Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 2 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 $f^{-1}(Q)=P$.

share|improve this answer
    
Thank you for your clarification! –  user3620 Nov 17 '10 at 17:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.