homomorphisms f ∗ : A(Y ) → A(X)([10, p.16]), i.e. we obtain a bijection{

morphisms X → Y}→ {k-algebra homomorphisms OY (Y )→ OX(X)

}.

To see this, let ϕ, ψ : X → Y be morphisms where Y = Z (p) ⊂ Am for some primeideal p ⊂ k[x1, . . . , xm]. Suppose that ϕ∗ = ψ∗. Then ϕi := xi ◦ ϕ = ϕ∗(xi) equalsψi = xi ◦ ψ = ψ∗xi for all i = 1, . . . , n. However as discussed before, ϕ : X → Y is themap given by P →

(ϕ1(P ), . . . , ϕn(P )

)and the same holds for ψ so that ϕ = ψ.

Conversly, if

g : OY (Y ) = k[y1, . . . , ym]/p1 → k[x1, . . . , xn]/p2 = OX(X)

is any k-algebra homomorphism between the coordinate rings of Y and X, let us defineϕi = g(xi), where the bar indicates the residue class modulo p. Now this determines apolynomial map ϕ(P ) =

(ϕ1(P ), . . . , ϕn(P )

). This map is well-defined for if f ∈ p then

f(ϕ(P )

)= g(f)

(P), as f = 0. Note here that this means that ϕi are regular functions!

Since ϕ∗ = g by construction, it remains to prove that ϕ indeed defines a morphisms.This follows from the following

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Lemma 3.29. Let Y ⊂ An be a quasi-affine variety, and let x1, . . . , xn be the coordinatefunctions on An. If X is a quasi-affine variety and ϕ : X → Y is a map, then ϕ is amorphisms if and only if the functions ϕi = xi ◦ ϕ for i = 1, . . . n are regular functionson X.

Proof. Note that ϕ is the map given by P →(ϕ1(P ), . . . , ϕn(P )

). Because coordinate

functions are regular we find that if ϕ is a morphism this implies that ϕi are regularfunctions. Now suppose that ϕi are regular functions for all i = 1, . . . , n. We willfirst prove that ϕ is continous after which we shall show that for f regular on an openV ⊂ Y , the function f ◦ ϕ is regular on ϕ−1(U). By definition this would imply that ϕis a morphism.

To prove continuity, let Z ⊂ Y be closed. Then there are polynomials g1 . . . , gr ∈k[x1, . . . , xn] such that Z = Y ∩ Z (g1, . . . , g4) so that ϕ−1(Z) ⊂ X is the intersectionof the sets (gj ◦ φ)−1(0). Since ϕ1, . . . ϕn are regular by assumption the functions gj ◦ ϕare regular too. Regular functions are continous and hence ϕ−1(Z) is closed in X. Thisproves that ϕ is a continous function.

For the second part, let P ∈ X be a point, setϕ(P ) = Q ∈ Y and let f be aregular function on an open set V ⊂ Y which contains Q. Then by definition of aregular function, there exists an open neighbourhood U ⊂ V of Q and polynomialsg, h ∈ k[x1, . . . , xn] with h nowhere zero on U , such that f |U = g/h. It follows thatf ◦ϕ = (g◦ϕ)/(h◦ϕ) as functions on ϕ−1(U). Now h◦ϕ is still nowhere zero on ϕ−1(U),hence f ◦ ϕ is a regular function on ϕ−1(U). This proves that ϕ is a morphism.

Hopefully you have not forgotten where we are going: the geometric interpretation ofNoether’s normalization lemma! We are almost there, the last three definitions we needare that of a dominant, proper and a finite morphism. The first one is rather easy!

Definition 3.30 (dominant morphism). A morphism ϕ : X → Y between two varietiesis said to be dominant if the image of ϕ is dense in Y .

Definition 3.31 (finite morphism). A morphism of varieties ϕ : X → Y is called finiteif one of three equivalent conditions holds:

i) There exists an affine open covering Y = ∪mj=1Vj such that for every index j theinverse image Uj = ϕ−1(Vj) is again affine, and such that A(Uj) is finitely generatedas a module over A(Vj).

ii) For every affine open V ⊂ Y , the pre-image U = ϕ−1(V ) is again affine, and A(U)is finitely generated as a module over A(V ).

iii) The function ϕ is a dominant morphism of affine varieties and k[X] is integral overf ∗(k[Y ]) ' k[Y ].

A proof of this equivalence can be found in [10, p.75]A morphism is said to be quasi-finite if for every y ∈ Y the fiber ϕ−1(y) is finite. That

this is a weaker property than finite follows from the next proposition, as stated andproved in [5, p.43].

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Proposition 3.32. Let f : X → Y be a finite morphism of affine varieties, then

i) for any P ∈ Y the set f−1(P ) is finite

ii) the map f is surjective if dimX = dimY .

iii) the map f is closed (even proper as we define next).

