Isomorphisms preserve the property of being a principal ideal ring 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? 
 A: $S$ 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?
A: 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.
