Noetherian local ring of depth $1$ whose localization at some prime ideal has higher depth?

Question

Let $(R,\mathfrak m)$ be a Noetherian local ring of depth $1$.

Then, is it possible that depth$(R_P)\ge 2$ for some prime ideal $P$ of $R$?

Of course such an example has to be non-Cohen-Macaulay, i.e. we would have to have $\dim R \ge ht(P)=\dim (R_P) \ge \operatorname{depth}(R_P) \ge 2>\operatorname{depth}(R)$.

Answer 1

Let $R=K[[X,Y,Z,T,W]]/(X^2,XY,XZ,XT,W^2)$.

We have $$(X^2,XY,XZ,XT,W^2)=(X,W^2)\cap(X^2,Y,Z,T,W^2),$$ so the height of this ideal is $2$.
It follows that $\dim R=3$.

Now notice that $x-w$ is a maximal $R$-sequence since $R/(x-w)\simeq K[[X,Y,Z,T]]/(X^2,XY,XZ,XT)$ and $\operatorname{depth}R/(x-w)=0$. It follows that $\operatorname{depth}R=1$.

Let $\mathfrak p=(x,y,z,w)$. Then $R_{\mathfrak p}\simeq K[[Y,Z,T,W]]_{(Y,Z,W)}/(W^2)$ is Cohen-Macaulay of dimension $2$.