# Is every ideal of Quotient Ring a PID?

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 ?

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.
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.