Part III Commutative algebra

Lecturer: Dr. C.J.B. Brookes

Michaelmas term 20131

1Transcribed by Stiofain Fordham

Contents

PageLecture 1 51.1. Introduction 5Lecture 2 62.1. §1: Noetherian rings: definitions, examples & ideal structure 7Lecture 3 93.1. Cohen’s theorem, the nilradical and Jacobsen radical 9Lecture 4 114.1. Weak and strong Nullstellensatz 11Lecture 5 135.1. Minimal primes, annihilators and associated primes 13Lecture 6 156.1. §2: Localisation 16Lecture 7 177.1. Modules 17Lecture 8 208.1. §3: Dimension 20Lecture 9 229.1. More about integrality 22Lecture 10 2410.1. The Going-up and Going-down theorems 24Lecture 11 2511.1. Proof of the Going-down theorem 25Lecture 12 2712.1. Noether normalisation lemma 27Lecture 13 2813.1. Remarks on the Noether normalisation lemma 29Lecture 14 3014.1. Proof of Krull’s Hauptidealsatz 31Lecture 15 3215.1. Filtrations 33Lecture 16 3516.1. The lemma of Artin-Rees and the theorem of Hilbert-Serre 35Lecture 17 3717.1. The Hilbert polynomial and the Samuel polynomial 37Lecture 18 3918.1. §4: Valuation rings and Dedekind domains 3918.2. Recipe for valuation rings 40Lecture 19 4119.1. Discrete valuation rings 41Lecture 20 4220.1. §5: Tensor products, homology and cohomology 43Lecture 21 44

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21.1. Restriction and extension of scalars 44Lecture 22 4622.1. Projective & injective modules; Tor and Ext 46Lecture 23 4923.1. Induced long exact sequences from Tor and Ext 49Lecture 24 5124.1. Hochschild cohomology (non-examinable) 5124.2. Hochschild dimension 52

Bibliography 53

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Lecture 1

Lecture 1

11th October 10:00

The prerequisites for this course are some experience in ring theory. On books:there is a very concise book by Atiyah & Macdonald [AM69]- it is a very goodintroduction, but there is material in there that I will not talk about. On the otherhand it is not all there - I want to say something about homological algebra. You’llfind quite a lot of things are exercises. If you want to see the details, you can lookat the book by Sharp [Sha00]. There is the book by Matsumura [Mat70]but thisis not an introduction but does include the homological algebra. There are the“weighty tomes”: Bourbaki [Bou74] and Zariski & Samuel [ZS60].

We start on chapter 0, an introduction

1.1. Introduction. A good place to start thinking about commutative alge-bra is with David Hilbert - he had a series of papers on invariant theory 1888-1893. The idea is, if k is a field and you have a polynomial ring in n variablesk[X1, . . . ,Xn], and if Σn is the symmetric group on {1, . . . , n}, then the symmetricgroup acts on the polynomial ring k[X1, . . . ,Xn] by permuting variables. We lookat the set of polynomials fixed under this action: the set (actually form a ring)

S = {f ∈ k[X1, . . . ,Xn]∶ g(f) = f for all g ∈ Σn}

called the ring of invariants. We have the elementary symmetric functions

f1(X1, . . . ,Xn) =X1 + ⋅ ⋅ ⋅ +Xn

f2(X1, . . . ,Xn) =∑i<j

XiXj

⋮

fn(X1, . . . ,Xn) =X1X2 . . .Xn

In fact S is generated as a ring by these fi and

S ≅ k[f1, . . . , fn]

as a polynomial ring i.e. no algebraic dependence. Hilbert showed that S is finitelygenerated, and for lots of other groups as well. Along the way he proved fourimportant theorems

(1) Basis theorem(2) Nullstellensatz (zeroes theorem)(3) polynomial nature of the Hilbert function(4) Syzygy theorem

The next person to mention is Emmy Noether - 1921, she abstracted fromHilbert’s work the fundamental property that made the Basis theorem work.

Definition 1.1. A (commutative) ring R is Noetherian if every ideal of R is finitelygenerated.

Remark. There are equivalent definitions - you should be reminding yourself ofthis for next time.

Noether developed the idealtheorie, the theory of ideals for Noetherian rings -e.g. primary decomposition which generalises factorisation into primes from numbertheory.

Next let me tell you or remind you of the link between commutative algebraand algebraic geometry. One way of viewing the fundamental theorem of algebra isthat a polynomial f ∈ C[X] is determined upto a scalar multiple by its zeroes upto

5

Lecture 2

multiplicity. If we go to n variables now, f ∈ C[X1, . . . ,Xn], there is a polynomialfunction

f ∶Cn → Cn

(a1, . . . , an)↦ f(a1, . . . , an)

Thus we get polynomial functions on affine n-space. Given I ⊂ C[X1, . . . ,Xn],define

Z(I) = {(a1, . . . , an) ∈ Cn∶ f(a1, . . . , an) = 0 for all f ∈ I}

the set of common zeroes. Such a subset of Cn is an algebraic set. Note that we canreplace I by the ideal generated by I without changing Z(I). For a subset S ⊂ Cn,define

I(S) = {f ∈ C[X1, . . . ,Xn]∶ f(a1, . . . , an) = 0 for all (a1, . . . , an) ∈ S}

This is an ideal of C[X1, . . . ,Xn] and moreover it is radical: if fr ∈ I(S) for somer ≥ 1 then f ∈ I(S).

So the Nullstellensatz is really a family of theorems but one way of looking atit, is that there is a 1-1 correspondance

{radical ideals in C[X1, . . . ,Xn]}←→ {algebraic subsets of Cn}where

I ↦ Z(I)

I(S)↤ S

in particular, the maximal ideals of C[X1, . . . ,Xn] correspond to points in Cn.

Remark. There is a topology on Cn under which the closed sets are the algebraicones - called the Zariski topology.

The Basis theorem says that

Theorem 1.1. If R is Noetherian, then R[X] is.

and there is a sort of obvious corollary

Corollary 1.2. If k is a field then k[X1, . . . ,Xn] is Noetherian

Quite a large section of the course is about dimension theory. There are atleast three ways of defining dimension:

(1) in terms of maximal length of chains of prime ideals,(2) in the geometric context, we look at growth rates in the vicinity of points

of a function (in the context of the Hilbert function),(3) if you are looking at an integral domain, then you can take a field of frac-

tions, and you can ask about algebraic dependence - transcendence degreeof the field of quotients, which says how many transcendence elements toyou need to take.

In the commutative case these all give you the same answer, and in fact thereis a fourth way using homological algebra which for “nice” Noetherian rings, givesyou the same answer again. So it really does make sense to spend a lot of time ondimension. The case of dimension-0 is field theory and the case of dimension-1 isnumber theory.

Most of the content of this course comes from around 1920-1950 so it is “ancienthistory”.

Lecture 2

14th October 10:00

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Lecture 2

2.1. §1: Noetherian rings: definitions, examples & ideal structure.This class, you will probably have seen all this before. As I suggested at the end ofthe last lecture, you should be looking over your previous notes. We take R to bea commutative ring with a 1. Most of you are probably comfortable with rings butmaybe not with modules, so we will start with modules. Let M be a left R-module.

Lemma 2.3 (1.3). The following are equivalent

(1) every submodule is finitely generated.(2) the ascending chain condition: there is no strictly ascending chain of sub-

modules.(3) every non-empty subset of submodules of M contains at least one maximal

member.

Proof. (1)→ (2) Suppose we have a chain

M1 ⫋M2 ⫋M3 ⫋ ⋯

Then take N = ⋃Mi. It is a submodule of M . Assuming (1), then N is finitelygenerated & a finite generating set lies in some Mj , so N =Mj , which contradictsthe strict ascent.

(2)→ (3) Assume the ascending chain condition (ACC). Let S be a non-emptysubset of submodules. Choose M1 ∈ S. If it’s a maximal member, then we’re done.If not, then choose M2 which is bigger etc. By the ACC, this process stops.

(3)→ (1) Suppose that (3) holds. Let N be a submodule of M . Let

S = set of all finitely generated submodules of N

then S is non-empty because it contains the zero submodule. So S has a maximalmember L, say. Then we just check that L = N , and thus N is finitely-generated.1

�

Definition 2.2. A module satisfying these conditions is called Noetherian.

Lemma 2.4 (1.4). Let N be a submodule of M . Then M is Noetherian iff N isNoetherian and M/N is Noetherian.

Proof. I’ll proove one direction; you can do “→” yourself.2

Suppose N and M/N are Noetherian and we have an ascending chain of sub-modules of M

L2 ⊆ L2 ⊆ L3 ⊆ ⋯

Now we set⎧⎪⎪⎨⎪⎪⎩

Qi/N = (Li +N)/N

Ni = Li ∩N

By the ACC there are r, s such that Qi/N = Qr/N for i ≥ r and Ni = Ns for i ≥ s.Now we set k = max(r, s). Then you can check for yourself that Li = Lk for i ≥ k.[pp. 5 of Wilkin’s notes, much more straightforward]. �

Lemma 2.5 (1.5). Suppose that we have modules with

M =M1 +⋯ +Mn

(not necessarily direct). Then M is Noetherian iff each Mi is Noetherian.

1Suppose L is generated by {x1, . . . , xr} and let x ∈ N , then the submodule K generated by{x1, . . . , xr, x} is finitely-generated, so by maximality we have L =K, so x ∈ L thus L = N .

2Suppose that M is Noetherian, then N is Noetherian, since any submodule of N is alsoa submodule of M and hence finitely-generated. Also any submodule L of M/N is of the form{N + x∶x ∈ J} for some submodule J of M satisfying N ⊂ J . But J is finitely-generated (since Mis Noetherian). Let x1, . . . , xr be a finite generating set for J . Then L + x1, . . . , L + xr is a finite

generating set for K. Thus M/N is Noetherian.

7

Lecture 2

Proof. The direction “→” is okay. For the “←” direction, if each Mi is Noe-therian then

M1 ⊕M2 ⊕⋯⊕Mn

is Noetherian and so M is, since it is an image of M1⊕⋯⊕Mn under the canonicalmap

M1 ⊕⋯⊕Mn →M1 +⋯ +Mn �

Definition 2.3. A ring R is Noetherian if it is Noetherian as a (left) R-module.

Remark. The submodules of R as an R-module are its ideals. The ACC formodules is then equivalent to the ACC for ideals.

Lemma 2.6 (1.6). Let R be a Noetherian ring. Then any finitely generated R-module is Noetherian.

Proof. Suppose that

M = Rm1 +⋯ +Rmn

We have R-module maps R → Rm1 given by r ↦ rm1. R is Noetherian and so Rm1

is. Hence M is Noetherian by lemma 2.5. �

Theorem 2.7 (1.7, Hilbert’s basis theorem). Let R be a Noetherian ring. ThenR[x] is Noetherian.

Proof: (sketch). We prove that every ideal of R[x] is finitely generated.Let I be an ideal. Define

I(n) = set of polynomials in I of degree ≤ n.

We have 0 ∈ I(n) and so I(n) is non-empty. We have

I(0) ⊆ I(1) ⊆ ⋯

LetR(n) = set of all leading coefficients of elements of I(n)

(i.e. the coefficients of xn) and we can check that

R(0) ⊆ R(1) ⊆ ⋯

is an ascending chain and each R(n) is an ideal in R. The ascending chain conditionthen yields that⋃R(n) = RN for someN . We know that eachR(0),R(1), . . . ,R(N)is finitely generated. Each R(j) is generated by

aj1, . . . , ajkj

say, and these are leading coefficients of

fj1, . . . , fjkj

in I(j). And then the claim (which is for you to check) is that the set

{fjk ∶ j ≤ N,1 ≤ k ≤ kj}

generate I.3 �

Remark. In practice one uses Grobner bases for ideals - one generally would liketo get a “nice” set of generators for the ideal that one is dealing with, and thesesatisfy this: they are generating sets with added properties that make algorithmsefficient.

3Let g ∈ I have degree n ≤ N and leading coefficient b. Then there are elements c1, . . . , ckn ∈ Rsuch that b = c1an1 + ⋅ ⋅ ⋅ + cknankn then g = c1fn1 + ⋅ ⋅ ⋅ + cknfnkn + r where deg r < deg g, so an

induction argument works. If g has degree n > N then the expression you need is g = c1xn−NfN1+

⋅ ⋅ ⋅ + ckNxn−NfNkN

+ r and an induction argument works.

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Lecture 3

Now we will look at some examples

(1) Fields are Noetherian.(2) Principal ideal domains e.g. k[X] and Z are Noetherian.(3) Localisations of Noetherian rings are necessarily Noetherian. Take

Z(p) = {q ∈ Q∶ q is of the form m/n with m,n ∈ Z, and p ∤ n}

(4) The case k[X1, . . . ,Xn, . . . ] with infinitely many indeterminates, is notNoetherian, consider

(X1) ⫋ (X1,X2) ⫋ ⋯

(5) k[X1, . . . ,Xn] is Noetherian - this is a corollary of the Basis theorem.Z[X1, . . . ,Xn] is Noetherian. Any finitely generated ring is Noetherian(since such a ring is an image of some Z[X1, . . . ,Xn]).

(6) k[[X]] the ring of formal power series, is Noetherian, things like

a1 + a2X + a3X2 + . . .

with ai ∈ k. There is a more general result

Lemma 2.8 (1.8). If R is Noetherian then R[[X]] is Noetherian.

Proof. Next time. Exercise: do it by an argument analogous to that for theBasis theorem, but by using “trailing coefficients”. I will prove it by a differentmethod next time. �

Lecture 3

16th October 10:00

3.1. Cohen’s theorem, the nilradical and Jacobsen radical.

Proposition 3.9 (1.8). If R is Noetherian then R[[X]] is Noetherian.

Proof. Either by using trailing coefficients by a method analogous to the Basistheorem, or use Cohen’s theorem.4

Theorem 3.10 (1.9, Cohen). If every prime ideal is finitely generated then R isNoetherian.

Remark. Should view this as: only need to look at the primes.

Proof. Suppose that R is not Noetherian, so there exists ideals that are notfinitely generated. By Zorn’s lemma, there is a maximal member I of the family ofnon-finitely generated ideals (recall that to apply Zorn’s lemma, one needs to checkthat the family is non-empty and that the ascending chain of non-finitely generatedideals has a union which is non-finitely generated).

Claim: I is prime (& hence we’re getting a contradiction if we’re assuming allprimes to be finitely-generated). Suppose now that there exists a ∉ I and b ∉ I butwith ab ∈ I. Then I + Ra is an ideal strictly containing I. The maximality of Iensures I +Ra is finitely generated by

i1 + r1a, . . . , in + rna

say. Let J = {s ∈ R∶ sa ∈ I} ≥ I +Rb ≩ I. So by maximality of I, we know that J isfinitely generated. We will show that

I = Ri1 + ⋅ ⋅ ⋅ +Rin + Ja

a finitely generated ideal, a contradiction: take t ∈ I ≤ I +Ra, so

t = u1(i1 + r1a) + ⋅ ⋅ ⋅ + un(in + rna)

4Lemma 0306 just seems simpler really.

9

Lecture 3

for some ui ∈ R. So u1r1 + ⋅ ⋅ ⋅ + unrn ∈ J and so it is of the required form. �

To use Cohen’s theorem, we apply

Proposition 3.11. Let Pc be a prime ideal of R[[X]] and θ is the map θ∶R[[X]]→R sending X to 0. Then P is a finitely generated ideal of R[[X]] iff θ(P ) is afinitely generated ideal of R.

Proof. Clearly if P is finitely generated then θ(P ) is. Conversely, supposethat θ(P ) is finitely generated with

θ(P ) = Ra1 + ⋅ ⋅ ⋅ +Ran

If X ∈ P then P is generated by a1, . . . , an,X and we are done. If X ∉ P , letf1, . . . , fn be power series with constant terms a1, . . . , an in P . Take g ∈ P with

g = b + . . .

(where b is the constant term). But b = ∑ biai, so

g −∑ bifi = g1X

for some g1 - note that g1X ∈ P . But P is prime and X ∉ P and so g1 ∈ P . Similarly

g1 =∑ cifi + g2X

with g2 ∈ P . Continuing we get h1, . . . , hn ∈ R[[X]] with

hi = bi + ciX + . . .

satisfying that

g = h1f1 + ⋅ ⋅ ⋅ + hnfn �

�

Proposition 3.12 (1.11). Let R be commutative, but not necessarily Noetherian.The set N(R) of all nilpotent elements of R is an ideal and R/N(R) has no non-zero nilpotent elements.

Proof. If x ∈ N(R) then xn = 0 for some n and so (rx)n = 0, thus rx ∈ N(R).If x, y ∈ N(R) then xn = 0 and ym = 0 for some m,n then you can convince yourselfthat (x + y)n+m+1 = 0, so x + y ∈ N(R). If s ∈ R/N(R) then s = x + N(R) andsn = xn +N(R). If sn = 0 then xn ∈ N(R) and so x ∈ N(R) so s = 0. �

Definition 3.4 (1.12). This ideal N(R) is called the nilradicaldfn:nilradical of R.

