Table Of ContentPart III Commutative algebra
Lecturer: Dr. C.J.B. Brookes
Michaelmas term 20131
1TranscribedbyStiofa´inFordham
Contents
Page
Lecture 1 5
1.1. Introduction 5
Lecture 2 6
2.1. §1: Noetherian rings: definitions, examples & ideal structure 7
Lecture 3 9
3.1. Cohen’s theorem, the nilradical and Jacobsen radical 9
Lecture 4 11
4.1. Weak and strong Nullstellensatz 11
Lecture 5 13
5.1. Minimal primes, annihilators and associated primes 13
Lecture 6 15
6.1. §2: Localisation 16
Lecture 7 17
7.1. Modules 17
Lecture 8 20
8.1. §3: Dimension 20
Lecture 9 22
9.1. More about integrality 22
Lecture 10 24
10.1. The Going-up and Going-down theorems 24
Lecture 11 25
11.1. Proof of the Going-down theorem 25
Lecture 12 27
12.1. Noether normalisation lemma 27
Lecture 13 28
13.1. Remarks on the Noether normalisation lemma 29
Lecture 14 30
14.1. Proof of Krull’s Hauptidealsatz 31
Lecture 15 32
15.1. Filtrations 33
Lecture 16 35
16.1. The lemma of Artin-Rees and the theorem of Hilbert-Serre 35
Lecture 17 37
17.1. The Hilbert polynomial and the Samuel polynomial 37
Lecture 18 39
18.1. §4: Valuation rings and Dedekind domains 39
18.2. Recipe for valuation rings 40
Lecture 19 41
19.1. Discrete valuation rings 41
Lecture 20 42
20.1. §5: Tensor products, homology and cohomology 43
Lecture 21 44
3
21.1. Restriction and extension of scalars 44
Lecture 22 46
22.1. Projective & injective modules; Tor and Ext 46
Lecture 23 49
23.1. Induced long exact sequences from Tor and Ext 49
Lecture 24 51
24.1. Hochschild cohomology (non-examinable) 51
24.2. Hochschild dimension 52
Bibliography 53
4
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 good
introduction, butthereismaterialintherethatIwillnottalkabout. Ontheother
handitisnotallthere-Iwanttosaysomethingabouthomologicalalgebra. You’ll
find quite a lot of things are exercises. If you want to see the details, you can look
at the book by Sharp [Sha00]. There is the book by Matsumura [Mat70]but this
is 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 variables
k[X ,...,X ], andifΣ isthesymmetricgroupon{1,...,n}, thenthesymmetric
1 n n
group acts on the polynomial ring k[X ,...,X ] by permuting variables. We look
1 n
at the set of polynomials fixed under this action: the set (actually form a ring)
S ={f ∈k[X ,...,X ]∶g(f)=f for all g∈Σ }
1 n n
called the ring of invariants. We have the elementary symmetric functions
f (X ,...,X )=X +⋅⋅⋅+X
1 1 n 1 n
f (X ,...,X )=∑X X
2 1 n i j
i<j
⋮
f (X ,...,X )=X X ...X
n 1 n 1 2 n
In fact S is generated as a ring by these f and
i
S ≅k[f ,...,f ]
1 n
as a polynomial ring i.e. no algebraic dependence. Hilbert showed that S is finitely
generated, and for lots of other groups as well. Along the way he proved four
important 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 from
Hilbert’s work the fundamental property that made the Basis theorem work.
Definition1.1. A(commutative)ringRisNoetherianifeveryidealofRisfinitely
generated.
Remark. There are equivalent definitions - you should be reminding yourself of
this for next time.
Noether developed the idealtheorie, the theory of ideals for Noetherian rings -
e.g.primarydecompositionwhichgeneralisesfactorisationintoprimesfromnumber
theory.
Next let me tell you or remind you of the link between commutative algebra
andalgebraicgeometry. Onewayofviewingthefundamentaltheoremofalgebrais
that 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[X ,...,X ], there is a polynomial
1 n
function
f∶Cn→Cn
(a ,...,a )↦f(a ,...,a )
1 n 1 n
Thus we get polynomial functions on affine n-space. Given I ⊂ C[X ,...,X ],
1 n
define
Z(I)={(a ,...,a )∈Cn∶f(a ,...,a )=0 for all f ∈I}
1 n 1 n
thesetofcommonzeroes. SuchasubsetofCn isanalgebraic set. Notethatwecan
replace I by the ideal generated by I without changing Z(I). For a subset S ⊂Cn,
define
I(S)={f ∈C[X ,...,X ]∶f(a ,...,a )=0 for all (a ,...,a )∈S}
1 n 1 n 1 n
This is an ideal of C[X ,...,X ] and moreover it is radical: if fr ∈I(S) for some
1 n
r≥1 then f ∈I(S).
