If $A$ is a Principal Ideal Domain, and $\mathfrak{a}$ its ideal. prove that $\frac{A}{\mathfrak{a}}$ is also a Principal Ideal Domain.

Question

Let $A$ be a Principal Ideal Domain, and $\mathfrak{a}$ its ideal. I have to prove that $\frac{A}{\mathfrak{a}}$ is also a Principal Ideal Domain.

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This is what I've done:

As $A$ is a Principal Ideal Domain and $\mathfrak{a}$ its ideal, $\mathfrak{a}$ is principal, I mean, $\mathfrak{a}=(a)$.

Now, using the correspondence theorem, I know that if $\mathfrak{a}$ is an ideal of $A$ and $\mathfrak{b}\subseteq\mathfrak{a}$, $\frac{\mathfrak{a}}{\mathfrak{b}}=\frac{(a)}{\mathfrak{b}}$ is going to be an ideal of $\frac{A}{\mathfrak{a}}$.

But how do I prove that $\frac{(a)}{\mathfrak{b}}$ is principal?

Answer 1

The fact that $\mathfrak{a}$ is principal is actually irrelevant.

Suppose that $f\colon A\to B$ is a surjective ring homomorphism. Then

1. every ideal of $B$ is of the form $f(I)$, where $I$ is an ideal of $A$ such that $I\supseteq\ker f$;

2. 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.