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 -module , which is always assumed to be finitely generated. Let be an ideal with . We know that if is a nonzerodivisor on , then is part of a maximal -sequence in , which has length necessarily. It follows that has a -sequence of length (because the initial is thrown out) which can be extended no further. In particular, we find

Proposition 15Hypotheses as above, let be a nonzerodivisor on . Then

This is strikingly analogous to the dimension of the module . Recall that is defined to be the Krull dimension of the topological space for the annihilator of . But the “generic points” of the topological space , or the smallest primes in , are precisely the associated primes of . So if is a nonzerodivisor on , we have that is not contained in any associated primes of , so that must have smaller dimension than . That is,

But I claim that we have in fact equality.

Lemma 16For any f.g. -module , we have .

*Proof:* If you use the interpretation of via systems of parameters, this is not very interesting. For consistency, I will assume that everyone thinks of as defined as the combinatorial (i.e. Krull) dimension of . In particular, I will give a proof using the principal ideal theorem. Then I claim that .

Indeed, this is easily seen by localization: if is such that , then and . Similarly, is a non-unit in and thus , so . 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 is at least the initial dimension minus one. So this gives the other inequality. In particular, we deduce:

Proposition 17Let be a f.g. module over the noetherian ring . Then

for any ideal with .

*Proof:* Indeed, if is a maximal -sequence in , then

by the above remarks. This implies that . That proves the result. This does not tell us much about how depends on , though; it just says something about how it depends on . In particular, it is not very helpful when trying to estimate . Nonetheless, there is a somewhat stronger result, which we will need in the future.

Proposition 18Hypotheses as above, is at most the length of every chain of primes in that starts at an associated prime of and ends at a prime containing .

*Proof:* Consider a chain of primes where is an associated prime and contains . The goal is to show that

By localization, we can assume that is the maximal ideal of ; recall that localization can only increase the depth. In this case, the argument has become:

Lemma 19Let be a noetherian local ring. Let be a finite -module. Then the depth of on is at most the dimension of for an associated prime of .

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 -regular (nonzerodivisor on ) . Then . Let be an associated prime of . I claim that is properly contained in an associated prime of . Indeed, , so cannot itself be an associated prime. However, I claim that annihilates a nonzero element of .

To see this, consider maximal principal submodule of annihilated by . Let this submodule be for some . Then if is a multiple of , say , then would be a larger submodule of annihilated by —here we are using the fact that is a nonzerodivisor on . So the image of this in is nonzero and is clearly annihilated by .

Thus is contained in an associated prime of . Call this prime . Now we know that . Also, by the inductive hypothesis, we know that . But the dimension of is strictly greater than that of , so at least . This proves the lemma.

November 6, 2010 at 10:18 pm

Are your hands sore because of the typing?

November 6, 2010 at 10:34 pm

No, I’ve just been busy thinking about other things, like the talk. And doing coursework.

November 7, 2010 at 12:29 am

I see. I thought it was literally because of sore hands. I know I get sore hands when I do a lot of typing. So you will be stopping the algebraic topology notes?

November 7, 2010 at 12:30 am

I meant on your harvard fas website. It said that you would no longer by typing the alg topology notes but continue with the commutative algebra notes.

November 7, 2010 at 3:36 pm

Yes. Someone else was going to take over, anyway.