I asked this question before but did not get an answer, I tried solving it myself and I think I'm heading somewhere, I just need a push in the right direction.
Let $\phi: R \to S$ be a homomorphism of commutative rings. Suppose that $\phi$ is surjective and that $R$ is a principal ideal ring (Meaning all the ideals are principal ideals, spanned only by 1 term).
Show that $S$ is also a principal ideal ring.
My "psuedo" solution
We know that $R/ker(\phi)$ is isomorphic to $Im(S) = S$ ($\phi$ is surjective).
We also know that $ker(\phi)$ is an ideal, and we know that if $R$ is a principal ideal ring and $I$ is a prime ideal then $R/I$ is also a principal ideal ring.
So I wanted to show that $ker(\phi)$ is a prime ideal, that would solve the question. But it doesn't have to be...Let's look at $ab \in ker(\phi)$:
$ab \in ker(\phi)$ implies $\phi(ab) = 0$ which implies $\phi(a)\phi(b)=0$ which implies $a \in ker(\phi)$ or $b \in ker(\phi)$ IF AND ONLY IF $S$ has no zero divisors. We we're not given that. it could be possible to find $\phi(a),\phi(b) \in S$ such that their product is zero and neither of them is zero.
And that is where I am stuck.
The ring S is a principal ideal ring, but not necessarily a principal ideal DOMAIN. If $J$ is an ideal in $S$, then $I=\phi^{-1}(J)$ is an ideal of $R$. So $I$ is principal, generated by some $x\in R$. Given $j\in J$, write $\phi^{-1}(j)=ax$ for some $a\in R$. Then $j=\phi(a)\phi(x$) showing that $J$ is generated by $\phi(x)$.