So the Nullstellensatz is really a family of theorems but one way of looking at
it, is that there is a 1-1 correspondance
{radical ideals in C[X ,...,X ]}←→{algebraic subsets of Cn}
1 n
where
I ↦Z(I)
I(S)↤S
in particular, the maximal ideals of C[X ,...,X ] correspond to points in Cn.
1 n
Remark. There is a topology on Cn under which the closed sets are the algebraic
ones - 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[X ,...,X ] is Noetherian
1 n
Quite a large section of the course is about dimension theory. There are at
least 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,andyoucanaskaboutalgebraicdependence-transcendencedegree
of the field of quotients, which says how many transcendence elements to
you need to take.
In the commutative case these all give you the same answer, and in fact there
is a fourth way using homological algebra which for “nice” Noetherian rings, gives
you the same answer again. So it really does make sense to spend a lot of time on
dimension. The case of dimension-0 is field theory and the case of dimension-1 is
number theory.
Mostofthecontentofthiscoursecomesfromaround1920-1950soitis“ancient
history”.
Lecture 2
14th October 10:00
6
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 of
the last lecture, you should be looking over your previous notes. We take R to be
a commutative ring with a 1. Most of you are probably comfortable with rings but
maybenotwithmodules,sowewillstartwithmodules. LetM bealeftR-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) everynon-emptysubsetofsubmodulesofM containsatleastonemaximal
member.
Proof. (1)→(2) Suppose we have a chain
M ⫋M ⫋M ⫋⋯
1 2 3
Then take N = ⋃M . It is a submodule of M. Assuming (1), then N is finitely
i
generated & a finite generating set lies in some M , so N =M , which contradicts
j j
the strict ascent.
(2)→(3)Assumetheascendingchaincondition(ACC).LetS beanon-empty
subset of submodules. Choose M ∈S. If it’s a maximal member, then we’re done.
1
If not, then choose M which is bigger etc. By the ACC, this process stops.
2
(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 maximal
member L, say. Then we just check that L=N, and thus N is finitely-generated.1
(cid:3)
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 is
Noetherian 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
L ⊆L ⊆L ⊆⋯
2 2 3
Now we set
⎧
⎪⎪Q /N =(L +N)/N
⎨ i i
⎪⎪N =L ∩N
⎩ i i
By the ACC there are r,s such that Q /N =Q /N for i≥r and N =N for i≥s.
i r i s
Now we set k =max(r,s). Then you can check for yourself that L =L for i≥k.
i k
[pp. 5 of Wilkin’s notes, much more straightforward]. (cid:3)
Lemma 2.5 (1.5). Suppose that we have modules with
M =M +⋯+M
1 n
(not necessarily direct). Then M is Noetherian iff each M is Noetherian.
i
1SupposeLisgeneratedby{x1,...,xr}andletx∈N,thenthesubmoduleK generatedby
{x1,...,xr,x}isfinitely-generated,sobymaximalitywehaveL=K,sox∈LthusL=N.
2Suppose that M is Noetherian, then N is Noetherian, since any submodule of N is also
a submodule of M and hence finitely-generated. Also any submodule L of M/N is of the form
{N+x∶x∈J}forsomesubmoduleJ ofM satisfyingN⊂J. ButJ isfinitely-generated(sinceM
isNoetherian). Letx1,...,xr beafinitegeneratingsetforJ. ThenL+x1,...,L+xr isafinite
generatingsetforK. ThusM/N isNoetherian.
7
Lecture 2
Proof. The direction “→” is okay. For the “←” direction, if each M is Noe-
i
therian then
M ⊕M ⊕⋯⊕M
1 2 n
isNoetherianandsoM is, sinceitisanimageofM ⊕⋯⊕M underthecanonical
1 n
map
M ⊕⋯⊕M →M +⋯+M (cid:3)
1 n 1 n
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 for
modules 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 =Rm +⋯+Rm
1 n
WehaveR-modulemapsR→Rm givenbyr↦rm . RisNoetherianandsoRm
1 1 1
is. Hence M is Noetherian by lemma 2.5. (cid:3)
Theorem 2.7 (1.7, Hilbert’s basis theorem). Let R be a Noetherian ring. Then
R[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)⊆⋯
Let
R(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)⊆⋯
isanascendingchainandeachR(n)isanidealinR. Theascendingchaincondition
thenyieldsthat⋃R(n)=R forsomeN. WeknowthateachR(0),R(1),...,R(N)
N
is finitely generated. Each R(j) is generated by
a ,...,a
j1 jkj
say, and these are leading coefficients of
f ,...,f
j1 jkj
in I(j). And then the claim (which is for you to check) is that the set
{f ∶j ≤N,1≤k≤k }
jk j
generate I.3 (cid:3)
Remark. In practice one uses Gr¨obner bases for ideals - one generally would like
to get a “nice” set of generators for the ideal that one is dealing with, and these
satisfy this: they are generating sets with added properties that make algorithms
efficient.
