Is $A_P$ a GCD domain? Is it true that if $A$ is a GCD domain, then $A_P$ (the localization at $P$) is a GCD domain for every prime $P$?
Thanks for you help :)
 A: Here is an "element-wise" solution. We'll prove the following more general result (this appears as exercise 15.10 in Pete L. Clark's notes on commutative algebra).
Theorem: Let $D$ be a GCD domain. If $S$ is a multiplicatively closed subset of $D$ such that $0\notin S$, then $S^{-1}D$ is a GCD domain.
Proof: Let $\alpha, \beta\in S^{-1}D$. If we write $\alpha=a/s$ and $\beta=b/t$, with $a,b\in D$ and $s,t\in S$, we claim that $\gcd(\alpha,\beta)=\gcd(a,b)/1$. Then it follows immediately that $S^{-1}D$ is a GCD domain.
Let's put $\gcd(a,b)=d$, so $d\mid a$ and $d\mid b$. This means there are $a',b'\in D$ such that $a=da'$ and $b=db'$. Then $$\frac{d}{1}\cdot \frac{a'}{s}=\frac{a}{s},$$ $$\frac{d}{1}\cdot \frac{b'}{t}=\frac{b}{t}.$$
Therefore $d/1\mid a/s$ and $d/1\mid b/t$. Now, let $\gamma\in S^{-1}D$ such that $\gamma\mid a/s$ and $\gamma\mid b/t$. If we write $\gamma=c/u$, with $u\in S$, then there are $a''/s', b''/t'\in S^{-1}D$ ($s',t'\in S$) such that $c/u\cdot a''/s'=a/s$ and $c/u\cdot b''/t'=b/t$. Thus $$\frac{a}{s}=\frac{ca''}{us'},$$ $$\frac{b}{t}=\frac{cb''}{ut'}.$$
This lead us to $aus'=ca''s$ and $but'=cb''t$, so if we set $us'=s_1$ and $ut'=s_2$ we have $s_1,s_2\in S$ and $c\mid as_1$, $c\mid bs_2$. Then $c\mid as_1s_2$ and $c\mid bs_1s_2$. Therefore $$c\mid \gcd(as_1s_2,bs_1s_2)=s_1s_2\gcd(a,b)=s_1s_2d.$$
Let's set $s_1s_2=s_3$, then $s_3\in S$ and $c\mid ds_3$. Thus there is $x\in D$ such that $ds_3=cx$ and hence $$\frac{c}{u}\cdot \frac{xu}{s_3}=\frac{d}{1}.$$
This means that $\gamma\mid d/1$ and so we conclude that $d/1=\gcd(\alpha, \beta)$.