We next define a proper morphism. With this we can link the definition of a finitemorphism to quasi-finite and proper morphism, so that we get a more intuitive feel ofwhat it means for a morphism to be finite.

Definition 3.33 (proper morphism). A morphism of varieties ϕ : X → Y is called aproper morphism if for all varieties Z the morphism ϕ× Idz : (X ×Z)]→ (Y ×Z) is aclosed map.

The next proposition links the above notions. A proof of this can be found in [10,p.76].

Proposition 3.34. A morphism ϕ : X → Y is finite if and only if it is quasi-finite andproper.

In conclusion we have that a finite morphism has finite fibres that are nonzero every-where.

3.3 Geometric interpretation of Noether’s

normalisation lemma

We are here! We now have all the tools we need for an understanding of Noether’slemma. The geometric meaning is as follows. Every affine variety is a finite cover ofsome affine space. In other words, it admits a finite dominant map to Ad. Moreover, if Zis an algebraic set in An a linear change of coordinates in An makes that the projectionof Z to the first d coordinates is a finite morphism. In the figure φ represents the finitemorphism.

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More precise, the geometric interpretation of Noether’s normalisation lemma is asfollows:

Theorem 3.35 (Noether’s normalisation lemma). Let X ⊂ An an affine variety. Thenthere exists an affine space Am such that the projection ϕ : X → Am is a finite morphism.

Proof. This proof follows the one given in [6, p. 33]. Let X ⊂ An be a variety andcosider the coordinate ring

A = k[a1, . . . , an] = k[x1, . . . , xn]/I (X),

where ai := xi mod I. Since this is a finitely generated k-algebra, the algebraic versionof Noether’s normalisation says that there exist algebraically independent linear formsy1, . . . , ym in a1, . . . , an such that A is a finite k[y1, . . . , ym]-algebra.

These linear forms lift (nonuniquely) to linear forms y1, . . . , ym in x1, . . . , xn, whereyi ≡ yi mod I. With these linear forms, we can define a projection to Am by

π := (y1, . . . , ym) : An → Am.

If we now restrict π to X we obtain a map

ϕ|X : X → Am.

Note here that ϕ is independent of the choice of lifts yi since on X the equality

yi|X = yi|X

holds.

Note that by proposition 3.34, to prove that ϕ is finite, it suffices to prove that forevery point P ∈ Am the fiber ϕ−1(P ) is finite and nonempty.

The tower laws imply that we can find an integer N and polynomials fi,0, . . . , fi,N−1

for 1 ≤ i ≤ n, such that

aNi + fi,N−1(y1, . . . , ym)aN−1i + . . .+ fi,0(y1, . . . , ym) = 0.

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This implies that there exists a gi ∈ I (X) such that

xNi + +fi,N−1(y1, . . . , ym)xN−1i + . . .+ fi,0(y1, . . . , ym) = gi(x1, . . . , xn).

If x = (x1, . . . , xn) ∈ X we find that g(x) = 0. If we define

fi := xN + +fi,N−1(y1, . . . , ym)xN−1 + . . .+ fi,0(y1, . . . , ym),

we find that fi(x) = 0.Since I (X) is a prime ideal, the coordinate ring is an integral domain and has field of

fractions k(a1, . . . , an). By the fundamental theorem of algebra, there are only finitelymany solutions to f i(x) = 0. This implies that the fibres of ϕ are finite.

What remains to show is that ϕ is surjective. As we explained earlier, for P =(b1, . . . , bn) a point in Am, we have that I (P ) = k[x1, . . . , xn] only if Z (I (P )) = ∅.Therefore it is sufficient to prove that I (P ) 6= k[x1, . . . , xn]. This is equivalent toshowing that (y1−b1, . . . , ym−bm) 6= k[a1, . . . , zn]. Since this is a maximal ideal, unequalto k[a1, . . . , an], the theorem follows from Nakayama’s lemma with B = k[y1, . . . , ym],A = k[a1, . . . , an] and m = (y1 − b1, . . . , ym − bm).

We will illustrate this theorem with two intuitive examples:

Example 3.36. Condiser A = k[X, Y ]/(XY −1). Then Y is algebraic, but not integralover k[X]. This corresponds geometrically to the fact that the hyperbola XY = 1 hasthe line X = 0 as an asymptotic line so that the projection to the X-axis has a ‘missing’root over X = 0. The solution is however rather simple. We tilt the hyperbola a littlebefore projecting it, as illustrated in figure 4.1, such that it no longer has a verticalasymptotic line. This gives us a projection which is both surjective and has finite fibres.Algebraically this corresponds to taking X ′ = X − εY . Note that X ′ is not algebraicover k[X]. Now look at the relation

(X ′ + εY )Y = 1.

