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Let $A$ be a ring and $p$ be a prime ideal. By elementary considerations we have that $\operatorname{depth}_p A_p \geq \operatorname{depth}_p A$. Is it true that the reverse holds if $p$ is maximal?

This question came up as I was trying to work through Problem III.3.5 in Hartshorne's Algebraic Geometry. Let $U$ be an open subset of a Noetherian scheme and $p \in U$ a closed point, i.e., $p$ is a maximal ideal in an affine open. In this problem one needs to characterize unique extendibility of sections of the structure sheaf from $U \setminus p$ to $U$ in terms of the depth of the local ring at $p$.

This question came out naturally from trying to solve the above problem.

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In Bruns and Herzog notation one knows that $\mathrm{grade}(\mathfrak p,A)\le\mathrm{depth}(A_{\mathfrak p})$ for every prime ideal $\mathfrak p$ since a regular sequence remains regular after localization.
In general, it is not true that $\mathrm{grade}(\mathfrak p,A)=\mathrm{depth}(A_{\mathfrak p})$, and a counterexample is the following: $A=K[X,Y,Z]/(X^2,XY,XZ)$ and $\mathfrak p=(x,y)$, where $x$, respectively $y$ denote the residue classes of $X$, respectively $Y$. We have $\mathrm{grade}(\mathfrak p,A)=0$ since $x\mathfrak p=0$, and $\mathrm{depth}(A_{\mathfrak p})=1$.
Such phenomenon occurs since $\mathfrak p$ is not an associated prime.
For maximal ideals this can not happen. That is, if $\mathfrak m$ is a maximal ideal with $\mathrm{grade}(\mathfrak m,A)=0$, then $\mathfrak m\in\mathrm{Ass}(A)$, and thus $\mathfrak mA_{\mathfrak m}\in\mathrm{Ass}(A_{\mathfrak m})$ which shows that $\mathrm{depth}(A_{\mathfrak m})=0$.
This leads immediately to the conclusion that $\mathrm{grade}(\mathfrak m,A)=\mathrm{depth}(A_{\mathfrak m})$ for every maximal ideal $\mathfrak m$.

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