pragmatic 2

C: The Homological Approach

Before launching into this discussion I want to say a few words about homomorphisms. We’ve been concentrating on graded structures (in fact, graded by either ℤ or ℕ) but I’ve not said much about the maps between these objects. Of course there is the map from R to R/I, when I is a homogeneous ideal and that map has the property that it takes forms in R of degree d to forms in R/I of degree d. That will be the type of homomorphism that we want to think about.

Let M = ⊕_{i∈ℤ}M_{i} and N = ⊕_{i∈ℤ}N_{i} be graded A-modules (where A = ⊕_{i∈ℕ}A_{i} is a standard graded k-algebra). An A-module homomorphism f : M → N is said to be graded of degree d if

f(M_{i}) ⊆ N_{i+d} for all i.

Clearly, if we replace N by N(d) then

f : M_{i} → (N(d))_{i}

is then of degree 0.

Now if M = ⊕_{i∈ℤ}M_{i} and M′ = ⊕_{i∈ℤ}M′_{i} then M ⊕ M′ has a natural gradation by

(M ⊕ M′)_{i} = M_{i} ⊕ M′_{i}.

However, when dealing with graded modules, certain maps arise which don’t seem obviously graded although we would like them to be so. Consider the following simple example which explains what I have in mind.

Example:

Let F and G be forms of degrees a and b respectively (a ≠ b) in R = k[x_{0}, …, x_{n}] and let I = (F, G). I is a homogeneous ideal, hence a graded module, as is R and R ⊕ R.

Let ϕ : R ⊕ R ⟶ I where ϕ(A, B) = AF + BG. It is clear that the kernel of ϕ is a submodule of R ⊕ R which is generated by pairs (M, N) where M and N are forms of R. But, these forms need not have the same degree and so the kernel of ϕ is not a graded submodule of R ⊕ R with the usual grading.

If however, we view ϕ as a map:

ϕ : R(−a) ⊕ R(−b) ⟶ R

all turns out well. E.g. the kernel element (G, −F) is now an element of degree a + b. In fact, the basis element e_{1} = (1, 0) of R(−a) ⊕ R(−b) is seen to be an element of degree a while the second basis vector e_{2} = (0, 1) is an element of degree b and ϕ maps e_{1} to F and e_{2} to G. In fact, ϕ is now a map of degree 0.

We will use this “trick” often in what follows.

We now turn to a quick look at the homological approach to the study of commutative algebra and algebraic geometry. As with so much in these areas, the study begins with an observation of Hilbert.

Let I = (F_{1}, …, F_{r}) be a homogeneous ideal in the polynomial ring R = k[x_{0}, …, x_{n}] where the F_{i} are forms of degree d_{i}.

Definition: A syzygy on {F_{1}, …, F_{r}} is an element of R^{r} (call it (G_{1}, …, G_{r}) ) such that

∑_{i=1}^{r} G_{i}F_{i} = 0.

So, syzygies are nothing more than the elements of the kernel of the map

ϕ : R^{r} ⟶ I given by ϕ(H_{1}, …, H_{r}) = ∑_{i=1}^{r} H_{i}F_{i}

Since the F_{i} are forms it is easy to see that the kernel of ϕ is generated by syzygies of the form (G_{1}, …, G_{r}) where the G_{i} are also forms in R. So, if we use the “trick” mentioned above, i.e. consider

ϕ : R(−d_{1}) ⊕ ⋯ ⊕ R(−d_{r}) ⟶ I where ϕ(e_{i}) = F_{i}

then Ker ϕ is actually a graded submodule of F_{0} = ⊕_{i=1}^{r}R(−d_{i}). That submodule is called the module of first syzygies of {F_{1}, …, F_{r}}, denoted Syz_{1}{F_{1}, …, F_{r}} ⊂ F_{0}.

Considered as a graded submodule of F_{0} the module of first syzygies has a set of homogeneous generators H_{1}, …, H_{ℓ} ( which we will write as column vectors) i.e.