Another way of looking at the nilradical (due to Krull) is the following

Proposition 3.13 (1.13, Krull). N(R) is the intersection of all the prime idealsof R.

Proof. Let

I = ⋂P prime

P

If x ∈ R is nilpotent, then xn = 0 ∈ P for any prime. So x ∈ P . Hence N(R) ≤ I.Now, suppose that x is not nilpotent. Set S to be the set of ideals J such thatfor n > 0, xn ∉ J . S is non-empty as (0) ∈ S, and we can apply Zorn’s lemmaagain to get a maximal member J1 of S (if R is Noetherian, we don’t have to thinkabout Zorn’s lemma).We claim that J1 is prime. For, suppose that yz ∈ J1 withy ∉ J1, z ∉ J1. So the ideals J1 + Ry and J1 + Rz strictly contain J1 and henceby the maximality of J1, we have xn ∈ J1 + Ry and xn ∈ J1 + Rz for some n. Sox2n ∈ J1 +Ryz. But since yz ∈ J1, this implies x2n ∈ J1, a contradiction. �

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Lecture 4

Definition 3.5 (1.14). The radical ideal√I of an ideal I of a ring R is defined by

√I/I = N(R/I)

An ideal is radical if I =√I. Note that

√I is radical, and

√I = ⋂

P prime,containing I

P

Definition 3.6 (1.15). The Jacobsen radical J(R) of R is the intersection of themaximal ideals of R.

Lemma 3.14 (1.16, Nakayama). If M is a finitely generated R-module with MJ =M (where J := J(R)) then M = 0.

Proof. If M is a non-zero finitely generated R-module, Zorn’s lemma (orNoetherian property, if R is Noetherian) yields maximal (proper) submodules. TakeM1 maximal in M . Thus M/M1 is an irreducible/simple R-module5 so take agenerator of M/M1, M1 + a say, then

(3.1) M/M1 ≅ R/I

with I a maximal ideal of R, e.g.

R →M/M1

r ↦ r(M1 + a)

is an R-module homomorphism and the kernel is necessarily a maximal ideal. Butby definition, J ≤ I. So we deduce that MJ ≤M1 ≨M (multiply (3.1) by J). So if

M ≠ 0 then MJ ≨M . �

Lecture 4

18th October 10:00

4.1. Weak and strong Nullstellensatz. For a commutative ring one has

N(R) = ⋂P prime

P ≤ J(R) = ⋂p maximal

P

These need not be equal, e.g. consider

R = {m/n ∈ Q∶ fixed prime p ∤ n}

then the unique maximal ideal is

P = {m/n ∈ Q∶p ∣m and p ∤ n}

and since this is happening inside of Q, this is an integral domain, so there are nonon-zero nilpotent elements, N(R) = (0) and J(R) = P . However for rings of theform

R = k[X1, . . . ,Xn]/I

for k algebraically closed with I any ideal, then N(R) = J(R). This is the contentof the Nullstellensatz (there are strong and weak versions of the Nullstellensatz,but bizarrely they are actually equivalent so they are in fact equally ‘strong’). Thisis what we are going to prove now - the proof I will present is due to Artin andTate.

5The simple modules over a ring R are the (left or right) modules over R which have nonon-zero proper submodules. If M1 is a maximal proper submodule of M , then M/M1 is simple

because any submodule of M/M1 is of the form {x +M1∶x ∈ N} where N is a submodule of M

with M1 ⊂ N .

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Lecture 4

Lemma 4.15 (1.17). Let R ⊆ S ⊆ T be commutative rings. Suppose that R isNoetherian and T is generated by R and finitely many elements t1, . . . , tn. Supposethat T is a finitely generated S-module. Then S is generated by R and finitely manyelements (in other words S ‘inherits’ the property that T had regarding generation).

Proof. Let T be generated by x1, . . . , xn ∈ T as an S-module, i.e.

T = Sx1 + ⋅ ⋅ ⋅ + Sxn

Then we have

(4.2) ti =∑j

sijxj

for some sij ∈ S. We write

(4.3) xixj =∑k

sijkxk

for some sijk ∈ S. Let S0 be the ring generated by R and the sij and sijk. ThusR ⊆ S0 ⊆ S. Any element of T is a ‘polynomial’ in the ti with coefficients in R.Using equation 4.2 and 4.3, we see that each element of T is a linear combinationof the xj with coefficients in S0. Thus T is a finitely generated S0-module. ButS0 is Noetherian, being generated as a ring by R and finitely many elements. Tis a Noetherian S0-module, S is an S0-module of T and hence is finitely generatedas an S0-module. But S0 is generated by R and finitely many elements. So S isgenerated by R and finitely many elements. �

Proposition 4.16 (1.18). Let k be a field, R a finitely generated k-algebra. If Ritself is a field then it is an algebraic extension of k.

Proof. Suppose that R is generated by k and x1, . . . , xn say and is a field.If R is not algebraic over k we can reorder the x1, . . . , xn so that x1, . . . , xmare algebraically independent (i.e. the ring generated by k and x1, . . . , xm as apolynomial algebra k[x1, . . . , xm]). Now xm+1, . . . , xn are algebraic over the fieldF = k(x1, . . . , xm) (the field of fractions of k[x1, . . . , xm]). Hence R is a finite alge-braic extension of F and hence a finitely generated F -module6 (R contains a copyof F since R is a field). Apply lemma 4.15 for k ⊆ F ⊆ R. If follows that F is afinitely generated k-algebra generated by k and q1, . . . , qt say with each

qi = fi/gi

with

fi ∈ k[x1, . . . , xm]

gi ∈ k[x1, . . . , xm]

gi ≠ 0

There exists a polynomial h ∈ F which is prime to each of the gi (e.g. g1 . . . gm +1) and the element 1/h cannot be in the ring generated by k and q1, . . . , qt, acontradiction. Hence R is algebraic over k. �

Theorem 4.17 (1.19, weak Nullstellensatz). Let k be a field, S be a finitely gener-ated k-algebra. Let P be a maximal ideal. Then S/P is a finite algebraic extensionof k. In particular, if k is algebraically closed, then S/P ≅ k.

Proof. Apply proposition 4.16 to R = S/P a field. �

6There is a result he is using here: if a finitely generated algebra is integral over the coefficientring, then it is a finitely generated module: for proof Hartshorne chpt. 1, theorem 3.9 (a.) quotes

the book of Matsumura [Mat70] - or see page 23 of J.S. Milne’s algebraic number theory bookhttp://www.jmilne.org/math/CourseNotes/ANT210.pdf

12

Lecture 5

Theorem 4.18 (1.20, strong Nullstellensatz). Let k be an algebraically closed fieldand S a finitely generated k-algebra. Let P be a prime ideal. Then

P =⋂{maximal ideals containing P}

Furthermore, any radical ideal is the intersection of the maximal ideals containingit.

Proof. Let s ∈ S/P . Let

s = image of s in R = S/P

and S/P is an integral domain (since P is prime) and is finitely generated as a k-algebra by elements r1, . . . , rm. Invert s to get T = ⟨R,s−1⟩ ⊆ fraction field of R = S/P .Take a maximal ideal Q of T . Then by theorem 4.17 we have T /Q ≅ k and so Q∩Rcontains elements ri − λi for some λi ∈ k. Hence Q ∩R is a maximal ideal of R notcontaining s. Thus there exists a maximal ideal of S containing P , but not s. Thus

⋂{maximal ideals containing P} = P �

Lemma 4.19 (1.21). If R is Noetherian then every ideal I contains a power of its

radical√I. In particular N(R) is nilpotent (i.e. N(R)m = 0 for some m)

Proof. Suppose x1, . . . , xm generate√I as an ideal. So xni

i ∈ I for some ni(for each i). Let n be suitably large, something like

n =∑ni + 1

should work. Then (√I)n is generated by products xr11 , . . . , x

rmm with ∑ ri = n. We

must have some ri ≥ ni. Hence all their products are in I. �

Lemma 4.20 (1.22). If R is Noetherian, then a radical ideal is the intersection offinitely many primes.

Proof. Next time. �

Lecture 5

21st October 10:00

5.1. Minimal primes, annihilators and associated primes.

Proposition 5.21 (1.22). If R is Noetherian, then a radical ideal is the intersectionof finitely many primes.

Proof. Suppose not, and take I to be a maximal member of the set of radicalideals not of this form (i.e. not the intersection of finitely many primes). Claim: Iis prime (which would yield a contradiction). Proof: suppose not, so there existsideals J1, J2 with J1 ≩ I and J2 ≩ I but with J1J2 ≤ I. So by maximality of I, we

have that√J1 = Q1 ∩ ⋅ ⋅ ⋅ ∩Qs and

√J2 = Q

′

1 ∩ ⋅ ⋅ ⋅ ∩Q′

t with each Qi,Q′

j prime. Nowset

J = Q1 ∩ ⋅ ⋅ ⋅ ∩Qs ∩Q′

1 ∩ ⋅ ⋅ ⋅ ∩Q′

t =√J1 ∩

√J2

So Jm1 ≤ J1 and Jm2 ≤ J2 for some m1,m2, and hence

Jm1+m2 ≤ J1J2 ≤ I

But I is radical and so J ≤ I. But all Qi,Q′

j ≥ I and so J ≥ I. Hence J = I, acontradiction. �

13

Lecture 5

Now suppose that√I = P1 ∩ ⋅ ⋅ ⋅ ∩ Pm, then we may remove any prime which

contains one of the others. So we may assume that Pi ≰ Pj for i ≠ j. If P is prime

with√I ≤ P , then

P1 . . . Pm ≤⋂Pi =√I ≤ P

and so some Pi ≤ P . This motivates a definition of minimal primes.

Definition 5.7 (1.23). The minimal primes P over an ideal I of a Noetherian ringare such that if P ′ is prime and I ≤ P ′ ≤ P then P ′ = P .

Now observe that the Pi above are minimal primes. We can show the following

Proposition 5.22 (1.24). Let I be an ideal of a Noetherian ring. Then√I is the

intersection of the minimal primes over I and I contains a finite product of theminimal primes over I.

Proof. If P is a prime minimal over I, and x ∈√I then there is n such that

xn ∈ I ⊆ P , so x ∈ P . Thus, each (minimal) prime over I contains√I, so the primes

minimal over I are exactly those minimal over√I. The above discussion shows

that√I is the intersection of these. Their product lies in their intersection and

thus in√I, so lemma 4.19 yields the final statement. �

Definition 5.8 (1.25). Let M be a finitely generated R-module where R is Noe-therian. A prime ideal P is an associated primedfn:associated-prime of M if it is the annihilator of an

element of M - the annihilator is Ann(m) = {r ∈ R∶ rm = 0} and we writedfn:annihilator

AssM = {P ∶P is prime, P = Ann(m) for some m ∈M}

and we have e.g. Ass(R/P ) = {P} for P prime.

Definition 5.9 (1.26). A submodule N of M is (P -)primary if Ass(M/N) = {P}for a prime ideal P . An ideal is P -primary if I is P -primary as a submodule of R.

Lemma 5.23 (1.27). If Ann(M) = P for a prime ideal P then P ∈ AssM .

Proof. Let m1, . . .mk generate M and Ij = Ann (mj). Then the product ∏ Ijannihilate each mj and so

∏ Ij ≤ AnnM = P

So Ij = P for some j and so P ∈ AssM . �

In fact we can always be sure that we always do have some associated primes- i.e. that AssM is non-empty.

Lemma 5.24 (1.28). Let Q be maximal among all annihilators of non-zero elementsm of M . Then Q is prime and so Q ∈ AssM .

Proof. It is by definition an annihilator, so write Q = Ann(m) and r1r2 ∈ Qwith r2 /∉ Q. We show that r1 ∈ Q1. So r1r2 ∈ Q implies that r1r2m = 0 sor1 ∈ Ann(r2m). Now r2 ∉ Q implies that r2m ≠ 0. But Q ≤ Ann(r2m). Hence Qand r1 lie in Ann(r2m). Maximality of Q among annihilators forces that r1 ∈ Q. �

Lemma 5.25 (1.29). For a finitely generated non-zero R-module M with R Noe-therian, there is a chain of submodules

0 ≨M1 ≨M2 ≨ ⋅ ⋅ ⋅ ≨Mt =M

with Mi/Mi−1 ≅ R/Pi for some prime ideal Pi.

Proof. By lemma 5.24, there is m1 ∈ M with Ann(m1) prime, P , say. SetM1 = Rm1. The map r ↦ rm1 from R →M1 will have kernel P1, thus M1 ≃ R/P1.Repeat this for M/M1 to find M2/M1 ≅ R/P2 for some prime P2. Repeat, then theNoetherian property of M forces the process to terminate. �

14

Lecture 6

Lemma 5.26 (1.30). If N ≤M then

Ass(M) ⊆ Ass(N) ∪Ass(M/N)

Proof. Suppose that P = Ann(m) for some m ∈M , P prime (using 5.24). LetM1 = Rm ≅ R/P . For any [m1] ∈ M1 and [m1] ≠ 0, then if r(m1 + P ) = 0 thenrm1 ∈ P so r ∈ P since P is prime, thus Ann(m1) = P . If M1 ∩N ≠ ∅ then thereexists m1 ∈M1 ∩N with Ann(m1) = P and so P ∈ Ass(N). If M1 ∩N = ∅ then theimage of M1 in M/N is isomorphic to R/P , thus P ∈ Ass(M/N). �

Proposition 5.27 (1.31). Ass(M) is finite for any finitely generated R-module(when R is Noetherian).

Proof. From 5.25 we have a chain

0 ≨M1 ≨M2 ≨ ⋅ ⋅ ⋅ ≨Mt =M

with Mi/Mi−1 ≅ R/Qi for some prime ideal Qi. So we have (by 5.26)

Ass(M) ⊆ Ass(Mt−1) ∪Ass(Mt/Mt−1) ≅ Ass(Mt−1) ∪Ass(R/Qt).

Since Ass(R/Qt) = {Qt}, then applying this inductively to the above we obtainAss(M) ⊆ {Q1, . . . ,Qt}. �

Next time I will state the theorem for primary decomposition but I will probablynot prove it.

Lecture 6

23rd October 10:00

Recall from last time: we met the associated primes Ass(m) of a finitely generatedR-module (for R Noetherian).

Proposition 6.28. Each minimal prime over an ideal I is an associated prime ofR/I.

Proof. By lemma 5.22 there is a product of minimal primes over I (possiblywith repetition) contained in I

P s11 . . . P snn ≤ I,

with Pi ≠ Pj if i ≠ j. Consider

Ann (P s22 . . . P snn + I/I) = J.

Certainly J ≥ P s11 . Also JP s22 . . . P snn ≤ I and I ≤ P1 (since Pi is minimal over I)and since P1 is prime (≠ P2, . . . Pn) we must have J ≤ P1. Let M = P s22 . . . P snn +I/I.Using lemma 5.25 there is a chain of submodules in M

0 ≨M1 ≨M2 ≨ ⋅ ⋅ ⋅ ≨Mt =M,

such that each factorMj/Mj−1 ≅ R/Qj for some prime idealQj . But P s11 annihilatesM and hence each Mj/Mj−1, then the primeness of Qj ensures P1 ≤ Qj for each j.Not all the P1 ≨ Qj since ∏Qj ≤ J ≤ P1 and hence some Qj ≤ P1. Pick the least jsuch that Qj = P1 and then

∏k<j

Qk ≰ Pj ,

for otherwise we would have to have some Qk ≤ Pj since the Qk’s are prime,which would contradict the minimality of j. Take x ∈ Mj/Mj−1 . If j = 1, thenAnn (x) = Qj = P1 and so P1 ∈ Ass(R/I). If j > 1, take r ∈ (∏k<j Qk)/P1. Notethat r(sx) = 0 for any s ∈ P1 = Qj . So s(rx) = 0 and so P1 ≤ Ann (rx). Howeverrx ∉ Mj−1 since Mj/Mj−1 ≅ R/Qj = R/P1, and r ∉ P1 because r ∈ (∏k<j Qk)/P1.Thus if t(sx) = 0 then t ∈ P1. So Ann (rx) ⊆ P1, so Ann (rx) = P1 and we haveshown P1 ∈ Ass(M) ⊆ Ass(R/I). �

15

Lecture 6

However the converse is not true.

Example. An example where there exists P ∈ Ass(R/I) with P not minimal overI: let R = k[x, y] and P = (x, y) > Q = (x) and let I = PQ = (x2, xy). ThenAss(R/I) = {P,Q}. But Q is the only minimal prime over I.

I stated before that I am not going to do primary decomposition, because inpractice Ass(M) is of more practical use than primary decomposition - I will stateit though.

Proposition 6.29 (Primary decomposition). Let M be a finitely generated R-module, where R is Noetherian and N ⊆M . Then there exist N1, . . . ,Nt with

N = N1 ∩ ⋅ ⋅ ⋅ ∩Nt

with Ass(M/Ni) = {Pi} for some distinct P1, . . . , Pt.

