So I’ve missed a few days of MaBloWriMo. But I do have a talk topic now (I was mistaken–it’s actually tomorrow)! I’ll be speaking about some applications of Sperner’s lemma. Notes will be up soon.

Today I want to talk about how depth (an “arithmetic” invariant) compares to dimension (a “geometric” invariant). It turns out that the geometric invariant wins out in size. When they turn out to be equal, then the relevant object is called Cohen-Macaulay. This is a condition I’d like to say more about in future posts.

0.5. Depth and dimension

Consider an ${R}$-module ${M}$, which is always assumed to be finitely generated. Let ${I \subset R}$ be an ideal with ${IM \neq M}$. We know that if ${x \in I}$ is a nonzerodivisor on ${M}$, then ${x}$ is part of a maximal ${M}$-sequence in ${I}$, which has length ${\mathrm{depth}_I M}$ necessarily. It follows that ${M/xM}$ has a ${M}$-sequence of length ${\mathrm{depth}_I M - 1}$ (because the initial ${x}$ is thrown out) which can be extended no further. In particular, we find

Proposition 15 Hypotheses as above, let ${x \in I}$ be a nonzerodivisor on ${M}$. Then$\displaystyle \mathrm{depth}_I (M/xM) = \mathrm{depth} M - 1.$

This is strikingly analogous to the dimension of the module ${M}$. Recall that ${\dim M}$ is defined to be the Krull dimension of the topological space ${\mathrm{Supp} M = V( \mathrm{Ann} M)}$ for ${\mathrm{Ann} M}$ the annihilator of ${M}$. But the “generic points” of the topological space ${V(\mathrm{Ann} M)}$, or the smallest primes in ${\mathrm{Supp} M}$, are precisely the associated primes of ${M}$. So if ${x}$ is a nonzerodivisor on ${M}$, we have that ${x}$ is not contained in any associated primes of ${M}$, so that ${\mathrm{Supp}(M/xM)}$ must have smaller dimension than ${\mathrm{Supp} M}$. That is,

$\displaystyle \dim M/xM \leq \dim M - 1.$

But I claim that we have in fact equality.

Lemma 16 For any f.g. ${R}$-module ${M}$, we have ${\dim M /xM \geq \dim M - 1}$.

Proof: If you use the interpretation of ${\dim }$ via systems of parameters, this is not very interesting. For consistency, I will assume that everyone thinks of ${\dim }$ as defined as the combinatorial (i.e. Krull) dimension of ${\mathrm{Supp} M=V(\mathrm{Ann} M)}$. In particular, I will give a proof using the principal ideal theorem. Then I claim that ${\mathrm{Supp}(M/xM) = \mathrm{Supp} M \cap V(x)}$.

Indeed, this is easily seen by localization: if ${\mathfrak{p}}$ is such that ${M_{\mathfrak{p}}/x M_{\mathfrak{p}} \neq 0}$, then ${M_{\mathfrak{p}} \neq 0}$ and ${\mathfrak{p} \in \mathrm{supp} M}$. Similarly, ${x}$ is a non-unit in ${R_\mathfrak{p}}$ and thus ${x \in \mathfrak{p}}$, so ${\mathfrak{p} \in V(x)}$. The converse is proved the same way using Nakayama’s lemma. But we know that Krull’s principal ideal theorem says that the dimension of a closed set intersected with a “hypersurface” like ${V(x)}$ is at least the initial dimension minus one. So this gives the other inequality. In particular, we deduce:

Proposition 17 Let ${M}$ be a f.g. module over the noetherian ring ${R}$. Then$\displaystyle \mathrm{depth}_I M \leq \dim M$

for any ideal ${I \subset R}$ with ${IM \neq M}$.

Proof: Indeed, if ${x_1, \dots, x_r}$ is a maximal ${M}$-sequence in ${I}$, then

$\displaystyle \dim M/(x_1 , \dots, x_r) M = \dim M - r$

by the above remarks. This implies that ${r \leq \dim M}$. That proves the result. This does not tell us much about how ${\mathrm{depth}_I M}$ depends on ${I}$, though; it just says something about how it depends on ${M}$. In particular, it is not very helpful when trying to estimate ${\mathrm{depth} I = \mathrm{depth}_I R}$. Nonetheless, there is a somewhat stronger result, which we will need in the future.

Proposition 18 Hypotheses as above, ${\mathrm{depth}_I M}$ is at most the length of every chain of primes in ${\mathrm{Spec} R}$ that starts at an associated prime of ${M}$ and ends at a prime containing ${I}$.

Proof: Consider a chain of primes ${\mathfrak{p}_0 \subset \dots \subset \mathfrak{p}_k}$ where ${\mathfrak{p}_0}$ is an associated prime and ${\mathfrak{p}_k}$ contains ${I}$. The goal is to show that

$\displaystyle k \leq \mathrm{depth}_I M.$

By localization, we can assume that ${\mathfrak{p}_k}$ is the maximal ideal of ${R}$; recall that localization can only increase the depth. In this case, the argument has become:

Lemma 19 Let ${(R,\mathfrak{m})}$ be a noetherian local ring. Let ${M}$ be a finite ${R}$-module. Then the depth of ${\mathfrak{m}}$ on ${M}$ is at most the dimension of ${R/\mathfrak{p}}$ for ${\mathfrak{p}}$ an associated prime of ${M}$.

To prove this, first assume that the depth is zero. In that case, the result is immediate. We shall now argue inductively. Assume that that this is true for modules of smaller depth. We will quotient out appropriately to shrink the support and change the associated primes.

Namely, choose a ${M}$-regular (nonzerodivisor on ${M}$) ${x \in R}$. Then ${\mathrm{depth}_I M/xM = \mathrm{depth}_I M -1}$. Let ${\mathfrak{p}_0}$ be an associated prime of ${R}$. I claim that ${\mathfrak{p}_0}$ is properly contained in an associated prime of ${M/xM}$. Indeed, ${x \notin \mathfrak{p}_0}$, so ${\mathfrak{p}_0}$ cannot itself be an associated prime. However, I claim that ${\mathfrak{p}_0}$ annihilates a nonzero element of ${M/xM}$.

To see this, consider maximal principal submodule of ${M}$ annihilated by ${\mathfrak{p}_0}$. Let this submodule be ${Rz}$ for some ${z \in M}$. Then if ${z}$ is a multiple of ${x}$, say ${z = xz'}$, then ${Rz'}$ would be a larger submodule of ${M}$ annihilated by ${\mathfrak{p}_0}$—here we are using the fact that ${x}$ is a nonzerodivisor on ${M}$. So the image of this ${z}$ in ${M/xM}$ is nonzero and is clearly annihilated by ${\mathfrak{p}_0}$.

Thus ${\mathfrak{p}_0}$ is contained in an associated prime of ${M/xM}$. Call this prime ${\mathfrak{q}_0}$. Now we know that ${\mathrm{depth}_I M/xM = \mathrm{depth}_I M -1}$. Also, by the inductive hypothesis, we know that ${\dim R/\mathfrak{q}_0 \geq \mathrm{depth}_I M/xM = \mathrm{depth}_I M -1}$. But the dimension of ${R/\mathfrak{q}_0}$ is strictly greater than that of ${R/\mathfrak{p}_0}$, so at least ${\dim R/\mathfrak{q}_0 +1 = \mathrm{depth}_I M}$. This proves the lemma.