H_{i} = ( G_{i1} ; ⋮ ; G_{ir} ) where deg H_{i} = δ_{i}

It’s clear that we can now repeat the process and define a map ϕ̄

ϕ̄: ⊕_{i=1}^{ℓ}R(−δ_{i}) ⟶ F_{0}

where

ϕ̄: ( S_{1} ; ⋮ ; S_{ℓ} ) → ( \begin{matrix}G_{11} & ⋯ & G_{ℓ1} \\ ⋮ & ⋮ & ⋮ \\ G_{1r} & ⋯ & G_{ℓr} \end{matrix} ) ( S_{1} ; ⋮ ; S_{ℓ} )

and then consider the kernel of this map.

That submodule of F_{1} = ⊕_{i=1}^{ℓ}R(−δ_{i}) will be called the module of second syzygies of {F_{1}, …, F_{r}} and denoted Syz_{2}{F_{1}, …, F_{r}}.

One continues in this fashion and unbelievably one has:

Theorem: (Hilbert’s Syzygy Theorem)

Syz_{n+1}(F_{1}, …, F_{r}) is always a free module.

Given the tremendous flexibility one has in choosing generators at each stage, one might wonder:

Does it always take as many as n + 1 steps to get a free module?

The answer to this question is obviously no. Consider I = (F) a principal ideal, where the degree of F is d. We have

0 → R(−d) ⟶^{ϕ} I → 0

where ϕ(e_{1}) = F and the kernel is 0, hence free.

A less trivial example is given by the following: let F_{1} and F_{2} be forms of degrees d_{1} and d_{2} in R = k[x_{0}, …, x_{n}]. Suppose (for the moment) that they have no common factor.

Define ϕ_{1} : R(−d_{1}) ⊕ R(−d_{2}) → I = (F_{1}, F_{2}) where ϕ_{1}(e_{i}) = F_{i}.

It is a simple matter to show that (F_{2}, −F_{1}) generates Ker ϕ and that this is an element of degree d_{1} + d_{2}. So Ker ϕ is a free module and we have the exact sequence of free modules

0 → R(−(d_{1} + d_{2})) ⟶^{ϕ_{2}} R(−d_{1}) ⊕ R(−d_{2}) ⟶^{ϕ_{1}} (F_{1}, F_{2}) → 0

where ϕ_{2}(ē_{1}) = (F_{2}, −F_{1}), and ē_{1} is the standard basis vector for R(−(d_{1} + d_{2})).

Another interesting (and simple) example to consider is the following:

Let I = (x^{2}, xy) ⊂ k[x, y, z] = R. We have ϕ_{1} : R(−2)^{2} → I by ϕ_{1}(e_{1}) = x^{2} and ϕ_{1}(e_{2}) = xy. In this case it is easy to see that ker ϕ_{1} is generated by (y, −x) which has degree 3.

So, we have an exact sequence of free modules:

0 → R(−3) ⟶ R(−2)^{2} ⟶ (x^{2}, xy) ⟶ 0

It is worth noting in these examples that both ideals have their module of first syzygies free despite the fact that the first ideal is generated by a regular sequence and the second is not. Note also that the second ideal has height 1 while the first ideal has height 2.

Another obvious question one could ask is:

does the first time a free module of syzygies appears depend on the choice of the generators at each step?

Again the answer is no, but this time the reasoning is a bit less obvious. (A proof, in the case of graded modules over rings like R, follows from, e.g. Schanuel’s lemma and the fact that a graded module which is a direct summand of a free graded module is itself free. This last uses the graded version of Nakayama’s Lemma.)

The fact that the answer is NO gives us a new integer invariant which we can attach to a f.g. graded R-module. To be more precise:

Definition: Let M be a f.g. graded R-module (R = k[x_{0}, …, x_{n}]). We say that the homological dimension of M is d, and write h.dim_{R} M = d, if:

1) There is an exact sequence of free modules

0 → F_{d} ⟶^{∂_{d}} ⋯ → F_{1} ⟶^{∂_{1}} F_{0} ⟶^{∂_{0}} M → 0  (∗)

2) There is no shorter such exact sequence.

Remarks:

1) An exact sequence like (∗) above is called a (graded) free resolution of M.