Remark. Recall that such an Ni is a Pi-primary submodule of M . So how doesthis fit in with the example we gave? In the example I is not Q-primary despite√I = Q. But I = Q ∩ P 2 is a primary decomposition.

Remark. If you look in the books you may think that I have defined my primaryideals in a strange way. But they are equivalent: an alternative definition of aprimary ideal I is that

(1) I is proper,(2) if ab ∈ I but a ∉ I then there exists n such that bn ∈ I.

If so,√I is a prime ideal P , and the ideal is P -primary (see examples sheet 2,

question 1).

That’s the end of chapter 1. I will give out the example sheet on Friday.

6.1. §2: Localisation.

Remark. A word of warning: my use of the word localisation may be a lot moregeneral than what most people are used to.

Let R be commutative with a 1. Let S be a multiplicatively closed subset of R(i.e. it is closed under multiplication and 1 ∈ S7). Define a relation ‘≡’ on R × S by

(r1, s1) ≡ (r2, s2)⇔ (r1s2 − r2s1)x = 0

for some x ∈ S. This is an equivalence relation, the only one worth justifying istransitivity: let (r1, s1) ≡ (r2, s2) and (r2, s2) ≡ (r3, s3), then there exists x, y ∈ Swith

(r1s2 − r2s1)x = 0 = (r2s3 − r3s2)y

Then (r1s3 − r3s1)s2xy = 0 (using r1s3s2xy = r2s3s1xy and r3s1s2xy = r2s1s3xy),and S is multiplicatively closed so s2xy ∈ S. Denote the equivalence class of (r, s)by r/s, and the set of equivalence classes by S−1R. Then S−1R can be made into aring with addition given by

(a1/s1) + (a2/s2) = (a1s2 + a2s1)/s1s2

and multiplication given by

(a1/s1)(a2s2) = a1a2/s1s2

There is a natural ring homomorphism

θ∶R → S−1R

r ↦ r/1.

The ring S−1R satisfies the following universal property.

7Not standard, but we will have this convention here for ease or something.

16

Lecture 7

Proposition 6.30 (2.1). Let φ∶R → T be a ring homomorphism where φ(s) is aunit in T for all s ∈ S. Then there exists a unique ring homomorphism α∶S−1R → Twith φ = α ○ θ.

Proof. Define α∶S−1R → T by s/t ↦ φ(s)φ(t)−1. This is a homomorphismbecause φ is a homomorphism, and it is clear that φ = α○θ. Suppose that s/t =m/nin S−1R then there exists u ∈ S such that u(sn − tm) = 0 so that

φ(u(sn − tm)) = φ(u)φ(sn − tm) = φ(s)φ(n) − φ(t)φ(m) = 0

and since φ(x) is a unit for all x ∈ S it follows that φ(s)φ(t)−1 = φ(m)φ(n)−1 sothe map is well-defined.

To show uniqueness, suppose that β is another map satisfying the propertiesof the statement, then β(r/s) = β(r/1)β(1/s) = φ(r)β(1/s) and since 1 = β(s/s) =β(s/1)β(1/s) = φ(s)β(1/s) so in fact β(r/s) = φ(r)φ(s)−1 so α = β as required. �

Example.(1) Integral domains: put S = R/{0}.(2) S−1R is the zero ring if and only if 0 ∈ S.(3) If I is any old ideal in our commutative ring R, then we can set S = 1+I =

{1 + x∶x ∈ I}.(4) If P is a prime ideal of R, then set S = R/P . We write RP for S−1R in

this case, and the process of passing from R to RP is called localising atP , or localisation (some authors only call this specific case localisation,rather than the general thing). The elements r/s with r ∈ P form an idealof RP , and this is a unique maximal ideal in RP : if r/s is such that r ∉ P ,then r ∈ S and so has an inverse in RP .8

Definition 6.10. A ring with a unique maximal ideal is called a local ring dfn:local.

Remark. Some authors require a local ring to be Noetherian.

Example.(1) Take R = Z and P = (p) a prime ideal, where p is prime. Then

RP = {m/n∶p ∤ n} ⊆ Q

and this is the example of a Noetherian ring that I gave in the introduction.(2) Take R = k[x1, . . . , xn] and P = (x1−a1, . . . , xn−an), then RP is a subring

of k(x1, . . . , xn) (the ring of rational functions) of rational functions de-fined at (a1, . . . , an) ∈ k

n and the unique maximal ideal consists of thosefunction which are zero at (a1, . . . , an).

Lecture 7

25th October 10:00

7.1. Modules. Given an R-module M , define an equivalence relation ‘≡’ onM ×S with respect to a multiplicatively closed set S by (m1, s1) ≡ (m2, s2) iff thereis an x ∈ S with

x(s1m2 − s2m1) = 0.

8Recall the results from Atiyah & Macdonald: every non-unit of a ring A is contained in amaximal ideal; let A be a ring and m ≠ (1) an ideal of A such that every x ∈ A/m is a unit in A,

then A is a local ring and m its maximal ideal; let A be a ring and m a maximal ideal of A, suchthat every element of 1 +m is a unit in A, then A is a local ring.

17

Lecture 7

This is an equivalence relation (exercise). An equivalence class (m,s) is denotedm/s and the set of equivalence classes is denoted S−1M . Then S−1M is an S−1R-module via

m1/s1 +m2/s2 = (s2m1 + s1m2)/s1s2

(r1/s1)(m2/s2) = (r1m2)/(s1s2).

Write MP for the case where we take S = R/P for a prime ideal P . If θ∶M1 →Mis an R-module map then

S−1θ∶S−1M1 → S−1M,

is an S−1R-module map given by

S−1θ∶m1/s↦ θ(m1)/s,

and if φ∶M →M2 is also an R-module map, then

S−1(φ ○ θ) = S−1φ ○ S−1θ.

A sequence of R-modules

M0 M2 . . . Mi . . . Mt→ → →θ →φ

→

is exact at Mi if Im θ = kerφ. A short exact sequence is of the form

0 M1 M M2 0→ →θ →φ

→

with exactness at M1,M and M2.

Lemma 7.31 (2.3). If M1,M,M2 are R-modules and

M1 M M2→θ →φ

is exact at M , then

S−1M1 S−1M S−1M2→S−1θ →S−1φ

is exact at S−1M

Proof. Since kerφ = Im θ we have φ ○ θ = 0 so (S−1φ) ○ (S−1θ) = S−10 = 0 andhence ImS−1θ ⊆ kerS−1φ. Now suppose m/s ∈ kerS−1φ ⊆ S−1M . So φ(m)/s = 0 inS−1M2, which means that there exists t ∈ S with tφ(m) = 0 in M2. But tφ(m) =φ(tm) since φ is an R-module map. So tm ∈ kerφ = Im θ thus tm = θ(m1) for somem1 ∈M1. Hence in S−1M we have

m/s = θ(m1)/ts = S−1θ(m1/ts) ∈ ImS−1θ

Thus kerS−1φ = ImS−1θ. �

Lemma 7.32 (2.4). Let N be a submodule of M . Then

S−1(M/N) ≅ S−1M/S−1N

Proof. Apply lemma 7.31 to the short exact sequence

0 N M M/N 0,→ →θ →φ

→

to get

0 S−1N S−1M S−1(M/N) 0,→ →S−1θ →S−1φ

→

is a short exact sequence. Note S−1θ is an embedding and so we obtain

S−1(M/N) ≅ S−1M/S−1N. �

18

Lecture 7

Let R be a ring with a multiplicatively closed subset S and define φ∶R → S−1Rby r ↦ r/1. If I is an ideal in R then S−1I is an ideal of S−1R.

Lemma 7.33 (2.5).(1) Every ideal in S−1R is of the form S−1I for some ideal I of R.(2) The prime ideals of S−1R are in 1-1 correspondence with the prime ideals

of R that do not meet S.

Proof.(1) Let J be an ideal of S−1R. If r/s ∈ J then r/1 ∈ J . Let

I = {r ∈ R∶ r/1 ∈ J}

Clearly J ⊆ S−1I. If r ∈ I then r/1 ∈ J and hence r/s ∈ J , so S−1I = J .(2) Let Q be a prime in S−1R, then set P = {r ∈ R∶ r/1 ∈ Q}. P is prime:

if xy ∈ P then xy/1 ∈ Q and so either x/1 or y/1 ∈ Q and so x ∈ P ory ∈ P . P does not meet S: if r ∈ S ∩P then r/1 ∈ Q and 1/r ∈ Q, so 1 ∈ Q,a contradiction. Conversely, let r/s, x/y ∈ S−1P then rx/sy ∈ S−1P . Soz(rx) ∈ P for some z ∈ S and therefore rx ∈ P or z ∈ P , but we cannothave z ∈ P because P cannot meet S by the previous argument, whichimplies that r ∈ P or x ∈ P which implies that r/s ∈ S−1P or x/y ∈ S−1P ,so S−1P is prime as required.

�

Lemma 7.34 (2.6). If R is Noetherian, then S−1R is Noetherian.

Proof. Any chain of ideals in S−1R is of the form

J1 ≤ J2 ≤ . . .

Consider Ik = {r ∈ R∶ r/1 ∈ Jk}. Then I1 ≤ I2 ≤ . . . is a chain, which must terminatesince R is Noetherian. So It = It+1 = . . . for some t. Then Jt = Jt+1 = . . . sinceJk = S

−1Ik for each k. �

Definition 7.11. A property P of a ring R (or an R-module M) is local if

{R (or M) has property P}⇔ {RP (or MP ) has property Pfor each prime ideal P of R

}

Lemma 7.35. Let M be an R-module. The following are equivalent

(1) M = 0,(2) MP = 0 for all prime ideals P of R,(3) MQ = 0 for all maximal ideals Q of R.

Proof. It’s clear that (1) → (2) → (3). To show that (3) → (1), suppose (3)holds and M ≠ 0. Take 0 ≠m ∈M . The annihilator of M is a proper ideal of R. Ittherefore is contained in a maximal ideal Q by Zorn’s lemma. Consider m/1 ∈MQ.By assumption MQ = 0 and so m/1 = 0. So sm = 0 for some s ∈ S where S = R/Q,a contradiction, since Q contains the annihilator of m. The other implications arestraightforward. �

Lemma 7.36 (2.9). Let φ∶M → N be an R-module map, then the following areequivalent

(1) φ is injective.(2) φP ∶MP → NP is injective for all prime ideals P of R.(3) φQ∶MQ → NQ is injective for all maximal ideals Q of R.

19

Lecture 8

Proof. (1) → (2) The sequence 0 → M → N is exact, hence by lemma 7.310→MP → NP is exact, so φP is injective.

(2)→ (3) A maximal ideal is prime.(3) → (1) Let M ′ = kerφ, then the sequence 0 →M ′ →M → N is exact, hence

0 → M ′

Q → MQ → NQ is exact by lemma 7.31, and therefore M ′

Q ≅ ker(φQ) = 0,

since φQ is injective. Hence M ′ = 0 by lemma 7.35. �

Lecture 8

28th October 10:00

Lemma 8.37 (2.10). Let P be a prime ideal of a ring R and S a multiplicativelyclosed subset of R with S ∩ P ≠ ∅. By lemma 7.33 (2), S−1P is a prime ideal ofS−1R. Then (S−1R)S−1P ≅ RP . In particular if Q is a prime ideal of R and P ≤ Qthen (RQ)PQ

≅ RP (taking S = R/Q).

Proof. We have ring homomorphisms θ1∶R → S−1R and θ2∶S−1R → (S−1R)S−1P .

Let φ = θ2 ○ θ1, then φ∶R → (S−1R)S−1P is a ring homomorphism with φ(s) a unitfor all s ∈ S′ = R/P . So we can apply our universal property

R RP

(S−1R)S−1P

→θP

→φ → α

to give a unique ring homomorphism α. The map α is surjective since all elementsof (S−1R)S−1P are of the form φ(r)φ(s′)−1 for r ∈ R. To show that α is injective:suppose that r/s′ ∈ kerα ≤ RP with s′ ∈ S′, then r/1 ∈ kerα and hence r ∈ kerφ.But if φ(r) = 0 then (r/1)(x/y) = 0 in S−1R for some x/y ∉ S−1P and so srx = 0for some s ∈ S and x ∉ P . But S ∩ P = ∅ and so sx ∉ P and therefore we havey = sx ∈ S′ = R/P such that ry = 0, hence r/s′ ∈ RP is zero in RP . �

8.1. §3: Dimension. All rings are commutative with a 1.

Definition 8.12 (3.1). The spectrum of the ring R is defined

Spec (R) = {P ∶P is a prime ideal of R}

Definition 8.13 (3.2). The length of a chain of prime ideals

P0 ≨ P1 ≨ P2 ≨ ⋅ ⋅ ⋅ ≨ Pn

is n (note the numbering started at 0).

Definition 8.14 (3.3). The (Krull) dimension dimR of R is defined to be

sup{n∶ there is a chain of prime ideals of length n}

and is denoted ∞ if the above does not exist.

Definition 8.15. The height ht(P ) of P ∈ Spec (R) is defined

sup{n∶ there is a chain of prime ideals P0 ≨ P1 ≨ P2 ≨ ⋅ ⋅ ⋅ ≨ Pn = P}

Remark. There is a 1-1 correspondence between primes that have empty intersec-tion with R/P and the primes of RP . This shows that ht(P ) = dimRP .

Example.

20

Lecture 8

(1) An Artinian ring (see example sheet - these are rings that satisfy the de-scending chain condition on ideals) have dimension 0, since all prime idealsare maximal (exercise). Conversely, any Noetherian ring of dimension 0is Artinian (exercise). Example of an Artinian ring: a field9.

(2) We have dimZ = 1 - any chain of maximal length is of the form (0) ≨ (p)where p is prime. Also one has dimk[X] = 1 where k is a field. Theseare both examples of Dedekind domains (integrally closed domains ofdimension 1) - we will say some more about these next term.

(3) One has k[X1, . . . ,Xn] ≥ n, since we have a chain of prime ideals

(0) ≨ (X1) ≨ (X1,X2) ≨ ⋅ ⋅ ⋅ ≨ (X1, . . . ,Xn)

In fact dimk[X1, . . . ,Xn] = n when k is a field. To prove this we need toprove some results about the relationship between chains of prime idealsin subrings of chains in the whole ring under some condition relating thesubring to the larger ring. First though observe the following

Lemma 8.38 (3.5). The height 1 primes of k[X1, . . . ,Xn] are precisely those ofthe form (f) where f is irreducible.

Proof. Cf. question 3 on example sheet 1. Height 1 primes in a unique fac-torisation domain are the principal ideals generated by an irreducible. �

Definition 8.16 (3.6). Let R ⊂ S be two rings, then x ∈ S is integral over R if itsatisfies some monic polynomial with coefficients in R, e.g. the elements of Q thatare integral over Z are precisely the elements of Z.

Lemma 8.39 (3.7). The following are equivalent

(1) x ∈ S is integral over R,(2) R[x] (subring generated by R and x) is a finitely generated R-module,(3) R[x] is contained in a subring T of S with T being a finitely generated

R-module.

Remark. Some authors say S is finite over R if S is a finitely generated R-moduleand a k-algebra R is of finite type if it is finitely generated as a k-algebra.

Proof. (1) → (2) Let x ∈ S be integral over R and f(y) ∈ R[y] a monicpolynomial of degree n ≥ 1 such that f(x) = 0. For an arbitrary polynomial g(y) ∈R[y] we may write

g(y) = q(y)f(y) + r(y),

with q(y), r(y) ∈ R[y] and deg(r(y)) < n, so that one has

g(x) = r(x) = r0 + r1x + ⋅ ⋅ ⋅ + rn−1xn−1,

with r0, . . . , rn−1 ∈ R. Thus R[x] is generated as an R-module by 1, x, . . . , xn−1.(2)→ (3) Trivial.(3) → (1) Consider multiplication by x in the ring T and take y1, . . . yn R-

module generators for T and let

xyi =∑j

rijyj

for each i. So

∑(xδij − rij)yj = 0

Now multiply on the left by the adjugate of the matrix Aij = (xδij−rij). We deducethat detAyj = 0 for all j. But 1 is an R-linear combination of the yj ’s. So detA = 0But detA is of the form

xn + rn−1xn−1 + ⋅ ⋅ ⋅ + r0 �

9Hurr durr.

21

Lecture 9

Remark. This proof is reminiscent of a proof often used for Nakayama’s lemmae.g. see Atiyah & MacDonald’s proof of Nakayama’s lemma.

Lecture 9

30th October 10:00

9.1. More about integrality.

Lemma 9.40 (3.8). If x1, . . . , xm ∈ S are integral over R then R[x1, . . . , xm], thesubring of S generated by R and x1, . . . , xm, is a finitely generated R-module.

Proof. By induction: the base case was done previously. Since xk is integralover A = R[x1, . . . , xk−1], then the previous proof shows that A[xk] = R[x1, . . . , xk]is finitely generated over R. �

Lemma 9.41 (3.9). The set T ⊆ S of elements integral over R forms a subring ofS.

