Thanks to all who responded to the bleg yesterday. I’m still haven’t completely decided on the topic owing to lack of time (I actually wrote this post last weekend), but the suggestions are interesting. My current plan is, following Omar’s comment, to look at Proofs from the Book tomorrow and pick something combinatorial or discrete-ish, like the marriage problem or Arrow’s theorem. I think this will be in the appropriate spirit and will make for a good one-hour talk.

4. ${\mathrm{Ext}}$ and depth

One of the first really nontrivial facts we need to prove is that the lengths of maximal ${M}$-sequences are all the same. This is a highly useful fact, and we shall constantly use it in arguments (we already have, actually). More precisely, let ${I \subset R}$ be an ideal, and ${M}$ a finitely generated module. Assume ${R}$ is noetherian.

Theorem 12 Suppose ${M}$ is a f.g. ${R}$-module and ${IM \neq M}$. All maximal ${M}$-sequences in ${I}$ have the same length. This length is the smallest value of ${r}$ such that ${\mathrm{Ext}^r(R/I, M) \neq 0}$.

I don’t really have time to define the ${\mathrm{Ext}}$ functors in any detail here beyond the fact that they are the derived functors of ${\hom}$. So for instance, ${\mathrm{Ext}(P, M)=0}$ if ${P}$ is projective, and ${\mathrm{Ext}(N, Q) = 0}$ if ${Q}$ is injective. These ${\mathrm{Ext}}$ functors can be defined in any abelian category, and measure the “extensions” in a certain technical sense (irrelevant for the present discussion).

So the goal is to prove this theorem. In the first case, let us suppose ${r = 0}$, that is there is a nontrivial ${R/I \rightarrow M}$. The image of this must be annihilated by ${I}$. Thus no element in ${I}$ can act as a zerodivisor on ${M}$. So when ${r = 0}$, there are no ${M}$-sequences (except the “empty” one of length zero).

Conversely, if all ${M}$-sequences are of length zero, then no element of ${I}$ can act as a nonzerodivisor on ${M}$. It follows that each ${x \in I}$ is contained in an associated prime of ${M}$, and hence by the prime avoidance lemma, that ${I}$ itself is contained in an associated prime ${\mathfrak{p}}$ of ${M}$. This prime avoidance argument will crop up quite frequently.

Then there is an injection ${R/\mathfrak{p} \rightarrowtail M}$, which when composed with reduction ${R/I \rightarrow R/\mathfrak{p}}$ shows that ${\hom(R/I, M) \neq 0}$. Note that we have used the fact that ${R}$ is noetherian and ${M}$ finitely generated here, or otherwise the whole business of associated primes wouldn’t work. Now I want to claim that if ${x_1, \dots, x_s}$ is any ${M}$-sequence in ${I}$, then ${\mathrm{Ext}^q( R/I, M) = 0}$ for ${q < s}$. We have shown this for ${s=1}$; the conclusion is then simply ${\hom(R/I, M)=0}$. Namely, I want to claim that the length of any maximal ${M}$-sequence is at most the minimal ${r}$ as in the theorem.

Lemma 13 If ${x_1, \dots, x_s}$ is any ${M}$-sequence in ${I}$, then ${\mathrm{Ext}^q( R/I, M) = 0}$ for ${q < s}$.

Proof: As I have said, this is true for ${s=0}$. We want to prove this by induction on ${s}$. So suppose it true for ${s-1}$. Then we know that ${\mathrm{Ext}^q(R/I, M)=0}$ for ${q < s-1}$. We are left to showing that $\displaystyle \mathrm{Ext}^{s-1} (R/I, M) = 0.$

Well, we can apply the inductive hypothesis again to conclude that ${\mathrm{Ext}^q(R/I, M/x_1 M) = 0}$ for ${q < s-1}$ as there is a ${s-1}$-length ${M/x_1 M}$ sequence ${x_2, \dots, x_s}$. There is an exact sequence $\displaystyle 0 \rightarrow M \stackrel{x_1}{\rightarrow} M \rightarrow M/x_1 M \rightarrow 0,$

by definition of regularity. The standard argument in these proofs is to get from here a long exact sequence and use induction. Here the long exact sequence looks like $\displaystyle \mathrm{Ext}^{s-2}(R/I, M/x_1 M) \rightarrow \mathrm{Ext}^{s-1}(R/I, M) \stackrel{x_1}{\rightarrow} \mathrm{Ext}^{s-1}(R/I, M) \rightarrow \dots$

The first term is zero by the inductive assumption. So multiplication by ${x_1}$ is injective on ${\mathrm{Ext}^{s-1}(R/I, M)}$. But ${x_1}$ acts by zero on ${R/I}$. Thus it acts by zero on ${\mathrm{Ext}^{s-1}(R/I, M)}$ by the interpretation as a derived functor. This means that ${\mathrm{Ext}^{s-1}(R/I, M) =0}$. And that proves the lemma.

So we have gotten one direction. We need to do the other. Namely, we need to show that if ${x_1, \dots, x_r}$ is an ${M}$-sequence which cannot be extended in ${I}$, then $\displaystyle \mathrm{Ext}^r(R/I, M) \neq 0.$

This will imply the other inequality and prove the theorem. For this we prove:

Lemma 14 If ${x_1, \dots, x_r}$ is an ${M}$-sequence, then $\displaystyle \mathrm{Ext}^r(R/I, M) \simeq \hom(R/I, M/(x_1, \dots, x_r)M).$

This is actually more general than the previous lemma. In that case, we knew that ${x_1 \dots x_r}$ could be extended, so there was an element of ${I}$ which was a nonzerodivisor on ${M/(x_1 \dots x_r) M}$, so the hom-set on the left was zero. Proof: Though this is really a generalization of the previous result, the argument is very similar (and there was basically no need for the previous result therefore, if we showed a slightly more general result). Anyway, let’s induct on ${r}$. We know it for ${r=0}$ by the initial discussion. Draw the exact sequence ${0 \rightarrow M \stackrel{x_1}{\rightarrow} M \rightarrow M/x_1 M\rightarrow 0}$. This leads to the exact sequence $\displaystyle \mathrm{Ext}^{r-1}(R/I, M/x_1 M) \rightarrow \mathrm{Ext}^r(R/I, M) \stackrel{x_1}{\rightarrow} \mathrm{Ext}^r(R/I, M).$

As before, the last multiplication is zero, so we find an isomorphism $\displaystyle \mathrm{Ext}^{r-1}(R/I, M/x_1 M) \simeq \mathrm{Ext}^r(R/I, M).$

Given the ${M/x_1 M}$-sequence ${x_2, \dots, x_r}$ which can be extended no further, we see that ${\mathrm{Ext}^{r-1}(R/I, M/x_1 M) \neq 0}$, which proves the result.

Next time, we will show that depth is bounded in terms of dimension.