math 746 notes topics in ring theory

Math 746 NotesTopics in Ring Theory

Lectures by Daniel ErmanNotes by Daniel Hast

Spring 2014

Contents

I Dimension theory 7I.1 2014-01-21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

I.1.1 Course information . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7I.1.2 Crash review on geometry . . . . . . . . . . . . . . . . . . . . . . . . 7I.1.3 Motivating questions . . . . . . . . . . . . . . . . . . . . . . . . . . . 8I.1.4 Axioms for dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . 8I.1.5 Krull dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

I.2 2014-01-23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9I.2.1 Dimension of ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9I.2.2 Dimension of modules . . . . . . . . . . . . . . . . . . . . . . . . . . 10I.2.3 Integral extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10I.2.4 Completions and flatness . . . . . . . . . . . . . . . . . . . . . . . . . 11

I.3 2014-01-28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12I.3.1 Completions and flatness, continued . . . . . . . . . . . . . . . . . . . 12I.3.2 Principal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12I.3.3 Proof of the Principal Ideal Theorem . . . . . . . . . . . . . . . . . . 13I.3.4 Consequences of the Principal Ideal Theorem . . . . . . . . . . . . . 13

I.4 2014-01-30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13I.4.1 Krull’s Principal Ideal Theorem, continued . . . . . . . . . . . . . . . 14I.4.2 Dimension of local rings . . . . . . . . . . . . . . . . . . . . . . . . . 14I.4.3 Restatement of the Principal Ideal Theorem . . . . . . . . . . . . . . 15I.4.4 Dimension of fibers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

I.5 2014-02-04 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16I.5.1 Dimension of base and fiber . . . . . . . . . . . . . . . . . . . . . . . 16I.5.2 Going down for flat extensions . . . . . . . . . . . . . . . . . . . . . . 16I.5.3 Corollaries of dimension theorem . . . . . . . . . . . . . . . . . . . . 17

I.6 2014-02-06 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18I.6.1 Nakayama’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18I.6.2 Regular local rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18I.6.3 Regular sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19I.6.4 Complete regular local rings . . . . . . . . . . . . . . . . . . . . . . . 19I.6.5 Affine rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

I.7 2014-02-13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20I.7.1 Dimension of polynomial rings . . . . . . . . . . . . . . . . . . . . . . 20

3

4 CONTENTS

I.7.2 Weak Noether normalization . . . . . . . . . . . . . . . . . . . . . . . 21I.7.3 Strong Noether normalization . . . . . . . . . . . . . . . . . . . . . . 21I.7.4 Transcendence degree and dimension . . . . . . . . . . . . . . . . . . 22

I.8 2014-02-18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22I.8.1 Transcendence degree and dimension, continued . . . . . . . . . . . . 22I.8.2 Going down . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23I.8.3 End of proof of Theorem A . . . . . . . . . . . . . . . . . . . . . . . 23I.8.4 Catenary rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23I.8.5 Remarks on past homework . . . . . . . . . . . . . . . . . . . . . . . 24I.8.6 Axioms of dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

I.9 2014-02-20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24I.9.1 Weak Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . . . . . . 24I.9.2 Finiteness of integral closure . . . . . . . . . . . . . . . . . . . . . . . 25I.9.3 Problems with Noetherian rings . . . . . . . . . . . . . . . . . . . . . 26

II Gröbner bases 27II.1 2014-02-25: Monomial orders and Gröbner bases . . . . . . . . . . . . . . . . 27

II.1.1 Graded rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27II.1.2 Graded modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28II.1.3 Monomial orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

II.2 2014-02-27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29II.2.1 Monomial ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29II.2.2 Division algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30II.2.3 Gröbner bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

II.3 2014-03-04: Buchberger’s algorithm . . . . . . . . . . . . . . . . . . . . . . . 31II.3.1 Reduced Gröbner bases . . . . . . . . . . . . . . . . . . . . . . . . . . 31II.3.2 Buchberger’s criterion . . . . . . . . . . . . . . . . . . . . . . . . . . 32II.3.3 Buchberger’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . 32

II.4 2014-03-06: Some algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . 33II.4.1 Buchberger’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . 33II.4.2 Organizing a proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34II.4.3 Ideal membership . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34II.4.4 Elimination of variables . . . . . . . . . . . . . . . . . . . . . . . . . 35II.4.5 Geometry of elimination . . . . . . . . . . . . . . . . . . . . . . . . . 36

II.5 2014-03-11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36II.5.1 Elimination theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36II.5.2 Solving polynomial equations . . . . . . . . . . . . . . . . . . . . . . 37II.5.3 Implicitization, i.e., computing kernels . . . . . . . . . . . . . . . . . 37II.5.4 Remarks on computation . . . . . . . . . . . . . . . . . . . . . . . . . 38II.5.5 Radical membership . . . . . . . . . . . . . . . . . . . . . . . . . . . 38II.5.6 Dimension of a ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

II.6 2014-03-13: Flatness and Gröbner bases [incomplete] . . . . . . . . . . . . . 39II.6.1 Project progress report . . . . . . . . . . . . . . . . . . . . . . . . . . 39II.6.2 Flatness and Gröbner bases . . . . . . . . . . . . . . . . . . . . . . . 40

II.7 2014-03-25: Flatness and Gröbner bases, continued . . . . . . . . . . . . . . 40

CONTENTS 5

II.8 2014-03-27: Hilbert polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 42II.8.1 Hilbert polynomials and short exact sequences . . . . . . . . . . . . . 42

IIIHomological methods 43III.1 2014-03-27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

III.1.1 Complexes of R-modules . . . . . . . . . . . . . . . . . . . . . . . . . 43III.2 2014-04-01: Minimal free resolutions . . . . . . . . . . . . . . . . . . . . . . 43

III.2.1 Minimal free resolutions over a local ring . . . . . . . . . . . . . . . . 43III.2.2 Conjectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44III.2.3 Minimal free resolutions in the graded case . . . . . . . . . . . . . . . 45

III.3 2014-04-08: Koszul complexes . . . . . . . . . . . . . . . . . . . . . . . . . . 45III.4 2014-04-10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45III.5 2014-04-17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

III.5.1 Projective dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . 46III.5.2 Global dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

III.6 2014-04-22: Cohen–Macaulay rings . . . . . . . . . . . . . . . . . . . . . . . 48III.6.1 Cohen–Macaulay rings . . . . . . . . . . . . . . . . . . . . . . . . . . 48

III.7 2014-04-29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50III.7.1 Easy step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50III.7.2 Homological proof, main step . . . . . . . . . . . . . . . . . . . . . . 51III.7.3 Gröbner basis proof, main step . . . . . . . . . . . . . . . . . . . . . 51

6 CONTENTS

Chapter I

Dimension theory

I.1 2014-01-21

I.1.1 Course information

Course website: http://www.math.wisc.edu/~derman/746.htmlBackground ≈ [E], 2–4.Useful program: Macaulay2Grading:

• 50% homework, due weekly.

• 50% final project.

I.1.2 Crash review on geometry

Consider I = (f1, . . . , fr) ⊆ C[x1, . . . , xn]. Write

V (I) = {f1 = f2 = · · · = fr = 0} ⊆ Cn.

This is an affine algebraic variety. The geometry of V (I) corresponds to algebraic propertiesof C[x1, . . . , xn]/I. The local geometry of V (I) near (a1, . . . , an) corresponds to algebraicproperties of the localization (

C[x1, . . . , xn]

I

)(x1−a1,...,xn−an)

.

A map of ringsC[x1, . . . , xn]

I→ C[x1, . . . , xn]

J

induces a map of varieties V (J)→ V (I).

7

http://www.math.wisc.edu/~derman/746.html

8 CHAPTER I. DIMENSION THEORY

I.1.3 Motivating questions

(1) What is the dimension of V (I)?

(2) How does the algebra of C[x1, . . . , xn]/I encode topological information (e.g., the genusof a Riemann surface) about V (I)?

(3) Which results from algebraic geometry hold over a field like Q,Fq, . . . ?

(4) Consider the generic q × p matrix (xij). Let I ⊆ Z[x11, . . . , xij, . . . , xpq] be the idealgenerated by the maximal minors of the matrix. Is I a prime ideal? (It is.)

(5) Consider the map Z⊕3 → Z⊕2 given by[1 2 34 5 6

], giving an exact sequence

0→ K → Z⊕3 → Z⊕2 → Q→ 0.

What are Q and K?

Let’s answer (5). By looking at ranks, K ∼= Z. Explicitly, K is generated by (1,−2, 1).It turns out that Q = Z/3Z. This is because, as it turns out,√

annQ =√

ideal of maximal minors of presentation matrix.

I.1.4 Axioms for dimension

We will first treat dimension axiomatically.

(Axiom 1) Dimension is a local property:

dimR = supP⊆R prime

dimRP

anddimRP = dim RP .

(Axiom 2) Nilpotent ideals do not affect dimension: if I ⊆ R is nilpotent, then

dimR = dimR/I.

(Axiom 3) Dimension is preserved by a “map with finite fibers”: if R ⊆ S is a finite extension,then

dimR = dimS.

Definition I.1.1. R ⊆ S is a finite extension of rings if S is a finitely generated R-module.

Axiom 3 is motivated by the following:

Theorem I.1.2 (Sard’s theorem). If f : M → N is a smooth map of smooth manifolds, andn ∈ N is a general point, then

dimM = dimN + dim f−1(n).

(Axiom 4) If k is a field, then dim k[[x1, . . . , xn]] = n. IfR is a DVR, then dimR[[x2, . . . , xn]] =n.

I.2. 2014-01-23 9

I.1.5 Krull dimension

Definition I.1.3. dimR is the supremum of the length of a chain of prime ideals in R,where

P0 $ P1 $ · · · $ Pr

has length r.

Example I.1.4. If dimR = 0, then every prime ideal is maximal.Example I.1.5. Let R be a DVR with uniformizer π. (For example, R = k[[t]] with π = t,or R = Zp with π = p.) Then dimR = 1 and the unique maximal chain of prime ideals is(0) ⊆ (π).Example I.1.6. If R = k[x1, . . . , xn], then we have

(0) ⊆ (x1) ⊆ (x1, x2) ⊆ · · · ⊆ (x1, . . . , xn),

so dimR ≥ n. (Later, we’ll see dimR = n.)

Definition I.1.7. Let I ⊆ R be an ideal. Define

dim I = dim(R/I),

which is the supremum of the length of a chain

I ⊆ P0 $ · · · $ Pn.

Definition I.1.8. Let I ⊆ R be an ideal. The codimension (or height) of I is

codim I = height I = minP⊇I

min. prime

dimRP .

I.2 2014-01-23

I.2.1 Dimension of ideals

Recall: Let R be a commutative ring, and let P ⊆ R be a prime and I ⊆ R be any ideal.Then there are natural bijections

{primes in RP} ←→ {primes in R contained in P} ,{primes in R/I} ←→ {primes in R containing I} .

Last time, we defined dim I := dim(R/I).

Lemma I.2.1. dim I is the sup of the length of a chain of primes ascending from I.

Proof. Given a chain of primes in R

I ⊆ P0 $ P1 $ · · · $ Pr,

we get a chain of primes in R/I

0 ⊆ P0 $ P1 $ · · · $ Pr,

where Pi = Pi(R/I), and conversely.


I.2.2 Dimension of modules

Definition I.2.2. Let M be an R-module. Then we define

dimM := dim(annM),

codimM := codim(annM),

where we recall thatannM =

{r ∈ R

∣∣ rm = 0 ∀m ∈M}.

Is it true that dim I + codim I = dimR? In general, no. However, it is true if R is anintegral domain.

Example I.2.3. Consider R =C[x, y, z]

(xz, yz). This is the union of the (x, y)-plane and the z-axis.

Let I = (z − 1)R = (z − 1, x, y)R. Then

dim I = dimR/I = dimC[x, y, z]

(z − 1, x, y)= 0.

Moreover,

RI =

(C[x, y, z]

(xz, yz)

)(z−1,x,y)

=

(C[x, y, z]

(x, y)

)(z−1)

= C[z](z−1)

is a DVR, socodim I = min

P⊇I minimalRP = dimRI = 1.

