# Does $\mathrm{grade}(J,M) = \mathrm{depth} M$?

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 $$\mathrm{grade}(J,M) = \mathrm{grade}(JR_p,M_p)$$ ?

• 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 $J=(0)$! – abx Mar 12 at 12:46
• Yes, $R$ is a local ring. If $J\neq 0$, then is the above equality true? – Tri Nguyen Mar 12 at 13:01
• Of course not. You can get anything between $0$ and $\operatorname{depth}(M)$. – abx Mar 12 at 22:16