2) If

0 → M_{1} ⟶^{ϕ_{0}} M_{2} ⟶^{χ_{0}} M_{3} → 0

is an exact sequence of f.g. graded R-modules with maps which are homogeneous of degree 0, then there are some standard ways to construct resolutions for one of these modules given resolutions for the other two.

E.g. the Horseshoe Lemma permits us to build, on the direct sum of the modules in a resolution for M_{1} and M_{3}, a resolution for M_{2}. The maps in this resolution are not quite what one might expect however.

Another construction, called the Mapping Cone Construction builds a resolution of M_{3} from those of M_{1} and M_{2}. The idea, briefly, is the following:

let F →^{{d}} M_{1} → 0 be a free resolution of M_{1} and G →^{{∂}} M_{2} → 0 be a free resolution of M_{2}, then one always can find a sequence of maps ϕ_{i} : F_{i} → G_{i} which are built over the map ϕ_{0} from M_{1} to M_{2} and which make all the appropriate diagrams commute.

Define H_{i} = F_{i−1} ⊕ G_{i} and δ_{i} : H_{i} → H_{i−1} where

δ_{i} = ( \begin{matrix}−d_{i−1} & 0 \\ ϕ_{i−1} & ∂_{i} \end{matrix} )

Then the H_{i} and the δ_{i} give us a free resolution of M_{3}.

3) With M_{1}, M_{2} and M_{3} as above, there are relations between their homological dimensions. More specifically,

a) If h.dim_{R}M_{2} > h.dim_{R}M_{1} then h.dim_{R}M_{3} = h.dim_{R}M_{2};
b) If h.dim_{R}M_{2} < h.dim_{R}M_{1} then h.dim_{R}M_{3} = h.dim_{R}M_{1} + 1;
c) If h.dim_{R}M_{2} = h.dim_{R}M_{1} then h.dim_{R}M_{3} ≤ h.dim_{R}M_{2} + 1.

It follows from this that:

Proposition: If I is an ideal of R then

h.dim_{R}(R/I) = h.dim_{R}(I) + 1.

(It’s enough to consider the exact sequence 0 → I → R → R/I → 0 and consider separately the cases where I = (F) is free and I is not free).

4) Using the graded version of Nakayama’s Lemma, we have a unique notion for the number of generators in a minimal set of generators for an R-module (the notion could have meant either of two things: “fewest members” of any generating set, or “a set of generators which cannot be shortened”).

So, given a f.g. R-module M, the graded R/m-vector space (M/mM), has a basis of vectors with certain degrees which are the degrees of a minimal set of generators for M. So, although a minimal generating set for M is not unique, the number of and the degrees of the generators in it are uniquely determined by M.

If we build a resolution for M like (∗) above, but choosing at each step a minimal set of generators for the resulting module of syzygies, we get what is called a minimal free resolution for M.

There is a way to recognize a minimal free resolution: The maps between the free modules in the resolutions can (after we choose the usual bases) be represented by matrices whose entries are forms of R. Then, a resolution like (∗) above is minimal if and only if all the entries in the matrices are in the (irrelevant) maximal ideal. I.e. no units of R can appear in the matrices.

5) In the case of a minimal free resolution, if we write each F_{i} = ⊕R(−j)^{β_{ij}}, then the exponents β_{ij} that arise are a very subtle set of invariants of the module M. These invariants are called the graded Betti numbers of M. Obviously, if one knows all the graded Betti numbers of M, one knows the Hilbert function of M and hence the graded Betti numbers encode all the invariants that come from the Hilbert polynomial, and more.

Of course, having invariants is a fine thing – but useless if you can never calculate them and modules of syzygies give the impression that they are unapproachable. However, nothing could be further from the truth, in fact (in my opinion) one of the most significant advances in commutative algebra and algebraic geometry of the last 50 years has been the advent of computational methods which, for the first time, actually permit us to calculate all these invariants. (You’ll have seen in the demos of CoCoA and Macaulay how this actually works out in practice.)

