(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 {I} is an ideal such that {IM \neq M}. Then we define the {I}-depth of {M} to be the maximum length of a maximal {M}-sequence contained in {I}. When {R} is a local ring and {I} the maximal ideal, then that number is simply called the depth of {M}.

The depth of a proper ideal {I \subset R} is its depth on {R}.

The definition is slightly awkward, but it turns out that all maximal {M}-sequences in {I} 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 {I}.

Proposition 9
Let {R} be a ring, {I \subset R} an ideal, and {M} an {R}-module with {IM \neq M}. Then {\mathrm{depth}_I M = \mathrm{depth}_{\mathrm{Rad}(I)} M}.

The inequality {\mathrm{depth}_I M \leq \mathrm{depth}_{\mathrm{Rad} I} M} is trivial, so we need only show that if {x_1, \dots, x_n} is an {M}-sequence in {\mathrm{Rad}(I)}, then there is an {M}-sequence of length {n} in {I}. For this we just take a high power \displaystyle  x_1^N, \dots, x_n^{N}

where {N} is large enough such that everything is in {I}. We can do this as powers of {M}-sequences are {M}-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 {IM = M}, we define

\displaystyle  \mathrm{depth}_I (M) = \infty.

Proposition 10
Let {R} be a noetherian ring, {I \subset R} an ideal, and {M} a f.g. {R}-module. Then

\displaystyle  \mathrm{depth}_I M = \min_{\mathfrak{p} \in V(I)} \mathrm{depth}_{\mathfrak{p}} M .

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

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

It is obvious that we have an inequality

\displaystyle  \mathrm{depth}_I \leq \min_{\mathfrak{p} \in V(I)} \mathrm{depth}_{\mathfrak{p}} M

as each of those primes contains {I}. We are to prove that there is a prime {\mathfrak{p}} containing {I} with

\displaystyle  \mathrm{depth}_I M = \mathrm{depth}_{\mathfrak{p}} M .

But we shall actually prove the stronger statement that there is {\mathfrak{p}} containing I with {\mathrm{depth}_{\mathfrak{p}} M_{\mathfrak{p}} = \mathrm{depth}_I M}. Note that localization at a prime can only increase depth because an {M}-sequence in {\mathfrak{p}} leads to an {M}-sequence in {M_{\mathfrak{p}}} thanks to Nakayama’s lemma and the flatness of localization.

So let {x_1, \dots, x_n \in I} be a {M}-sequence of maximum length. Then {I}
acts by zerodivisors on {M/(x_1 , \dots, x_n) M} or we could extend the sequence further. In particular, {I} is contained in an associated prime of {M/(x_1, \dots, x_n)M} by elementary commutative algebra (basically, prime avoidance).

Call this associated prime {\mathfrak{p} \in V(I)}. Then {\mathfrak{p}} is an
associated prime of {M_{\mathfrak{p}}/(x_1, \dots, x_n) M_{\mathfrak{p}}}, and in particular acts only by zerodivisors on this module. Thus the {M_{\mathfrak{p}}}-sequence {x_1, \dots, x_n} can be extended no
further in {\mathfrak{p}}. In particular, since as we will see soon, the depth
can be computed as the length of any maximal {M_{\mathfrak{p}}}-sequence,

\displaystyle  \mathrm{depth}_{\mathfrak{p}} M_{\mathfrak{p}} = \mathrm{depth}_I M.

Perhaps we should note a corollary of the argument above:

Corollary 11
Hypotheses as above, we have {\mathrm{depth}_I M \leq \mathrm{depth}_\mathfrak{p} M_{\mathfrak{p}}} for
any prime {\mathfrak{p} \supset I}. However, there is at least one {\mathfrak{p}} containing I where equality holds.