Let $\varphi: A\to B$ be a commutative ring homomorphism and $P$ a prime ideal of $A$. The expansion of an ideal $I\subset A$ is the ideal generated by $\varphi(I)$ in $B$, and the contraction of an ideal $J\subset B$ is $\varphi^{-1}(J)$, and they are denoted by $I^e$ and $J^c$, respectively.
I want to show that there exists a prime ideal $Q\subset B$ such that $Q^c=P$ iff $(P^e)^c=P$.
I don't really have any ideas on how to begin this problem: there's a hint to localize $B$ at $\varphi(A-P)$, but I'm not sure what I'm supposed to do after that. Additional hints would be appreciated!
Since $P$ is a prime ideal of $A$, the set $S = A \setminus P$ is multiplicatively closed. Homomorphic images of multiplicatively closed sets remain multiplicatively closed, hence $T := \varphi(S)$ is multiplicatively closed and does not intersect $P^c$. It follows that the extension of $P^c$ in the localization $T^{-1} B$, call it $R$, is a proper ideal. Hence there is a maximal ideal $N \subset T^{-1} B$ containing $R$. It follows that the contraction of $N$ inside $B$ under the localization homomorphism is prime. Call this ideal $Q$. Then you can show that $Q$ does the job.