So, in a real sense, these invariants are “new” ones and their geometric interpretations have yet to be found (although M. Green and D. Eisenbud and S. Goto have made a number of interesting conjectures about them). This has been the new research of the latter half of the 20th century (naturally with roots reaching back for over a 100 years).

There are actually a few cases where it is possible to write down the minimal free resolution for an ideal I without recourse to a computer.

Example: (The Koszul Resolution)

Let R = k[x_{0}, …, x_{n}] and I = (G_{1}, …, G_{r}) be forms with deg G_{i} = d_{i} such that the G_{i} are a regular sequence. In this case we have the following minimal free resolution for I:

0 → F_{r−1} ⟶^{ϕ_{r−1}} ⋯ → F_{1} ⟶^{ϕ_{1}} F_{0} ⟶^{ϕ_{0}} I → 0

where, if we use e_{i} to denote the usual basis vectors for R(−d_{i}), then

\begin{aligned}F_{0} = ⊕_{i=1}^{r}R(−d_{i}) & where & ϕ_{0}(e_{i}) = G_{i} \\ F_{1} = ⋀^{2}(F_{0}) & = & ⊕_{1≤i_{1}<i_{2}≤r}R(−(d_{i_{1}} + d_{i_{2}}) \\ F_{2} = ⋀^{3}(F_{0}) & = & ⊕_{1≤i_{1}<i_{2}<i_{3}≤r}R(−(d_{i_{1}} + d_{i_{2}} + d_{i_{3}})) \\ ⋮ & ⋮ & ⋮ \\ F_{r−1} = ⋀^{r}(F_{0}) & = & R(− ∑_{i=1}^{r} d_{i}) \end{aligned}

The maps in the resolution are not very hard to describe and you will find them in any book on homological algebra. I’ll just write down ϕ_{1} and ϕ_{r−1} and explain what they do to decomposable vectors (which is enough to describe them in general):

ϕ_{1}(F e_{i} ∧ e_{j}) = F [ϕ_{0}(e_{i})e_{j} − ϕ_{0}(e_{j})e_{i}] = F [G_{i}e_{j} − G_{j}e_{i}]

and

ϕ_{r−1}(F e_{1} ∧ ⋯ ∧ e_{r}) = ∑_{i=1}^{r}(−1)^{i+1}G_{i}F e_{1} ∧ ⋯ ∧ ē_{i} ∧ ⋯ e_{r}.

It follows from this that if I is generated by a regular sequence of r forms then

h.dim_{R}(I) = r − 1 and h.dim_{R}(R/I) = r

The connection between homological dimension and what we have seen before is provided by the fantastic theorem of M. Auslander and D. Buchsbaum. I’ll state it only for the polynomial ring R although they proved it in much more generality for arbitrary standard k-algebras and even for local rings.

Theorem: (Auslander and Buchsbaum) Let R = k[x_{0}, …, x_{n}] and let I be a homogeneous ideal of R. Then

h.dim_{R}(R/I) + depth (R/I) = n + 1.

Remarks: There are many consequences of this theorem. Let me mention a few.

1) Since depth (R/I) ≤ K-dim(R/I) = (n + 1) − ht.I we have that

n + 1 ≤ h.dim_{R}(R/I) + K-dim(R/I) = h.dim_{R}(R/I) + (n + 1) − ht.I .

Thus

ht. I ≤ h.dim_{R}(R/I) = h.dim_{R}I + 1

2) If we take the equality in the last remark, i.e. if we have

ht. I = h.dim_{R}I + 1 i.e. ht. I = h.dim_{R}(R/I)

then

ht. I + depth (R/I) = n + 1.

But, ht. I = (n + 1) − K-dim(R/I). Thus we have depth (R/I) = K-dim(R/I), i.e. R/I is a CM ring.

Since all the steps are reversible in this last argument we have:

Theorem: With the hypotheses above:

R/I is a CM ring ⇔ ht. I = h.dim_{R}I + 1 = h.dim_{R}(R/I).

3) We can now see very clearly the difference between the two examples we looked at before.