But we have a chain of primes (z) ⊆ (x, z) ⊆ (x, y, z) in R, so

dimR ≥ 2 > 0 + 1 = dim I + codim I.

I.2.3 Integral extensions

Theorem I.2.4 ([E], §4). Let R ⊆ S be an integral extension of rings. Then:

(1) Lying over: If P ⊆ R is prime, then there exists Q ⊆ S prime with Q ∩R = P .

(2) Going up: Let P1 ⊆ P2 ⊆ R and Q1 ⊆ S be primes with Q1 ∩ R = P1. Then thereexists Q2 ⊆ S prime with Q1 ⊆ Q2 and Q2 ∩R = P2.

(3) Incomparability: If Q ⊆ Q′ ⊆ S are prime with Q ∩R = Q′ ∩R, then Q = Q′.

This lets us prove (Axiom 3): If R ⊆ S is a finite extension, then dimR = dimS.

Proposition I.2.5 ([E], Prop. 9.2). Let ψ : R→ S be a map making S integral over R. LetI ⊆ S be any ideal. Then dim I = dimψ−1I.

I.2. 2014-01-23 11

Proof. Replace R by its image and assume kerψ = 0, and hence R ⊆ S integral. Considera chain in R:

ψ−1I ⊆ P0 $ P1 $ · · · $ Pr.

By going up, we get a chain

I ⊆ Q0 $ Q1 $ · · · $ Qr

in S with Qi ∩R = Pi for i = 0, 1, . . . , r. Hence dim I ≥ dimψ−1I.On the other hand, given a chain

I ⊆ Q0 $ Q1 $ · · · $ Qr ⊆ S,

contracting to R yields a chain

I ∩R ⊆ Q0 ∩R ⊆ Q1 ∩R ⊆ · · · ⊆ Qr ∩R ⊆ R,

and by incomparability, this chain has the same length.

Corollary I.2.6. Krull dimension satisfies (Axiom 3).

Proof. R ⊆ S is finite, hence integral. (Indeed, for s ∈ S, let µs ∈ HomR(S, S) denote mul-tiplication by s. By the Cayley–Hamilton theorem, µs satisfies its characteristic polynomial.[E, §2]) By the above proposition,

dimS = dim 0S = dim 0R = dimR.

Exercise I.2.7. If R ⊆ S is a finite extension of Noetherian rings, then Spec(S) → Spec(R)has finite fibers.

I.2.4 Completions and flatness

The most important fact about completion is that it’s flat.Given an R-module N and a short exact sequence

0→M ′ →M →M ′′ → 0,

tensoring yields an exact sequence

M ′ ⊗R N →M ⊗R N →M ′′ ⊗R N → 0.

Definition I.2.8. N is a flat module if −⊗R N is exact, i.e., for any exact sequence

0→M ′ →M →M ′′ → 0,

we obtain a short exact sequence

0→M ′ ⊗R N →M ⊗R N →M ′′ ⊗R N → 0.

Lemma I.2.9. Let P ⊆ R be prime. Then RP is a flat R-module.

Proof. [AM, Prop. 3.3] or [E, Prop. 2.5].

Definition I.2.10. Let (R,m) be a local ring. The completion of R is

R := lim←−R/mi =

{(f1 + m1, f2 + m2, . . . ) ∈

∏R/mi

∣∣ fj − fi ∈ mi ∀i < j}.


I.3 2014-01-28

I.3.1 Completions and flatness, continued

Let R be a ring and m ⊆ R a maximal ideal. We defined

R = lim←−R/mi =

{(x0, x1, x2, . . . )

∣∣ xi ∈ R/mi, xi = xi−1 mod mi−1}.

Example I.3.1. (1) Let R = Z and m = (p), where p is a prime number. Then R = Zp isthe ring of p-adic integers.

(2) If R = k[x1, . . . , xn] and m = (x1, . . . , xn), then R = k[[x1, . . . , xn]].

Theorem I.3.2. R is a flat R-module.

(We’ll prove this later.)E.g., if M is an R-module, then M ∼= M ⊗R R.

Remark I.3.3. Geometrically, completion corresponds to a “much smaller neighborhood” ofthe point. For example, if

R =k[x, y]

(y2 − x3 − x2),

then any localization of R is an integral domain, but if we complete R at the node m = (x, y),then R has zero divisors.

I.3.2 Principal ideals

Theorem I.3.4. If (x) is a proper principal ideal, then any prime p $ (x) has codimension0.

Proof. Suppose there exist Q $ P $ (x) with Q,P prime. Mod out by Q: we are in theintegral domain S = R/Q, where we still have 0 6= P $ (x) in S.

Let y ∈ P . Then y = ax for some a ∈ R. Since x /∈ P and P is prime, it follows thata ∈ P . Hence, P = xP . By Nakayama’s lemma, there exists b ∈ (x) such that (1− b)P = 0.But P has nonzero elements and S is an integral domain, so 1 − b = 0; hence, b = 1 and(x) = S, a contradiction.

Krull proved a stronger statement:

Theorem I.3.5 (Krull’s Principal Ideal Theorem). Let R be a Noetherian ring, and letP ⊆ R be a prime ideal minimal over (x). Then codimP ≤ 1.

Let us first recall some auxiliary results.

Proposition I.3.6. The following are equivalent for I ⊆ P with P prime and I any ideal:

(1) P is minimal over I.

(2) RP/IP is Artinian.

I.4. 2014-01-30 13

(3) In RP , for all n� 0, we have P nP ⊆ IP .

Lemma I.3.7 (Nakayama’s lemma). If I ⊆ Jac(R) and M is a finitely-generated R-module,then IM = M implies M = 0.

Definition I.3.8 (Symbolic powers). Say Q ⊆ R is prime. Consider the map Rπ−→ RQ.

DefineQ(n) = π−1

((QQ)n

)={r ∈ R

∣∣ ∃s ∈ R \Q such that sr ∈ Qn}.

Note that Qn ⊆ Q(n) ⊆ Q for all n ≥ 1.

Remark I.3.9. Inside RQ, we have(Q(n)

)Q

= (QQ)n.

I.3.3 Proof of the Principal Ideal Theorem

We want to show that, if Q $ P with Q prime, then RQ has dimension 0.Replace R by RP , so that P is the unique maximal ideal. We have the following inclusions:

(x) ⊆ Q(n) + (x) ⊆ P ⊆ R.

Note that, in R/(x), there is only one prime ideal, i.e., all primes are maximal, i.e., R/(x) isArtinian. Thus, Q(n) + (x) ⊆ Q(n+1) + (x) for n� 0; in particular, Q(n) ⊆ Q(n+1) + (x). Sofor all f ∈ Q(n), we can write f = ax+ g for some a ∈ R and g ∈ Q(n+1).

Hence ax = f − g ∈ Q(n), since f ∈ Q(n) and g ∈ Q(n+1) ⊆ Q(n). But x /∈ Q (otherwise(x) ⊆ Q, and P would not be minimal), so a ∈ Q(n).

Note I.3.10. This is due to a general property of symbolic powers of prime ideals: If ax ∈ Q(n)

and x /∈ Q, then a ∈ Q(n).

So far: Q(n) = (x) ·Q(n) +Q(n+1). Hence, Q(n) = (x) ·Q(n) mod Q(n+1), so by Nakayama’slemma, Q(n) = 0 mod Q(n+1). Thus, Q(n) = Q(n+1) for all n� 0.

Localizing at Q, we get QnQ = Qn+1

Q ⊆ RQ, i.e., Q ·QnQ = Qn

Q. So by Nakayama’s lemma,QnQ = 0, whence RQ is Artinian.

I.3.4 Consequences of the Principal Ideal Theorem

Corollary I.3.11 (Another version). Let R be a Noetherian ring, and x ∈ R not a unit ora zero divisor. Then any minimal prime P ⊇ (x) has codimP = 1.

Example I.3.12. Let R = k[x, y]/(xy). Then codim(x) = 0.

Corollary I.3.13. Let x1, . . . , xc ∈ R, and let P be minimal among the primes which contain(x1, . . . , xc). Then codimP ≤ c, with equality if x1, . . . , xc is a regular sequence.

Corollary I.3.14. In k[x1, . . . , xn], we have codim(x1, . . . , xc) = c.

I.4 2014-01-30Guest lecture by Dima Arinkin.


I.4.1 Krull’s Principal Ideal Theorem, continued

Recall (10.1): if x ∈ R and p ⊂ R is minimal among primes p 3 x, then codim p ≤ 1.

Proposition I.4.1 (10.2). If x1, . . . , xc ∈ R and p ⊂ R is minimal among primes containingx1, . . . , xc, then codim p ≤ c.

Proof. Step 1: Localize at p: without loss of generality, p is maximal and R is local.

Step 2: p/(x1, . . . , xc) ⊂ R/(x1, . . . , xc) is both maximal and minimal in a local ring, henceis nilpotent.

Step 3: Let p1 $ p be a prime ideal such that there are no primes in between. WOLOG,x1 /∈ p1. Then p is minimal among primes containing p1 + (x1).

Step 4: Therefore: p/(p1 + (x1)) is nilpotent in R/(p1 + (x1)), so we have

xn2 = a2x1 + y2,

xn3 = a3x1 + y3,

...

with y2, y3, · · · ∈ p1. A prime ideals contains x1, . . . , xc iff it contains {x1, y2, . . . , yc}.Now,

p1 ⊃ (y2, . . . , yc) $ p

and p ⊇ (x1, . . . , xc), (x1, y2, . . . , yc).It remains to show p1 is minimal among primes containing (y2, . . . , yc). But this is clear,

because if there were p % p1 % p2 ⊃ (y2, . . . , yc), then p/(y2, . . . , yc) would have codimension≥ 2 in R/(y2, . . . , yc), contradicting the Principal Ideal Theorem.

Example I.4.2. p = (x1, . . . , xc) ⊂ k[x1, . . . , xn] is prime, and codim p = c.

Corollary I.4.3 (Converse, 10.5). If codim p = c, then there exist x1, . . . , xc ∈ p such thatp is a minimal prime containing (x1, . . . , xc).

Proof. Pick x1 ∈ p so that it is not in any minimal prime. (This is possible because thereare only finitely many minimal primes (because R is Noetherian), so we can use primeavoidance.)

Any prime containing x1 has codim ≥ 1. On the other hand, minimal primes with thisproperty have codim = 1. The result follows by induction.

I.4.2 Dimension of local rings

Suppose R is local and m is its maximal ideal. Then

dimR = codimm = min{c∣∣ ∃x1, . . . , xc ∈ m : m/(x1, . . . , xc) is nilpotent

}is the smallest number c such that for some x1, . . . , xc ∈ m, m is minimal among primescontaining x1, . . . , xc. Equivalently, mk ⊆ (x1, . . . , xc) for k � 0, i.e., m =

√(x1, . . . , xc).

I.4. 2014-01-30 15

Definition I.4.4. We say x1, . . . , xc are a system of parameters for R provided that c =dimR and m =

√(x1, . . . , xc).

Example I.4.5. If R = C[[x, y]] or C[x, y](x,y), then x, y is a system of parameters. There areother systems of parameters, e.g., (x2, y2) or (x, y2 + x3). Geometrically, being a system ofparameters for R means that the corresponding algebraic curves intersect “properly” at theorigin, i.e., the point of intersection has the right dimension. (The curves are not requiredto intersect transversely.)

Consider the maximal ideal m ⊂ S = R/(y2 + x3). Note that m is generated by x and y,but m2 = (x2, xy) ⊂ (x) $ m, so dimS = 1.Remark I.4.6. There is a section of [E] on parameters for modules ; we will skip this.

I.4.3 Restatement of the Principal Ideal Theorem

A choice of elements x1, . . . , xc ∈ S is equivalent to a morphism Z[x1, . . . , xc] → S. Theprime ideal p = (x1, . . . , xc) ⊂ Z[x1, . . . , xc] has image mS = (x1, . . . , xc) ⊆ S.

The Principal Ideal Theorem says that codim(mS) ≤ c. This is a statement about thefibers of the map SpecS → SpecZ[x1, . . . , xc].Proposition I.4.7. Suppose R is a local ring and m ⊂ R is maximal. Then for any R-algebra S,

codimS mS ≤ codimRm = dimR.