Proof. According to the previous results, if s1, . . . , sn ∈ S are integral over R,then so is any element s of R[s1, . . . , sn], because R[s1, . . . , sn, s] = R[s1, . . . , sn]is a finitely generated R-module. So in particular, given two integral elementss1, s2 ∈ S, then s1 + s2 and s1s2 are also integral over R. �

Definition 9.17. T is called the integral closure of R in S. If T = R, then Ris integrally closed in S. If T = S, then S is integral over R. If R is an integraldomains then we just say that R is integrally closed if it is integrally closed in itsfield of fractions.

Example.(1) Z is integrally closed.(2) k[X1, . . . ,Xn] is integrally closed. In an algebraic number field K with

[K ∶Q] < ∞ then the integral closure of Z in K is the ring of integers inK denoted OK .

Lemma 9.42 (3.11). If R ⊆ S ⊆ T (rings) with T integral over S and S integralover R, then T is integral over R.

Proof. Take t ∈ T , and let tn + s1tn−1 + ⋅ ⋅ ⋅ + sn = 0 be an equation with

coefficients in S. Write A = R[s1, . . . , sn]. Then A[t] is a finitely generated A-module. If S is integral over R, then A[t] is even finitely generated over R. Thust is integral over R. �

Lemma 9.43 (3.12). Let R ⊆ T be rings with T integral over R, then

(1) if J is an ideal of T then T /J is integral over R/(J ∩ R) (identifyingR/(J ∩R) with R + J/J as a subring of T /J),

(2) if S is a multiplicatively closed subset of R then S−1T is integral overS−1R.

Proof. If x ∈ T then

(9.4) xn + rn−1xn−1 + ⋅ ⋅ ⋅ + r0 = 0

for some ri ∈ R. Then working modulo J (writing bars over elements for images inT /J) then

xn + rn−1xn−1 + ⋅ ⋅ ⋅ + r0 = 0

in T /J with ri ∈ R + J/J .(2) Suppose that x/s ∈ S−1T , then x satisfies an equation of the form (9.4).

Then(x/s)n + (rn−1/s)(x/s)

n−1 + ⋅ ⋅ ⋅ + (r0/sn) = 0

in S−1T . So x/s is integral over S−1R. �

22

Lecture 10

Lemma 9.44 (3.13). Suppose that R ⊆ T are integral domains with T integral overR. Then T is a field iff R is a field.

Proof. Suppose R is a field. Let 0 ≠ t ∈ T and choose a monic equation ofleast degree of the form

tn + rn−1tn−1 + ⋅ ⋅ ⋅ + r0 = 0

with ri ∈ R. T is an integral domain and so r0 ≠ 0 (otherwise we have t(tn−1 + ⋅ ⋅ ⋅ +r1) = 0 and tn−1 + ⋅ ⋅ ⋅ + r1 = 0 a contradiction by minimality of degree). So t has aninverse

−r−10 (tn−1 + rn−1tn−2 + ⋅ ⋅ ⋅ + r1) ∈ T

So T is a field.Conversely suppose that T is a field. Let 0 ≠ x ∈ R. Then x has inverse x−1 in

T . So x−1 satisfies a monic equation

x−m + r′m−1x−m+1 + ⋅ ⋅ ⋅ + r′0 = 0,

with coefficients in R. Then rearrange for a formula for x−1 by multiplying theabove by xm−1 to get

x−1 = −m

∑i=1

rm−ixi−1,

and note that it is in R. Thus R is a field. �

Lemma 9.45 (3.14). Let R ⊆ T be rings with T integral over R. Let Q be a primeideal of T and set P = R ∩Q. Then Q is maximal iff P is maximal.

Proof. By lemma 9.43 (1), T /Q is integral over R/P and since P and Q areprime, T /Q and R/P are integral domains. So by lemma 9.44 we have T /Q is afield iff R/P is a field. So Q is maximal iff P is maximal. �

Theorem 9.46 (3.15, Incomparability theorem). Let R ⊆ T be rings with T integralover R. Let Q ≤ Q1 be prime ideals of T . Suppose Q∩R = P = Q1∩R. Then Q = Q1.

Proof. Applying lemma 9.43 (2) with S = R/P we have TP integral over RP(with abuse of notation TP = S−1T ). From chapter 2 we have that there is a primeS−1P in RP which is the unique maximal ideal of RP . Also there are S−1Q andS−1Q1 in TP which are also prime and

⎧⎪⎪⎨⎪⎪⎩

S−1Q ∩ S−1R = S−1P

S−1Q1 ∩ S−1R = S−1P

By lemma 9.45, S−1Q and S−1Q1 are maximal, since S−1P is. But S−1Q ≤ S−1Q1

and so S−1Q = S−1Q1. But the 1-1 correspondence between prime ideals of S−1Tand those of T that do not meet S yields that Q = Q1. �

Theorem 9.47 (3.16, Lying over theorem). Let R ⊆ T be rings with T integral overR. Let P be a prime ideal of R. Then there is a prime ideal Q of T with Q∩R = P(we say that Q ‘lies over’ P ).

Proof. By lemma 9.43 (2) TP is integral over RP . Let S = R/P . Takea maximal ideal of TP . It must be of the form S−1Q for some ideal Q of T ,necessarily prime (because primality is preserved under the 1-1 correspondence).Then S−1Q∩S−1R

(=RP )

is maximal by lemma 9.45. But RP has a unique maximal ideal

namely S−1P and so S−1Q ∩ S−1R = S−1P .Therefore Q ∩R = P . �

The next two results are quite fundamental, they are due to Cohen and Sei-denberg called the ‘going-up theorem’ and the ‘going-down theorem’.

23

Lecture 10

Lecture 10

1st November 10:00

10.1. The Going-up and Going-down theorems. We next have two the-orems due to Cohen and Seidenberg (1947) that allow us to move from chains ofprime ideals in R to chains of prime ideals in T with R ⊂ T rings with T integralover R. However, the second one requires stronger conditions as you will now see.

Theorem 10.48 (3.17, Going-up theorem). Let R ⊆ T be rings with T integralover R. Let P1 ≤ ⋅ ⋅ ⋅ ≤ Pn be a chain of prime ideals of R and Q1 ≤ ⋅ ⋅ ⋅ ≤ Qm (withm < n) be a chain of prime ideals of T with Qi ∩R = Pi for 1 ≤ i ≤ m. Then thechain of Qi’s extends to a chain Q1 ≤ ⋅ ⋅ ⋅ ≤ Qn with Qi ∩R = Pi.

Theorem 10.49 (3.18, Going-down theorem). Let R ⊂ T be integral domains, Rintegrally closed and T integral over R. Let P1 ≥ ⋅ ⋅ ⋅ ≥ Pn be a chain of primeideals of R and let Q1 ≥ ⋅ ⋅ ⋅ ≥ Qm with m < n be a chain of prime ideals of T withQi ∩R = Pi for 1 ≤ i ≤m. Then we can extend the chain of Qi’s to give

Q1 ≥ ⋅ ⋅ ⋅ ≥ Qn,

with Qi ∩R = Pi for 1 ≤ i ≤ n.

Our major application of these is in the context of finitely generated k-algebrasT (assume T is an integral domain). Noether normalisation yields R ⊂ T with Tintegral over R and R isomorphic to a polynomial algebra and hence integrallyclosed.

Proof: (of 10.48). By induction. It is enough to consider the case n = 2 andm = 1. Write R = R/P1 and T = T /Q1. Then R ↪ T with T integral over R (usingQ1 ∩R = P1 in 9.43 (1)). By the Lying over theorem 9.47, there is a prime Q2 of Tsuch that Q2 ∩ R = P2. Lifting back gives a prime ideal Q2 of T with Q2 ≥ Q1 andQ2 ∩R = P2. �

Corollary 10.50 (3.19, corollary of 10.48). Let R ⊆ T with T integral over R.Then dimR = dimT .

Proof. Take a chain of primes in T

Q0 ≨ Q1 ≨ ⋅ ⋅ ⋅ ≨ Qn.

By the incomparability theorem 9.46 we have a chain

P0 ≨ P1 ≨ ⋅ ⋅ ⋅ ≨ Pn

of prime ideals of R with Qi ∩ R = Pi. Thus dimR ≥ dimT . Conversely, if P0 ≨

P1 ≨ ⋅ ⋅ ⋅ ≨ Pn is a chain of primes of R, then there is a prime Q0 lying over P0 bytheorem 9.47 and then by the going-up theorem 10.48 we have a chain

Q0 ≤ Q1 ≤ ⋅ ⋅ ⋅ ≤ Qn,

with Qi ∩R = Pi. Note that we have strict inequalities, so dimR ≤ dimT . �

Corollary 10.51 (3.20, corollary of 10.49). Let R ⊂ T be integral domains with Rintegrally closed and T integral over R. Let Q be a prime of T . Then ht(Q ∩R) =ht(Q).

Proof. Take a chain Q0 ≨ Q1 ≨ ⋅ ⋅ ⋅ ≨ Qn = Q in Spec (T ). As above, there is

P0 ≨ P1 ≨ ⋅ ⋅ ⋅ ≨ Pn = Q ∩R with Qi ∩R = Pi. So ht(Q ∩R) ≥ ht(Q). Conversely if

P0 ≨ P1 ≨ ⋅ ⋅ ⋅ ≨ Pn = Q ∩R then the Going-down theorem 10.49 yields

Q0 ≨ Q1 ≨ ⋅ ⋅ ⋅ ≨ Qn = Q,

with Qi ∩R = Pi. So ht(R ∩Q) ≤ ht(Q). �

24

Lecture 11

The examples classes are on the 4th,7th,8th at 4-5pm in MR15, MR15, MR13respectively.

The proof of the Going-down theorem 10.49 requires a couple of lemmas and abit of knowledge of field theory (Galois theory).

Definition 10.18. If I is an ideal of R with R ⊂ T then x ∈ T is said to be integralover I if x satisfies a monic equation

(10.5) xn + rn−1xn−1 + ⋅ ⋅ ⋅ + r0 = 0

with ri ∈ I. The integral closure of I in T is the set of such x.

Lemma 10.52 (3.22). Let R ⊂ T be rings with T integral over R. Let I be an ideal

of R, then the integral closure of I in T is the radical√

(TI) (observe that TI isan ideal of T ) and is thus closed under addition and multiplication. In particular,

if R = T then we get that the integral closure of I in R is the radical√I.

Proof. If x ∈ T is integral over I, then 10.5 implies that xn ∈ TI and thus

x ∈√

(TI). Conversely if x ∈√

(TI) then

xn =k

∑i

tiri

for some ri ∈ I, ti ∈ T . But each ti is integral over R and so lemma 9.40 shows thatM = R[t1, . . . , tk] is a finitely generated R-module. Also xnR[t1, . . . , tk] ⊆ IM . Lety1, . . . , ys be a generating set of M qua R-module. Then we have

xnyj =∑l

rjlyl

with rjl ∈ I. As in the proof of lemma 8.39 we have

∑(xnδjl − rjl)yl = 0

and we deduce that xn satisfies a monic equation

(xn)s + ⋅ ⋅ ⋅ + r′0 = 0

namely detA = 0. Note that all but the top coefficient is in I (the top coefficient is1 so if I is not a trivial ideal then 1 ∉ I). Thus x is integral over I. �

Lecture 11

4th November 10:00

11.1. Proof of the Going-down theorem.

Lemma 11.53 (3.23). Let R ⊂ T be integral domains with R integrally closed andlet x ∈ T be integral over an ideal I of R. Then x is algebraic over the field offractions K of R and its minimal polynomial over K

(11.6) Xn + rn−1Xn−1 + ⋅ ⋅ ⋅ + r0,

has its coefficients rn−1, . . . , r0 in√I.

Proof. Certainly x is algebraic over K (from the integrality of x over R). Weclaim that the coefficients in (11.6) are integral over I. Take an extension field L ofK containing all the conjugates x1, . . . , xn of x e.g. a splitting field of the minimalpolynomial of x over K. There is a K-automorphism of L sending x↦ xi and so if

xm + rm−1xm−1 + ⋅ ⋅ ⋅ + r0 = 0,

where ri ∈ I, then

xmi + rm−1xm−1i + ⋅ ⋅ ⋅ + r0 = 0.

25

Lecture 11

Thus each xi is integral over I and in particular lies in the integral closure T of Rin L. Lemma 9.42 implies that a polynomial with coefficients in Z in the xi’s willalso be integral over I. But the coefficients of the minimal polynomial of x over Kare of this form by the usual theory linking coefficients to roots of polynomials. Sowe’ve established the claim. �

Proof: (of theorem 10.49). By induction it is enough to look at the casewhere n = 2, m = 1 with P1 ≩ P2 and Q1∩R = P1. We want to find Q2 with Q1 ≥ Q2

and Q2 ∩R = P2. Let S2 = R/P2 and S1 = T /Q1. Set S = S1S2 = {rt∶ r ∈ S2, t ∈ S1}.Thus S is multiplicatively closed and contains both S1 and S2. We’ll show thatTP2 ∩ S = ∅. Assume this for now. TP2 is an ideal of T and so S−1(TP2) is anideal of S−1T . It is proper since TP2 ∩ S = ∅ (our assumption). Hence S−1(TP2)lies in a maximal ideal of S−1T , which is necessarily of the form S−1Q2 for someprime ideal Q2 of T with Q2 ∩S = ∅ (1-1 bijection with primes that don’t meet S)and TP2 ≤ Q2 (since S−1(TP2) ≤ S

−1Q2). Hence P2 ≤ TP2 ∩R ≤ Q2 ∩R. Similarly,suppose that x ∈ R and x ∉ P2, then x ∈ S2 = R/P2 ⊂ S, but Q2 ∩S = ∅ so x ∉ Q2 sox ∉ Q2 ∩R ⊂ Q2, hence Q2 ∩R ≤ P2, so Q2 ∩R = P2.

It remains to prove that TP2 ∩ S = ∅. Take x ∈ TP2 ∩ S. By lemma 10.52,x is integral over P2 in T (using lemma 10.52 with I = P2). So by lemma 11.53,it is algebraic over the field of fractions K of R and its minimal polynomial Xn +rn−1X

n−1 + ⋅ ⋅ ⋅ + r0 over K has coefficients in√P 2 = P2. But x ∈ S and so x is of

the form rt with r ∈ S2 and t ∈ S1 so t = x/r has minimal polynomial over K

Xn +rn−1rXn−1 +

rn−2r2

Xn−2 + ⋅ ⋅ ⋅ +r0rn

and these coefficients are in R (using lemma 10.52 with I = R) since t ∈ T is integralover R. Write these coefficients as r′i (i = 0, . . . , n − 1). But ri ∈ P2, r ∉ P2 and so

r′i ∈ P2. So by definition t is integral over P2 and so by lemma 10.52, t is in√

(TP2),

a contradiction since t ∈ S1 = T /Q1 and TP2 ≤ Q1 (and hence√

(TP2) ≤ Q1). �

We now concentrate on finitely generated k-algebras where k is a field - warning,some authors call these affine algebras.

Theorem 11.54 (3.24). Let T be a finitely generated k-algebra, which is an integraldomain with fraction field L. Then dimT = tr degk(L) where tr degk(L) is thetranscendence degree of L (over k).

Remark. What is transcendence degree? If x1, . . . , xn are algebraically indepen-dent over k, that is the map

k[X1, . . . ,Xn]→ k[x1, . . . , xn]

Xi ↦ xi

is injective, then k[x1, . . . , xn] may be regarded as a polynomial algebra.As in linear algebra, one wants to consider maximal, algebraically independent

sets - they all have the same cardinality. Such a set is a transcendence basis of L overk, and the transcendence degree is the cardinality. There is a rough correspondencehere between the linear algebra case and what we are doing now

linearly independent sets←→ algebraically independent sets

span ⟨S⟩←→ algebraic closure of S

⎛⎜⎜⎝

the set of elements in L which arealgebraically dependent over the field

generated by K and S

⎞⎟⎟⎠

spanning set←→ S whose algebraic closure is L

26

Lecture 12

Example. Let L = k(X1, . . . ,Xn) be the fraction field of k[X1, . . . ,Xn] and f anirreducible in k[X1, . . . ,Xn], and let K be

K = Frac(k[X1, . . . ,Xn]/(f)).

Then tr degk(L) = n and tr degk(K) = n − 1 since K is an algebraic extension ofk(X1, . . . ,Xi−1,Xi+1, . . . ,Xn) where Xi appears in some term of f .

Lecture 12

6th November 10:00

12.1. Noether normalisation lemma. The key result that we need to provetheorem 11.54 is the following

Theorem 12.55 (3.25, Noether’s normalisation lemma). Let T be a finitely gen-erated k-algebra. Then there exists a subring R = k[X1, . . . ,Xr] with X1, . . . ,Xr

algebraically independent with T integral over R.