When I = (F, G) is generated by two forms which form a regular sequence we showed that h.dim_{R}I = 1 and this then implies that h.dim_{R}(R/I) = 2. Since ht. I = 2 we get that the K-dim(R/I) = (n + 1) − 2 = n − 1. By the A-B theorem we get that depth (R/I) = (n + 1) − 2 also. Thus R/I is a CM ring.

In the other example we considered earlier i.e. where I = (x^{2}, xy), we also have h.dim_{R}I = 1 and so h.dim_{R}(R/I) = 2. But this time ht.I = 1 and so K-dim(R/I) = 3 − 1 = 2 and not 3 − 2 = 1. Thus k[x, y, z]/(x^{2}, xy) is not a CM ring.

Tensor Products and Tor Modules

Recall the definition of M ⊗_{R} N where M and N are f.g. R-modules. When M and N are graded then so is M ⊗_{R} N. In fact, (M ⊗ N)_{t} is generated, as an abelian group, by products m ⊗ n where m and n are homogeneous and deg(m) + deg(n) = t.

Some important things to recall are:

(R/I) ⊗_{R} M ≃ M/IM

and

R(−j) ⊗_{R} M ≃ M(−j)

Now, let

F : ⋯ → F_{i} ⟶^{ϕ_{i}} F_{i−1} → ⋯ → F_{1} ⟶^{ϕ_{1}} F_{0} → M → 0

be a minimal free resolution for M and let N be any graded R-module. Then the complex F ⊗ N is defined to be:

F ⊗ N := ⋯ → F_{i} ⊗ N ⟶^{ϕ_{i}⊗id} F_{i−1} ⊗ N → ⋯ → F_{1} ⊗ N ⟶^{ϕ_{1}⊗id} F_{0} ⊗ N → 0.

Definition: The cohomology modules of this complex, i.e.

Ker(ϕ_{i} ⊗ id)/Im(ϕ_{i+1} ⊗ id.)

are called T or_{i}^{R}(M, N).

Remarks:

1) We could just as well have started the procedure above with a free resolution of N (rather than M), tensored that resolution with M and then taken cohomology. It is not at all clear, but it is one of the important theorems of the subject, that the resulting cohomology modules don’t change. This means, in practice, that we have two different ways to calculate the module T or_{i}^{R}(M, N) and that is sometimes extremely useful.

2) It is an easy to see that

T or_{0}^{R}(M, N) ≃ M ⊗_{R} N.

3) If f : M → M′ then f induces a map f_{i} : T or_{i}(M, N) → T or_{i}(M′, N) for each i. If f is multiplication by a form of R, then f_{i} is also multiplication by that same form.

4) (Long Exact Sequence of Tor’s) If we are given a short exact sequence of graded Rmodules with homomorphisms graded of degree 0 we obtain a long exact sequence of Tor’s. More precisely, if

0 → M_{1} ⟶^{f} M_{2} ⟶^{g} M_{3} → 0

is a s.e.s. of graded R-modules with graded homomorphisms of degree 0 then we have a long exact sequence

⋯ → T or_{i}^{R}(M, M_{1}) ⟶^{f_{i}} T or_{i}^{R}(M, M_{2}) ⟶^{g_{i}} T or_{i}^{R}(M, M_{3}) ⟶^{∂_{i}} T or_{i−1}^{R}(M, M_{1}) → ⋯

⋯ → T or_{1}^{R}(M, M_{3}) ⟶^{∂_{1}} M ⊗ M_{1} ⟶ M ⊗ M_{2} ⟶ M ⊗ M_{3} → 0

5) Notice that if we carry out the construction for T or_{i}^{R}(M, N) with N = R/m ≃ k then the maps in the complex F ⊗ k all become the zero map (all the entries in all the matrices were in m) and all that remains are the graded vector spaces F_{i} ⊗ k.

I.e. T or_{i}^{R}(M, k) = F_{i} ⊗ k and dim_{k}[T or_{i}^{R}(M, k)]_{ℓ} = β_{iℓ} (as defined earlier). So, the graded betti numbers are simply the dimensions of certain Tor modules.

