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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 :)

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    What's up, Alopiso, and @Xam, asking or answering a PSQ (problem, likely from homework, that adds no context whatsoever)? – 2017-11-22
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    @amWhy at that time I wasn't really interested in handling PSQ posts or similar. My thoughts about such posts have changed along the time since I became an active member of this site. Certainly now I care about these low quality questions, but I don't think it's necessary to delete my answer because I actually use it as a reference for such result (which is not very popular in commutative algebra textbooks). Btw, I also don't think my answer deserves a downvote. And finally, since it'd be kind of weird to close a question that I answered, I won't vote to close it. – 2017-11-22
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    It's natural to came up with this kind of questions when one is studying some special integral domains like UFDs, Bezout domains, GCD domains, etc. – 2017-11-22

1 Answers 1

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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 $d/1=\gcd(\alpha, \beta)$.