For ε 6= 0 this equation is monic in Y .In general this means that for the unlucky choice where yn is not integral over

k[y1, . . . , yn−1] we can change the choice of y1, . . . , yn−1 to make yn integral over k[y1, . . . , yn−1].

Example 3.37. Define f as f := x1x2 + x2x3 + x3x1 ∈ k[x1, x2, x3] and let S :=Z (f) ⊂ A2

k. We have A = k[x1, x2, x3]/(f) as the coordinate ring and we use thenotation xi := xi mod (f) for i = 1, 2, 3.. We search for a map ϕ : S → A2 thathas finite fibres. For this we have to make a change of coordinates. For example, letz = x2 + x1. Then we have f = z(x1 + x3)− x2

1. Now A is algebraic over k[a1 + a2, a3],and the corresponding map ϕ is given by

ϕ : S → A2k,

(x1, z, x3)→ (z, x3).

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Figure 3.1: Projecting the hyperbola XY = 1.

We see that the fibres are ϕ−1(a, b) = {(x, a, b) |x2 − ax − ab} and that they consistof at most two points. Note that S contains the coordinate axis so that the projectionto the xi, xj plane

S → A2

(x1, x2, x3)→ (xi, xj),

has an infinite fiber over (0, 0). For (a, b, c) ∈ S something similar holds: S contains theline L; = {(λa, λb, λc) |λ ∈ k} which is mapped to (0, 0) by the projection (x1, x2, x3)→(bx1 − ax2, cx1 − ax3). However, there is a dense set of hyperplanes such that thecorresponding projection has finite fibres.

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Reflection

The aim of this bachelor thesis was to discuss Noether’s normalisation lemma from boththe algebraic and the geometric point of view. To obtain a full understanding of thislemma, I discuss related concepts in commutative algebra as well as algebraic geometry.The first chapter is an introduction to commutative algebra, where I explain notions ofthe spectrum of a ring, modules, finite algebras and the normalisation of a ring.

In chapter 2, a comprehensive proof of Noether’s normalisation lemma is given, fol-lowing the outline of the proof given in [7, p.63].

The third chapter then explains a bridge between algebraic geometry and commutativealgebra: This links polynomial rings to geometric objects by a one-to-one correspondencebetween prime ideals of the coordinate ring of an affine variety, and the closed ireeduciblesubsets of Y . The chapter concludes with the geometric interpretation of Noether’slemma.

Noether’s normalisation lemma is important for many further results. Some are dis-cussed in chapter 4, where we use the theorem to prove Hilbert’s weak Nullstellensatzand consequently, the strong version of the Nullstellensatz. However, there are manymore application which would be interesting to consider. Furthermore, it would be

The concepts discussed in this thesis should give an intuitive feeling of how algebraicgeometry and commutative algebra are intertwined. It was very interesting to considerone theorem in two worlds. At first, it seemed to be two totally different statements,but after reading my thesis I hope you agree that they are not!

Although I was not able to obtain the perfect overview I would have liked, this thesishas given me a lot of new insights. It also made me realise how much fun it can be totry to completely understand a small part of mathematics.

In conclusion, I would like to give a word of thanks to my supervisor, Arie Peter-son, for his enthausiastic feedback and his good teaching skills, which really helped meunderstand the intuitions behind the definitions and theorems.

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Populaire samenvatting

‘Het normalisatie lemma van Noether’, een titel die misschien nog niet gelijk tot deverbeelding spreekt. Echter, Emmy Noether is een van de belangrijkste in de geschiedenisvan de wiskunde en haar normalisatie lemma is een resultaat uit het (naar mijn mening)mooiste deelgebied uit de wiskunde.

Allereerst, wie was Emmy Noether? Amalie Emmy Noether werd geboren in 1882en is bekend geworden om haar baanbrekende bijdrage aan de commutatieve algebra entheoretische natuurkunde. Zelfs Einstein noemde haar een genie. Noether’s normalisatielemma is een belangrijk resultaat uit de commutatieve algebra en staat centraal in mijnscriptie.

Het vakgebied. In commutatieve algebra staat de studie van commutatieve ringencentraal. Ringen zijn verzamelingen, bijvoorbeeld getallen of polynomen, waarvoor wede elementen kunnen optelllen en vermenigvuldigen. Hierbij moet de uitkomst ook altijdweer een element uit de verzameling zijnt. Voor commutatieve ringen geldt dat voor devermenigvuldiging van twee elementen a en b uit de ring,

a · b = b · a.

Typisch, vragen uit de commutatieve algebra gaan over eigenschappen van ringen, of overeigenschappen van objecten die aan de ring gerelateerd zijn, zoald idealen en modules.Zo kun je jezelf bijvoorbeeld afvragen of er een unieke manier bestaat om een element uiteen ring te schrijven als het product van irreducibele elementen. Hierbij is een elementp irreducibel als het niet geschreven kan worden als het product twee elementen, allebeiongelijk aan p.

