PID which is not a Prüfer domain

Question

I am wondering if there exists a commutative ring which is not a Prüfer domain, i.e. intersection does not distribute over addition of ideals, but which is also a PID. The reason I ask is because all of the examples of domains which are not Prüfer that I can come up with ($k[x,y]$ for some field $k$ and $\mathbb{z}[x]$ as two) seem not to be Prüfer precisely because there are ideals which are not generated by a single element. I cannot, however, prove that every PID is Prüfer. Do such rings exist and if not why?

Answer 1

No, there doesn't. Prüfer domains are domains in which non-zero finitely generated ideals are invertible, and in a P.I.D., all ideal are principal, hence free since it is a domain. If $R$ is such a P.I.D., and if $I=Ra$, then its inverse is the fractionary ideal $$I^{-1}=R\cdot\frac1a. $$

Answer 2

Another way to skin the cat is knowing that a domain is Prüfer iff its finite generated ideals are projective. In a PID, all the ideals are free (isomorphic to $R$ or else $\{0\}$), and free modules are all projective.

Since a commutative domain is Dedekind iff all of its ideals are projective, this also shows a PID is a Dedekind domain. (If you know Dedekind domains are Prüfer, you have another way to see it.)