Lecture 25.
Conclude the Hilbert Syzygies theorem by
> mention regular local rings,
> extending the theorem to regular local rings,
> the AuslanderBuchsbaum strengthening of the result.
I give a quick and amateurish discussion of this. You can find
higher powered versions in [Ma] and [Ei].
Direction for the last 5 lectures.
Memorable point: Write S = = k[x1,..xN]. A quotient ring A = S/I of
dimension d is CohenMacaulay <=> depth d.
AuslanderBuchsbaum says that A has a free resolution of
length <= codim c = Nd
Moreover A is Gorenstein <=> CohenMacaulay plus the free
resolution ends with a free module of rank 1 S(k).
A couple of examples.
codim 1. I an ideal in A = k[x1,..xn] with dim A/I = n1.
if A/I is CohenMacaulay, then I = (f), so its resolution is
of length 1, so is
A <f A < 0
On the other hand, A/I may have embedded primes, so saying
dim A/I = n1 without CohenMacaulay doesn't given anything  if
A/I has a 0dimensional associated prime (so A/I has a submodule
isomorphic to A/m) then it can only have a resolution of length n.
HilbertBurch
Suppose that a quotient ring A = k[x1,..xn]/I
(1) has codimension 2, and
(2) has a free resolution of length 2
Then the free resolution is necessarily of the form
A <F (d+1)*A <M d*A < 0

v
A/I
where M is a (d+1) x d matrix and F is the vector \Wedge^d M made up by
its d x d minors (in appropriate order and with appropriate signs).
The proof is also memorable: the module we are resolving has rank 1, and
the sequence is exact outside V(I). In particular it is exact after
tensoring with the function field k(x1.. xn). Therefore the ranks are
d+1 and d (so that the alternating sum of 1,d+1,d comes to 0). Next, still
arguing over k(x1.. xn), the cokernel of a (d+1) x d matrix of rank d is
given by its dxd minors (Cramer's rule) up to proportionality. If the
constant of proportionality is not a unit, then A/I has a associate prime
of dimension n1, which contradicts the assumption that dim A/I = n2.
====
CohenMacaulay and Gorenstein rings
Preliminary section introducing Ext_A^r(M,N)
Q. Why the name Ext? Ext is about the Hom functor
Hom_A(M,?) is a covariant left exact functor, as we have seen:
If 0 > N1 > N2 > N3 > 0 is a ses then
0 > Hom_A(M,N1) i> Hom_A(M,N2) p> Hom_A(M,N3)
is exact. The cokernel of p at the right end measures
_extensions_ of M by N1.
I describe this briefly. Given the ses and a homomorphism
f: M > N3, consider the set theoretic fibre product
Ef in N2 + M defined by Ef = { n,m  p(n) = f(m) }.
E has projections into the two summands, and fits into
a cartesian square. One sees that the kernel of E > M
is a copy of N1 in E fitting into an extended ses
0 > N1 > Ef > M > 0 (*)
  f
v v
0 > N1 > N2 > N3 > 0
called an _extension_ of M by N1. The construction continues:
one sees that
> a lifting of f to N2 gives a splitting of the upper ses: Ef = M + N1.
> the upper sequence is split if and only if f lifts to N2.
> the set of extensions 0 > N1 > E > M > 0 (up to isomorphism of
exact sequences) can be given the structure of an Amodule Ext^1(N1,M),
by multiplying the homomorphisms by elements of A and taking pullbacks.
> taking f in Hom_A(M,N3) to the extension Ef defines a continuation
of the exact sequence
0 > Hom_A(M,N1) > Hom_A(M,N2) > Hom_A(M,N3)
> Ext^1_A(M,N1)
and this continues.
Continuing this construction in explicit terms is possible, but
would be long and tedious. Instead, we give a more systematic
treatment in homological algebra.
====
Q. What are Ext^i good for?
The functors Ext^i limit the depth  e.g. as we have seen
Hom(A/m, M) <> 0 implies depth = 0.
As first consequence we get that lots of Ext^i = 0 is equivalent
to depth > something.
