(~~I don’t know why these <br >’s appear below. If anyone with better HTML knowledge than I could explain what I’m doing wrong, I’d appreciate it!~~)

Today, we will continue with our goal of understanding some aspects of commutative algebra, by defining depth.

** 0.3. Depth **

Constructing regular sequences sequences is a useful task. We often want to ask how long we can make them subject to some constraint. For instance,

**Definition 8**

Suppose is an ideal such that . Then we define the **-depth of ** to be the maximum length of a maximal -sequence contained in . When is a local ring and the maximal ideal, then that number is simply called the **depth** of .

* The depth of a proper ideal is its depth on .
*

The definition is slightly awkward, but it turns out that all maximal -sequences in have the same length. So we can use any of them to compute the depth.

The first thing we can prove using the above machinery is that depth is really a “geometric” invariant, in that it depends only on the radical of .

Proposition 9

Let be a ring, an ideal, and an -module with . Then .

*Proof:*

The inequality is trivial, so we need only show that if is an -sequence in , then there is an -sequence of length in . For this we just take a high power

where is large enough such that everything is in . We can do this as powers of -sequences are -sequences.

This was a fairly easy consequence of the above result on powers of regular sequences. On the other hand, we want to give another proof, because it will

let us do more. Namely, we will show that depth is really a function of prime ideals.

For convenience, we set the following condition: if , we define

Proposition 10

Let be a noetherian ring, an ideal, and a f.g. -module. Then

So the depth of on can be calculated if you know the depths at each prime containing . In this sense, it is clear that depends only on (and the depths on those primes), so clearly it depends only on *up to radical*.

*Proof:*

In this proof, we shall **use the fact that the length of every maximal -sequence is the same**, something which we will prove below.

It is obvious that we have an inequality

as each of those primes contains . We are to prove that there is a prime containing with

But we shall actually prove the stronger statement that there is containing with . Note that localization at a prime can only increase depth because an -sequence in leads to an -sequence in thanks to Nakayama’s lemma and the flatness of localization.

So let be a -sequence of maximum length. Then

acts by zerodivisors on or we could extend the sequence further. In particular, is contained in an associated prime of by elementary commutative algebra (basically, prime avoidance).

Call this associated prime . Then is an

associated prime of , and in particular acts only by zerodivisors on this module. Thus the -sequence can be extended no

further in . In particular, since as we will see soon, the depth

can be computed as the length of any maximal -sequence,

Perhaps we should note a corollary of the argument above:

Corollary 11

Hypotheses as above, we have for

any prime . However, there is at least one containing where equality holds.

November 1, 2010 at 9:13 pm

I think you probably have a linebreak after \mathfrak{p} in your formula.

November 1, 2010 at 9:19 pm

Yes, you’re right. Thanks!