Proof. Let T = k[a1, . . . , an], the proof is by induction on n, the numberof generators. Let r = the maximum number of algebraically independent ele-ments. Observe, we may assume that r ≥ 1 - otherwise T is a finite dimensionalk-vector space. There is nothing to do if a1, . . . , an are algebraically independent. Sorenumber the ai’s so that a1, . . . , ar are algebraically independent and ar+1, . . . , anare algebraically dependent on a1, . . . , ar. Take 0 ≠ f ∈ k[X1, . . . ,Xr,Xn] with

f(a1, . . . , ar, an) = 0. Thus f(X1, . . . ,Xr,Xn) is a sum of terms λlXl11 . . .X lr

r Xlnn

where l = (l1, . . . , lr, ln) an (r + 1)-tuple. Claim: there exists positive integers

m1, . . . ,mr such that φ∶ l ↦m1l1 + ⋅ ⋅ ⋅ +mrlr + ln is 1-1 for those l with λl ≠ 0. Proof

of claim: there are finitely many possibilities for differences d = l− l′ with λl ≠ 0 ≠ λl′(because a polynomial has finitely many monomials). Write d = (d1, . . . , dr, dn) andconsider the finitely many non-zero (d1, . . . , dr) ∈ Zr determined. Vectors in Qrorthogonal to one of these lie in finitely many (r − 1)-dimensional subspaces. Pick(q1, . . . , qr) ∈ Qr with each qi > 0 so that ∑ qidi ≠ 0 for all of the finitely many non-zero (d1, . . . , dr). Multiplying by a positive integer to get (m1, . . . ,mr) (mi ≥ Z>0)

so that ∣∑midi∣ > ∣dn∣ for all of the finitely many differences d with (d1, . . . , dr) ≠ 0

and so if φ(l) = φ(l′) then d1 = d2 = ⋅ ⋅ ⋅ = dr = 0 and thus ln = l′

n and so l = l′. Now,for these m1, . . . ,mr, set

g(X1, . . . ,Xr,Xn) = f(X1 +Xm1n , . . . ,Xr +X

mrn ,Xn)

This is a sum

∑l∶λl≠0

λl(X1 +Xm1n )l1⋯(Xr +X

mrn )lrX ln

n

Different terms have different powers of Xn, and so there will be a single term withthe highest power of Xn. As a polynomial in Xn, the leading coefficient is thereforeone of the λl and is therefore in k. Put bi = ai − a

min for 1 ≤ i ≤ r and

h(Xn) = g(b1, . . . , br,Xn)

This has leading coefficient in k and all coefficients are in k[b1, . . . , br]. Moreover,

h(an) = g(b1, . . . , br, an) = f(a1, . . . , ar, an) = 0

Dividing through by the leading coefficient shows that an is integral over k[b1, . . . , br].So for each i with 1 ≤ i ≤ r, ai = bi+a

min is also integral over k[b1, . . . , br]. Hence T is

integral over k[b1, . . . , br, ar+1, . . . , an−1] and we may apply the induction hypothesisto this subring with fewer generators. �

So now we use it to go back and prove what we wanted to about dimension.

27

Lecture 13

Proof: (of theorem 11.54). Let T be a finitely generated k-algebra. Ap-ply theorem 12.55 to get x1, . . . , xr algebraically independent with T integral overk[x1, . . . , xr] (≅ a polynomial algebra). By corollary 10.50, we have dimT = dimk[x1, . . . , xr].Thus any finitely generated k-algebra has dimension equal to the dimension of apolynomial algebra with r variables where tr degk(Frac(T )) = r. So we need toshow that

dimk[x1, . . . , xr] = r,

- recall from an earlier example that dimk[x1, . . . , xr] ≥ r). We prove equality byinduction on r. The case r = 0 is okay. Assume that r ≥ 1. If P0 ≨ P1 ≨ ⋯ ≨ Ps is achain of prime ideals - we may assume that P0 = 0, and since P1 contains f , for someirreducible (f) (k[x1, . . . xr] is a unique factorisation domain - see question 3 on ex-ample sheet 1) we may assume that P1 = (f). But tr degk(Frac(k[x1, . . . , xr]/(f)) =r − 1. So we have

dimk[x1, . . . , xr]/(f) = dimk[Y1, . . . , Yr−1]

for some polynomial algebra with r−1 variables, and so is equal to r−1 by induction.But P1 = (f) and P1/P1 ≨ P2/P1 ≨ ⋯ ≨ Ps/P1 is a chain of length s−1. So s−1 ≤ r−1and so s ≤ r. Thus dimk[x1, . . . , xr] = r �

Proposition 12.56 (3.26). Let Q be a prime ideal of T , a finitely generated k-algebra which is an integral domain, with dimT = n. Then

ht(Q) + dim(T /Q) = n

Proof. Denote m = ht(Q) and pick a prime chain

Q0 ≨ Q1 ≨ ⋯ ≨ Qm = Q

By the Noether normalisation theorem 12.55, there is a sub-algebra R ≅ a polyno-mial algebra, with T integral over R. By corollary 10.50 we have dimT = dimRand by theorem 11.54, n = dimR = tr degR = number of variables in the polyno-mial algebra. Write Pi = Qi ∩R. Observe that ht(Q1) = 1 as otherwise we couldfind a longer chain. So by the corollary 10.51, we know that since R is integrallyclosed (as it is a polynomial algebra, it is integrally closed in its fraction field), thenht(P1) = 1. So P1 = (f) (as R is a unique factorisation domain) (where f is anirreducible) and so tr deg(Frac(R/P1)) = n− 1. Hence we have dimR/P1 = n− 1 bytheorem 11.54. Now we want to apply induction to the prime Q/Q1 in T /Q1, andwe observe that

(1) ht(Q/Q1) =m − 1,(2) dim(T /Q1) = dim(R/P1) = n−1 (since R/P1 embeds in T /Q1 as R+Q1/Q1

and T /Q1 is integral over it, using corollary 10.50 and lemma 9.43),

(3) dim( T /Q1

Q/Q1) = dim(T /Q).

So induction gives that

m − 1 + dim(T /Q) = n − 1,

as required. �

Lecture 13

8th November 10:00

28

Lecture 13

13.1. Remarks on the Noether normalisation lemma. We celebratedbeing half-way through the course by proving the Noether normalisation theorem.Now we will make some remarks.

Remark.(1) If k = C then the maximal ideals of k[X1, . . . ,Xr](= T?) are of the form

P = (X1 − a1, . . . ,Xr − ar)

by the nullstellensatz so they correspond to points of Cr. But T /TPis finite dimensional over C since T is a finitely-generated k[x1, . . . , xr]-module (integrality of T over R). So T /TP is Artinian and hence onlyhas finitely many primes, which are all maximal. They correspond to themaximal ideals of T lying over P . Thus there is a map

f ∶{maximal ideals

of T}Ð→ {

maximal idealsof k[x1, . . . , xr]

} ≅ Cr

Qz→ P

with each fibre f−1(P ) being non-empty and finite.(2) In fact if T is a finitely generated k-algebra which is an integral domain

then its integral closure T1 in its fraction field L is a finitely generatedT -module (and hence Noetherian). Take k = C then we have a map

g∶{maximal ideals

of T1}Ð→ {

maximal idealsof T

}

and we have a normal variety (if you are dealing with curves, then beingnormal is equivalent to being non-singular) and the fibres g−1(Q) are finitenon-empty.

Proposition 13.57 (3.27). Let R be a Noetherian integral domain that is integrallyclosed in its field of fractions K. Let L be a finite degree separable field extensionof K. Let T1 be the integral closure of R in L. Then T1 is finitely generated as anR-module.

Corollary 13.58. If R = Z, then the integral closure of Z in an algebraic numberfield L is a finitely generated Z-module.

A field is separable if it has Hochschild cohomological dimension zero - we mightdiscuss this towards the end of the course.

Proof: (of 13.57) (sketch). The proof uses the trace function, defined

TrL/K(x) = −[L ∶K(x)] ⋅ (next to top coefficient in

minimal polynomial of x over K)

for any finite degree field extension L of K. Equivalently, if L is Galois over K thenwe have

TrL/K(x) = ∑g∈Gal(L/K)

g(x)

a sum of conjugates but potentially with repetitions (and hence getting a multipleof the relevant coefficient of the minimal polynomial). We quote from Miles Reid’sbook on commutative algebra: if L is separable then

L ×L→K

(x, y)→ Tr(xy)

is a non-degenerate symmetric bilinear form.

29

Lecture 14

Pick a K-vector space basis of L, y1, . . . , yn. By multiplying by suitable ele-ments of K, we may assume that yi ∈ T1 (if the minimal polynomial of yi is

xm + rm−1/sm−1xm−1 + ⋅ ⋅ ⋅ + r0/s0

with ri/si ∈ K then the minimal polynomial of yi/(Πsj) has coefficients in R).Since Tr(xy) yields a non-degenerate symmetric bilinear form, there is a basisx1, . . . , xn so that Tr(xiyj) = δij . Let z ∈ T1. Then z = ∑λixi with λi ∈ K. SoTr(zyj) = Tr(∑λixiyj) = ∑λiδij = λj . But z and yj are in T1 and hence zyj ∈ T1.By lemma 11.53 (with I = R) the coefficients of the minimal polynomial of zyj lie inR (using that R is integrally closed) and so Tr(zyj) ∈ R. So λj ∈ R for each j. HenceT ⊆ ∑Rxi, a Noetherian R-module. So T is a finitely generated R-module. �

Now we move on to Krull’s principal ideal theorem for 1931 that tells us aboutthe minimal primes over principal ideals.

Theorem 13.59 (3.28, Principal ideal theorem). Let R be a Noetherian ring anda ∈ R a non-unit. Let P be a minimal prime over (a). Then ht(P ) ≤ 1.

The result above provides the inductive step for the following.

Theorem 13.60 (3.29, Generalised principal ideal theorem). Let R be a Noetherianring, I a proper ideal generated by n elements. Then ht(P ) ≤ n for any minimalprime P over I.

Corollary 13.61 (3.30).(a) Each prime of a Noetherian ring has finite height.(b) Every Noetherian local ring has finite dimension (namely the height of

the unique maximal ideal) and this is ≤ the minimal number of generatorsof the unique maximal ideal P and this is equal to the R/P -vector spacedimension of P /P 2.

Proof: (of 13.61 from 13.60). (a) Any prime ideal is a maximal ideal overitself and is finitely generated if the ring is Noetherian.

(b) For a local ring dimR = htP where P is the unique maximal ideal. Apply(a) to get that dimR is finite. Theorem 13.60 implies that ht(P ) ≤ the minimalnumber of generators of P . The final equality follows from Nakayama’s lemma. Pis generated by x1, . . . , xn iff P /P 2 is generated by x1, . . . , xn where an over-linedenotes the image of an element of P in P /P 2, which implies the result. For the ‘←’case: suppose that x1, . . . , xn generated P /P 2, then consider I = (x1, . . . , xn) ≤ P ,then clearly I +P 2 = P and so P (P /I) = P /I, then Nakayama’s lemma implies thatP /I = 0. �

Corollary 13.62 (3.31). A Noetherian ring satisfies the descending chain conditionon prime ideals.

Proof. If we have a strictly descending chain

P = P1 ≩ P2 ≩ P3 ≩ . . . ,

of prime ideals then the chain can have length at most ht(P ). Use corollary 13.61(a). �

Lecture 14

11th November 10:00

30

Lecture 14

14.1. Proof of Krull’s Hauptidealsatz.

Theorem 14.63 (3.29, Generalised principal ideal theorem). Let R be a Noetherianring, I a proper ideal generated by n elements. Then htP ≤ n for each minimalprime P over I.

Definition 14.19. A regular local ring is one where dimR = dimP /P 210 where Pis the unique maximal ideal.

In fact, regular local rings are integral domains. In geometry they correspondto localisations at non-singular points.

In corollary 13.61, we have that the dimension of a Noetherian local ring is ≤the minimal number of generators of the maximal ideal P . This inequality can bestrengthened for Noetherian local domains: the dimension of the ring is = to theminimal number of generators of some ideal I with

√I = P .

Lemma 14.64. If m is a maximal ideal of a Noetherian ring R, then R/mn is anArtinian R-module, for every n > 0.

Proof. Let M = mn−1/mn (and M = R/m, if n = 1). We show that M is anArtinian R-module. The action of R on M induces a module action of R/m on Mgiven by (r + m)x = rx which is well-defined because mM = 0. Then M has thesame submodules as an R-module and as an R/m-module. Since R is Noetherian,mn−1 is a finitely generated R-module, and M is finitely generated as an R-module,and as an R/m-module. But m is maximal so R/m is a field, hence M is a finite-dimensional vector space over R/m and M is Artinian as an R/m-module and asan R-module. Since (R/mn)/(mn−1/mn) ≅ R/mn−1, the result then follows fromexamples sheet 1, question 12, and induction (the base case of induction is donebecause M = R/m when n = 1). �

Now we can prove the principal ideal theorem, often called Krull’s Hauptide-alsatz or Krull’s intersection theorem.

Proof: (of theorem 13.59). Let P be a minimal prime over (a), wherea ∈ R is a non-unit. First, we localise at P to get RP which has a unique maximalideal PP = S−1P where S = R/P . We observe that S−1P is a minimal prime overS−1(a) (this follows from the correspondence between ideals in R and ideals in thelocalisation RP ). So we may assume that R is local with unique maximal ideal P .

Suppose that htP > 1 and so there is a chain of primes Q′ ≨ Q ≨ P . ConsiderR/(a). This has a unique maximal ideal P /(a). Moreover it is also a minimalprime. So it is the only prime. So N(R/(a)) = P /(a) is nilpotent. So Pn ≤ (a) forsome n. By the previous result it follows that R/Pn and hence R/(a) is Artinian.Now consider In = {r ∈ R ∶ r/1 ∈ S−1Qn} where S = R/Q. Clearly

(14.7) Q = I1 ≥ I2 ≥ . . . ,

and hence(I1 + (a))/(a) ≥ (I2 + (a))/(a) ≥ . . . ,

is a descending chain in R/(a) which necessarily terminates

Im + (a) = Im+1 + (a),

for some m.Next, we show that (14.7) terminates. Let r ∈ Im. Then r = t + xa for some

t ∈ Im+1 (x ∈ R). So xa = r − t ∈ Im. But a ∉ Q (since P is the minimal prime over(a) and Q = I1 ≥ Im ≥ Qm). So x ∈ Im since we have

S−1R ≥ S−1Q ≥ . . . ≥ S−1Qm

10The second dim here means the R/P -vector space dimension.

31

Lecture 15

and if x/1 ∉ S−1Qm then xa/1 ∉ S−1Qm. So Im = Im+1 + Ima and hence

Im/Im+1 = P (Im/Im+1)

since a ∈ P . Nakayama’s lemma then implies that Im/Im+1 = 0 and thus Im = Im+1.Now

(S−1Q)m = S−1(Qm) = S−1Im

(S−1Q)m+1 = S−1(Qm+1) = S−1Im+1

So (S−1Q)m = (S−1Q)m+1. Then using Nakayama’s lemma for the maximal idealS−1Q of RQ gives that (S−1Q)m = 0 in RQ. Then since S−1Q is prime we have

S−1Q =√

(S−1Q)m =√

(0). Thus S−1Q = N(S−1R), and by Krull’s theorem(proposition 3.13) the nilradical is the intersection of all the prime ideals but thecorrespondence between primes under localisation then gives that S−1Q′ is a prime≨ S−1Q, a contradiction. �

Proof: (of theorem 14.63). Let R be Noetherian, and let I ≨ R generatedby n elements. We’re aiming to show htP ≤ n for each minimal prime P over I.We argue by induction on n. The case n = 1 is the principal ideal theorem. Soassume that n > 1. We may assume by passing to RP that R is local with uniquemaximal ideal P . Pick any prime Q maximal subject to Q ≨ P and thus P is theonly prime strictly containing Q. We show that htQ ≤ n − 1, it’s enough to dothis for all such Q as we then deduce that htP ≤ n. Since P is minimal over I,Q /≧ I. By assumption, there are generators a1, . . . , an for I and we may assumethat an ∉ Q. Now since P is the only prime containing Q+(an) so as in the proof of(3.28), R/(Q+ (an)) is Artinian and we note that the maximal ideal of an Artinianlocal ring is nilpotent11. So there is m such that

ami ∈ Q + (an),

for all 1 ≤ i ≤ n − 1. So ami = xi + rian for some xi ∈ Q, and ri ∈ R. Any prime ofR containing x1, . . . , xn−1 and an contains a1, . . . , an. Note that (x1, . . . , xn−1) ⊆ Q

since xi ∈ Q. Claim: Q is a minimal prime of R where R = R/(x1, . . . , xn−1) (we

write bars for images in R). Note that the unique maximal ideal P of R is a minimal

prime over (an). Proof: apply the principal ideal theorem to P . So ht(P ) ≤ 1 and

thus Q must be of height 0. Thus Q is minimal prime of R. From the claim we canapply the inductive hypothesis to Q to get that ht(Q) ≤ n − 1. �

Next, we consider filtrations by powers of an ideal I

R > I ≥ I2 ≥ I3 ≥ . . .

