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Let $(R,\mathfrak{m})$ be a Cohen–Macaulay ring, $J$ an ideal of $R$ such that $\dim R/J >0$ and $M$ a finitely generated $R$-module.

Is $\mathrm{grade}(J,M) =\mathrm{depth} M$ true?

Here $\mathrm{grade}(J,M)= \inf \{ i : Ext^i_R(R/J,M)\neq 0\}$. If it is true, then is there any $p\in V(J+\mathrm{Ann}_R M)\setminus\{ \mathfrak{m}\}$ such that $\mathrm{grade}(J,M) = \mathrm{grade}(JR_p,M_p)$ ?

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  • $\begingroup$ You probably mean that $R$ is local, with maximal ideal $\mathfrak{m}$? Then what you say is false. What do you call $\operatorname{grade}(J,M) $ is usually called $\operatorname{depth}(J,M) $, the depth of $M$ w.r.t. $J$ — the maximal length of a $M$-regular sequence contained in $J$; while $\operatorname{depth}(M):= \operatorname{depth}(\mathfrak{m},M)$ is the maximal length of a $M$-regular sequence contained in $\mathfrak{m}$. Obviously $\operatorname{depth}(J,M) \leq \operatorname{depth}(M) $, but it is in general smaller — take $J=(0)$! $\endgroup$ – abx Mar 12 at 12:46
  • $\begingroup$ Yes, $R$ is a local ring. If $J\neq 0$, then is the above equality true? $\endgroup$ – Tri Nguyen Mar 12 at 13:01
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    $\begingroup$ Of course not. You can get anything between $0$ and $\operatorname{depth}(M) $. $\endgroup$ – abx Mar 12 at 22:16

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