Question: show that homomorphic image of a principal ideal ring is a principal ideal ring.
My attempt : let $R$ be PIR (principal ideal ring) and $f$ be homomorphism from $R$ to a ring $S$ then by properties of ring homomorphism we know that $f(R)$ is subring of $S$. Now we consider an ideal $I$ of $f(R)$ then by properties of homomorphism we know its pullback that is $f^{-1}(I)$ is an ideal of $R$. But $R$ is PIR and hence there must exists some $a\in R$ such that $f^{-1}(I)=<a>$.
Hence to prove the result we just need to show that $I$ is principal ideal. My intension is that $I=<f(a)>$. But how to prove it?
Let $s\in <f(a)>$ then by definition of principal ideal, $s=s'f(a)$ for some $s'\in f(R)$. How to show $s\in I$? If $f(a)\in I$ then as $I$ is an ideal of $f(R)$ hence we have $s=s' f(a)\in I$ and we are done! But how $f(a)\in I$? I didnt get this! (Since $R$ my not have unity and hence $a$ does not belongs to $<a>=f^{-1}(I)$ and so we can't say $f(a)\in I$) and also how to show other direction that is, $I\subseteq <f(a)>$
Please help.