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 .