$A$ 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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$A=k[[x,y]]/(x^2,xy)$ 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. 