This is an example of a more general situation where one filters a ring R by Risatisfying RiRj ≤ Ri+j . We write negative powers: R−j = I

j . We can form thegraded ring grR

⊕i∈ZRi/Ri−1

I will talk about this later.

Lecture 15

13th November 10:00

11Take the chain P ⊇ P 2⊇ P 3

⊇ . . . which stabilises so Pm= P (Pm

) for some m, then sincewe’re in a local ring, Nakayama’s lemma gives the result.

32

Lecture 15

15.1. Filtrations.

Definition 15.20 (3.33). A (Z)-filtered ring R is one whose additive group isfiltered by subgroups

. . . ⊆ R−1 ⊆ R0 ⊆ R1 ⊆ R2 ⊆ . . .

where the Ri are additive subgroups with

⎧⎪⎪⎨⎪⎪⎩

RiRj ⊆ Ri+j for i, j ∈ Z1 ∈ R0

Note that ⋃Ri is a subring and ⋂Ri is an ideal of ⋃Ri. We shall assume as usualthat ⋃Ri = R (called ‘exhaustive’) and ⋂Ri = {0} (called ‘separated’). Also notethat Ri for i < 0 are ideals of R0.

Example.(1) Let I be an ideal of R and let Ri = R for i ≥ 0 and Ri = I

−i for i < 0. Thisis called the I-adic filtration.

(2) Let R be a k-algebra generated by x1, . . . , xn and set Ri = 0 for i < 0,R0 = k ⋅ 1 and in the case of i > 0, let Ri to be the subspace spanned bypolynomial expressions in the xj ’s of total degree ≤ i.

Definition 15.21 (3.34). The associated graded ring is

grR =⊕Ri/Ri−1

as an additive group with multiplication given by

(r +Ri−1)(s +Rj−1) = rs +Ri+j−1

for r ∈ Ri and s ∈ Rj .

Remark. Notation: often textbooks refer to ‘the symbol’ of r ∈ Ri/Ri−1 as

σ(r) = r +Ri−1

Definition 15.22 (3.35). A Z-graded ring is a ring S with additive subgroups Siso that S =⊕Si with SiSj ⊆ Si+j for i, j ∈ Z. Si is the ith homogeneous component,S0 is a subring and the Si are S0-modules.

Definition 15.23. A graded ideal I is of the form ⊕ Ii with Ii ⊆ Si.

Remark. Note that for such an I, if it is finitely generated then it can be generatedby a finite set of homogeneous elements.

Definition 15.24 (3.36). Let R be a filtered ring with filtration {Ri}, and M anR-module. Then M is a filtered R-module if there is a compatible filtration {Mi}of M of additive subgroups such that RjMi ⊆Mi+j .

Definition 15.25 (3.37). The associated graded module of a filtered R-module is

grM =⊕Mi/Mi−1

regarded as an additive group, and it is a graded grR-module via

(r +Ri−1)(m +Mj−1) = rm +Mi+j−1

If S =⊕Si is a graded ring then a graded S-module V is of the form ⊕Vj withSiVj ⊆ Vi+j .

Given a filtered R-module M with filtration {Mi} and N an R-submodule ofM , then there are induced filtrations {N ∩Mi} of N and {(N +Mi)/N} of M/N .The inclusion N ⊆M allows the definition

φi∶ (N ∩Mi)/(N ∩Mi−1)Ð→Mi/Mi−1

33

Lecture 15

Putting these together gives a map of additive groups

φ∶⊕(N ∩Mi)/(N ∩Mi−1)Ð→⊕Mi/Mi−1

grN Ð→ grM

This is a grR-module homomorphism. Consider N +Mi/N ≅ Mi/(N ∩Mi), thenwe factor in the induced filtration in the quotient

(N+Mi)

N(N+Mi−1)

N

≅Mi/(Mi−1 + (N ∩Mi))

and there is a canonical map

Mi/Mi−1 Ð→Mi/(Mi−1 + (N ∩Mi))

yielding

πi∶Mi/Mi−1 Ð→

(N+Mi)

N(N+Mi−1)

N

Then putting these together gives

π∶grM Ð→ grM/N

You should check that this is a grR-module homomorphism.

Lemma 15.65 (3.38). If N ≤M , and M is a filtered R-module, then

0 grN grM gr (M/N) 0→ →φ

→π →

is exact when N and M/N are endowed with the filtrations induced from that ofM .

Proof. One has

kerπi = (Mi−1 + (N ∩Mi))/Mi−1 ≅ (N ∩Mi)/(N ∩Mi−1).

So we have

0 (N ∩Mi)/(N ∩Mi−1) Mi/Mi−1

(N+Mi)

N(N+Mi−1)

N

0,→ →φi

→πi →

is exact. Put these together and the result follows. �

Definition 15.26 (3.39). Let R be a filtered ring, with filtration {Ri}. The Reesring of the filtration is the subring E = ⊕i∈ZRiT

i of the Laurent polynomial ringR[T,T −1]. Since RiRj ⊆ Ri+j , E is a graded ring and its homogeneous componentsare RiT

i of degree i.

Observe that

E/(T ) ≅ grR

E/(1 − T ) ≅ R

where (T ) is the ideal generated by T , and the second follows because (1 − T ) isthe kernel of the map

E → R

∑ riTi ↦∑ ri

Thus if E is Noetherian, then R and grR are.

34

Lecture 16

Example. Let R be Noetherian, and let Rj = I−j for j > 0, the I-adic filtration.

Then I is finitely generated by x1, . . . , xn say. Then E = ⊕RiTi is generated as a

ring by R0 = R and T,x1T−1, . . . , xnT

−1. So E is a ring image of R[Z0, Z1, . . . , Zn]and so E is Noetherian.

Definition 15.27 (3.40). The associated Rees module is Re(M) = ⊕T iMi for afiltered R-module M . It is an E-module via

(∑ rjTj)(T imi) = T

i+j(rjmi)

ForN ⊆M , given the induced filtrations, then lemma 15.65 implies that Re(M/N) =Re(M)/Re(N)

Definition 15.28 (3.41). A filtration of M is good if Re(M) is a finitely generatedE-module.

Lemma 15.66 (3.42). Let N ⊆M be R-modules with {Mi} a good filtration of M .If the Rees ring E of the filtration is Noetherian then the induced filtration of Nand M/N are also good.

Proof. Re(N) is an E-submodule of Re(M). But Re(M) is finitely generatedE-module and hence Noetherian. So Re(N) is finitely generated and

Re(M/N) ≅ Re(M)/Re(N)

is also finitely generated. So the induced filtrations of N and M/N are both good.�

Lecture 16

15th November 10:00

16.1. The lemma of Artin-Rees and the theorem of Hilbert-Serre.Let R be a filtered ring, M an R-module with a compatible filtration {Mi}, andlet E denote the associated Rees ring. If we have a good filtration i.e. Re(M) isa finitely generated E-module, then it is generated by a finite set of homogeneouselements

T k1mk1 , . . . , Tknmkn

say, with mki ∈Mki . So the ith homogeneous component is

T iMi = Ri−k1Ti−k1(T k1mk1) + ⋅ ⋅ ⋅ +Ri−knT

i−kn(T knmkn)

and so

Mi = Ri−k1mk1 + ⋅ ⋅ ⋅ +Ri−knmkn

for these mki ∈Mki .

Example. Let M be a finitely generated R-module, where R is Noetherian, andMi = I−iM for i < 0. Take N ≤ M . We deduce from lemma 15.66 that theinduced filtration {N ∩I−iM} is a good filtration of N . So there is a generating setnk1 , . . . , nkn of N (negative integers ki) with nkj ∈ N ∩ I−kjM and

N ∩ I−iM = I−i+k1nk1 + ⋅ ⋅ ⋅ + I−i+knnkn

So for i ≤ min{kj} = k one has

N ∩ I−iM = I−i+k(N ∩ I−kM)

for i ≤ k ≤ 0. Set a = −i and c = −k then we have the following result.

35

Lecture 16

Lemma 16.67 (3.43, Artin-Rees, 1956). Let R be Noetherian, N,M finitely-generated R-modules with N ≤ M and I an ideal of R. Then there exists c ≥ 0such that

N ∩ IaM = Ia−c(N ∩ IcM)

for a ≥ c.

Some more notes on dimension now. Suppose that R is a finitely generatedk-algebra which is an integral domain and I an ideal. Form the I-adic filtrationand its Rees ring E. Then E is a finitely generated k-algebra which is an integraldomain. The principal ideal theorem implies that the minimal primes over theideals (1 − T ) and (T ) in E are of height 1 and proposition 12.56 (the catenaryproperty) yields

dimE = 1 + dim(E/(T ))

= 1 + dim(E/(1 − T ))

and then using that R ≅ E/(1−T ) and grR ≅ E/(T ), thus dimR = dim grR. So it’suseful to consider the dimensions of graded rings. We’ll consider positively gradedrings S =⊕∞

i=0 Si and finitely generated graded S-modules V =⊕∞

i=0 Vi.

Remark. This all applies for negatively graded rings as arising from I-adic fil-trations - once one has formed the graded ring, one can remember to change theindexing to be positive.

Suppose that S is Noetherian, generated by S0 and homogeneous elementsx1, . . . , xm of degree k1, . . . , km. Let λ be an additive function, taking integralvalues on the finitely generated S0-modules, i.e. if

0 U1 U2 U3 0→ → → →

is a short exact sequence of finitely generated S0-modules then λ(U2) = λ(U1) +λ(U3), e.g.

(1) if S0 is a field k, then we can take λ = k-vector space dimension,(2) more generally if S0 is local Artinian with maximal ideal P then each

finitely generated S0-module U has a chain of submodules

U ≥ U1 ≥ U2 ≥ ⋅ ⋅ ⋅ ≥ Us = 0

with each factor ≅ S0/P . The number of terms in the chain is the com-position length of Ui. Exercise: check this is independent of the choice ofchain. We can take λ to be the composition length.

Definition 16.29 (3.44). The Poincare series of V is the power series

P (V, t) =∑λ(Vi)ti ∈ Z[[t]]

Theorem 16.68 (3.45, Hilbert-Serre). Suppose that S is Noetherian, generated byS0 and homogeneous elements x1, . . . , xm of degree k1, . . . , km. The Poincare seriesP (V, t) is a rational function in t of the form

f(t)

∏mi=1(1 − t

ki)

where f(t) ∈ Z[t] and ki is the degree of generator xi

Proof. By induction on the number of generators m. If m = 0 then S = S0

and V is a finitely generated S0-module, so Vi = 0 for large enough i, and clearly

36

Lecture 17

P (V, t) is a polynomial. Now suppose that m > 0 and assume that it is true for thecase of m − 1 generators. Now consider a map consisting of multiplication by xm

Vi Vi+km→xm

and so we can get an exact sequence

(16.8) 0 Ki Vi Vi+km Li+km 0→ → →xm → →

where Ki is the kernel, and Li+km is the cokernel of the map

Vi Vi+km→xm

Let K =⊕Ki and L =⊕Li, then K is a graded submodule of V =⊕Vi and hencea finitely generated S-module, and L is the quotient submodule of V = ⊕Vi andhence a finitely generated S-module. Both K and L (≅ V /xmV ) are annihilated byxm and so are S0[x1, . . . , xm−1]-modules. Apply λ to (16.8) to obtain

λ(Ki) − λ(Vi) + λ(Vi+km) − λ(Li+km) = 0

Multiplying by ti+km and sum up to obtain

tkmP (K, t) − tkmP (V, t) + P (V, t) − P (L, t) = g(t)

where g(t) ∈ Z[t] polynomial arising from the first few terms. Apply the inductivehypothesis to P (K, t) and P (L, t) to obtain the result. �

Corollary 16.69 (3.46). Suppose that in the situation of the previous theorem thateach ki is 1 then for large enough i λ(Vi) = φ(i) where φ(t) ∈ Q[t] of degree d − 1where d = the order of the pole of P (V, t) at t = 1. Moreover

i

∑j=0

λ(Vj) = χ(i)

where χ(t) ∈ Q[t] of degree d.

Definition 16.30 (3.47). In the above corollary, φ(t) is called the Hilbert polyno-mial and χ(t) is called the Samuel polynomial.

Lecture 17

18th November 10:00

17.1. The Hilbert polynomial and the Samuel polynomial. We provethe result stated at the end of the last class - corollary (3.46).

Proof. P (V, t) = f(t)(1−t)d

for some d with f(1) ≠ 0 and f(t) ∈ Z[t]. Since

(1 − t)−1 = 1 + t + t2 + . . . then repeated differentiation yields

(1 − t)−d =∑(d + i − 1

d − 1)ti

If f(t) = a0 + a1t + ⋅ ⋅ ⋅ + asts say, then

(17.9) λ(Vi) = a0(d + i − 1

d − 1) + a1(

d + i − 2

d − 1) + ⋅ ⋅ ⋅ + as(

d + i − s − 1

d − 1)

Setting ( rd−1

) = 0 for r < d − 1. The right hand side of (17.9) can be rearranged togive φ(i) for a polynomial φ(t) ∈ Q[t] valid for d + i − s − 1 ≥ d − 1

φ(t) =f(1)

(d − 1)!td−1 + terms of lower order

37

Lecture 17

So degree of φ(t) = d− 1 since f(1) ≠ 0. Using (17.9) we can produce an expression

for ∑ij=0 λ(Vj): the formula

(m

n) = (

m − 1

n − 1) + (

m − 1

n)

yields thati

∑j=0

(d + j − 1

d − 1) = (

d + i

d)

and soi

∑j=0

λ(Vj) = a0(d + i

d) + a1(

d + i − 1

d) + ⋅ ⋅ ⋅ + as(

d + i − s

d)

for i ≥ s, thus is equal to χ(i) for χ(t) ∈ Q[t] of degree d. �

Now if we return to R a finitely generated k-algebra negatively filtered e.g.I-adic filtration for some ideal I. Let M be a finitely generated R-module witha good (negative) filtration {Mi}. Form V = grM and S = grR and renumber sothat S is positively graded. We can apply our Hilbert-Serre analysis of dimensionsusing e.g. λ = k-vector space dimension. From our analysis just done (3.46), thereis the Samuel polynomial χ(t) ∈ Q[t] where for large enough i

χ(i) =0

∑i

dimk(Mj/Mj−1) (i < 0)

= dimk(M0/M−i)

Remark. In fact, the degree is independent of which graded filtration of M wepick (for a particular filtration R).

Definition 17.31 (3.48). d(M) = degree of χ(t)

Theorem 17.70 (3.49). For a finitely generated k-algebra R that is an integraldomain, then

dimR = tr deg (Frac(R)) = d(R)

using the P -adic filtration for any maximal ideal P of R.

Remark. Note that this implies that d(R) is independent of the choice of P - if youtake two P1 and P2 and go through all this then you end up with two polynomialsof the same degree.

Proof: (rather sketchy). We’ve established the first equality already in(3.24) and seen that dimR = dim grR with respect to the P -adic filtration. Soit remains to show that for finitely generated graded k-algebras S, dimS = d(S).Prove this by induction - S is a finite dimensional k-vector space iff dimS = d(S) = 0.The induction step comes from considering S/xS where x is a homogeneous elementwhich is not a zero divisor. The principal ideal theorem (3.28?) and the catenaryproperty (3.24) imply that dimS/xS = dimS−1. But also observe from the proof ofthe Hilbert-Serre theorem - in (16.8) replace xm with x. Then K = 0 since x is nota zero divisor and we deduce from the equation involving g(t) that d(L) = d(M)−1where L = S/xS. So d(S/xS) = d(S) − 1, M = S. Apply the inductive hypothesisto S/xS to get that dimS = d(S) in general. �

Remark. This is sketchy, because when we quotient out we may not have anintegral domain any more. But this is the general strategy needed to prove it ingeneral - if you worry about when you do and do not have an integral domain, thenyou can prove it.

38

Lecture 18

Example. Take R = k[X1, . . . ,Xn] a polynomial algebra, then the number of

monomials of total degree n is (n+m−1m−1

) for all n ≥ 0. Thus

φ(t) =1

(m − 1)!(t +m − 1) . . . (t + 1)

is the Hilbert polynomial of degree m − 1.

Exercise: d(M) = max{d(M1), d(M2)} where

0 M1 M M2 0→ → → →

is a short exact sequence.

Theorem 17.71. Let R be a Noetherian local ring that is an integral domain, thend(R) = dimR, where d(R) is taken with respect to the P -adic filtration , where Pis the unique maximal ideal of R.

Next time we will start a new chapter on valuations - discrete valuation ringsand so on.

Lecture 18

20th November 10:00

The examples classes are on Monday, Tuesday and probably Thursday.

18.1. §4: Valuation rings and Dedekind domains.

Definition 18.32 (4.1). Let A be an integral domain with field of fractions K,then A is a valuation ring of K if for each 0 ≠ x ∈ K either x ∈ A or x−1 ∈ A (orboth).

Example. An example of this, take K = Q and A = Z(p) the localisation of Z at aprime ideal (p) with p ≠ 0.

