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I'm trying to prove the following result:

Let $R$ be a principal ideal domain, $S$ an integral domain and $f: R\to S$ a surjective morphism. Prove that if $f$ is not an isomorphism, then $S$ is a field.

I have done the following: if $f$ is surjective, then $Im f=S$ and by the first theorem of isomorphism I know that $R/\ker f$ is isomorphic to $S$. So, $S$ is a field if and only if $R/\ker f$ is a field. Then I noticed that $\ker f$ is an ideal of $R$ and since $R$ is a principle ideal domain, there exists $a\in R$ such that $\ker f = <a>$. Besides, $a\neq 0$, because we assume that $f$ is not an isomorphism; since $f$ is surjective, it can't be injective and its kernel is not trivial.

I don't know how to proceed, now. I've tried to prove that $\ker f$ is maximal, and this would finish the proof, but I couldn't do it.

Many thanks!

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  • $\begingroup$ Every non-zero prime ideal is maximal in a PID. $\endgroup$ – Crostul Apr 27 '15 at 21:29
  • $\begingroup$ In fact I have no idea of what Krull dimension is :) but thank you for your comments! @user26857 what you said is true because $R/<a>$ integral domain $<=>$ a irreducible $<=>$ a prime (since R PID) $<=>$ $R/<a>$ field, right? $\endgroup$ – user194469 Apr 27 '15 at 21:31
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As $\ker f\ne 0$ there exists $a\ne 0$ with $\ker f =\langle a\rangle$. Assume $\ker f\subseteq I\subsetneq R$. Then $I=\langle b\rangle$ and $b=ra$ for some $r\in R$. Then $f(r)f(a)=0$ in $S$, so one of $f(r), f(a)$ is $0$, i.e., one of $r,a$ is in $\ker f$. Then also $b\in \ker f$, i.e., $I=\ker f$. Hence $\ker f $ i smaximal.

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In a PID, any nonzero prime ideal is maximal. Now:

  • either $\ker f =0$, which means $f$ is injective, hence is an isomorphism since it is surjective by hypothesis,
  • or $\ker f$ is a non-zero prime ideal since $S$ is an integral domain, so that $S\simeq R/\ker f$ is a field since $\ker f$ is a maximal ideal.
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