Commutatieve algebra is nodig voor het ontwikkelen van getaltheorie en meetkunde,allebei deelgebieden van de wiskunde met veel ‘echte wereld’ toepassingen, zoals coder-ingssystemen en het besturen van robots.

Het normalisatie lemma van Noether. Het lemma van Noether geeft een structuurgeeft aan een bepaald soort ringen. Dit zijn de ringen die eindig gegenereerd zijn alsk-algebra.

Een k-algebra mag worden gezien als een uitbreiding van het begrip vectorruimte uitde lineaire algebra, waarbij in een algebra, naast optelling en scalaire vermenigvuldiging,ook de vermenigvuldiging van elementen (vectoren) onderling mogelijk is. De k staathierbij voor de scalars die bij de vermenigvuldiging horen, waarbij k een lichaam is. Ditis een ring waarin elk element een inverse heeft.

48

Een k-algebra A is eindig gegenereerd, als alle elementen geschreven kunnen wordenals lineaire combinaties van een eindig aantal elementen y1, . . . yn.

Voor deze eindig gegenereerde k-algebra, geeft het lemma van Noether een struc-tuur. Deze houdt in dat de k-algebra een eindige (ring) uitbreiding is van de ringB = k[z1, . . . , zn]. Hoe dit precies in zijn werk gaat wordt uitgelegd in 1.33.

De stelling is een belangrijk resultaat in de commutative algebra omdat het essentieelis voor het bewijs van veel andere stellingen, zoals onder andere Hilbert’s zwakke ensterke Nullstellensatz.

Een tweede interpretatie. Een ander gebied in de wiskunde dat ons interesse heeftin deze scriptie is meetkunde. We zouden graag het normalisatie lemma van Noethernogmaals bekijken, maar dan nu met een meetkundige bril op. Om dit te kunnen doen,moeten we een manier vinden om wiskundige dingen uit de ene (commutatieve algebra)wereld te vertalen naar dingen uit de andere (meetkundige) wereld. We bouwen dus eenbrug tussen de algebra en de meetkunde.

De reden dat dit kan, is omdat de algebraısche meetkunde en commutatieve algebranauw met elkaar verwoven zijn. Dit verband gaat terug tot de zeventiende eeuw waarDescartes het Cartesisch coordinaten stelsel introduceerde. Dit gaf de eerste mogeli-jkheid voor een brug tussen de meetkunde en algebra. Nadat ook complexe getallenwaren ingevoerd en er was bewezen dat elk polynoom ongelijk aan nul een nulpuntheeft, kon deze brug precies gemaakt worden. Zo kunnen we voor een verzameling vanpolynomen, die we even S noemen, kijken op welke punten deze allemaal de uitkomst 0geven. Dit noemen we de zero locus van de verzameling S, wat we noteren als Z (S).We krijgen dus de verzameling punten

Z (I) = {(a1, . . . , an) ∈ An | f(a1, . . . , an) = 0, zodat f in S zit},

waarbij An = kn een affiene ruimte is. Intuitief is een affiene ruimte een vector ruimtewaarin 0 geen speciale rol vervult. We kunnen Z zien als een afbeelding van een verza-meling polynomen uit een polynoomring naar een meetkundige ruimte, zoals bijvoorbeeldhet vlak.

Op een zelfde manier kunnen we een afbeelding I definieren. Deze afbeelding bekijktvoor een verzameling punten X ⊂ An, welke polynomen in n variabelen er zijn, die vooralle punten uit X de uitkomst nul geven. Dus,

I (X) = {f ∈ k[x1, . . . , xn] | f(x1, . . . , xn) = 0 voor alle punten (a1, . . . , an) ∈ X}.

We vinden de eerste correspondentie tussen de twee werelden:

49

In hoofdstuk 4 wordt dit verder uitgewerkt zodat we aan gesloten verzameling uit demeetkundige ruimte, uniek een priem ideaal kunnen koppelen.

Een tweede interpretatie van Noether’s normalisatie lemma. Met deze brugkunnen we nu het resultaat van Noether uit de commutatieve algebra, vertalen naareen zinvol resultaat in de algebraısche meetkunde. Noether’s normalisatie lemma geeftdan dat we een gesloten verzameling X uit de affiene ruimte An kunnen projecterenop een vlak met dezelfde dimentie als X, zodat deze projectie een eindige afbeeldingis. Een eigenschap van een eindige afbeelding is bijvoorbeeld dat het maar eindig veelpunten hetzelfde beeld hebben onder een functie. Dit heet eindige vezels in de wiskunde.Schematisch geeft dit het volgende plaatje:

50

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