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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 .

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  • $\begingroup$ 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. $\endgroup$ – walkar Jun 3 at 15:30
  • $\begingroup$ @Louis The only thing you need in order to move from one element to many is the following: $\frac{R/I}{(a_1,\dots,a_i)(R/I)}=\frac{R/I}{(a_1,\dots,a_i)+I/I}\simeq\frac{R}{(a_1,\dots,a_i)+I}$. $\endgroup$ – user26857 Jun 4 at 7:38
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I decided to expand my comment into an answer. My reference is to Hochster's Local Cohomology notes (http://dept.math.lsa.umich.edu/~hochster/615W11/loc.pdf).

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. $\operatorname{depth}_{\mathfrak{m}}(R/I)$.

This proof method might not be easiest, since one must go through the route of Koszul complexes to prove it. I expect there's an easier method only using regular sequences somewhere out there.

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