Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PIR.

Question

I think I have proved this:

Let $J$ be an ideal of $S$. Then $f^{-1}(J)=(a)$ is a principal ideal of $R$. Hence $J=f (f^{-1}(J))=f((a))=(f(a))$ is a principal ideal of $S$. Is my proof correct?

But I also think there are some counterexamples by considering the quotient rings of polynomial rings $\mathbb{Z}[x,y]/f(x,y)$. I am confused...

The proof is correct, and $\mathbb{Z}[x,y]$ is not a PID, if that is what is confusing you. — Prahlad Vaidyanathan, Dec 29 '13 at 03:42
The proof is "correct", but the claim is obviously false! Take $R=\Bbb Z$. — , Dec 29 '13 at 22:04
The proof is correct but $Z[x,y]$ is not a PID, see
http://math.stackexchange.com/questions/697204/is-any-ufd-also-a-pid/701581#701581 — user 1, Mar 12 '14 at 16:54

score 3 · Answer 1 · answered Dec 25 '14 at 00:42

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This is a community wiki answer designed to help eliminate this question from the unanswered queue

The proof has been verified to be correct.

The example given is not a counterexample since $\mathbb Z[x, y]$ is not a principal ideal domain.

answered Dec 25 '14 at 00:42

Robert Cardona

9,271

Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PIR.

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