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 $f:R\rightarrow S$ 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?

share|improve this question
    
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
show 1 more comment

2 Answers 2

$S$ 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?

share|improve this answer
add comment

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.

share|improve this answer
add comment

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.