$A$ 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 $\mbox{dim }M\leq\mbox{depth }A$?

1$\begingroup$ Noetherian, both here and in the other question. $\endgroup$ – Victor Protsak Aug 6 '10 at 4:51

1$\begingroup$ AuslanderBuchsbaum'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. $\endgroup$ – ashpool Aug 6 '10 at 13:38

$\begingroup$ I retagged, since this was on the front page anyway $\endgroup$ – 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 $M$.
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 "CohenMacaulayRings". In that chaper there is more material on rings that admit a finitely generated module of finite injective dimension.