Q. Definition of Ext_A^r(M,N) via projective resolution of M
More systematically, the functors Ext^i are right derived functor
of Hom_A(M,?)
Partly answers the old question: why do all that business about
complexes and free resolutions?
The connection with projective resolutions: if P is projective, a
surjective map P > M turns the composite with P > M > N3 into a
morphism with a splitting P > N2, together with worries about the
kernel of where the kernel of P > M goes. If we replace M with a
projective resolution up to d terms, we get the derived functors
Ext_A^i(M, ?) for i0.
(3) Ext^i_A(M,N) is contravariant in M, and covariant in N.
(4) s.e.s. of Ns gives rise to long exact cohomology seq.
P. > M a projective resolution (up to n terms, no need
to assume finite for the moment)
Define Ext^i_A(M,N) := H_i(Hom_A(P.,N)).
because it Hom is contravariant in M, the indices increase
Hom(Pi, N) > Hom(P_{i+1}, N) > ..
so we write H_i.
1. Ext^0 = Hom
2. If M is already projective Ext^i(M,N) = 0 for all i.
If M is projective, we can just take P. to be P0 = M and
nothing more, so all Ext^i(M,N) = 0 for all i.
More formally, we need to know that the definition is independent of
the choice of P. Any two projective resolutions are homotopy
equivalent. (Exercise, omit discussion.)
3. Nothing to prove
4. Let 0 > N1 > N2 > N3 > 0 be a s.e.s.
Then each Pi is projective, so the sequence
0 > Hom(Pi, N3) > Hom(Pi, N2) > Hom(Pi, N1) > 0
is exact. Therefore the 3 complexes Hom(P.,Ni) fit together as a
s.e.s. of complexes, and the l.e.s. comes by the usual snake lemma
argument.
====
There is a very similar treatment of Ext^r(M,N) using _injective_
resolutions of N. The definition is "dual" to projective (in the
categorical sense "just reverse the arrows").
Definition: An Amodule I is injective if the contravariant functor
Hom_A(?,I) is exact.
A restatement is that whenever M1 into M2 is injective, every
Ahomomorphism f: M1 > I is the restriction of some M2 > I, or
f _extends to_ M2.
An Appendix below discusses what injective modules are, and gives a
proof that every module M over a ring is a submodule of an injective
module. However, you should treat this as an existence proof. (It is
not comparable to the process of finding a free resolution by writing
generators of a module and relations between them.)
For an Amodule N, there exists an injective resolution: a complex
I.: 0 > I0 > I1 > .. > In .. (not assumed to be finite) such that
> each In is an injective Amodule,
> the complex is exact at each i > 0, and
> H^0(I.) = N.
In other words, there is an inclusion N into injective I0 then an
inclusion of (I0/N) into injective I1, and so on indefinitely
Now Ext^r(M,N) is the homology H_r(M,I.). This construction has a list
of properties similar to the above, including a long exact sequence for
s.e.s. in the M variable (Hom(?,N) covariant). That the 2 constructions
give the same groups is left as an exercise in [Ma], Appendix B, p. 277.
====
There is an interplay between
depth M.
number of initial Ext^i(A/I,M) = 0.
number of final H_{ni}(K.(y1,..yn,M)) = 0 in Koszul complex.
The basic point to remember is that a finite module M over a Noetherian
ring A has Idepth zero if and only every f in I is a zerodivisor, and
this happens iff there is a prime P in V(I) with P in Ass M. This holds
because every zerodivisor of M is contained in an associated prime of M,
of which there are finitely many, and I contained in a finite union of
primes implies I is contained in one of the primes. So
Idepth M = 0 <=> Hom(A/P,M) <> 0 for some P in Ass M.
If A,m is local then
mdepth M = 0 <=> M contains a submodule isomorphic to the field A/m.
[Ma] Theorems 16.616.7 characterise depth of a finite module in terms
of Ext^i:
The Idepth of M is defined as the maximal length of a Mregular
sequence (x1,.. xd) contained in I.
This is equal to the number of initial Ext^i(A/I, M) that vanish
(for i = 0,.. d1).
