About the localization of a UFD 
I was wondering, is the localization of a UFD also a UFD? 

How would one go about proving this? It seems like it would be kind of messy to prove if it is true.
If it is not true, what about localizing at a prime? Or what if the UFD is Noetherian?

As far as I can tell, the localization of a UFD is always a UFD. Let $R$ be a UFD and $S \subseteq R$ multiplicatively closed.
A ring is a UFD if and only if every height 1 prime ideal is principal. So, let $P$ be a height 1 prime ideal of $S^{-1}R$. Then there is a prime ideal $I \lhd R$ such that $P = S^{-1}I$. Now, localization does not change height, so $I$ has height 1, hence it is principal as $R$ is a UFD, say $I = \langle a \rangle$. Then $P = S^{-1}I = S^{-1}\langle a \rangle = \langle \frac{a}{1} \rangle$, so $P$ is principal. Hence, all height 1 prime ideals of $S^{-1}R$ are principal, hence $S^{-1}R$ is a UFD.
Edit: As Arturo mentions, this requires $0 \not\in S$. Also, it requires that $R$ is noetherian.
