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Let $f: A \rightarrow B$ be a ring homomorphism between commutative rings with identity. Then there exists an induced map $f' : Spec(B) \rightarrow Spec(A)$. If $f'$ is surjective, then clearly every prime ideal of $A$ is a contracted ideal. Now my question is, is the converse true?

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up vote 4 down vote accepted

Let $P$ be a prime ideal of $A$, and suppose that $P=f^{-1}(I)$ where $I$ is some ideal of $B$. Let $S=A-P$, a multiplicatively closed subset of $A$, and $T=f(S)$. Then $T$ is a multiplicatively closed subset of $B$ and is disjoint from $I$. By Zorn's lemma, there is an ideal $Q$ of $B$ maximal with respect to the conditions that $Q$ contains $I$ and $Q$ is disjoint from $T$. By a standard argument, $Q$ is prime. Moreover $f^{-1}(Q)=P$.

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Thank you for your clarification! – user3620 Nov 17 '10 at 17:22

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