# Depth of $R/I$ as an $R$-module versus as a ring

Let $$(R, \mathfrak m,k)$$ be a Noetherian local ring. Let $$I\subseteq \mathfrak m$$ be an ideal of $$R$$. Then $$(R/I, \mathfrak m/I, k)$$ is a Noetherian local ring but also $$R/I$$ is a finitely generated $$R$$-module.

Is it true that the depth of $$R/I$$ as an $$R$$-module is equal to the depth of the ring $$R/I$$ ?

I can see that $$x\in \mathfrak m$$ is a non zero divisor on the module $$R/I$$ if and only if $$x+I \in \mathfrak m/I$$ is a non zero divisor in the ring $$R/I$$, but I'm not sure if that is enough to conclude what I want or not. Please help .

• If one knows the characterization of depth in terms of vanishing of local cohomology, then you can use that it doesn't matter if you compute the local cohomology of $R/I$ with respect to $\mathfrak{m}$ as an R module or $R/I$-module. – walkar Jun 3 at 15:30
• @Louis The only thing you need in order to move from one element to many is the following: $\frac{R/I}{(a_1,\dots,a_i)(R/I)}=\frac{R/I}{(a_1,\dots,a_i)+I/I}\simeq\frac{R}{(a_1,\dots,a_i)+I}$. – user26857 Jun 4 at 7:38

By Theorem 6.9, one can test depth by checking the first nonvanishing index of local cohomology. Then, we can use Corollary 7.11 with $$R \rightarrow R/I$$, $$M=R/I$$. This says $$H^i_{\mathfrak{m}}(R/I) \simeq H^i_{\mathfrak{m}/I}(R/I)$$, so they must vanish together. This gives the desired result, since $$\operatorname{depth}_{\mathfrak{m}/I}(R/I)$$ is the minimum index $$i$$ where $$H^i_{\mathfrak{m}/I}(R/I) \neq 0$$, which is the same as the minimum index where $$H^i_{\mathfrak{m}}(R/I) \neq 0$$, i.e. $$\operatorname{depth}_{\mathfrak{m}}(R/I)$$.