# Does maximal Cohen-Macaulay modules localize?

Let $A$ be a Noetherian local ring and $M$ a finitely generated $A$-module such that $$\operatorname{depth}M= \dim M=\dim A.$$ I can prove that $$\operatorname{depth}M_{\mathfrak{p}}= \dim M_{\mathfrak{p}}\leq\dim A_{\mathfrak{p}}\quad\forall\mathfrak{p}\in\operatorname{Supp}M.$$ I guess the inequality is in fact an equality, but can't seem to be able to prove it. Does anyone have any idea?

P.S. If $A$ is Cohen-Macaulay and pd$(M)<\infty$ (i.e., if $M$ is perfect), then it can be shown using $$\operatorname{pd}M=\operatorname{depth}A-\operatorname{depth}M.$$

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I cleaned up the TeX in this posting. Notice that writing \operatorname{depth} not only prevents italicization, but also provides proper spacing that was lacking. And \dim is an operatorname in its own right. And using \mbox in this context is an abomination: en.wikipedia.org/wiki/Wikipedia:Manual_of_Style/Mathematics/… (What it says on the linked page does not apply only to Wikipedia. I do think this will mislead people about the proper use of \mbox when TeX is used normally as opposed to the way it's used on websites.) – Michael Hardy Sep 29 '11 at 20:49
If the suppport of $M$ is not equal to $\mathrm{Spec}(A)$, then the equality is not true: just take a prime $\mathfrak p$ not in the support, then $M_{\mathfrak p}=0$ but $A_{\mathfrak p}$ has no reason to has zero dimension. Example: let $A$ be $\mathbb C[x,y]/(x,y)$ localized at $(x,y)$ and let $M=A/xA$. – user18119 Apr 24 '13 at 6:00

It is true if $A$ is CM. I'm not sure if $A$ is not CM. Here is a proof:
Let $\mathfrak{p}\in\operatorname{Supp} M$, and $\operatorname{ht}\mathfrak{p}=n$. Since $A$ is CM, there is an $A$-sequence $a_1,\ldots,a_n\in\mathfrak{p}$ (see Matsumura, Commutative Ring Theory, Theorem 17.4). Since $M$ is maximal, it is also an $M$-sequence (see Eisenbud, Commutative Algebra, Proposition 21.9). Then $a_1/1,\ldots,a_n/1\in\mathfrak{p}A_{\mathfrak{\mathfrak{p}}}$ is an $M_{\mathfrak{\mathfrak{p}}}$-sequence, so $\operatorname{depth} M_{\mathfrak{\mathfrak{p}}}\geq n$. On the other hand, $$n=\dim A_{\mathfrak{\mathfrak{p}}}\geq\dim M_{\mathfrak{\mathfrak{p}}}\geq\operatorname{depth} M_{\mathfrak{\mathfrak{p}}},$$ so they are all equal.