(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
Letbe 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
Letbe 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 havefor
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!