UFD not implies PID

Statement
A unique factorization domain (UFD) need not be a principal ideal domain (PID).

More general results
There are many ways of seeing this -- one approach is to observe that a polynomial ring over a UFD continues to remain a UFD; however, the polynomial ring over a PID is not necessarily a PID (in fact, a polynomial ring over a ring is a PID iff the base ring is a field). This gives examples of UFDs which are not PIDs:


 * Polynomial ring in two or more variables over a field
 * Polynomial ring over the integers

Partial truth
It is true that under certain circumstances, being a UFD is the same as being a PID. An example of such circumstances is a Dedekind domain. In particular, all rings of integers in number fields are Dedekind domains, and hence any such ring is a PID iff it is a UFD.