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Let $f:R\rightarrow S$ be an isomorphism. Prove that if we let $R$ be a principal ideal ring it follows that $S$ is a principal ideal ring too.

How should I begin the proof?

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What is isomorphic? – Phira May 11 '11 at 21:35
If the map is an isomorphism, then everything is obvious (the ideals are the same, the generators are the same, everything you want is preserved by isomorphisms). What you probably want is for the map to be surjective. – Aaron May 11 '11 at 21:41
@Zev Chonoles: I do not think that it is certain that the OP wanted to talk about isomorphisms. – Phira May 11 '11 at 21:41
@user9325: The original text was "Let R --> S be an isomorphic". It would make sense to have "Let $R$ and $S$ be isomorphic", but the existence of both the word "an" and the arrow notation, indicates to me that the isomorphism was intended to be an explicit part of this question. – Zev Chonoles May 11 '11 at 21:49
Also I have no compunctions about changing the title to something that is descriptive of what the question is about. – Zev Chonoles May 11 '11 at 21:50

$S$ is a principal ideal ring, by definition, when every ideal $I\subseteq S$ is principal, that is, $$I=(s)=\{sx\mid x\in S\}$$ for some $s\in S$.

You should know from class (or, if not, prove on your own) that, for any homomorphism $g:A\rightarrow B$ where $A$ and $B$ are rings, then if $J\subseteq B$ is an ideal of $B$, then $g^{-1}(J)$ is an ideal of A.

Let $I$ be any ideal of $S$. We want to show $I=(s)$ for some $s\in S$. Using the above fact, we know that $f^{-1}(I)$ is an ideal of $R$. Do you see how to use what we've assumed to be true about $R$ (that $R$ is a principal ideal ring), combined with the ability to send elements of $R$ to elements of $S$ by $r\mapsto f(r)$ and vice versa by $s\mapsto f^{-1}(s)$ (because $f$ is an isomorphism), to prove what we want?

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Isomorphism is a very strong condition. A ring $R$ has every property an isomorphic ring has as a ring, such as being principal ideal domain or being a field.

I think the condition of this problem can be weakened a little bit.

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