One of the more important relationships between free resolutions occurs when one considers what Juan has called “hyperplane sections”. The algebraic version of what Juan has used is:

Theorem: (Change of Rings) Let M be a f.g. graded R-module and let L be an element of R_{1} which is not a zero divisor on M (i.e. L : M(−1) → M is 1-1). Set S = R/LR.

If F → M → 0 is a minimal free resolution of M as an R-module then F ⊗_{R}(R/LR) = F ⊗_{R} S is a minimal free resolution of M/ℓM as an S-module i.e.

T or_{i}^{R}(M, k) ≃ T or_{i}^{S}(M/LM, k).

In particular, β_{iℓ}(M) (as R-module) = β_{iℓ}(M/LM) (as S-module).

The theorem will be particularly useful when M = R/I and L ∈ R_{1}. In this case

M/LM ≃ (R/I)/((L) + I)/I) ≃ R/(L, I).

Since k is an infinite field (and so one can always find a linear form L as in the theorem) then this process allows us to move from considering I in R to considering another ideal J, having the same height as I, but now in a polynomial ring in ht. I variables.

Let me be more precise:

Definition: Let A = R/I be a CM ring and L_{1}, …, L_{s} be a maximal R-sequence in R_{1} on A. The artinian ring B = A/(L̄_{1}, …, L̄_{s}) is called an artinian reduction of A.

Notice that B ≃ R/(I, L_{1}, …, L_{s}) and that, with a change of variables, we could just as well consider the L_{i} as “variables” of R and then factor them out. That is the sense in which we get that B can be considered as a quotient of a polynomial ring in ht. I variables. From an algebraic point of view such ideals are somewhat easier to manage.

Hom and Ext Modules

Very similar to the way we constructed the modules T or_{i}^{R}(M, N), starting with the tensor product, we can construct other cohomology modules starting with the functors Hom_{R}(M, −) and Hom_{R}(−, N). Unlike the two functors −⊗_{R}N and M ⊗_{R}− (which were both covariant and left exact) the Hom functors have different variances. This involves us, in the case of Hom_{R}(M, −), in considering injective, rather than free, resolutions. There is no time to enter into that complication in these lectures so I will just mention what goes on with the functor Hom_{R}(−, N).

If we fix N and let M be any f.g. graded R-module we can consider a free resolution of M as earlier, i.e

F : 0 → F_{r} ⟶^{ϕ_{r}} ⋯ → F_{1} ⟶^{ϕ_{1}} F_{0} ⟶^{ϕ_{0}} M → 0

We define the complex Hom_{R}(F, N) by:

Hom_{R}(F, N) := 0 → Hom_{R}(F_{0}, N) ⟶^{ϕ_{1}^{t}} Hom_{R}(F_{1}, N) ⟶^{ϕ_{2}^{t}} ⋯ ⟶^{ϕ_{r}^{t}} Hom_{R}(F_{r}, N) → 0

and its cohomology modules as,

Ext_{R}^{i}(M, N) ≃ (Ker ϕ_{i+1}^{t})/(Im ϕ_{i}^{t})

Remarks:

1) It’s easy to see that Ext_{R}^{0}(M, N) ≃ Hom_{R}(M, N).

2) Naturally one has to prove the independence of this definition from the choice of a resolution of M, but that is the standard fare of homological algebra. The apparent imbalance in the way of defining these cohomology modules, as I mentioned above, is more apparent than real. You should consult any good book on Homological Algebra (e.g. the book of Weibel) to find out how to fix this apparent “imbalance”.

3) If M is free then Ext_{R}^{i}(M, N) = 0 for all i > 0 and for all N. More generally,

h.dim_{R}M = n ⇒ Ext_{R}^{i}(M, N) = 0 for all i > n and, for all N.

4) If f : M_{1} → M_{2} is a homomorphism of degree 0 then there are induced homomorphisms (of degree 0)

f_{i}^{t} : Ext_{R}^{i}(M_{2}, N) → Ext_{R}^{i}(M_{1}, N)

for each i.

Similarly if g : N_{1} → N_{2} we have maps

g_{i} : Ext_{R}^{i}(M, N_{1}) → Ext_{R}^{i}(M, N_{2}).

