Depth and dimension

$A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. Then is it true that $\mbox{depth }M\le\mbox{depth }A$ just like $\mbox{dim }M\le\mbox{dim }A$? I don't see any relation between an $M$-sequence and an $A$-sequence. At least I know it is true when $\mbox{inj.dim }M<\infty$, from the relation $\mbox{depth }M\leq\mbox{dim }M\leq\mbox{inj. dim }M=\mbox{depth }A\leq\mbox{dim }A$. But what happens when $\mbox{inj.dim }M=\infty$? Another inequality I'm not quite sure about when $\mbox{inj.dim }M=\infty\$: is it true that $\mbox{dim }M\leq\mbox{depth }A$?

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Noetherian, both here and in the other question. –  Victor Protsak Aug 6 '10 at 4:51
Auslander-Buchsbaum's theorem says when $\mbox{proj.dim }M<\infty$ $\mbox{depth }A-\mbox{depth }M=\mbox{proj.dim }M$ so it appears the inequality holds when either projective dimension or injective dimension is finite. –  ashpool Aug 6 '10 at 13:38
I retagged, since this was on the front page anyway –  David White Jan 11 '12 at 18:48

$A=k[[x,y]]/(x^2,xy)$ then depth$(A)=0$. Let $M=R/(x)=k[[y]]$ then $y$ is a nonzerodivisor on $M$.
Did you mean $M=A/(x)$? –  Sándor Kovács Jan 11 '12 at 16:28
In the paper "Eine Dualität zwischen den Funktoren Ext und Tor" (J. Algebra 11, 510–531) Ischebeck shows that if $A$ admits a finitely generated module $N$ of finite injective dimension, then the answer is affirmative. More precisely, for any finitely generated module $M$ one has $\text{depth}\ A - \text{depth}\ M = \sup\left\lbrace i : \text{Ext}^i_A(M,N) \neq 0 \right\rbrace$. This is Excercise 3.1.24 in Bruns/Herzog "Cohen-Macaulay-Rings". In that chaper there is more material on rings that admit a finitely generated module of finite injective dimension.