# Prime ideal is contraction of prime ideal iff it's saturated

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!

• Have you tried doing the $\Rightarrow$ implication? That doesn't seem as hard. Forget the hint for a moment. – Hoot Jul 1 '14 at 2:21
• That direction is "If $Q$ exists, then $P$ is saturated," right? – Nishant Jul 1 '14 at 2:24
• Right. One always has $P^{ec} \supset P$, of course, so you just need to get the opposite inclusion somehow. – Hoot Jul 1 '14 at 2:26
• Hmm, so if $P=Q^c$, then I need to show that $(P^e)^c\subseteq Q^c$, which is accomplished if I can show that $P^e\subseteq Q$. That is true, because if $Q$ contracts to $P$, it must contain the image of $P$, and thus the ideal generated by that image, right? – Nishant Jul 1 '14 at 2:45
• Hmm wait, it might not hold if $\phi$ is not injective? – Nishant Jul 1 '14 at 2:48

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.