Proof. Since m =√

(x1, . . . , xc), any prime ideal containing mS also contains (x1, . . . , xc)S.By the PIT, codim(x1, . . . , xc)S ≤ c.

I.4.4 Dimension of fibers

Theorem I.4.8 ([E], 10.10). Let (R,m)→ (S, n) be a map of local rings. Then

dimS ≤ dimR + dimS/mS.

Proof. Let d = dimR, and choose x1, . . . , xd ∈ R such that m =√

(x1, . . . , xd). If e =dimS/mS, then we can choose y1, . . . , ye ∈ S such that√

(y1, . . . , ye) + mS = n/mS ⊂ S/mS.

For k � 0, we have nk ⊆ (y1, . . . , ye) + mS. Hence, for j � k,

nj ⊆ (y1, . . . , ye) + (x1, . . . , xd)S = (y1, . . . , ye, x1, . . . , xd)S ⊂ S.

By the PIT, dimS = codim n ≤ e+ d.

Remark I.4.9. How can the inequality be strict? Consider chains of primes

0 ⊆ pd $ · · · $ p1 $ p0 = m ⊂ R,

mS ⊆ qe $ · · · $ q1 $ q0 = n ⊂ S.

A priori, we cannot necessarily lift the primes pi to primes p′i ⊂ mS.To get this, we need a “going down” theorem, which does not work in general, but works

if S is flat over R, in which case

dimS = dimR + dimS/mS.


I.5 2014-02-04

I.5.1 Dimension of base and fiber

Theorem I.5.1 ([E], 10.10). Let (R,m) → (S, n) be a map of local rings (sending m inton). Then

dimS ≤ dimS + dimS/mS,

with equality if S is a flat R-module.

Example I.5.2. Consider k ⊆ k[[x]]. We have

dim k[[x]] = 1 = 0 + 1 = dim k + dim k[[x]] /(0) = dim k + dim k[[x]] .

Example I.5.3. Consider R = k[[x]] ⊆ k[[x, y]] /(xy) = S. This is not flat. We have dimS = 1because S is finite over k[[x− y]], and so

dimS = 1 < 2 = 1 + dim k[[y]] = dimR + dimS/(x).

Example I.5.4. Blowing up and normalization are generally not flat. For example, let R =k[[t2, t3]] ⊆ k[[t]] = S. This is a finite extension of rings, but it is not flat.

I.5.2 Going down for flat extensions

Lemma I.5.5 ([E] 10.11). Let R ⊆ S be a flat extension of rings. Fix primes P ′ ⊆ P ⊆ Rand Q ⊆ S with Q ∩R = P . Then there exists a prime Q′ ⊆ Q ⊆ S such that Q′ ∩R = P ′.

Remark I.5.6. As a prelude to the proof, we make a few observations.

• If Q′ ⊆ S is a minimal prime, then Q′ − {0} consists entirely of zero divisors on S.

• Let M be a flat R-module, and let x ∈ R be a non-zerodivisor. Then x is a non-zerodivisor on M .1

Note that a nonzero element x ∈ R is a non-zerodivisor on M iff the map M·x−→

M, m 7→ xm is injective. If x is a non-zerodivisor on R, then 0 → R·x−→ R is

injective. Since M is flat, tensoring by M preserves exactness, so 0 → M·x−→ M is

injective. Hence, x is a non-zerodivisor on M .

Proof of going down. Observe that P ′S ⊆ PS = (Q ∩ R)S ⊆ Q. Choose Q′ ⊆ Q minimalover P ′S. Tensoring preserves flat ring maps; hence,

R/P ′ → S ⊗R/P ′ = S/P ′S

is flat. Now replace R by R/P ′, and S by the image of that map. Now R ⊆ S is flat, P ′ = 0,and Q′ is a minimal prime of S.

Since S is a flat R-module, every non-zerodivisor on R is a non-zerodivisor on S. But Q′is a minimal prime of S, so Q′ − {0} consists of zero-divisors. Thus, Q′ ∩R = (0) = P ′.

1An element x ∈ R−{0} is a zero divisor on R (resp. on M) if there exists r ∈ R−{0} (resp. m ∈M−{0})such that xr = 0 (resp. xm = 0).

I.5. 2014-02-04 17

I.5.3 Corollaries of dimension theorem

Corollary I.5.7. If (R,m) is a local ring, then dimR = dim R.

Proof. Since (R,m)→ (R, m) is flat,

dim R = dimR + dim R/mR = dimR + 0.

Corollary I.5.8.

(i) Let k be a field. Then dim k[x1, . . . , xn] = n.

(ii) Let R be a Noetherian ring. Then dimR[x] = dimR + 1.2

(iii) If P ⊆ R is prime, then there exists Q ⊆ R[x] such that Q ∩ R = P . If Q is maximalwith respect to this property, then dimR[x]Q = 1 + dimRP .

Proof. Clearly, (ii) implies (i). To show (ii), given a chain P1 $ · · · $ Pd ⊂ R, we get a chain

P1R[x] $ · · · $ PdR[x] $ PdR[x] + (x) ⊂ R[x],

so dimR[x] ≥ dimR + 1. The other inequality follows from (iii): dimR[x] = dimR[x]Q forsome maximal Q, whence (ii) because Q is maximal among all primes of R[x] contracting toP := P ∩R.

We prove (iii) in cases: If R be a field, then P = (0). Every prime of R[x] contracts to(0). Let Q ⊆ R[x] be any maximal ideal. Then R[x]Q is a DVR of dimension 1.

In general, fix a prime P ⊆ R, and replace R by RP . Let Q ⊆ R[x] be any maximalideal. Note that PR[x] $ Q. One inequality is now clear:

dimR[x]Q ≥ 1 + dimRP

because a chain P1 $ · · · $ Pd ⊂ RP extends to P1R[x] $ · · · ⊆ PdR[x] $ Q. To get theother inequality, we use the theorem on the dimension of base and fiber:

dimR[x]Q ≤ dimRP + dimR[x]QPR[x]Q

= dimRP + 1.

We can now prove (Axiom 4) for Krull dimension:

dim k[[x1, . . . , xn]] = dim k[x1, . . . , xn](x1,...,xn) = dim k[x1, . . . , xn] = n.

2The Noetherian hypothesis is necessary; Nagata constructed an example of a finite-dimensional non-Noetherian ring R such that dimR[x] = dimR+ 2.


I.6 2014-02-06

I.6.1 Nakayama’s lemma

Let (R,m) be a local ring, M a finitely-generated R-module.

Lemma I.6.1 (Nakayama’s lemma, standard version). If mM = M , then M = 0.

Lemma I.6.2 (Nakayama’s lemma, better version). Fix m1, . . . ,ms ∈ M . The followingare equivalent:

(1) m1, . . . ,ms are a basis for M/mM .

(2) m1, . . . ,ms are an irredundant set of generators for M .

Proof. Consider the map⊕s

i=1R → M sending the i-th generator to mi. This induces anexact sequence

s⊕i=1

R→M → Q→ 0.

Tensoring with R/m is right exact, so we obtain an exact sequences⊕i=1

R/m→M/mM → Q/mQ→ 0.

So Q = 0 ⇐⇒ Q/mQ = 0. (Independence and irredundancy are left as exercises to thereader.)

I.6.2 Regular local rings

Let (R,m) be a Noetherian local ring. Last week, we noted that

dimR = min{d∣∣ ∃f1, . . . , fd :

√(f1, . . . , fd) = m

}.

Consequently, dimR ≤ min {d | ∃f1, . . . , fd : (f1, . . . , fd) = m}. We have a name for whenthis is an equality.

Definition I.6.3. A local ring (R,m) is a regular local ring if m = (x1, . . . , xd), whered = dimR. In this case, x1, . . . , xd is called a regular system of parameters .

If A is any ring, then A is regular if AP is a regular local ring for all P maximal.

Remark I.6.4. Question: If R is regular local and P ⊂ R is prime, is RP regular local? Thiswas a major open question for a long time, until ultimately being resolved in the affirmativein the 1950s by Serre using homological methods.Example I.6.5. R = Q[[x1, . . . , xr]] is a regular local ring, and x1, . . . , xr is a regular systemof parameters.Example I.6.6. Zp[[t]] is a regular local ring with regular system of parameters p, t.Example I.6.7. Q[[x, y]] /(xy) is not regular.Remark I.6.8 (“Meta-theorem”). Everything you want to be true about a regular local ringis true.

Corollary I.6.9 ([E], 10.14). If R is regular local, then R is an integral domain.

I.6. 2014-02-06 19

I.6.3 Regular sequences

Definition I.6.10. Let R be any ring and M an R-module. A sequence of elementsf1, . . . , fn ∈ R is a regular sequence on M provided that:

(1) (f1, . . . , fn)M 6= M .

(2) For i = 1, . . . , n, fi is a non-zerodivisor on M/(f1, . . . , fi−1).

Proposition I.6.11. If f1, . . . , fn is a regular sequence on R, then

dimR/(f1, . . . , fn) = dimR− n.

Proof. Apply the Principal Ideal Theorem inductively.

Proposition I.6.12 ([E], 10.15). Let (R,m) be a local ring, and x1, . . . , xd a regular systemof parameters. Then x1, . . . , xd is a regular sequence on R.

Proof. For each i, R/(x1, . . . , xi) is a regular local ring with regular system of parametersxi+1, . . . , xd. Hence R/(x1, . . . , xi) is an integral domain, and xi+1 6= 0 on R/(x1, . . . , xi). Soxi+1 is a non-zerodivisor on R/(x1, . . . , xi). Moreover, (x1, . . . , xd)R = mR 6= R.

I.6.4 Complete regular local rings

Let k be an algebraically closed field. Let (R,m) be a complete, regular, local k-algebra ofdimension d.

Theorem I.6.13. (R,m) ∼= k[[x1, . . . , xd]].

Proof. This follows from the Cohen structure theorem; see [E, Prop. 10.16].

Recall the following fact in differential geometry: “Locally, every smooth, connectedmanifold of dimension d looks the same.” (More precisely, if X is a smooth, connected,d-dimensional manifold, and x ∈ X, then there is a local neighborhood of x which is diffeo-morphic to some local neighborhood of 0 ∈ Rd.)

Proposition I.6.14 (Algebraic analogue). Let k be an algebraically closed field, let R bea regular integral k-algebra of dimension d, and let P ⊂ R be a maximal ideal. Then thecompletion of R at P is isomorphic to k[[x1, . . . , xd]].3

I.6.5 Affine rings

We will now jump to [E, chapter 13] and consider affine rings , i.e., rings of the formk[y1, . . . , yn]/I, where k is an arbitrary field and I is an arbitrary ideal.4

Theorem I.6.15. If k is a field, then dim k[x1, . . . , xr] = r.

3There is also an étale version of this local isomorphism.4This is equivalent to studying closed subschemes of An

k = Spec k[y1, . . . , yn].


Lemma I.6.16 ([E], 13.3(c)). Let S = k[x1, . . . , xr] and f ∈ S. Fix ai ∈ k, and letx′i = xi− air. There is a polynomial g(s1, . . . , sr−1), depending only on f , with the followingproperty: If g(a1, . . . , ar−1) 6= 0, then

k[x′1, . . . , x′r−1] ⊆ S/(f)

is a finite extension.

Proof. Observe that d := deg f(x1, . . . , xr) = deg f(x′1, . . . , x′r−1, xr). The coefficient of xdr in

f(x′1, . . . , x′r−1, xr) is a polynomial combination g of a1, . . . , ar−1. If g(a1, . . . , ar−1) 6= 0, then

f(x′1, . . . , x′r−1, xr) = g(a1, . . . , ar−1) · xdr + f1x

d−1r + · · ·+ fd−1xr + fd,

where fi ∈ k[x′1, . . . , x′r−1]. So if g(a1, . . . , ar−1) 6= 0, then S/(f) is generated by 1, xr, . . . , x

d−1r

as a module over k[x′1, . . . , x′r−1].

I.7 2014-02-13

I.7.1 Dimension of polynomial rings

Recall the following lemma:

Lemma I.7.1 ([E], Lemma 13.3). Let S = k[x1, . . . , xr] and f ∈ S. Fix ai ∈ k, and letx′i = xi − aixr. Then there exists g ∈ k[T1, . . . , Tr−1] where g(a1, . . . , ar−1) 6= 0 impliesk[x′1, . . . , x

′r−1] ⊆ S/(f) is a finite extension.

