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.