3Letg∈Ihavedegreen≤Nandleadingcoefficientb. Thenthereareelementsc1,...,ckn ∈R
such that b=c1an1+⋅⋅⋅+cknankn then g =c1fn1+⋅⋅⋅+cknfnkn+r where degr<degg, so an
inductionargumentworks. Ifghasdegreen>N thentheexpressionyouneedisg=c1xn−NfN1+
⋅⋅⋅+c xn−Nf +r andaninductionargumentworks.
kN NkN
8
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 ={q∈Q∶q is of the form m/n with m,n∈Z, and p∤n}
(p)
(4) The case k[X ,...,X ,...] with infinitely many indeterminates, is not
1 n
Noetherian, consider
(X )⫋(X ,X )⫋⋯
1 1 2
(5) k[X ,...,X ] is Noetherian - this is a corollary of the Basis theorem.
1 n
Z[X ,...,X ] is Noetherian. Any finitely generated ring is Noetherian
1 n
(since such a ring is an image of some Z[X ,...,X ]).
1 n
(6) k[[X]] the ring of formal power series, is Noetherian, things like
a +a X+a X2+...
1 2 3
with a ∈k. There is a more general result
i
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 the
Basis theorem, but by using “trailing coefficients”. I will prove it by a different
method next time. (cid:3)
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. EitherbyusingtrailingcoefficientsbyamethodanalogoustotheBasis
theorem, or use Cohen’s theorem.4
Theorem 3.10 (1.9, Cohen). If every prime ideal is finitely generated then R is
Noetherian.
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 not
finitely generated. By Zorn’s lemma, there is a maximal member I of the family of
non-finitelygeneratedideals(recallthattoapplyZorn’slemma,oneneedstocheck
thatthefamilyisnon-emptyandthattheascendingchainofnon-finitelygenerated
ideals has a union which is non-finitely generated).
Claim: I is prime (& hence we’re getting a contradiction if we’re assuming all
primes to be finitely-generated). Suppose now that there exists a∉I and b∉I but
with ab ∈ I. Then I +Ra is an ideal strictly containing I. The maximality of I
ensures I+Ra is finitely generated by
i +r a,...,i +r a
1 1 n n
say. Let J ={s∈R∶sa∈I}≥I+Rb≩I. So by maximality of I, we know that J is
finitely generated. We will show that
I =Ri +⋅⋅⋅+Ri +Ja
1 n
a finitely generated ideal, a contradiction: take t∈I ≤I+Ra, so
t=u (i +r a)+⋅⋅⋅+u (i +r a)
1 1 1 n n n
4Lemma0306justseemssimplerreally.
9
Lecture 3
for some u ∈R. So u r +⋅⋅⋅+u r ∈J and so it is of the required form. (cid:3)
i 1 1 n n
To use Cohen’s theorem, we apply
c Proposition 3.11. LetP beaprimeidealofR[[X]]andθ isthemapθ∶R[[X]]→
R sending X to 0. Then P is a finitely generated ideal of R[[X]] iff θ(P) is a
finitely generated ideal of R.
Proof. Clearly if P is finitely generated then θ(P) is. Conversely, suppose
that θ(P) is finitely generated with
θ(P)=Ra +⋅⋅⋅+Ra
1 n
If X ∈ P then P is generated by a ,...,a ,X and we are done. If X ∉ P, let
1 n
f ,...,f be power series with constant terms a ,...,a in P. Take g∈P with
1 n 1 n
g=b+...
(where b is the constant term). But b=∑b a , so
i i
g−∑b f =g X
i i 1
forsomeg -notethatg X ∈P. ButP isprimeandX ∉P andsog ∈P. Similarly
1 1 1
g =∑c f +g X
1 i i 2
with g ∈P. Continuing we get h ,...,h ∈R[[X]] with
2 1 n
h =b +c X+...
i i i
satisfying that
g=h f +⋅⋅⋅+h f (cid:3)
1 1 n n
(cid:3)
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=0and ym=0for somem,n thenyou canconvince yourself
that (x+y)n+m+1 = 0, so x+y ∈ N(R). If s ∈ R/N(R) then s = x+N(R) and
sn=xn+N(R). If sn=0 then xn∈N(R) and so x∈N(R) so s=0. (cid:3)
dfn:nilradical Definition 3.4 (1.12). This ideal N(R) is called the 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 ideals
of R.
Proof. Let
I = ⋂ P
P prime
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 that
for n > 0, xn ∉ J. S is non-empty as (0) ∈ S, and we can apply Zorn’s lemma
againtogetamaximalmemberJ ofS (ifR isNoetherian, wedon’thavetothink
1
about Zorn’s lemma).We claim that J is prime. For, suppose that yz ∈ J with
1 1
y ∉ J ,z ∉ J . So the ideals J +Ry and J +Rz strictly contain J and hence
1 1 1 1 1
by the maximality of J , we have xn ∈ J +Ry and xn ∈ J +Rz for some n. So
1 1 1
x2n∈J +Ryz. But since yz∈J , this implies x2n∈J , a contradiction. (cid:3)
1 1 1
10