A ring and I an ideal. M a finite Amodule.
For given d, equivalent conditions:
[Ma] Theorem 16.6
A Noetherian ring, I its ideal and M a finite Amodule.
Given d > 0, equivalent conditions:
(1) Ext^i(N,M) = 0 for i = 0,..d1
for EVERY finite Amodule N with Supp N contained in V(I).
(2) Ext^i(A/I,M) = 0 for i = 0,..d1.
(2') Ext^i(N,M) = 0 for i = 0,..d1
for SOME Amodule N with Supp N = V(I)
(3) M has Idepth M >= d
(<=> exists an Mregular sequence (y1,..yd) in I).
See below for reminder on Supp and Ass.
Assume N has property (2'). The main claim is i=0, Hom(N,M) = 0
Idepth M > 0. If Idepth M = 0 then every element of I is a
zerodivisor of M. Every zerodivisor of M is contained in an
associated prime P in Ass M. This is a finite set { Pi }, so
I contained in Union Pi, which implies I contained in some P.
Now the N has N_P <> 0, and a bit of localisation that I omit
([Ma] p. 129130) shows eventually that Hom_A(N,M) <> 0,
contradicting (2'). Hence there is an Mregular element x1 in I.
The result holds for M/x1M by induction, so (2') implies (3).
For the easier converse (3) implies (1), see [Ma] p. 130.
====
Summarise results in [Ma] about depth versus Exts.
[not finished]
A,m is local and N,M are finite Amodules
Suppose dim N = r and Ext^s(N,M) <> 0.
Then depth M <= r+s.
Step 1.
The basic case is r = 0 and s = 0, when statement reduces to
N is 0dimensional and phi: N > M is nonzero
then phi(N) subset Ass M, so depth M = 0.
Step 2. In the statement, we can assume N = A/P with P
a prime.
By primary decomposition N has a finite descending filtration
with Nj/N_{j+1} = A/Pj so dim A/Pj <= r. [Ma], Theorem 6.4 or
[UCA], Theorem 7.6.
Now by the long exact sequence of Exts, if Ext^s(N,M) <> 0
one of the A/Pj has Ext^s(A/Pj, M) <> 0.
Step 3. If N = A/P and dim N = r > 0, choose x in P notin m,
and do 0 > N x> N > Nbar > 0. Then dim Nbar <= r1.
The exact sequence of Exts (remember, contravariant and
coboundary does i > i+1) etc. by induction.
====
Intro to CM and Gor
Let A, M be local. One of the characterisations of dimension:
a _system of parameters_ (s.o.p.) is a sequence x1,.. xn in m
that generate an mprimary submodule. This means that
M/(x1,.. xn)M is an Artinian module, so of finite length or
zero dimensional. Then dim M = min length of a s.o.p.
We define A is CohenMacaulay if it has mdepth A = n = dim A.
Thus A has a s.o.p. of length n = dim A that is a regular
sequence. In geometric terms, we can cut A down by a regular
sequence to an Artinian quotient ring, with each step the
quotient by a principal ideal.
The context:
S polynomial ring k[x1,..xN] with maximal ideal m = (x1,..xN),
either graded or localised at m. We can think of S as functions
on the ambient space Spec S = AA^N.
Much the same applies to S a regular local ring of dimension N.
The ring A = S/I is the main object of study. In geometric situations
we can think of it as functions on a subvariety V(I) in AA^N.
If A has dim n, we can think of cutting it down by a sequence of
parameters A/(x1,.. xn) to get to the Artinian quotient A/(x1,.. xn),
or V(I) intersect (x1=.. xn=0) is zero dimensional.
In this context, CohenMacaulay means x1,..xn is a regular sequence.
[Gorenstein means that plus a bit.]
The module A/(x1,..xn) of course depends on the regular s.o.p. you
choose  for example, we should be able to do the exercise of
proving that
l( A/(x1^s,x2,..xn) ) = s*l( A/(x1,x2,..xn) )
has length s times l(A/x1,..xn). However, the point is that the
condition that the s.o.p. is a regular sequence is independent of
the choice.