Proposition 18.72 (4.2). Let A be a valuation ring with fraction field K, then

(1) A is a local ring,(2) if A ⊆ B ⊆K, then B is a valuation ring,(3) A is integrally closed.

Proof.(1) Let P be the set of non-units of A. Then x ∈ P iff x = 0 or x−1 ∉ A. We

see that P is an ideal:(a) if a ∈ A, and x ∈ P then ax ∈ P since otherwise (ax)−1 ∈ A and so

x−1 = (ax)−1a ∈ A, contradicting x ∈ P ,(b) if x, y ∈ P then x + y ∈ P : either xy−1 ∈ A or x−1y ∈ A. If xy−1 ∈ A

then x+ y = (1+xy−1)y which is of the form ay and so is in P (using(a)). Similarly x−1y ∈ A.

(2) ✓ This is clear.(3) Let x ∈K be integral over A. Thus

xn + an−1xn−1 + ⋅ ⋅ ⋅ + a0 = 0

for some ai ∈ A. Suppose that x ∉ A, then

x = −(an−1 + am−2x−1 + ⋅ ⋅ ⋅ + a0x

−n+1)

and this lies in A since x−1 does, a contradiction. �

39

Lecture 18

The definition of a valuation ring is not very enlightening - the reason for callingit a valuation ring is that one may associate a non-Archimedean valuation

v∶K× → Γ

where Γ is a (well-chosen) ordered Abelian group (i.e. every pair of elements satisfiesγ1 ≤ γ2 or γ2 ≤ γ1 and we only get both if γ1 = γ2, and γ1 ≤ γ2 ⇒ γ + γ1 ≤ γ + γ2)satisfying

(1) v(xy) = v(x) + v(y),(2) v(x + y) ≥ min(v(x), v(y)) the ultra-metric inequality,

so that if we define

A = {x ∈K ∶x = 0 or v(x) ≥ 0}

given such a v, then A is a valuation ring.

Definition 18.33. Using the notation as above, if Γ ∈ Z we say that A is a discretevaluation ring.

Example.(1) vp∶Q× → Z given by pn a

b↦ n where a, b ∈ Z are coprime, is the p-adic

valuation on Q with discrete valuation ring Z(p).(2) vf ∶k(X)× → Z given by fn g

h↦ n for f some irreducible polynomial k[X],

and g, h are coprime to f . This has a discrete valuation ring k[X](f), thelocalisation of k[X] at the prime ideal (f).

18.2. Recipe for valuation rings. Given R an integral domain with fieldof fractions K, take an algebraically closed field F . Consider pairs (R′, φ′) whereR′ is a subring of K, and φ′∶R′ → F is a ring homomorphism. We partially orderthese pairs by

(R1, φ1) ≤ (R2, φ2)⇔

⎧⎪⎪⎨⎪⎪⎩

R1 ≤ R2

φ2∣R1 = φ1

An ascending chain of such pairs has an upper bound (R0, φ0) such that R0 isthe union of all subrings appearing in the pairs in the chain, and φ0 is the mapthat when restricted to each subring, yields the corresponding φi. Applying Zorn’slemma, there is a maximal such pair (A, θ) say.

Claim: Such an A is a valuation ring of K. Proof: step 1: A is a local ringwith ker θ = P the unique maximal ideal (we proved that these are local rings inproposition 18.72. θ(A) is a subring of F1 a field, and so is an integral domain.So P = ker θ is a prime ideal. Extend θ to a ring homomorphism φ∶AP → F anda/s ↦ θ(a)/θ(s) and let S = A/P . The maximality of the pair (A, θ) ensures thatA = AP ⊆K, and hence A is a local ring with maximal ideal P .

Take x ≠ 0 in K. We must show that either x ∈ A or x−1 ∈ A i.e. A[x] or A[x−1]is equal to A. Step 2: first we show that either PA[x] ≨ A[x] or PA[x−1] ≨ A[x−1].

Proof: suppose that PA[x] = A[x] and PA[x−1] = A[x−1], so 1 ∈ PA[x] and1 ∈ PA[x−1] so

1 = anxm + ⋅ ⋅ ⋅ + a0(18.10)

1 = bnx−n + ⋅ ⋅ ⋅ + b0(18.11)

where ai ∈ P and bi ∈ P . Pick m and n minimal and assume that m ≥ n. Multiply(18.11) by xn to obtain

(1 − b0)xn = bn + ⋅ ⋅ ⋅ + b1x

n−1(18.12)

But b0 ∈ P and so 1 − b0 ∉ P and so 1 − b0 is a unit. So (18.12) gives that

xn = cn + . . . c1xn−1

40

Lecture 19

with ci ∈ P and so xm = cnxn−1 + ⋅ ⋅ ⋅ + c1x

m−1. Substituting in (18.10) gives anequation contradicting the maximality of n.

Step 3: We may assume I = PA[x] ≨ A[x]. Let B = A[x]. We show that B = Aand hence x ∈ A. Let Q be a maximal ideal of B containing I. Thus Q∩A = P sinceA ∩A ≤ A and P is in it. Regard A/P as a subring of B/Q. Both are fields, k1, k2say. k1 = k[x] where x is the image of x in B/Q. Thus k1 is an algebraic extension

of k. But θ induces a map θ∶k = A/P → F and thus extends to a map φ∶k1 → F

since F is algebraically closed. φ lifts back to a map B → F . The maximality ofthe pair (A, θ) ensures that A = B.

Lecture 19

22nd November 10:00

The example class is on Thursday in MR15 at 4-5pm.

Theorem 19.73. Let R be an integral domain with fraction field K. Then theintegral closure T of R in K is the intersection of all the valuation rings of Kcontaining R.

Remark. An example of this is Z = ⋂p primeZ(p).

Proof. Let A be a valuation ring containing R. But A is integrally closed byproposition 18.72 and hence T ⊆ A. Conversely if x ∉ T then x ∉ R[x−1] = R1. Sox−1 is not a unit of R1 and is therefore contained in a maximal ideal P1 of R1. LetF be the algebraic closure of the field R1/P1. The canonical map R1 → R1/P1 ⊆ Frestricts to give a map φ∶R → F . So in our recipe for valuation rings. By Zorn’slemma12 there is a maximal pair (A, θ) with A being a valuation ring. Since θextends θ(x−1) = φ(x−1) = 0. So x ∉ A. �

19.1. Discrete valuation rings. A valuation ring A with fraction field K.Then it is a local ring by proposition 18.72(1) and it is integrally closed by propo-sition 18.72(3). Denote the maximal ideal by P . If we have a (surjective) discretevaluation v∶K× → Z so that

A = {x ∈K ∶x = 0 or v(x) ≥ 0}

P = {x ∈K ∶x = 0 or v(x) ≥ 1}

(remark: people often set v(0) = ∞). If v(a) = v(b) then v(ab−1) = 0 and so ab−1

is a unit in A. So (a) = (b). If I is a non-zero ideal of A, there is a least k suchthat v(a) = k for some a ∈ I. So I contains every b with v(b) ≥ k (since b = a(b/a)and v(b/a) ≥ 0 and so b/a ∈ A and hence b ∈ (a)). Hence I = Ik = {x ∈ A∶ v(x) ≥ k}.Thus there is only one chain of ideals

A ≩ P = I1 ≩ I2 ≩ . . .

and therefore A is Noetherian. Thus P is the only non-zero prime, so dimA = 1.

Lemma 19.74 (4.5). Let A be a Noetherian local integral domain. Set k = A/Pwhere P is the unique maximal ideal. The following are equivalent

(1) A is a discrete valuation ring,(2) A is integrally closed,(3) P is principal,(4) dim(P /P 2) = 1,(5) every non-zero ideal (≠ A) is a power of P ,

12Brookes: I look after the graduate applications in my college. We received one from somefellow whose surname was Zorn. Naturally, we admitted him - axiom of choice and all that.

41

Lecture 20

(6) there exists x ∈ P such that every non-zero ideal ≠ A is of the form (xn)for some n ≥ 1.

Proof. (1)→ (2) is proposition 18.72 (iii).(2)→ (3). Let 0 ≠ a ∈ P , then dimension of A =1 and A is local. P is the only

minimal prime over (a) and we know Pn ≤ (a) for some n. Pick n minimal - soPn−1 /≤ (a) and we may pick b ∈ Pn−1 with b ∉ (a). Set x = a/b. Claim: P = (x).Note that x−1 ∉ A since b ∉ (a) and by the integrally closed property of A. x−1

is not integral over A. If x−1P ≤ P then P would be a A[x−1]-module, finitelygenerated as an A-module. Any A[x−1]-submodule will also be finitely generatedas an A-module (since A Noetherian). But any non-zero cyclic A[x−1]-submoduleof A[x−1] is isomorphic to A[x−1] and this gives a contradiction as A[x−1] is not afinitely generated A-module (x−1 is not integral). So x−1P /≤ P . Thus x−1P ≤ A byconstruction and so x−1P = A. Hence P = (x).

(3)→ (4). P is principal so P /P 2 is principal and thus dimk P /P 2 ≤ 1 (P ≠ P 2

by Nakayama’s lemma).(4) → (5). If 0 ≠ I ≠ A, ideal of A, then Pn ≤ I (as in (2) → (3)) But

dimk(P /P 2) = 1 thus P is principal (application of Nakayama’s lemma) P = (x)say. There exists r such that I ⊆ P r, I /⊆ P r−1 and hence there exists y ∈ I suchthat y = axr with y ∉ P r−1. So a ∉ P and hence a is a unit of A. So xr ∈ I andP r ⊆ I. So I = P r.

(5) → (6) By Nakayama’s lemma P ≠ P 2. Take x ∈ P /P 2. But then P = (x).Every ideal I is of the form P r for some r. So I = (xr).

(6) → (1). Assuming (6), we have P = (xr) for some r. But P is prime andso r = 1, i.e. P = (x). By Nakayama’s lemma P k ≠ P k+1 for any k. Then A is avaluation ring: if y ∈ K, y ∉ A then consider {x ∈ A∶xy ∈ A} an ideal of A andso is (xk) for some k. So yxk ∈ A/P (otherwise yxk ∈ P = (x) and yxk−1 ∈ A andxk−1 ∈ (xk) a contradiction). Thus yxk is a unit of A and we deduce that y−1 isin A. If a ∈ A then (a) = (xk) for exactly one value of k. Define v(a) = k. Extendto K× by v(ab−1) = v(a) − v(b). Check: this gives a well defined discrete valuationv∶K× → Z. �

Lecture 20

25th November 10:00

Definition 20.34 (4.6). An integrally closed Noetherian integral domain of di-mension 1 is a Dedekind domain.

Example.(1) Integral closure of Z in a finite field extension of Q.(2) Coordinate rings of (smooth) normal curves.

Remark. If S is integrally closed then S−1R will be integrally closed for anymultiplicatively closed subset S of R. In particular RQ will be integrally closed forany maximal ideal Q, and so RQ is a discrete valuation ring (DVR) using lemma19.74.

Lemma 20.75 (4.7). In a Dedekind domain, every ideal I with√I = Q a maximal

ideal, is a power of Q.

Proof. Let I be such that√I = Q, maximal. Then S−1I is a non-zero ideal

of RQ where S = R/Q. By lemma 19.74 applied to the DVR RQ, we have

S−1I = (S−1Q)r = S−1(Qr)

for some r. The bijective correspondence between ideals that don’t meet S and theideals of the localisation gives that I = Qr. �

42

Lecture 20

Theorem 20.76 (4.8, Dedekind). In a Dedekind domain R, every non-zero idealI has a unique factorisation as a product of prime ideals.

Proof. Given a non-zero ideal I, then R/I only has finitely many primes allof which are maximal and is Artinian since dimR = 1. So R/I is a direct product

of Artinian rings (examples sheet 1). Thus I = ⋂ Ij with√Ij = maximal ideal

Qj . But by lemma 20.75 we have Ij = Qmj

j for some mj . Thus I = ⋂Qmj

j But

for coprime ideals, products are the same as intersections (a small lemma, provedby induction on the number of ideals - see example sheets hint/solutions). SoI = ∏Q

mj

j . Uniqueness: the Qj appearing ar the minimal primes over I. In anyother similar expression for I, the same Q’s must appear and the primes must bethe same because the powers are unique in lemma 19.74. �

Definition 20.35 (4.9). Given an integral domain R, with fraction field K, anyR-submodule M of K is a fractional ideal of R if there exists 0 ≠ x ∈ R such thatxM ⊆ R

Remark.(1) Every finitely generated R-submodule M of K is a fractional ideal.(2) In a Dedekind domain, the factional ideals form a group under multipli-

cation - the class group.

20.1. §5: Tensor products, homology and cohomology. Roughly, youget homology and cohomology theories when you are dealing with functors that arenot exact.

Let L,M,N be R-modules.

Definition 20.36 (5.1). φ∶M ×N → L is R-bilinear if

(1) φ(r1m1 + r2m2, n) = r1φ(m1, n) + r2φ(m2, n),(2) φ(m,r1n1 + r2n2) = r1φ(m,n1) + r2φ(m,n2).

The idea of tensor products is to reduce the discussion of multilinear maps toa discussion of linear maps. If φ∶M ×N → T is bilinear and θ∶T → L is linear thenthe composition is bilinear. Thus composition with φ gives a well-defined function

φ∗∶{R-module maps

T → L}Ð→ {

bilinear mapsM ×N → L

} .

The map φ is universal if φ∗ is a 1-1 correspondence for all L.

Lemma 20.77 (5.2).(1) Given M,N there is an R-module T and a universal map φ∶M ×N → T .(2) Given two such map φ1∶M ×N → T1 and φ2∶M ×N → T2, then there is a

unique isomorphism β∶T1 → T2 with β ○ φ1 = φ2.

Proof.(1) Let F be the free module on generators e(m,n) indexed by pairs (m,n) ∈

M ×N . Let X be the R-submodule generated by all elements of the form

e(r1m1+r2m2,n) − r1e(m1,n) − r2e(m2,n),

and

e(m,r1n1+r2n2)− r1e(m,n1)

− r2e(m,n2).

Set T = F /X and write m ⊗ n for the image of the basis elements e(m,n)in T . Define a map

φ∶M ×N → T,

(m,n)↦m⊗ n.

43

Lecture 21

Note that T is generated by the elements m ⊗ n and φ is bilinear. Anymap α∶M ×N → L extends to a map α∶F → L by e(m,n) ↦ α(m,n). If αis bilinear then α vanishes on X and we have an induced map α′∶T → Lwith α′(m⊗ n) = α(m,n) and α′ is uniquely defined by this.

(2) This just follows from universality. �

Definition 20.37 (5.3). T (as above) is written M ⊗RN called the tensor productof M and N over R (often we drop the ‘R’ if it is clear what ring we are using).

Warning: not all elements of M ⊗N are of the form m⊗ n, a general elementis of the form ∑(mi ⊗ ni).

Example. If R is a field k, and M,N are k-vector spaces of dimensions s and t,then M ⊗k N is a k-vector space of dimension st.

Lemma 20.78 (5.4). There are unique isomorphisms

(1) M ⊗N → N ⊗M

(m⊗ n)↦ (n⊗m)

(2) M ⊗ (N ⊗L)→ (M ⊗N)⊗L

m⊗ (n⊗ l)↦ (m⊗ n)⊗ l

(3) (M ⊕N)⊗L→ (M ⊗N)⊕ (N ⊗L)

(m⊕ n)⊗ l ↦ (m⊗ l)⊕ (n⊗ l)

(4) R⊗M →M

r ⊗m↦ rm

Proof. Exercise. �

Lecture 21

27th November 10:00

21.1. Restriction and extension of scalars. If φ∶R → T is a ring homomor-phism and N is a T -module, it may be regarded as an R-module via rm = φ(r)m.Thus T itself may be regarded as an R-module - we call this restriction of scalars.Extension of scalars is: given an R-module M we can form T ⊗RM . This can beviewed as a T -module via t1(t2 ⊗m) = t1t2 ⊗m.

Example. In localisation we had a map R → S−1R. Given an R-module M and themultiplicatively closed set S, there is a unique isomorphism S−1R ⊗RM → S−1M .The map S−1R×M → S−1M is R-bilinear and universality yields an R-module map

S−1R⊗M → S−1M.

Check yourself that it is an isomorphism.

Definition 21.38 (5.5). Given R-module maps θ∶M1 →M2 and φ∶N1 → N2, thenwe define the tensor product θ ⊗ φ by

M1 ⊗N1 →M2 ⊗N2,

m1 ⊗ n2 ↦ θ(m1)⊗ φ(n1).

Note that the map

M1 ×N1 →M2 ⊗N2,

(m1, n2)↦ θ(m1)⊗ φ(n1),

is bilinear and so universality yields the map

θ ⊗ φ∶M1 ⊗N1 →M2 ⊗N2.

44

Lecture 21

Lemma 21.79 (5.6). Let R be a ring and let M,N,L be R-modules, then one hasan isomorphism

Hom(M ⊗N,L) ≅ Hom(M,Hom(N,L)).

