Showing that a module has depth zero in the proof of the Buchsbaum-Eisenbud criterion of exactness of finite free resolutions

Question

Let $R$ be a Noetherian ring with a maximal ideal $P$. Assume $\operatorname{depth} R_p=0$. Let $\varphi : F\to G$ be a homomorphism of free modules of finite ranks. Let $M:=\operatorname{coker}\varphi$. Then how do we show that $\operatorname{depth}(P_p,M_p)=0$?

One of my attempts is, using https://stacks.math.columbia.edu/tag/00LW, showing that $\operatorname{Ext}^{0}_{R_p}(R_p/P_p,M_p) = \operatorname{Hom}_{R_p}(R_p/P_p,M_p) \neq 0$. And I'm stuck at constructing a well defined nonzero $R_p$-module homomorphism $R_p/P_p \to M_p$.

This question originates from following proof (David Eisenbud, Commutative Algebra, Theorem 20.9) :

I'm now trying to understand the underlined statement. Here, the Auslander-Buchsbaum formula is,

Let $(R,P)$ be a local ring. If $M$ is a finitely generated $R$-module of finite projective dimension, then $ \operatorname{pd}M= \operatorname{depth}(P,R) - \operatorname{depth}(P,M)$.

So, if above our question is true, then it means that $\operatorname{pd}M_p$ is zero, so $M_p$ is a projective module over local ring $R_p$. So it is free over $R_p$.

Can anyone help?

Edit : For the underlined statement, if we can show that $P \in \operatorname{Ass}(M:=\operatorname{coker}(\varphi_1))$, then by the https://stacks.math.columbia.edu/tag/0310, $P_P \in \operatorname{Ass}(M_P)$ and as the proof(first paragraph) of the https://stacks.math.columbia.edu/tag/0BK4, we are done. And can we deduce $P \in \operatorname{Ass}(M)$ (from $P \in \operatorname{Ass}(R))$ ?

Answer 1

As stated your statement is false. Take $R=k[x,y]/(x^2,xy)$. $\phi:R\to R$ be multiplication by $x$. Then the cokernel $R/xR$ has depth one.

In the proof you quote, the module in question has finite projective dimension, so that Auslander-Buchsbaum applies. In the above example, $R/xR$ does not have finite projective dimension.