Nor does the length, or dimension over k = A/m of the final
Hom_A(k,A/(x1,x2,..xn) = Ext_A^n(k,A).
CohenMacaulay ring: dimension n, and there exists a s.o.p. y1,.. yn
for A that is a regular sequence, so that also
=> depth A = n
The main result is that if this holds, every s.o.p. is a regular
sequence.
CohenMacaulay module: either M = 0, or dim M = n there exists a
s.o.p. y1,.. yn for M that is a regular sequence
The Gorenstein condition is the additional requirement that
the Artinian quotient N = A/(x1,..xn) has socle of length 1.
Here the socle of a module M is the set of elements g in M
with Ann g = m. This is the same thing as Hom_M(A/m, M), or
the biggest possible kvector subspace of M.
Consider A = S/I or M some Amodule.
how I is generated, free resolution of A, compared to codimension
c = Nn
system of parameters of A is defined as
(y1,.. yn) in m that generate mprimary ideal.
(one of the equivalent definitions of n = dim A).
Note the two different steps: from S to quotient ring A, then from
A to the Artinian quotient A/(x1,..xn). At the start, m is generated
by the regular sequence x1,.. xN
"unmixed" ideals. Take an ideal I = (x1,.. xr), and suppose that its
height is r. (That is, every minimal prime of I has height r, or
equivalently, no P in Ass A/I has dim A/P < r  compare the
counterexamples in Lect1822, p.68.) CohenMac or exactness of
complexes only depends on dimension or height.
Aim: Macaulay's unmixedness theorem
depth is welldefined, independent of the choice of regular sequence
Duality: Socle of the Artinian quotient is independent of regular sequence
CohenMacaulay comes with a duality theory based on Ext_A^n(A,?)
Gorenstein
Artinian quotient Abar = A/(y1,.. yn)
Socle of Abar: Hom_A(A/m, Abar) = { a in Abar  ma = 0 }
= biggest kvector space in Abar.
============
Appendix on Supp M and Ass M
Intended as reminder/clarification. See [UCA] Chap 7 or [Ma] 6.4
A module M gives subsets of Spec A
Supp M = { P  M_P <> 0 }
That is, some a in M survives multiplication by every s notin P.
Recall V(I) = { P  P contains I } in Spec A (any Zariski closed set).
1. For a cyclic module, Supp(A/I) = V(I).
2. For a finite module Supp N = V(Ann N) = V(rad(Ann N)).
3. Supp(N) contained in V(I) iff I contained in rad(Ann(N)).
Proof of 1. Obviously, I = Ann(1 in A/I).
If P contains I then s notin P implies s notin I, so multiplication
by s takes 1 in A/I to a nonzero element of A/I. Localisation makes
S = AP invertible, so 1 survives multiplication by every s notin P,
that is, (A/I)_P = S^1(A/I) <> 0.
Conversely, if P does not contain I then some s in I notin P
and multiplication by s kills A/I.
Proof of 2.
Write N = sum A*n_i with generators ni.
Then Supp N = Union Supp A*ni = Union V(Ann n_i)
= V( Intersection Ann ni) = V(Ann N).
Proof of 3.
If J = Ann(N) then Supp(N) = V(J). Now V(J) contained in V(I)
is equivalent to I contained in rad(J).
Ass M is defined as the set of primes P such that M contains a copy of
the integral domain A/P. If M is finite and <> 0, this is a finite and
nonempty set. Every element a in A that is a zerodivisor for M is
contained in one of the P in Ass M.
============
Appendix on injective modules
Some details on their properties and existence
Summary: For ZZmodules, the inverse ptorsion modules (ZZ[1/p])/ZZ
are injective, and every ZZmodule M embeds into a product of these
(usually infinite). View any ring R as a ZZalgebra; then for an
injective ZZmodule I, the ZZmodule Hom_ZZ(R, I) becomes an
Rmodule under premultiplication, and is an injective Rmodule. For
an Rmodule M, view M as a ZZmodule and embed it into an injective
ZZmodule I. An inclusion of M into an injective Rmodule is then
provided by the tautological identity
Hom_ZZ(M, I) = Hom_R(M, Hom_ZZ(R, I)).