5) (Long Exact Sequences of Exts) If

0 → M_{1} ⟶^{f} M_{2} ⟶^{g} M_{3} → 0

is a short exact sequence of graded modules with graded homomorphisms of degree 0, then there is a long exact sequence of Ext modules

0 → Hom_{R}(M_{3}, N) ⟶^{g^{t}} Hom_{R}(M_{2}, N) ⟶^{f^{t}} Hom_{R}(M_{1}, N) ⟶^{δ_{0}} Ext_{R}^{1}(M_{3}, N) → ⋯

⋯ → Ext_{R}^{i}(M_{3}, N) ⟶^{g_{i}^{t}} Ext_{R}^{i}(M_{2}, N) ⟶^{f_{i}^{t}} Ext_{R}^{i}(M_{1}, N) ⟶^{δ_{i}} Ext_{R}^{i+1}(M_{3}, N) → ⋯

Analogously, if

0 → N_{1} ⟶^{f} N_{2} ⟶^{g} N_{3} → 0

is a s.e.s. of graded modules with homomorphisms of degree 0 then we have a long exact sequence

0 → Hom_{R}(M, N_{1}) ⟶^{∘f} Hom_{R}(M, N_{2}) ⟶^{∘g} Hom_{R}(M, N_{3}) ⟶^{∂_{0}} Ext_{R}^{1}(M, N_{1}) → ⋯

⋯ → Ext_{R}^{1}(M, N_{1}) ⟶^{f_{i}} Ext_{R}^{i}(M, N_{2}) ⟶^{g_{i}} Ext_{R}^{i}(M, N_{3}) ⟶^{∂_{i}} Ext^{i+1}(M, N_{1}) → ⋯

6) If f : M → M is multiplication by a form F of R then the induced map on

Ext_{R}^{i}(M, N) → Ext_{R}^{i}(M, N)

is also multiplication by F. The same is also true if we consider multiplication on N by F.

There is an important connection between the Ext modules and what we have described before.

The key observation is the following:

Lemma: Let M ≠ 0 be a f.g. R-module and J an ideal of R.

Hom_{R}(R/J, M) = 0 ⇔ there is an a ∈ J which is not a zero divisor on M.

(The proof of this Lemma is not difficult but involves some simple observations about associated primes of modules (not only those of the form R/I) which we have not discussed.)

One can then proceed by induction, using (if a is not a zero-divisor on M) the s.e.s.

0 → M ⟶^{×a} M → M/aM → 0

and the long exact sequence of Ext and Remark 6 above to prove:

Theorem: Let M be a f.g. R-module, J an ideal of R and a_{1}, …, a_{r} ∈ J a regular sequence on M, then

Ext_{R}^{r}(R/J, M) ≃ Hom_{R}(R/J, M/(a_{1}, …, a_{r})M).

We derive two interesting corollaries from this Theorem.

If we first apply the theorem to the special case in which M = R = k[x_{0}, …, x_{n}] and observe that if the height of J is r we can always find a regular sequence in J of length r (but not longer) since R is a CM ring we obtain:

Theorem: Under the hypotheses just stated:

\begin{matrix}Ext_{R}^{i}(R/J, R) = 0 for i < r \\ Ext_{R}^{r}(R/J, R) ≠ 0. \end{matrix}

(As Juan will mention, this non-zero module , up to a twist, is what is called the canonical module of R/J. )

If we now apply the theorem to the special case where R = k[x_{0}, …, x_{n}] and let J = m = (x_{0}, …, x_{n}) and set M = R/I, we find

Corollary: Let R = k[x_{0}, …, x_{n}], I an ideal of R. Let F_{1}, …, F_{r} be a maximal regular sequence on R/I. Then

1) Ext_{R}^{r}(R/m, R/I) ≃ Hom_{R}(R/m, R/(F_{1}, …, F_{r}, I)) ≠ 0;

and

2) Ext_{R}^{i}(R/m, R/I) = 0 for i < r.
I.e. depth (R/I) = minimum integer r such that Ext_{R}^{r}(R/m, R/I) ≠ 0.