Theorem I.7.2. Let k be a field. Then dim k[x1, . . . , xr] = r.

Proof. Let k be an algebraic closure of k. Since k ⊆ k is integral, it follows that S ⊆ S ⊗k kis also integral. Thus, dimS = dimS ⊗k k. Hence, we may assume k is infinite.

Now induct on r. If r = 0 or 1, the claim is obvious. We have a chain

0 $ (x1) $ (x1, x2) $ · · · $ (x1, . . . , xr).

So dimS ≥ r. To show the other direction, let 0 $ P1 $ · · · $ Pm be a chain of distinctprimes. Let f ∈ P1 \ {0}. Note that f ∈ Pi for all i. This yields a chain of length m− 1,

P1 $ P2 $ · · · $ Pm,

in S/(f). Let x′i = xi − aixr for some a1, . . . , ar−1 such that

S ′ ⊆ k[x′1, . . . , x′r−1] ⊆ S/(f)

is a finite extension. Incomparability yields a chain of distinct primes

S ′ ∩ P1 $ S ′ ∩ P2 $ · · · $ S ′ ∩ Pm.

By induction, m− 1 ≤ r − 1.

I.7. 2014-02-13 21

I.7.2 Weak Noether normalization

Theorem I.7.3 (Weak Noether normalization). Let k be a field, and let R = k[Y1, . . . , Yr]/Jbe a finitely-generated k-algebra.5 There exist x1, . . . , xd ∈ R such that S = k[x1, . . . , xd] ⊆ Ris a finite extension and S is a polynomial subring of R.

Note I.7.4. When k is infinite, each xi may be chosen as a k-linear combination of the Yj.

Proof. Write yi for the image of Yi in R, so R = k[y1, . . . , yr] (not a polynomial ring).Choose d maximal such that, after reordering, y1, . . . , yd are algebraically independent andyd+1, . . . , yr are algebraic over k[y1, . . . , yd].

For a given d, we induct over r. If d = r, there is nothing to prove. Let r > d and dothe induction step. So, yr is algebraic over k[y1, . . . , yr]. In particular, there is a polynomialf 6= 0 such that f(y1, . . . , yr) = 0. By the Lemma, choose y′i = yi−aiyr (and Y ′i = Yi−arYr)such that

k[Y ′1 , . . . , Y′r−1] ⊆ k[Y1, . . . , Yr]

(f)

is a finite extension, hence

k[y′1, . . . , y′r] ⊆ k[y′1, . . . , y

′r−1, yr] = R

is a finite extension. By the induction hypothesis, there exist x1, . . . , xd (each a k-linearcombination of the yj) such that k[x1, . . . , xd] ⊆ k[y′1, . . . , y

′r−1] is a finite extension and

k[x1, . . . , xd] is a polynomial ring. The composition of finite extensions is finite, so we aredone.

Example I.7.5. Even if R is actually a polynomial ring, there are many different Noethernormalizations. For example, k[xa11 , . . . , x

add ] ⊆ k[x1, . . . , xd] is a Noether normalization.

I.7.3 Strong Noether normalization

Theorem I.7.6 ([E], 13.3). Let k be a field, R = k[Y1, . . . , Yr]/J of dimension d, andI1 ⊆ · · · ⊆ Im ideals in R, where dim Ij = dimR/Ij = dj and d1 > d2 > · · · > dm ≥ 0. Thenthere exist x1, . . . , xd ∈ R such that

(1) S = k[x1, . . . , xd] is a polynomial subring of R.

(2) Ij ∩ S = (xdj+1, . . . , xd) for j = 1, . . . ,m.

This is a simultaneous Noether normalization of R/I1, . . . , R/Im: for j = 1, . . . ,m,

k[x1, . . . , xdj ] ⊆ R/Ij

is a finite extension.

Proof. See [E, Thm. 13.3].

5In [E], this is called an affine ring .


I.7.4 Transcendence degree and dimension

Theorem I.7.7 (Theorem A). Let R be a finitely-generated k-algebra that is also a domain.Let Q(R) be the field of fractions of R. Then

dimR = tr. degkQ(R),

the transcendence degree of Q(R) over k, and this is the length of every maximal chain ofprimes in R.

Example I.7.8. Assume char k 6= 2. Let R = k[x, y, z]/(x2 + y2 + z2). Then dimR = 2, andtr. degkQ(R) = 2. For instance, we have the chain of purely transcendental extensions

k ⊂ k(x) ⊂ k(x, y),

but k(x, y) ⊂ Q(R) is algebraic.

Remark I.7.9. This is the older version of dimension theory, prior to Krull’s notion of di-mension.

Corollary I.7.10. Let R be a finitely-generated k-algebra and a domain. If I ⊆ R is anyideal, then

dim I + codim I = dimR.

Proof. This follows immediately from the definition of dimension and codimension of anideal, along with the fact that every maximal chain of primes in R has length dimR.

Corollary I.7.11 (Dimension of base and fiber). Let R ⊆ T be an inclusion of finitely-generated k-algebras, both domains. Then

dimT = dimR + dimQ(R)⊗R T.

Proof. Replace dimension with transcendence degree, and note that transcendence degreefor fields is additive.

I.8 2014-02-18

I.8.1 Transcendence degree and dimension, continued

Theorem I.8.1 (Theorem A). Let R be a finitely-generated k-algebra and a domain. Then

dimR = tr. degkQ(R),

and this number is the length of every maximal chain of primes in R.

Let S = k[x1, . . . , xn] ⊆ R be a Noether normalization. Then k(x1, . . . , xn) = Q(S) ⊆Q(R) is a finite field extension, so

dimR = dimS = n = tr. degkQ(S) = tr. degkQ(R).

I.8. 2014-02-18 23

Lemma I.8.2 ([E], Prop. 13.??). Let S be a normal domain and Q(S) ⊆ L a finite, normalfield extension. Let T be the integral closure of S in L. Let P ⊂ S be prime. Then theprimes of T that lie over P are conjugate under the action of Gal(L/Q(S)).

Proof. Let P ′ and P1 be primes in T over P . Let P1, . . . , Pn be the Galois conjugates ofP1. Assume P ′ 6= Pi for all i. By incomparability, P ′ * Pi for any i. By prime avoidance,P ′ *

⋃i Pi. Hence, there exists a ∈ P ′ \

⋃i Pi. No conjugate of a can lie in any Pi, so

NL/Q(S)(a) /∈ Pi for any i. But since S is normal, NL/Q(S)(a) ∈ P ′ ∩ S = P = Pi ∩ S, acontradiction.

I.8.2 Going down

Theorem I.8.3 (Going down for integral extensions). Let S be a normal domain, andR ⊇ S a finite extension of domains. Then going down holds for R ⊇ S, i.e., given primesQ ⊆ Q1 ⊂ S and P1 ⊂ R with P1 ∩ S = Q1, there is a prime P ⊆ P1 with P ∩ S = Q.

Proof. Let Q(R) and Q(R) be the quotient fields, and let L ⊇ Q(R) ⊇ Q(S) be the normalclosure of Q(R) ⊇ Q(S). Let T be the integral closure of S in L. Then a big diagram ofintegral extensions [E, Figure 13.2] completes the proof.

I.8.3 End of proof of Theorem A

We now complete the proof of Theorem A (I.8.1). Let d := dimR. Consider a chain ofprimes

P0 $ P1 $ · · · $ Pm ⊂ R,

where Pm is maximal. If m < d, then we can wedge an additional prime P into this chain:Let dj := dimPj. Note d0 > d1 > · · · > dm. If m < d, then there exists ` where d`−1−d` ≥ 2.It suffices to construct a prime P between P`−1 $ P $ P`. Factor out P`−1 and relabel:

(0) $ P1 $ · · · $ Pm ⊂ R,

where d0 − d1 = d− d1 ≥ 2. Choose a strong Noether normalization S ⊆ R so that

P1 ∩ S = (xd1+1, . . . , xd) ⊇ (xd−1, xd).

Let Q = (xd), so that (0) $ Q $ Q1 = P1 ∩ S. By going down, there exists P $ Pi ⊂ Rcontracting to Q.

I.8.4 Catenary rings

Definition I.8.4. A ring R is catenary if, for all primes P,Q in R, all maximal chains ofdistinct primes connecting P and Q have the same length.

Definition I.8.5. A ring R is universally catenary if every finitely-generated R-algebra iscatenary.

Corollary I.8.6 (Cor. of Theorem A). Every field (equivalently, every finitely-generatedk-algebra) is universally catenary.


Proof. Fix P,Q in k[x1, . . . , xn]/I. Let R := k[x1, . . . , xn]/P . Then the result follows imme-diately from Theorem A.

Remark I.8.7. Bold claim: Every interesting ring is catenary.

I.8.5 Remarks on past homework

Remark I.8.8. A regular system of parameters (a minimal generating set for m that is also asystem of parameters) is a regular sequence and a system of parameters, but not conversely.

Example I.8.9. If R = k[[x, y]] /(y2), then X is a regular sequence and a system of parameters,but not a regular system of parameters.

Remark I.8.10. In exercise 13.4:

• It’s okay to only prove the “weak” version of Noether normalization.

• Hint: k[[x1, . . . , xn]] ⊆ S is a finite extension iff x1, . . . , xn is a system of parameters.

I.8.6 Axioms of dimension

Why is dimR determined by the axioms? By the local axiom, it suffices to let R be acomplete local ring.

By Noether normalization, if R is a complete local ring, then there is a finite extensionk[[x1, . . . , xn]] ⊆ R, so

dimR = dim k[[x1, . . . , xn]] = n.

I.9 2014-02-20

Today: End of “dimension theory.”Next:

• Graded rings

• Gröbner bases and computational commutative algebra

I.9.1 Weak Nullstellensatz

Corollary I.9.1 (Weak Nullstellensatz). Let R be a finitely-generated R-algebra, and P ⊂ Ra prime.

(1) If P is maximal, then R/P is a finite field extension of k.

(2) P =⋂

m⊇P m, where m runs over the set of maximal ideals containing P .

Proof. (1) R/P is a field, hence 0-dimensional. By Noether normalization, k ⊆ R/P isfinite.

I.9. 2014-02-20 25

(2) Clearly, P ⊆⋂

m⊇P m. It suffices to show that, if f /∈ P , then there exists m ⊇ Pwhere f /∈ m.

Factor out P , and write R = R/P . Then f ∈ R is nonzero. If f is a unit, then we aredone. Otherwise, f is a non-unit non-zerodivisor, so by the Principal Ideal Theorem,codim(f)R = 1. Use strong Noether normalization for R to get a finite extension

S = k[x1, . . . , xd] ⊆ R

such that (f) ∩ S = (x1). Let m0 := 〈x1 − 1, x2, x3, . . . , xd〉 ⊆ S. By lying over, thereexists a prime Q ⊆ R where Q∩S = m0. Since R/m0R is 0-dimensional, Q is maximal.Since m0 contains x1 − 1, it does not contain x1, so f /∈ Q.

I.9.2 Finiteness of integral closure

Slogan: “Integral closure is finite for finitely-generated k-algebras.”

Proposition I.9.2 ([E], 13.13). Let R be a finitely-generated k-algebra which is also a do-main, and let K = Q(R) be its field of fractions. Let L be a finite extension of K. Let T bethe integral closure of R in L. The R ⊆ T is a finite extension of rings.

Proof. By Noether normalization, there is a finite extension k[x1, . . . , xd] ⊆ R. Since Ris Noetherian, submodules of a finite R-module are finite, so we can make the followingsimplifications:

(1) Replace R with k[x1, . . . , xd].

(2) Replace L with the normal closure of L/K.

Since L/K is now a normal extension, we can split it into K ⊆ L′ ⊆ L, where L′/K is purelyinseparable and L/L′ is Galois.

