3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Noetherian, both here and in the other question. $\endgroup$ – Victor Protsak Aug 6 '10 at 4:51
  • 1
    $\begingroup$ 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. $\endgroup$ – 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
11
$\begingroup$

$A=k[[x,y]]/(x^2,xy)$ then depth$(A)=0$. Let $M=R/(x)=k[[y]]$ then $y$ is a nonzerodivisor on $M$.

$\endgroup$
  • $\begingroup$ Did you mean $M=A/(x)$? $\endgroup$ – Sándor Kovács Jan 11 '12 at 16:28
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.