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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?

Maria
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2 Answers2

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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 $f^{-1}(Q)=P$.

Robin Chapman
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By atiyah-macdonald thm 3.15 If there is an homomorphism from A to B and p^{ec}=p, then p is a contraction of a prime ideal of B.

In your question p is contracted ideal, so p satisfies p^{ec}=p

Hong
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