Purely inseparable case Without loss of generality, L′ 6= K. Let p be the characteristicof K. For some q = pe, L′ is obtained by adding q-th roots of rational functions.Enlarge L′ by adjoining q-th roots of all coefficients of those rational functions. Then

L′ = k′(x1/q1 , . . . , x

1/qd ),

where k′/k is obtained by adjoining q-th roots of the coefficients. SoR′ := k′[x1/q1 , . . . , x

1/qd ]

is the integral closure of R = k[x1, . . . , xd] in L′.

Galois case Assume L/K is Galois with Galois group G = {σ1, . . . , σn}. Let T be theintegral closure of R in L. Let b1, . . . , bn ∈ T form a k-basis of L/K. Let

Φ =(σi(bj)

)∈ Matn×n(T ).

Let ∆ := det(Φ).

Claim. ∆ 6= 0, and R ⊆ T ⊆ 1

∆2· (Rb1 + · · ·+Rbn).

See [E, Prop. 13.14] for the remainder of the proof.


I.9.3 Problems with Noetherian rings

In the class of arbitrary Noetherian rings, there can be problems with finiteness of integralclosure:

• It’s not necessarily true that the integral closure of a Noetherian ring is finite.

• Worse, it may not even be Noetherian!

There are several problems with Noetherian rings:

• They may not be catenary.

• Integral closure may not be finite, or even Noetherian.

• Completion may not preserve normality.

Possible responses:

• Focus on finitely-generated k-algebras.

• Enlarge the class of Noetherian rings. This leads to the notion of a Krull ring .

• Add axioms. This approach has become popular largely to the influence of Grothendieck.

The last approach leads to the notions of Nagata ring (or Japanese ring), universallyNagata ring , and excellent ring . Excellent rings, for which none of the above mentionedproblems arise, are the most commonly studied.

Chapter II

Gröbner bases

II.1 2014-02-25: Monomial orders and Gröbner bases

II.1.1 Graded rings

Definition II.1.1. A graded k-algebra is a ring R with a decomposition as an abelian group

R =⊕d∈N

Rd

satisfying:

(1) R0 = k

(2) If ri ∈ Ri and rj ∈ Rj, when rirj ∈ Ri+j.

Example II.1.2. Let S = k[x1, . . . , xn] with deg(xi) = 1. Then Sd consists of all homogeneous,degree d polynomials.

An element r ∈ R is homogeneous if r ∈ Rd for some d. In this case, d := deg(r).

Definition II.1.3. An ideal I ⊆ R is graded (or ideal) if I = 〈f1, . . . , fs〉 for some homoge-neous elements f1, . . . , fs ∈ R.Example II.1.4. Let R = k[x1, . . . , x5] with deg(xi) = 1. Then I = 〈x1x2 − x2

3, x104 − x10

5 〉 isa graded ideal.Example II.1.5. Let R = k[x, y, z] with degrees 1, 1, 3 Then I = 〈z2 − x4y2〉 is graded.Lemma II.1.6. Let R be graded and I ⊆ R a graded ideal. Then R/I is graded.

Proof. We will show thatR/I =

⊕d∈N

Rd/(I ∩Rd).

It suffices to show that I =⊕

d I ∩ Rd. Let I = 〈f1, . . . , fs〉 with fi homogeneous. Letg =

∑si=1 aifi ∈ I. Write [h]α for the degree α part of the element h. The claim follows from

the fact that

[g]α =s∑i=1

[ai]α−deg fi · fi ∈ I.

Hence, [g]α ∈ I ∩Rα.

27

28 CHAPTER II. GRÖBNER BASES

Remark II.1.7. The following will be our standard assumptions:

(1) R is a graded k-algebra.

(2) m =⊕

d>0Rd is the unique homogeneous maximal ideal in R.

(3) Rd is a finite-dimensional k-vector space.

Remark II.1.8 (“Graded/local principle”). Basically any theorem about a local ring holds fora graded k-algebra (R,m).

II.1.2 Graded modules

Definition II.1.9. Let R be a graded k-algebra. A graded R-module is an R-module Mwith a decomposition

M =⊕d∈Z

Md

of abelian groups such that, if ri ∈ Ri and mj ∈Mj, then rimj ∈Mi+j.

Lemma II.1.10 (Graded Nakayama’s lemma). Let (R,m) be a graded k-algebra, and let Mbe a finitely-generated graded R-module. If M = mM , then M = 0.

Proof. Assume M 6= 0 with homogeneous generators m1, . . . ,ms. Let d := min {deg(mi)}.Since every element in m has degree > 0,

0 6= Md = (M)d = (mM)d = 0,

a contradiction.

Remark II.1.11. If R is a graded, finitely-generated k-algebra, then

(1) There is a finite subring k[x1, . . . , xn] ⊆ R where each xi ∈ R is homogeneous.

(2) dimR is the longest chain of homogeneous prime ideals in R.

Definition II.1.12. The Hilbert function of a graded R-module M is

HFM : Z→ N ∪ {∞} ,d 7→ dimkMd.

Remark II.1.13. Motivation (think R = k[x1, . . . , xn]):

• Containment: Given f ∈ R and I ⊆ R, does f ∈ I?

• Equality: Given I, J ⊆ R, does I = J?

• Dimension: What is dimR/I?

II.2. 2014-02-27 29

II.1.3 Monomial orders

Let S = k[x1, . . . , xn]. For a monomial xα with α = (α1, . . . , αn) ∈ Nn, call α the multidegreeof xα.

Definition II.1.14. A (global) monomial order is a total order on the set of monomials ofS, satisfying:

(1) If xα > xβ, then xαxγ > xβxγ.

(2) The total order is a well-ordering, i.e., every nonempty subset has a minimal element.

Remark II.1.15. Every descending chain in a well-ordering stabilizes.

Example II.1.16. Consider xα versus xβ:

(1) Lexicographic order: higher xi-exponent wins, where i is the smallest index such thatαi 6= βi.

(2) Graded-lexicographic: first compare∑αi to

∑βi; larger wins. If tied, then use the

lexicographic order.

(3) Graded reverse lexicographic:

• first compare degrees; highest wins.

• if tied, compare the xn-exponent; highest loses.

• if tied, repeat for xn−1, and so on.

(4) Weight order w = (w1, . . . , wn) ∈ Qn≥0:

• Compare α · w and β · w; highest wins.• Break ties via some other method.

II.2 2014-02-27

II.2.1 Monomial ideals

Problem: Given f ∈ S and I ⊆ S, decide if f ∈ I.Example II.2.1. Let f = 2x2y2w2 − y5w − y4z2. We want to know whether f is in:

I =⟨xz − y2, xw − yz, yw − z2

⟩,

J = 〈xz, yz, yw〉 .

Since J ⊆ 〈z〉, we can easily see that f /∈ J .

Definition II.2.2. I ⊆ S is a monomial ideal if I = 〈f1, . . . , fs〉 for some monomials fi.

Example II.2.3. If charK 6= 2, then I = 〈x2 + y2, x2 − y2〉 is a monomial ideal.


Example II.2.4. 〈x2 + xy, y2〉 is not a monomial ideal.

Remark II.2.5. Big idea:

• Computing with monomial ideals is “easy”.

• Monomial orders reduce arbitrary computations to monomial computations.

Problem: Determine when an ideal I ⊆ S is:

• a monomial ideal;

• a homogeneous ideal.

II.2.2 Division algorithm

Recall that a monomial order is a well-ordering on the set of monomials of S that respectsmultiplication.

Also, recall division for univariate polynomials: Given f, g ∈ Q[t], there exist q, r ∈ Q[t]such that f = q · g + r and deg(r) < deg(g).

Let us generalize this to a division algorithm to a general monomial order:

Input f, g ∈ S and a monomial order > on S.

Output q, r ∈ S such that f = q · g + r and no term of r is divisible by LT>(g).

Definition II.2.6. The leading term LT>(g) is the maximal monomial xα that appears witha nonzero coefficient in g.

Example II.2.7. If g = yz − w2 and > is the lexicographic order, then LT>(g) = −w2.

Here is the division algorithm:

• Let p := 0 and f ′0 := f .

• While some term of f ′p is divisible by LT>(g), do:

– Let mp be the maximal term of fp divisible by LT>(g).

– Let qp :=mp

LT>(g)and f ′p+1 := f ′p − qpg.

– Let p 7→ p+ 1.

• Let q :=

p−1∑j=0

qj and r = f ′p.

II.3. 2014-03-04: BUCHBERGER’S ALGORITHM 31

Example II.2.8. Let f = x2y2w2 + xyzw3− y4z2 + z6 and g = yz−w2 with the lex order forx > y > z > w. Note that LT>(g) = yz. Applying the algorithm:

f ′0 = f

f ′1 = f ′0 −xyzw3

yz· g = f ′0 − xw3g = x2y2w2 + xw5 − y4z2 + z6

f ′2 = f ′1 − y3z · g = x2y2w2 + xw5 − y3w2z + z6

f ′3 = f ′2 − y2w2 · g = x2y2w2 + xw5 − y2w4 + z6

r = f ′3

Proof of the division algorithm. The maximal term of f ′p that is divisible by LT>(g) strictlydecreases at each step, so the algorithm terminates. Correctness is just a matter of unravelingnotation.

II.2.3 Gröbner bases

Definition II.2.9. Given I ⊆ S and a monomial order >, the initial ideal of I is

in>(I)def=⟨LT>(f)

∣∣ f ∈ I⟩ .Definition II.2.10. Given I ⊆ S and a monomial order >, a finite set G = {g1, . . . , gr} isa Gröbner basis for I if gi ∈ I and

in>(I) = 〈LT>(g1), . . . ,LT>(gr)〉 .

II.3 2014-03-04: Buchberger’s algorithmThroughout today’s lecture, let S = k[x1, . . . , xn], and let > be a fixed monomial order.

II.3.1 Reduced Gröbner bases

Definition II.3.1. Let I ⊆ S be an ideal. A set G = {g1, . . . , gs} ⊆ I is a Gröbner basis forI if

〈LT>(g1), . . . ,LT>(gs)〉 = in>(I).

G is a reduced Gröbner basis if:

• The coefficient of LT(gi) is 1 for all i.

• No monomial of gi is divisible by LT>(gj) for i 6= j.

Theorem II.3.2. Fix an ideal I ⊆ S and an order >. Then:

(1) I admits a Gröbner basis.

(2) I admits a unique reduced Gröbner basis.


(3) Every Gröbner basis of I generates I.

Corollary II.3.3. I ⊆ S is a monomial ideal iff every (or any) reduced GB of I consists ofmonomials.

Example II.3.4. I ⊆ S is homogeneous iff any reduced GB consists entirely of homogeneouselements.

Corollary II.3.5. I = J iff the reduced GB for I equals the reduced GB for J .

Proposition II.3.6. Let G = {g1, . . . , gr} be a GB for I ⊆ S. Let f ∈ S. Then there is aunique r = f % G ∈ S such that:

(1) No term of r is divisible by LT(gi) for any i.

(2) f − i ∈ I.

Corollary II.3.7. f ∈ I iff r = f % G = 0.

Write f % I = f % G, where G is any Gröbner basis of I.

II.3.2 Buchberger’s criterion

Definition II.3.8. Given f, g ∈ S, let xγ = lcm(LT(f),LT(g)). The S-polynomial of f andg is

S(f, g) :=xγ

LT(f)· f − xγ

LT(g)· g.

Example II.3.9. Let f = 2x2 + z2 and g = 5xy + y2 with the lex order x > y > z. Then

S(f, g) =y

2· f − x

5· g =

1

2yz2 − 1

5xy2.

Theorem II.3.10 (Buchberger’s criterion). G = {g1, . . . , gr} is a Gröbner basis for I iffS(gi, gj) % G = 0 for all i, j.

Proof. If G is a GB, then since S(gi, gj) ∈ I, the previous proposition implies S(gi, gj) %G = 0. For the converse, see [CLO, §2.6, Thm. 6] or [E, Thm. 15.8]. Sketch (proof bycontradiction):

• Assume S(gi, gj) % G = 0, but G is not a GB.

• Choose a minimal witness xγ ∈ in \ 〈LT(g1), . . . ,LT(gk)〉.

• Find a more minimal witness.

II.3.3 Buchberger’s algorithm

Input A list {f1, . . . , fs} ⊆ S and an order >.

Output A Gröbner basis {g1, . . . , gr} for I = 〈f1, . . . , fs〉 with respect to >.