This is mostly cribbed from Charles A. Weibel, An introduction to
homological algebra, CUP 1994.
====
Definition An Rmodule I is injective if Hom_R(, I) is an exact
functor. This means that if
0 > M1 > M2 is exact (that is, M1 embeds in M2)
then any homomorphism e: M1 > I has some extension f: M2 > I.
Extension means that f gives the same value as e on M1 in M2.
Example. A kvector space V is an injective kmodule. This means
that for U in W vector spaces, a klinear map U > V extends to W.
(This requires Zorn's lemma.)
Proposition To guarantee that a ZZmodule I is injective, it is
enough to prove that any homomorphism e from an ideal J in R to I
extends to a homomorphism f: R > I. (This needs Zorn's lemma.)
Proof Let M in N be an inclusion of Rmodules and e: M > I. Suppose
that it has been extended to e': M' > I for some M' with M in M' in
N (start from M = M').
If b in N \ M', the extension from ideals allows me to
extend e' to e": (M' + bR) > I. For this, define
J = { r in R s.t. br in M'}. Then J is an ideal of R.
The homomorphism J > I given by
r > br > e'(br)
is defined on J, so extends to R as a homomorphism
f: R > I. Then e": (M' + bR) > I is defined as
e' on M' and b > f(1). In more detail,
e": (m+br) > e'(m) + f(r).
[To check welldefined: If some different m1, r1 has
m+br = m1+br1 then
b(rr1) = mm1 in M', so rr1 in J where the
extension f started, so f(rr1) = e'(mm1).] QED
Corollary An Abelian group (a ZZmodule) is
injective iff it is divisible. The modules
QQ and (ZZ[1/p])/ZZ are injective, and any
injective is a direct product of these. I refer to
(ZZ[1/p])/ZZ as the inverse ptorsion module, by
analogy with Macaulay's inverse monomials. Check that
QQ/ZZ is the direct product of (ZZ[1/p])/ZZ taken over
all primes p. An analogous construction works for a
PID.
Lemma For a ZZmodule M and nonzero m in M, there
exists a prime p and a homomorphism
f: M > (ZZ[1/p])/ZZ with f(m) <> 0.
Proof The annihilator of m in M is an ideal (n) in ZZ
so mZZ iso ZZ/n. Choose any prime p  n and a
surjective map ZZ/n >> ZZ/p (if n = 0 then any prime p
works). Compose with an embedding ZZ/p in (ZZ[1/p])/ZZ
and extend from mZZ to the whole of M using injectivity
of the module. QED
Corollary Consider the set of all homomorphisms from M
to the injective modules (ZZ[1/p])/ZZ. The direct sum
of all these homomorphisms in an embedding of M into an
injective ZZmodule.
Now for modules over a general ring R (you need a
little care about left and right modules if R is
noncommutative). R is a ZZalgebra. If A is a
ZZmodule, the ZZmodule Hom_ZZ(R, A) becomes an
Rmodule under premultiplication. Namely, r acts on
Hom_ZZ(R, A) by f > fr, where fr is the map
s > f(sr) for s in R. That, multiply before applying
the map (IMPORTANT). One checks the following points:
1. The identity
Hom_ZZ(M, A) = Hom_R(M, Hom_ZZ(R, A)).
holds for an Rmodule M and ZZmodule A.
2. If I is an injective ZZmodule then Hom_ZZ(R, I)
is an injective Rmodule.
3. It follows that for any Rmodule M, the product of
all homomorphisms M > Hom_ZZ(R, (ZZ[1/p])/ZZ) is an
embedding of M into an injective Rmodule.
Finally, let F be a sheaf of OXmodules over a ringed
space X. For P in X, the stalk F_P is an OX_Pmodule.
Embed each stalk F_P into an injective OX_Pmodule I_P.
This defines an OXhomomorphism of F into the sheaf of
discontinuous sections of DisjointUnion I_P, which is
an injective sheaf.