How to determine whether a unique factorization domain is a principal ideal domain? Could someone please provide an example of a unique factorization domain that is not a principal ideal domain?
Furthermore, is there some way to determine whether a UFD is a PID?
 A: There is a theorem which says that if $R$ is a UFD then $R[x]$ is a UFD. The question then is whether $R$ is a PID implies that $R[x]$ is a PID, and the answer is no. In fact, it's a common fact that $R[x]$ is a PID if and only if $R$ is a field, in which case $R[x]$ is actually a Euclidean domain. So, if $R$ is a UFD which is not a field (e.g. $\mathbb{Z}$) then $R[x]$ is a UFD which is not a PID.
If you are asking about what extra conditions need to be imposed on a UFD to make it a PID the answer is the notion of a Dedekind domain. In other words, a ring $R$ is a PID iff it's a UFD which is also a Dedekind domain.
A: $\rm PID\:$s are precisely the $\rm UFD\:$s having dimension $\le 1\:,\ $ i.e. such that prime ideals $\ne 0$ are maximal. Below is a sketch of a proof of this and closely related results.
THEOREM $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$ 
$(1)\ \ $ prime ideals are maximal if nonzero
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$ 
$(5)\ \ $ $\rm D$ is Bezout
$(6)\ \ $ $\rm D$ is a $\rm PID$  
Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)
$(1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$ 
$(2\Rightarrow 3)$ $\ \: $ Clear.
$(3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$ 
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$ 
$(5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors
$(6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$
A: Alex Youcis's answer essentially deals with your question, but one can unwrap the Dedekind domain condition slightly (since UFDs are automatically integrally closed in their quotient fields).  
So what one finds is that a UFD is a PID precisely if it is Noetherian and of Krull dimension one (i.e. if all non-zero prime ideals are maximal).
One way to understand the Krull dimension condition is to consider the case of polymomial rings over UFDs (already discussed in Alex's answer):
If $R$ is a Noetherian ring of finite Krull dimension $d$, then $R[x]$ is of Krull dimension $d + 1$, so (as Alex notes) polynomial rings over UFDs are never
PIDs unless the UFD $R$ is of Krull dimension zero, i.e. is a field.