II.4. 2014-03-06: SOME ALGORITHMS 33

II.4 2014-03-06: Some algorithms

II.4.1 Buchberger’s algorithm

Input

• f1, . . . , fs ∈ S• a monomial order > on S

Output

• a Gröbner basis for 〈f1, . . . , fs〉

Steps

• G := {f1, . . . , fs}• P := {(p, q) | p, q ∈ G}• while P 6= 0:

– choose (p, q) ∈ P , and set P := P \ {(p, q)}.– h := S(p, q)

– while LT(h) is divisible by LT(gi) for some i:∗ h := h % gi.

– if h 6= 0, then:∗ P := P ∪ {(h, f) | f ∈ G}∗ G := G ∪ {h}

• return G

Correctness follows from Buchberger’s criterion.

Termination it suffices to show that G stabilizes. Observe that G gets larger when LT(h) /∈〈LT(g) | g ∈ G〉. Each time we add h, we make 〈LT(g) | g ∈ G〉 strictly larger. Thisyields a strictly ascending chain of ideals, which must be finite.

Remark II.4.1. In the linear case, finding a reduced Gröbner basis is equivalent to Gaussianelimination.

Example II.4.2. Let g1 = x+y+z+w and g2 = x+2y+3z+5w in the lex order x > y > z > w.Then

S(g2, g1) = g2 − g1 = y + 2z + 4w.

Here is a Gröbner basis for 〈g1, g2〉:

G = {x+ y + z + w, x+ 2y + 3z + 5w, y + 2z + 4w} .

The reduced Gröbner basis is:

{x− z − 3w, y + 2z + 4w}


Remark II.4.3. In the one-variable case, finding a reduced Gröbner basis is equivalent to theEuclidean algorithm.

Example II.4.4. Find a GB of 〈t3 − 1, t2 − 2t+ 1〉:

S(t3 − 1, t2 − 2t+ 1) = 2t2 − t− 1,

so h = (2t2 − t− 1) % g2 = 3t− 3 = 3(t− 1). The reduced GB is thus {t− 1}.

II.4.2 Organizing a proof

Problem: Compute the kernel of

φ : k[x, y, z, w]→ k[s, t]

x 7→ s3

y 7→ s2t

z 7→ st2

w 7→ t3

This is the twisted cubic:

P1 → P3

[s : t] 7→[s3 : s2t : st2 : t3

]Claim: the entire kernel is

I =⟨yz − xw, y2 − xz, z2 − yw

⟩.

Let f ∈ kerφ, and let > be the rev-lex order. Without loss of generality, we can assumeno term of f is divisible by yz, y2, or z2. Write

f =∑

(a,b)∈Z2

αa,bxawb + βa,bx

aywb + γa,bxazwb.

Thenφ(f) =

∑(a,b)

αa,bs3at3b + βa,bs

3a+2t3b+1 + γa,bs3a+1t3b+2.

Each term of φ(f) has a different multidegree, so there is no cancellation; hence,

αa,b = βa,b = γa,b = 0 ∀(a, b) ∈ Z2.

II.4.3 Ideal membership

Input

• f ∈ S• {g1, . . . , gr} ⊆ S

II.4. 2014-03-06: SOME ALGORITHMS 35

Output

• True if f ∈ 〈g1, . . . , gr〉, else False.• Optionally: a “witness”, i.e., an expression f =

∑aigi.

Steps

• Compute a Gröbner basis G of the ideal 〈g1, . . . , gr〉. [To get a “witness”, trackcoefficients in the division algorithm to get an expression for the elements of G interms of the gi.]

• Compute r = f % G. [To get a “witness”, track coefficients in use of the divisionalgorithm.]

• If r = 0, return True; else, return False.

II.4.4 Elimination of variables

Input

• a finite list f1, . . . , fs ∈ S;• an integer s ∈ {1, . . . , n}.

Output generators for 〈f1, . . . , fs〉 ∩ k[xs+1, . . . , xn].

Definition II.4.5 (elimination order). A monomial order > is an elimination order forxs+1, . . . , xn if

LT>(f) ∈ k[xs+1, . . . , xn] =⇒ f ∈ k[xs+1, . . . , xn]

for all f ∈ S.

Example II.4.6. The lex order on k[x, y] is an elimination order for y.

Definition II.4.7 (product order). Fix s ∈ {1, . . . , n}. Fix monomial orders>1 on k[x1, . . . , xs]and >2 on k[xs+1, . . . , xn]. The product order of >1 and >2 is the order >, where xα > xβ iff

• xα11 · · ·xαs

s >1 xβ11 · · ·xβss , or

• xα11 · · ·xαs

s = xβ11 · · ·xβss and xαs+1

s+1 · · ·xαnn >2 x

βs+1

s+1 · · ·xβnn .

Fact II.4.8. Every such order is an elimination order for xs+1, . . . , xn.

Steps

• Let > be an elimination order for xs+1, . . . , xN .

• Compute a Gröbner basis G for 〈f1, . . . , fr〉.• return G ∩ k[xs+1, . . . , xn].

Termination follows from termination of computation of the Gröbner basis.


Correctness Let J ⊆ k[xs+1, . . . , xn] be the ideal spanned by the output. Clearly, J ⊆I ∩ k[xs+1, . . . , xn].

Let h ∈ I ∩ k[xs+1, . . . , xn]. A term of h is divisible by LT(gi) only if LT(gi) ∈k[xs+1, . . . , xn]. Note that h % G = 0. In the computation of h % G, we only usegi ∈ G ∩ k[xs+1, . . . , xn], hence h ∈ J .

II.4.5 Geometry of elimination

Geometrically, elimination corresponds to projection.

Example II.4.9. Let f = ax2 + bx+ c and

I =

⟨f,∂f

∂x

⟩=⟨ax2 + bx+ c, 2ax+ b

⟩⊆ k[x, a, b, c].

What is I ∩ k[a, b, c]? In k[a, b, c], this is 〈b2 − 4ac〉. We can think of I as the set of f suchthat f and ∂f

∂xhave a common root.

Example II.4.10. Let f = y2 − x3 − ax − b. Let I =⟨f, ∂f

∂x, ∂f∂y

⟩⊆ k[x, y, a, b]. What is

I ∩ k[a, b]?

II.5 2014-03-11

“Homework”: work on projects, and compute something (in Macaulay2).

II.5.1 Elimination theory

Consider the following: F = y2 − x3 − ax− b ∈ C[x, y, a, b], and I =⟨F, ∂F

∂y, ∂F∂x

⟩. Then the

vanishing set V (F ) is a cubic curve in A2. (We sometimes call F the “universal curve”.)What is J := I ∩ k[a, b]? Last time, we saw we can answer this using Gröbner bases.

Note that

V (I) ={

(x, y, a, b) ∈ A2 × A2∣∣ (x, y) is a singular point of V (Fa,b)

}.

We have (a, b) ∈ V (J) ⇐⇒ V (Fa,b) has a singular point (so J corresponds to singular planecubics).

Claim II.5.1. J = 〈4a3 + 27b2〉.

Remark II.5.2. The same sort of procedure works to find the singular locus of other maps.This is the subject of elimination theory .

II.5. 2014-03-11 37

II.5.2 Solving polynomial equations

Rather than explaining in full generality, we consider an example: Solve

x2 + 2y2 − 3 = 0,

x2 + xy + y2 − 3 = 0.

Computing a Gröbner basis with respect to the lex order >, we obtain the reduced Gröbnerbasis

G ={y3 − y, xy − y2, x2 + 2y2 − 3

}.

Looking at y3 − y = 0, we see that y ∈ {0,±1}. If y = 0, then x2 − 3 = 0, so x = ±√

3. Ify = 1, then x2 +2−3 = 0, so x = ±1; however, (1,−1) does not satisfy the second equation.So the solutions are {

(−√

3, 0), (1, 1), (−1,−1), (√

3, 0)}.

II.5.3 Implicitization, i.e., computing kernels

We want to compute the image of the following map:

Ar → An

y = (y1, . . . , yr) 7→(f1(y), . . . , fn(y)

).

This is equivalent to computing the kernel of the map of rings

k[x1, . . . , xn]ϕ−→ k[y1, . . . , yr],

xi 7→ fi(y).

This can be done by the following algorithm:

Input {f1, . . . , fn} ⊆ k[y1, . . . , yr]

Output generators for ker(ϕ)

Steps

• choose an elimination order > for x1, . . . , xn on k[y1, . . . , yr, x1, . . . , xn]

• let I := 〈x1 − f1(y), . . . , xn − fn(y)〉• compute a Gröbner basis G of I• return G ∩ k[x1, . . . , xn]

Termination Each of the steps is already known to terminate.

Correctness Let Ar → Γϕ ⊆ Ar × Ar be the graph of ϕ. Then the diagram

Ar //

Γϕ

projection��

An

commutes, and we saw last time that projection corresponds to elimination of variables.


II.5.4 Remarks on computation

There’s a philosophy (perhaps not corresponding to a known theorem) due to Bayer andMumford (1984): the runtime for computing a Gröbner basis for I is approximately on theorder of the Castelnuovo–Mumford regularity of I.

Question: Given an ideal I ⊆ k[x1, . . . , xn] generated in degree ≤ d, can we bound theCM-regularity of I in temrs of n and d?

Answer: There’s a doubly exponential upper and lower bound.

Theorem II.5.3 (Galligo?). The CM-regularity of I is at most (2d)2n−2.

Theorem II.5.4 (Mayr–Meyer). There exists a family with regularity(d2

)2n−2

.

This leads to a mystery: Why do GB-computations actually terminate in a reasonableamount of time?

The answer (Bayer–Mumford): The ideals we care about do not achieve the lower bound.Some more remarks:

• Question: How does the monomial order affect runtime?

General principle: “lex is bad” (because lex is an elimination order for xs+1, . . . , xn forany s).

• Speedups:

– Prove criteria for knowing that an S-polynomial S(g1, g2) will reduce to zerowithout doing the reduction explicitly.

– Change the order for considering S-pairs or for reduction steps.

– Gröbner bases make no use of symmetry! In fact, there’s an inherent asymmetrydue to the term order. (Fixing a “generic initial ideal” is one approach to takingadvantage of symmetry.)

II.5.5 Radical membership

Does f ∈√I?

Lemma II.5.5 (Rabinovitch’s trick). Given an ideal I ⊆ S, let I := I + 〈1− tf〉 ⊆ S[t].Then f ∈

√I ⇐⇒ 1 ∈ I.

Proof. By definition, f ∈√I ⇐⇒ fm ∈ I for some m, which implies tmfm ∈ I. Then

1 = tmfm + (1− tmfm) = tmfm + (1− tf)(1 + tf + t2f 2 + · · ·+ tm−1fm−1) ∈ I .

Conversely, suppose 1 ∈ I. Note that S[t]/(1 − tf) ∼= S[f−1]. So 1 ∈ I · S[f−1], hence1 =

∑i aigi for some ai ∈ S[f−1] and gi ∈ I. Clearing denominators, fm =

∑i(f

mai)gi,where fmai ∈ S as long as m is sufficiently large. Thus, f ∈

√I.

Hence, we have an algorithm:

II.6. 2014-03-13: FLATNESS AND GRÖBNER BASES [INCOMPLETE] 39

Input f1 ∈ S and {g1, . . . , gs} ⊆ S

Output True if f ∈√〈g1, . . . , gs〉, otherwise False.

Steps

• let I := 〈g1, . . . , gs〉• let I := I + 〈1− tf〉 ⊆ S[t]

• compute a reduced GB G for I

• return True if G = {1}, otherwise return False

II.5.6 Dimension of a ring

Next time, we’ll consider how to compute the dimension of a quotient ring S/I, using thefollowing algorithm:

Input I = 〈g1, . . . , gs〉 ⊆ S

Output dim(S/I)

Steps

• compute G a GB of I

• note that in(I) = 〈LT(gi) | gi ∈ G〉• compute dim(S/ in(I))

• return dim(S/ in(I))

This relies on the following theorem:

Theorem II.5.6. dim(S/I) = dim(S/ in(I)).

