Is every ideal of Quotient Ring a PID?

Question

We are given a principal ideal domain $R$ and an ideal of this domain, $I$. Is it true that every proper ideal of $\displaystyle\frac{R}{I}$ a principal ideal domain ?

I have proven that every proper ideal, say, $\overline{K}$ of $\displaystyle\frac{R}{I}$ is an ideal generated by an element of the $\displaystyle\frac{R}{I}$. Is it true that $\overline{K}$ is an integral domain ?

Answer 1

No, in general $K$ can have zero divisors, and it will not always have an identity.

For example: $\mathbb Z/12\mathbb Z$ and its ideal $6\mathbb Z/12\mathbb Z$ are examples for both.

$K$ will not have zero divisors if $I$ is prime, but in a PID this would mean that $I=\{0\}$ or else a maximal ideal, which are both pretty uninteresting cases.

The best you can say is, as you concluded, that $R/I$ is always a principal ideal ring.

Answer 2

Consider $\mathbb{Z}$ and $p,q,r$, three prime integers. $\mathbb{Z}/pqr\mathbb{Z}$ is isomorphic to $\mathbb{Z}/p\times\mathbb{Z}/q\times \mathbb{Z}/r$. $\mathbb{Z}/p\times\mathbb{Z}/q\times 0$ is a proper ideal of $\mathbb{Z}/p\times\mathbb{Z}/q\times \mathbb{Z}/r$ and is not integral.