Proof. Given a bilinear φ∶M ×N → L we have a map

θ∶M → Hom(N,L),

m↦ {θm∶N → L},

where θm∶N → L is given by n ↦ φ(m,n). Conversely given θ∶M → Hom(N,L)we have a bilinear M × N → L given by (m,n) ↦ θ(m)(n). Thus there is anisomorphism

{bilinear mapsM ×N → L

}←→ {linear maps

M → Hom(N,L)} .

But the left-hand side corresponds to the linear maps M ⊗N → L. �

Definition 21.39 (5.7). Given φ1∶R → T1 (and thus T1 is an R-algebra) andφ2∶R → T2, the tensor product of the two R-algebras is T1 ⊗R T2 defined as anR-module.

T1 ⊗R T2 can be endowed with a product

(t1 ⊗ t2)(t′

1 ⊗ t′

2) = t1t′

1 ⊗ t2t′

2

Check that (T1 ⊗T2)(T1 ⊗T2)→ T1 ⊗T2 is well-defined - 1⊗ 1 is the multiplicativeidentity and

R → T1 ⊗ T2,

r ↦ φ1(r)⊗ 1 = 1⊗ φ2(r),

is a ring homomorphism (check this).

Example.● Let k be a field, so k[X] is a k-algebra then

k[X1]⊗k k[X2] ≅ k[X1,X2]

● We have

Q[X]/(X2 + 1)⊗Q C ≅ C[X]/(X2 + 1)

● One has

k[X1]/(f(X1))⊗k k[X2]/(g(X2)) ≅ k[X1,X2]/(f(X1), g(X2))

Lemma 21.80 (5.8). If M1 → M → M2 → 0 is an exact sequence, and N is anR-module then

M1 ⊗N M ⊗N M2 ⊗N 0

N ⊗M1 N ⊗M N ⊗M2 0

→ → →

→ → →

is exact.

Proof. We can prove this directly but often use knowledge of exactness forHom and then use lemma 21.79 - see next lecture. �

Remark. The sequences in the lemma above are not necessarily short exact se-quences - given a short exact sequence, applying (− ⊗ N) does not necessarilypreserve the injectivity of the left hand map - an example illustrating this is

0 2Z Z Z/2Z 0→ → → →

45

Lecture 22

then if we tensor with N = Z/2Z we lose injectivity, so exactness is not preserved.

Definition 21.40 (5.9). N is a flat R-module if given any short exact sequence

0 M1 M M2 0,→ → → →

then

0 M1 ⊗N M ⊗N M2 ⊗N 0,→ → → →

is exact.

Example.(1) R is itself a flat R-module.(2) Rn (free module on n generators) is a flat R-module.(3) If R = Z then Q is a flat Z-module. In fact any torsion-free Abelian group

is a flat Z-module.

If we consider Hom(−,N), we have an analogous situation, only things are nowcontravariant rather than covariant.

Lecture 22

29th November 10:00

22.1. Projective & injective modules; Tor and Ext.

Proposition 22.81 (5.10).

(1) M1 M M2 0→θ →φ

→ (�) is exact iff

0 Hom(M2,N) Hom(M,N) Hom(M1,N)→ →φ

→θ

is exact for all N ,

(2) 0 M1 M M2→ →θ →φ

is exact iff

0 Hom(N,M1) Hom(N,M) Hom(N,M2)→ →θ →φ

is exact for all N .

Proof. (i) Suppose

0 Hom(M2,N) Hom(M,N) Hom(M1,N)→ →φ

→θ

is exact for all N . Since Hom(M2,N)φÐ→ Hom(M,N) is injective for all N , the

map M →M2 is surjective. So we have exactness at M2 in (�). We need to checkexactness at M . First we show Im θ ⊆ kerφ. Take N =M2, f = identity map fromM2 →M2 then θ○φ(f) = 0. So f ○φ○θ = 0 and so φ○θ = 0. Finally take N =M/Im θ

and π∶M → N the usual quotient map. Then π ∈ ker θ and hence there exists ψsuch that π = φ(ψ) ∈ Hom(M2,N). So Im θ = kerπ ⊆ kerφ. Hence kerφ = Im θ.Rest of proof: exercise. �

Now we prove lemma 21.80 using lemma 21.79 and proposition 22.81.

Proof. Given the sequence M1 → M → M2 → 0 which is exact, we wantM1 ⊗N →M ⊗N →M2 ⊗N → 0 exact. Let P be any R module. The sequence

46

Lecture 22

0 Hom(M2,Hom(N,P )) Hom(M,Hom(N,P ))

Hom(M1,Hom(N,P ))

→ →φ

→

θ

is exact by proposition 22.81. Hence the sequence

0 Hom(M2 ⊗N,P ) Hom(M ⊗N,P ) Hom(M1 ⊗N,P ),→ →φ

→θ

is exact for any P using lemma 21.79. So proposition 22.81 implies that

M1 ⊗N →M ⊗N →M2 ⊗N → 0,

is exact. �

Observe that in general given a short exact sequence

0 M1 M M2 0,→ → → →

then the sequence

0 Hom(M2,N) Hom(M,N) Hom(M1,N) 0,→ → → →

is not necessarily exact.

Definition 22.41 (5.10). A module P is projective if when we have a surjectivemodule homomorphism M ↠M1 and a module homomorphism P →M1 then wecan complete the diagram with a map P →M .

P

M M1 0

→→

→ →

so that it commutes . Similarly we can define injective modules E by being able tocomplete diagrams of the form

0 M1 M

E

→ →

→ →

Example.(1) Free modules are projective.(2) For an integral domain R with fraction field K, then K is an injective

R-module, e.g. Q is an injective Z-module.

Lemma 22.82 (5.12). For an R-module P , the following are equivalent

(1) P is projective.(2) for every short exact sequence 0 → M1 → M → M2 → 0 the induced

sequence

0 Hom(P,M1) Hom(P,M) Hom(P,M2) 0→ → → →

is exact.(3) If ε∶M ↠ P is surjective then there exists a map β∶P → M such that

ε ○ β = id.(4) P is a direct summand in every module of which it is a quotient.(5) P is a direct summand of a free module.

47

Lecture 22

Proof. ‘(1)→ (2)’ - follows from the definition of a projective P .

‘(2)→ (3)’ Choose an exact sequence 0→ ker ε→Mε→ P → 0, then the induced

sequence

0 Hom(P,ker ε) Hom(P,M) Hom(P,P ) 0→ → → →

is exact and so there exists β∶P →M such that ε ○ β = id.

‘(3) → (4)’ Let P ≃ M/M1 a quotient of M and so we have 0 → M1 → Mα→

P → 0 a short exact sequence. By (3) there exists β∶P → M such that α ○ β = id,therefore P is a direct summand of M .

‘(4) → (5)’ P is a quotient of a free module: take a generating set X for P ,form the free module with basis {ei} indexed by X and a map of basis elementsex ↦ x, for x ∈X.

’(5) → (1)’ By (5), F = P ⊕Q. Since free modules are projective and we havegood behaviour under ⊕. P is projective. �

Remark.(1) Projectives are direct summands of free modules and free modules are

flat and tensor products behave well under ⊕, so we get that projectivemodules are flat.

(2) In a principal ideal domain we have from the structure theorem for mod-ules that direct summands of free modules are free - projectives are free.

Lemma 22.83 (5.13). For an R-module E, the following are equivalent

(1) E is injective,(2) for every short exact sequence 0→M1 →M →M2 → 0, the sequence

0 Hom(M2,E) Hom(M1,E) Hom(M1,E) 0→ → → →

is exact,(3) if µ∶E → M is a monomorphism, then there exists β∶M → E such that

β ○ µ = id.(4) E is a direct summand of every module which contains it as a submodule.

Proof. Exercise. �

Given an R-module M , there is certainly a free module F with F ↠M surjec-tively.

Definition 22.42 (5.14). A projective presentation of M is a short exact sequence

0 K P M 0→ → → →

with P projective. We say that it is a free presentation in the case where P is free.

Definition 22.43 (5.15). Given a projective presentation of M , then apply −⊗RNto get

K ⊗N P ⊗N M ⊗N 0→ → →

Define TorR(M,N) = ker(K ⊗N → P ⊗N). Apply Hom(−,N) to get

0 Hom(M,N) Hom(P,N) Hom(K,N)→ → →

and define Ext(M,N) = coker(Hom(P,N)→ Hom(K,N)).

Remark.(1) This is actually independent of the choice of presentation.

48

Lecture 23

(2) One may also take a projective presentation for N and apply M ⊗R − andthis gives the same kernel.

(3) One may take a short exact sequence

0 N R C 0→ → → →

with E injective and apply Hom(M,−) and the cokernel arising is thesame as Ext(M,N).

(4) Ext denotes ‘extensions’ - there is an alternative description in terms ofequivalence classes of extensions. Tor denotes ‘torsion’ - this is moreobscure - it comes from the idea that torsion free Abelian groups are flat.

Lecture 23

2nd December 10:00

23.1. Induced long exact sequences from Tor and Ext.

Example. We met the free presentation of Z/(R) (where R = 2)

0 Z Z Z/(2) 0→ →×2 → →

Apply − ⊗Z/(2) then we have

Tor(Z/(2),Z/(2)) = ker(Z⊗Z/(2)→ Z⊗Z/(2))

So Tor(Z/(2),Z/(2)) = Z/(2).Apply Hom(−,N) to our presentation, then

Ext(Z/(2),N) = coker(Hom(Z,N)→ Hom(Z,N))

induced by multiplication by 2 (note that Hom(Z,N) ≅ N).

Remark. For any principal ideal domain, we have a projective presentation

0 K P M 0→ → → →

for any finitely generated R-module M in which K is also projective (projectivesare free).

Definition 23.44 (5.16). A projective resolution of M is an exact sequence

. . . Pn . . . P1 P0 M 0→ → → → → →

with Pi projective.

Remark. If R is Noetherian and M is a finitely generated R-module we can pro-duce a projective resolution with all the Pi finitely generated projective modules.From the presentation

0 K0 P0 M 0→ → → →

and then take a presentation for K0

0 K1 P1 K0 0→ → → →

etc. ensuring at each stage that P1 is finitely generated, and hence K1 is finitelygenerated. Applying −⊗RN to a projective resolution for M yields a chain complex

. . . Pn ⊗N . . . P1 ⊗N P0 ⊗N M ⊗N 0→ → → → → →

(A chain complex is one where the image of one arrow is in the kernel of thenext)

At Pn ⊗N we have

49

Lecture 23

. . . Pn ⊗N . . .→θn →

θn−1

and ker θn/Im θn is an R-module - it is known as the homology of the chain complexPn ⊗N .

Definition 23.45 (5.17). TorRn (M,N) is the homology group at Pn ⊗N . Thus

Tor0(M,N) =M ⊗N

Tor1(M,N) = Tor(M,N)

(Use the chain complex

. . . Pn ⊗N . . . Pn ⊗N P0 ⊗N 0→ → → → →

with the homology of P0 ⊗N being M ⊗N).Similarly given a projective resolution for M , apply Hom(−,N) and get a

cochain complex

0 Hom(P0,N) Hom(P1,N) . . .→ → →

and we define ExtnR(M,N) to the (co-)homology group at Hom(Pn,N). Thus

Ext0(M,N) = Hom(M,N)

Ext1(M,N) = Ext(M,N)

Remark. If fact this is all independent of the choice of projective resolution. More-over, one can obtain Extn(M,N) by considering an injective resolution of N

0 N E0 E1 . . .→ → → →

is exact, with Ei injective R-modules. Apply Hom(M,−) to this and consideringthe homology groups of the resulting complex yields the same thing.

Lemma 23.84 (5.18). The following are equivalent

(1) Extn+1(M,N) = 0 for all R-modules N ,(2) M has a projective resolution of length n,

0 Pn Pn−1 . . . P0 M 0→ → → → → →

Proof. Omitted. �

Definition 23.46. The homological dimension of M is n if Extn+1(M,N) = 0 forall N and there is some N for which Extn(M,N) ≠ 0. The global dimension of Ris the supremum of all the homological dimensions of R-modules M .

Example.(1) For a field k, all modules are free and the global dimension is 0.(2) The global dimension Z is 1. In fact, this is obviously the case for any

principal ideal domain which isn’t a field e.g. k[X].(3) The condition that gl dimR = 0 is equivalent to saying that all submod-

ules of R are direct summands. In other words R is semisimple (butyou could complain here that I have veered off into the non-commutativecase, but you could just replace with a finite Abelian group) cf. the com-plex representation theory of finite groups G - the group algebra CG issemisimple.

Proposition 23.85 (5.20). Given a short exact sequence

0 M1 M M2 0→ → → →

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Lecture 24

there is a long exact sequence

. . . Tor2(M2,N)

Tor1(M1,N) Tor1(M,N) Tor1(M2,N)

Tor0(M1,N) Tor0(M,N) Tor0(M2,N) 0

→

→ → →

→ → → →

and

0 Ext0(M1,N) Ext0(M,N) Ext0(M2,N)

Ext1(M1,N) Ext1(M,N) Ext1(M2,N)

Ext2(M1,N) . . .

→ → →

→ → →

→ →

Proof. Omitted. �

Corollary 23.86 (Dimension shifting). Given a projective presentation

0 K P M 0→ → → →

then we have

Torn(M,N) = Torn−1(K,N) for n > 0,

Extn(M,N) = Extn−1(K,N).

Proof. Apply proposition 23.85 to our presentation and observe that for aprojective P we have

Extn(P,N) = 0

Torn(P,N) = 0

for n > 0. �

For a polynomial algebra R = k[X1, . . . ,Xn], there is a well chosen free resolu-tion of the trivial module k with Xi acting like 0, known as the Koszul complex,based on the exterior algebra on n-generators in each degree m, we have the freeR-module with generators being the extension products Xi1 ∧Xi2 ∧ ⋅ ⋅ ⋅ ∧Xin of mof the n variables. In general Hilbert’s syzygy theorem says that any ideal of R hasa projective module resolution of lengths ≤ n. So gl dimR = n.

Lecture 24

4th December 10:00

24.1. Hochschild cohomology (non-examinable). This is a bimodule co-homology. Let R be an R −R-bimodule, and suppose for this lecture that we dropthe commutativity condition. A k-algebra R is an R − R-bimodule - we need aprojective resolution for R as a bimodule. An R−R-bimodule may be viewed whenconvenient as a left R⊗k R

op-module. In Rop we have

x ⋅ y = yx

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Lecture 24

where the multiplication on the left is in Rop and on the left is multiplication in R.For R commutative we have Rop ≅ R

R⊗R → R

x⊗ y ↦ xy

where the thing on the right is a free R⊗R-module. The Hochschild resoltuion ofR is

. . . R⊗R⊗R⊗R R⊗R⊗R R⊗R R→ → → →

where R⊗R⊗R is a free R⊗R module and so on. The map R⊗n → R⊗(n−1) involvealternate sums e.g.

x⊗ y ⊗ z → xy ⊗ z − x⊗ yz

where we are using multiplication in R for the xy etc. For Hochschild cohomology,one tensors this resolution with R. If M is an R−R-bimodule, we can form M ⊗kRwhich is a bimodule and consider the homology groups in the arising chain complexHHi(R,R), and if we apply hom(−,R), the cohomology in the arising cocomplexis Hochschild cohomology HHi(R,R)

HH0(R,R) = centre of R

HH1(R,R) =derivations of R

inner derivations of R

A derivation is a map d satisfying d(xy) = xdy + (dx)y. The inner derivations arethose arising from ring commutators x ↦ [x, y] = xy − yx where the multiplicationis in R. For a commutative algebra we have HHi(R,R) ≅ derivations. The deriva-tions from a Lie algebra - they can be regarded as infinitesimal automorphisms.Hochschild cohomology has a product defined in it - it forms a ring, but in practiceit is hard to work with a particular examples. HH2 has a meaning to do withdeformations of the algebra.

24.2. Hochschild dimension. A k-algebra has Hochschild dimension n ifand only if R has a free/projective resolution of length n and no shorter (analoguesto global dimension definition).

Hochschild dimension 0 is equivalent to R being a projective R −R-bimodule,which is equivalent R being a direct summand of R⊗Rop. This is called a separablek-algebra.

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Bibliography

[AM69] M. F. Atiyah and I. G. Macdonald. Introduction to commutative algebra. Addison-WesleyPublishing Co., Reading, Mass.-London-Don Mills, Ont., 1969.

[Bou74] Nicolas Bourbaki. Algebre. Elements de mathematiques. Springer; 2eme ed., 1974.

[Mat70] Hideyuki Matsumura. Commutative algebra. W. A. Benjamin, Inc., New York, 1970.[Sha00] R. Y. Sharp. Steps in commutative algebra, volume 51 of London Mathematical Society

Student Texts. Cambridge University Press, Cambridge, second edition, 2000.

[ZS60] Oscar Zariski and Pierre Samuel. Commutative algebra. Vol. II. The University Series inHigher Mathematics. D. Van Nostrand Co., Inc., Princeton, N. J.-Toronto-London-New

York, 1960.

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