II.6 2014-03-13: Flatness and Gröbner bases [incomplete]

II.6.1 Project progress report

March 27:

• Progress report

• Approx. 1 page

• Don’t just rehash your source! Include something like:

– original examples

– Macaulay2 code

– your own big picture summary


II.6.2 Flatness and Gröbner bases

The best treatment of this is [E, §15.8].

II.7 2014-03-25: Flatness and Gröbner bases, continuedThroughout: S = k[x1, . . . , xn], > is any monomial order, and I is any ideal of S.

Lemma II.7.1. The set of monomials in S but not in in>(I) forms a k-basis for S/I.

Proof. If not, then there is a relation∑an · n ∈ I, where n ranges over monomials not in

in>(I).

Proposition II.7.2 (Universality of weight orders). There is a weight vector w ∈ Zn≥0 suchthat in>(I) = in>w(I).

Theorem II.7.3 (Flatness of Gröbner bases). There is an ideal I ⊆ S[t] such that S[t]/I isa flat k[t]-module, and

S[t]

I + 〈t− α〉∼=

{S/I if α 6= 0,

S/ in>(I) if α = 0.

Proof. By universality of weight orders, > = >w for some weight vector w = (w1, . . . , wn).Given g ∈ S, define

g := tdegw(LT>(g)) · g(t−w1x1, . . . , t−wnxn) = LT>(g) + t · (other stuff) ∈ S[t],

and let I := 〈g | g ∈ I〉 ⊆ S[t]. It is clear from the above that I + (t) = in>(I) + (t). Ifα 6= 0, it suffices to show that

S[t, t−1]

I

'−→ S[t, t−1]

I=S

I⊗ k[t, t−1].

Consider the map

ϕ : S[t, t−1]→ S[t, t−1],

(t, x1, . . . , xn) 7→ (t, tw1x1, . . . , twnxn).

This is an isomorphism on S[t, t−1]. Note that ϕ(g) = tdegw(LT>(g)) · g, hence ϕ(I) = I.To prove flatness, let B be the set of monomials in S but not in in>(I). We will show

that S[t]/I is free with k[t]-basis B. There is a natural map of k[t]-modules⊕b∈B

k[t]→ S[t]

I

b 7→ b.

Let K be the kernel and Q the cokernel of this map. Note that ϕ(b) = tdegw(b) · b. SinceB is a k[t, t−1]-basis for S[t, t−1]/I, it then follows that B is a k[t, t−1]-basis for S[t, t−1]/I.Hence, K[t−1] = 0, and so K = 0. Also, Q[t−1] = 0, hence some power of t annihilates Q.But

⊕b∈B k[t]→ S[t]/I is an isomorphism mod t, so Q/t = 0, whence Q = 0.

II.7. 2014-03-25: FLATNESS AND GRÖBNER BASES, CONTINUED 41

See [E, Thm. 15.17] for an alternate argument.

Corollary II.7.4. dimS/I = dimS/ in>(I).

(Requires stuff to be proved today.)Exercise II.7.5. If in>(I) =

√in>(I), then I =

√I. (Hint: if I 6=

√I, then there exists f /∈ I

such that f 2 ∈ I.)Example II.7.6 (Twisted cubic). Let S = Q[x, y, z, w] with the lex order >, and let

I =⟨yw − z2, xw − yz, xz − y2

⟩.

Thenin>(I) = 〈yw, xw, xz〉 = 〈x, y〉 ∩ 〈w, z〉 ∩ 〈w, x〉 .

To prove the corollary, we’ll need to use Hilbert functions. Let M be a graded S-module.The Hilbert function of M is the function

HFm : Z→ N ∪ {∞} ,d 7→ dimkMd.

The Hilbert series is the power series

HSM(z) =∑d∈Z

HFM(d)zd.

If M is finitely-generated, then HSM(z) ∈ Z[[z]] [z−1].If M is a graded S-module and a ∈ Z, define M(a) to be the S-module where M(a)d =

Ma+d.

Corollary II.7.7. If I ⊆ S is a graded ideal, then HFS/I = HFS/ in> I .

Proof. Define I ⊆ S[t] and g ∈ S[t] as before. Since I is homogeneous, I is homogeneouswith respect to the weights deg(xi) = 1, deg(t) = 0. We have

S[t]

I=⊕d∈Z

S[t]d

Id.

For each d, S[t]d/Id is a flat, finitely-generated k[t]-module, and so for all d, S[t]d/Id is a freek[t]-module. Thus,

HFS/ in>(I)(d) = dimSd

in>(I)d= rank

S[t]d

Id= dimK

SdId

= HFS/I(d).

Theorem II.7.8 (Hilbert). LetM be a finitely-generated graded S-module. There is a uniquepolynomial PM ∈ Q[t], called the Hilbert polynomial of M , such that PM(d) = HFM(d) forall d� 0.

Theorem II.7.9. Let M be a finitely-generated graded S-module. Then

deg(PM) + 1 = dimM = dimS/ ann(M).

Exercise II.7.10. The Hilbert polynomial of the twisted cubic is 3t+ 1.


II.8 2014-03-27: Hilbert polynomialsLet f : Z→ Z be a function. We say that f is of polynomial type of degree d if there existsP (t) ∈ Q[t] of degree d such that f(e) = P (e) for e� 0.

Lemma II.8.1 (Newton). Let (∆f)(e) = f(e) − f(e − 1). If ∆f is of polynomial type ofdegree d, then f is of polynomial type of degree d+ 1.

Theorem II.8.2. Let M be a finitely-generated graded S-module. Then there is a uniquePM(t) ∈ Q[t] such that PM(e) = HFM(e) for all e� 0.

Theorem II.8.3. degPM(t) + 1 = dimM = dimS/ ann(M).

We’ll prove this via the following claims:

Claim II.8.4. There is a chain 0 = N0 ( N1 ( · · · ( Nr = M , where Ni+1/Ni∼= S/Pi(ai)

for some Pi prime and aI ∈ Z.

Claim II.8.5. degPM = maxi{

degPS/Pi

}.

Claim II.8.6. dimM = maxi {dimS/Pi}.

See [BH] for the proof.

II.8.1 Hilbert polynomials and short exact sequences

Consider a short exact sequence

0→ A→ B → C → 0

of graded S-modules. Looking in degree e, we have a short exact sequence of S0 = k-modules

0→ Ae → Be → Ce → 0.

By additivity of dimension, dimk Be = dimk Ae + dimk Ce, so HFB(e) = HFA(e) + HFc(e) forall e. Thus, HFB = HFA + HFC , so the Hilbert polynomial is also additive on short exactsequences.

Chapter III

Homological methods

III.1 2014-03-27Next, we’ll start our study of homological commutative algebra. The main source is [E],starting with the chapter on the Koszul complex.Note III.1.1. For now, R will generally denote a Noetherian ring. We continue to reserve Sfor the ring k[x1, . . . , xn], where k is a field and deg(xi) = 1.

III.1.1 Complexes of R-modules

III.2 2014-04-01: Minimal free resolutionsA resolution is a complex

· · · ← Fi∂i+1← Fi+1 ← · · ·

where Fj = 0 for j � 0 and where Hj(F•) = 0 unless j = 0.

III.2.1 Minimal free resolutions over a local ring

Let (R,m) be a local ring, and let M be a finitely-generated R-module. A minimal freeresolution of M is a resolution

F0 F1φ1oo F2

φ2oo · · · ,oo

where

(1) each Fi is a free R-module of finite rank, and

(2) φi(Fi) ⊆ mFi−1.

Example III.2.1. Let R = C[[x, y]] and M = R/m = C. Then the minimal free resolution isgiven by the Koszul resolution

R/m R1oooo R2[x,y]oo R2

[ y−x ]oo 0.oo

43

44 CHAPTER III. HOMOLOGICAL METHODS

Theorem III.2.2. Every finitely-generated R-module M has a minimal free resolutions thatis unique up to (non-unique) isomorphism.

(Proof in [E].)

Example III.2.3 (infinite minimal free resolution). Let R = C[x]/x2 and M = R/x. Thenthe minimal free resolution is

M R1oo R1xoo R1xoo · · ·xoo

Lemma III.2.4. Let M be a finitely-generated R-module, and let m1, . . . ,mr ∈ M whoseimages form a basis for M/m. Then:

(1) m1, . . . ,mr generate M .

(2) Let m′1, . . . ,m′r ∈M whose images also form a basis of M/m. Then there is an invert-ible r × r matrix A = (aij) ∈ Matr×r(R) such that the following diagram commutes:

⊕ri=1Rεi

εi 7→mi// //

' A��

M

⊕ri=1 Rδi δi 7→mi

// //M

Proof. Let N := M/ 〈m1, . . . ,mr〉. Then N/m = 0, hence N = 0. This proves (1).To show (2), write mi =

∑j aijm

′j. Set A = (aij) : Rr → Rr. By construction, the

induced map A : (R/m)r → (R/m)r is an isomorphism. Hence det(A) ∈ (R/m) \ {0}, sodet(A) ∈ R \m is a unit in R.

Upshot: if F• is a minimal free resolution of M , then rankFi is an invariant of M .

Example III.2.5. Let R = C[[x, y]] and M = R/m. Then rankF0 = 1, rankF1 = 2, andrankF2 = 1.

III.2.2 Conjectures

Conjecture III.2.6 (Buchsbaum–Eisenbud–Horrocks). For i = 0, . . . , codimM , rankFi ≥(codimM

i

).

Remark III.2.7. There is no known counterexample for any Noetherian local ring. Moreover,a similar thing should be true for any free resolution, even dropping Noetherian or localhypotheses.

Conjecture III.2.8 (Arramov). In an infinite minimal free resolution F•, the rank mapi 7→ rankFi is nondecreasing for all i� 0.

III.3. 2014-04-08: KOSZUL COMPLEXES 45

III.2.3 Minimal free resolutions in the graded case

Let R be a graded k-algebra. Every graded free R-module has the form F0 =⊕

j∈ZR(−j)β0,jfor some β0,j ∈ N. (For Fi a graded free R-module, βi,j is the number of degree-j minimalgenerators of Fi.)

A graded map of R-modules is an R-module homomorphism that respects the grading.Example III.2.9. If deg(xy) = 2, then multiplication by xy is a graded map R(−2)

xy−−→ R.

Theorem III.2.10. Every finitely-generated graded R-module has a minimal free resolutionthat is unique up to (non-unique) isomorphism.

III.3 2014-04-08: Koszul complexes

III.4 2014-04-10Let f : F • → G• be a map of chain complexes (indexed cohomologically), i.e., df = fd.(Motivation: chain maps f∗ : C∗(X) → C∗(Y ) for continuous maps of topological spacesf : X → Y .)

Given a chain map f : F → G, we get induced maps H if : H iF • → H iG•.

Definition III.4.1. A chain map f is null-homotopic if there exists h = (hi) with hi : F i →Gi−1 such that f = dh+ hd.

Remark III.4.2. If f is null-homotopic, then f is a chain map. Indeed, f = dh+ hd, so

df − fd = d(dh+ hd)− (dh+ hd)d = d2h+ dhd− dhd− hd2 = 0.

Definition III.4.3. We say two chain maps f, g are homotopic to each other, denoted f ∼ g,if f − g is null-homotopic. This is an equivalence relation.

Proposition III.4.4. If f ∼ g, then H i(f) = H i(g) for all i.

Proposition III.4.5. If f is a chain map, we get a long exact sequence of cohomology

. . .→ H i(F •)→ H i(G•)→ H i(C•(f))→ H i+1(F •)→ . . .

Proposition III.4.6. Let F •, G• be complexes. Assume F i = Gi = 0 for i > 0, F i isprojective for all i, and H i(G•) = 0 unless i = 0. Then any map H0(F )→ H0(G) lifts to achain map, unique up to homotopy.

Corollary III.4.7. Projective resolutions are unique up to homotopy.

III.5 2014-04-17Let (R,m) be a local ring, and let k = R/m.

Corollary III.5.1. f1, . . . , fn ∈ m form a regular sequence on M iff K(f ,M) is a freeresolution of M/(f1, . . . , fn)M .

Let M,N be R-modules with projective resolutions P• and Q•. Then

Torp(M,N) = Hp(P• ⊗N) = Hp(M ⊗Q•) = Hp tot(P• ⊗Q•).


