If $A$ is a Principal Ideal Domain, and $\mathfrak{a}$ its ideal. prove that $\frac{A}{\mathfrak{a}}$ is also a Principal Ideal Domain.
The fact that $\mathfrak{a}$ is principal is actually irrelevant.
Suppose that $f\colon A\to B$ is a surjective ring homomorphism. Then
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every ideal of $B$ is of the form $f(I)$, where $I$ is an ideal of $A$ such that $I\supseteq\ker f$;
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if $I$ is a principal ideal in $A$, then $f(I)$ is a principal ideal in $B$.
Prove the two statements above and then apply them to the canonical map $A\to A/\mathfrak{a}$.
Hint: 1 is in the homomorphism theorems; for 2, take a generator $a$ of $I$ and conclude.
Caveat. You can conclude that $A/\mathfrak{a}$ is a principal ideal ring. It will be a principal ideal domain if and only if $\mathfrak{a}$ is a prime ideal.