# Is any UFD also a PID?

Is there any counterexample that will disprove that every unique factorization domain (UFD) is also a principal ideal domain (PID)? I mean, any PID is a UFD, does the converse hold?

• $\mathbb C[x, y]$ is a UFD but not a PID. – Dustan Levenstein Mar 3 '14 at 4:48
• $\mathbb{Z}[x]$ is another common example of a UFD that is not a PID. – BW. Mar 3 '14 at 4:51
• This Q is surely a duplicate. Perhaps someone can find a link (including the standard example $K[x,y]$). – Martin Brandenburg Mar 3 '14 at 11:55
A PID is always of dim $1$. So you can find a lot of UFD's that are not PID; for example:
Every regular local ring of dim greater than $1$ (i.e the maximal ideal can not generated with one element).