III.5.1 Projective dimension

Definition III.5.2. The projective dimension of M , denoted pdim(M), is the length of theshortest projective resolution of M .

Theorem III.5.3 (Hilbert’s syzygy theorem). Let (R,m) be a regular local ring, and let Mbe a finitely-generated R-module. Then pdim(M) ≤ dim(R).

Proof. Let F• be a minimal free resolution of M :

F• =[0→ Fp

ϕp−−→ . . .ϕ2−−→ F1

ϕ1−−→ F0

].

We have ϕIFi ⊆ mFi−1, and

Torj(M,k) = Hj(F• ⊗R R/m)

= Hj

(0→ Fp ⊗ k

0−→ . . .0−→ F1 ⊗ k

0−→ F0 ⊗ k)

= Fj ⊗R k = krankFj .

Let p be the length of a minimal free resolution of M . Then

p = max{j∣∣ Torj(M,k 6= 0

}≥ pdim(M).

But also Torj(M,k) = Hj(M ⊗K(x1, . . . , xn)), where (x1, . . . , xn) = m, so Torj(M,k) = 0 ifj > n = dimR. Thus,

dimR ≥ p ≥ pdim(M).

Corollary III.5.4. In fact, p = pdim(M).

III.5.2 Global dimension

Define the global dimension of a Noetherian ring R by

gl. dim(R)def= max

{pdim(M)

∣∣M ∈ R-Mod}.

By a theorem of Auslander [E, Thm. A3.18], it’s enough to look at finitely-generated R-modules.

Theorem III.5.5. Let (R,m) be a local ring. Then

gl. dim(R) =

{dimR if R is regular,∞ otherwise.

Lemma III.5.6. gl. dim(R) = pdim(R/m).

Proof. By definition gl. dim(R) ≥ pdim(k). Moreover, pdim(M) ≥ i ⇐⇒ Tori(M,k) 6= 0.If F• is a minimal free resolution of M , then

Tori(M,k) = Hi(F• ⊗ k) = krankFi = Hi(M ⊗G•),

where G• is a minimal free resolution of R/m. If Gi 6= 0, then pdim(M) ≥ i.

III.5. 2014-04-17 47

Theorem III.5.7 (Auslander–Buchsbaum formula). Let (R,m) be a local ring, and let Mbe a finitely-generated R-module. If pdim(M) <∞, then

pdim(M) = depth(m, R)− depth(m,M).

Proof. Outline:

• Exploit finiteness of pdim(M) via induction.

• Use Koszul complex to test depth.

If pdim(M) = 0, then M is free, and the statement is immediate. Fix a minimal freeresolution F• of M . If pdim(M) > 0, then we have

0→ Nφ−→ F0 →M → 0, (*)

and pdim(M) = pdim(N) + 1. By induction,

d := depth(m, N) = depth(m, R)− pdim(N).

So it suffices to show that depth(m,M) = d− 1.Let x1, . . . , xn be minimal generators of m. Tensor (*) by K(x1, . . . , xn) to get a long

exact sequence

. . .→ H i−1(K(x, F0))→ H i−1(K(x,M))→ H i(K(x,N))→ H i(K(x, F0))→ . . .

We have d = depth(N) ≤ depth(R). For i < d, we get vanishing for F0 and N , soH i−1(K(x,M)) = 0. Thus,

0→ Hd−1(K(x,M))→ Hd(K(x,N))f−→ Hd(K(x, F0))→ . . .

is exact. It suffices to prove that the above map f is zero.

Remark III.5.8. p = length of a minimal free resolution of M = pdim(M).

Theorem III.5.9 (Auslander–Buchsbaum, Serre). Any localization of a regular local ringat a prime is a regular local ring. Moreover, any localization of k[x1, . . . , xn] at a prime isregular.

Proof. Let (R,m) be regular local and P ⊂ R prime. By the lemma, it suffices to showpdim(RP/mP ) is finite. But if F• is a minimal free resolution of R/m, then (F•)P is a freeresolution of (R/m)P = RP/mP .


III.6 2014-04-22: Cohen–Macaulay ringsRecall:

Theorem III.6.1 (Auslander–Buchsbaum formula). Let (R,m) be a local ring, and M afinitely-generated R-module of finite projective dimension. Then

pdim(M) = depth(m, R)− depth(m,M).

The proof uses the Koszul complex to measure depth.We also had the following theorem:

Theorem III.6.2. Let (R,m) be a local ring. Then

gl. dim(R) = pdim(R/m) =

{dimR if R is regular,∞ otherwise.

Say R is non-regular. Let x1, . . . , xn be minimal generators for m. Then dimR < n, anddepth(m, R) < n by considering H i(K(x)). By Auslander–Buchsbaum, it suffices to showpdim(R/m) ≥ n.

Lemma III.6.3 ([E], 19.13). K(x) is a split subcomplex of the minimal free resolution ofR/m.

III.6.1 Cohen–Macaulay rings

“Life is really worth living in a Noetherian ring in which every system of parameters is aregular sequence. Such a ring is called Cohen–Macaulay.” — Mel Hochster

Lemma III.6.4. Let (R,m) be a local ring. Then dim(R) ≥ depth(m, R).

Proof. Let y1, . . . , ys be a maximal regular sequence. Then by the principal ideal theorem,

0 ≤ dimR/(y1, . . . , ys) = dimR− s = dimR− depth(R).

Definition III.6.5. A local ring (R,m) is called Cohen–Macaulay if

dimR = codim(m, R) = depth(m, R).

A Noetherian ring Q (not necessarily local) is Cohen–Macaulay (CM) if QP is CM for allprimes P .

Theorem III.6.6 (Alternate definition). Let R be a Noetherian ring. The following areequivalent:

(1) R is Cohen–Macaulay.

(2) depth(P,R) = codim(P,R) for all primes P .

(3) depth(I, R) = codim(I, R) for all ideals I ⊆ R.

III.6. 2014-04-22: COHEN–MACAULAY RINGS 49

Example III.6.7. Consider the ring

R =k[x, y, z](x,y,z)

(xz, yz).

Even though x is a zero-divisor, dimR/(x) = 1 = dimR − 1. This can’t happen in a CMring.

Theorem III.6.8. A ring R is CM iff R[x] is CM.

Theorem III.6.9. Let (R,m) be a local ring. Then (R,m) is CM iff (R, m) is CM.

Proof. We have dimR = dim R. Let x1, . . . , xn generate the maximal ideal. Since R is a flatR-module,

Hs(K(x)⊗ R

) ∼= Hs(K(x)

)⊗ R ∼= lim

j

Hs(K(x))

mj.

Theorem III.6.10. Let R be a CM ring. Then:

(1) R is universally catenary.

(2) R is equidimensional.

(3) R has no embedded primes.

(4) SpecR is connected in codimension 1.

Definition III.6.11. A ring R is equidimensional if:

(i) all minimal primes have the same dimension, and

(ii) all maximal primes have the same codimension.

Definition III.6.12. SpecR is connected in codimension 1 if, for all incomparable ideals Iand J with SpecR = Spec(R/I) ∪ Spec(R/J), we have codim(I + J) ≥ 1.

Definition III.6.13. Let (R,m) be a local ring, and M an R-module.

• M is Cohen–Macaulay if dimM = depth(m,M).

• M is maximal Cohen–Macaulay if M is CM and dimM = dimR.

If M is maximal CM, then M is free or pdim(M) =∞.

Example III.6.14. Let R = C[x, y, z]/(xz− y2) and M = coker

(x yy z

). Then M is maximal

Cohen–Macaulay, and we have a resolution

. . .→ R2

x yy z

−−−−−−→ R2

z −y−y x

−−−−−−−−−→ R2

x yy z

−−−−−−→ R2.


III.7 2014-04-29This is the last day of lectures.

Let k be a field, and let S = k[xij]. Let

Φ =

x11 . . . x1q... . . . ...xp1 . . . xpq

be a generic p× q matrix (where p ≤ q), and let I be the ideal generated by the p×p minorsof Φ.

The main theorem for today is:

Theorem III.7.1. I is prime of codimension q − p+ 1.

Remark III.7.2. There are several proofs of this theorem:

• Homological proof:

– Show S/I is Cohen–Macaulay (Main)– This implies I is prime (Easy)

• Gröbner basis proof:

– Compute a Gröbner basis of I (Main)– Conclude I is prime (Easy)

• Spectral sequence proof (which we won’t discuss)

III.7.1 Easy step

Lemma III.7.3.√I is prime of codimension q − p+ 1.

Corollary III.7.4. I is prime ⇐⇒ I =√I.

Proof of lemma. Let B = k[v1, . . . , vp]. Consider the incidence variety

V (J) ={

(M, v)∣∣M · v = 0

}⊆ Spec(S)× Proj(B) = Apq × Pp−1,

where we think of Apq as p × q matrices and Pp−1 as column vectors modulo scaling. Letπ : V (J)→ Apq be the restriction of the natural projection Apq × Pp−1 � Apq. Note that

π(V (J)) = V (I)

because M ∈ V (I) iff rankM ≤ p − 1 iff Mv = 0 for some v 6= 0 iff (M, v) ∈ V (J). Soπ : V (J) � V (I) is surjective, inducing an inclusion

S√I↪→ S[v1, . . . , vp]√

J.

Thus, if√J is prime, then so is

√I.

It remains to show that J is prime, dimV (J) = dimV (I), and codim I = q − p+ 1.

It remains to show I is radical, which is the hard step.

III.7. 2014-04-29 51

III.7.2 Homological proof, main step

Theorem III.7.5. S/I is Cohen–Macaulay, and hence I has no embedded primes.

In particular, this means that S/I is either reduced or everywhere non-reduced, so themain theorem will follow if we can find just one smooth point. We do this with the following:

Lemma III.7.6. Let P ⊆ S be the maximal ideal corresponding to the p × q matrix whichhas a (p−1)× (p−1) identity matrix in the upper left and is zero elsewhere. More explicitly,

P =⟨xi,i − 1

∣∣ i = 1, . . . , p− 1⟩

+⟨xi,j

∣∣ i 6= j or i = j = p⟩.

Then (S/I)P is regular local.

Sketch of proof of theorem. Construct a free resolution of S/I as an S-module, and use thisto show that

pdim(S/I) = codim(I).

By Auslander–Buchsbaum, it follows that S/I is Cohen–Macaulay. The free resolution ofS/I used in the proof is the “Eagon–Northcott complex”.

III.7.3 Gröbner basis proof, main step

We use the term order> which is reverse lex for the “antidiagonal” ordering x11, x21, . . . , xp1, x12, x22, . . . , xp2, . . . . . . , xpq.Fix J = {j1 < · · · < jp} ⊆ 1, . . . , q and ∆J the corresponding p× p minor of Φ.

Bibliography

[AM] Atiyah–Macdonald, Introduction to Commutative Algebra.

[BH] Bruns–Herzog, Cohen–Macaulay Rings.

[E] D. Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.

53

Index

affine ring, 19, 21

Castelnuovo–Mumford regularity, 38catenary, 23codimension, 9Cohen–Macaulay, 48

module, 49completion, 11connected in codimension 1, 49

dimensionof a module, 10of a ring, 9of an ideal, 9

elimination order, 35elimination theory, 36equidimensional, 49excellent ring, 26

finite extension of rings, 8flat, 11

global dimension, 46Gröbner basis, 31

reduced, 31graded k-algebra, 27graded ideal, 27graded map of R-modules, 45graded module, 28

height, 9Hilbert function, 28, 41Hilbert polynomial, 41Hilbert series, 41homogeneous

element, 27ideal, 27

homotopic, 45

initial ideal, 31

Japanese ring, 26

Krull ring, 26

leading term, 30

maximal Cohen–Macaulay, 49minimal free resolution, 43monomial ideal, 29monomial order, 29multidegree, 29

Nagata ring, 26Nakayama’s lemma, 13null-homotopic, 45

polynomial type, 42Principal Ideal Theorem, 12product order, 35projective dimension, 46

regular local ring, 18regular ring, 18regular sequence, 19regular system of parameters, 18resolution, 43

S-polynomial, 32symbolic power, 13system of parameters, 15

transcendence degree, 22

universally catenary, 23universally Nagata ring, 26

54

math 746 notes topics in ring theory

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