Let $\phi: (R,m) \rightarrow (S,n)$ be a morphism of local Noetherian rings. Let $M$ be an $S$-module which is finitely generated as $R$-module. Let $p \in \operatorname{Ass}_S M$ and let $x \in M$ be such that $\operatorname{ann}_S x = p$.
I want to prove that if $p \neq n$ then $p \cap R \neq m$.
My proof: Since $M$ is finite over $R$ and $R$ Noetherian, $M$ will be Noetherian. So every submodule of $M$ is finite over $R$. Hence $S/p \cong Sx$ is finite over $R$. Also, the kernel of $R \rightarrow S/p$ is $p \cap R$ and so we have an embedding $R/p \cap R \hookrightarrow S/p$ and so $S/p$ is finitely generated as $R/p\cap R$-module. If $p\cap R=m$ then $S/p \cong (R/m)^s$ for some positive integer $s$. Now if $n \neq p$, the Krull dimension of $S/p$ is greater than zero, whereas the Krull dimension of $(R/mR)^s$ is the Krull dimension of $R/mR$ which is zero since $R/mR$ is a field, contradiction.
I would appreciate if someone could verify that my proof is correct. In particular, i am not very confident regarding my last argument using the Krull dimension. Any instructive comment on that is highly welcome.
Remark: This is problem 1.2.26(a) in Bruns and Herzog, Cohen-Macaulay Rings.
Unfortunately your argument starting from $S/p\cong (R/m)^s$ is wrong. This is an isomorphism of $R/m$-vector spaces, and definitely not a ring isomorphism (in the end $S/p$ is a domain, right?). Instead you can finish the proof with ease: since $R/m$ is a field and the ring extension is finite, then $S/p$ is a field, a contradiction.