Prove that a ring of fractions of $\mathbb Q[x]$ is a PID

Question

I'm trying to prove the following:

Let $$ R = \left\{\frac{f(x)}{g(x)}\mid f(x),g(x)\in\mathbb{Q}[x], g(x)\not=0\right\}\subseteq\mathbb{Q}(x). $$ Show that $R$ is a principal ideal domain.

My approach was to first prove that for any integral domain $S$ we have $$ \begin{equation} \text{Frac}(S[x])\cong\text{Frac}(S)(x) \end{equation} $$ and so $R\cong\mathbb{Q}(x)$, hence it's a field and so a PID. But I couldn't come up with a proof of this statement. Is this the right approach or is there a better way?

Why didn't try to prove that $R$ is a field directly? – user26857 Mar 19 '21 at 20:18 — user26857, Mar 19 '21 at 20:18

score 1 · Accepted Answer · answered Dec 09 '20 at 16:51

1

$\Bbb Q$ is a field $\implies \Bbb Q[X]$ is a PID. Then as $R=S^{-1}(\Bbb Q[X])$ where $S=\Bbb Q[X] \setminus \{0\}$ and localization of a PID is a PID, hence, $R$ is also a PID.

answered Dec 09 '20 at 16:51

Brozovic

2,264

Hey, thank you! Do you have a link for a proof that a localization of PID is a PID? I haven't seen this result before... – 14159 Dec 10 '20 at 11:03
@14159 It's trivial to prove, anyway can see a proof here – Brozovic Dec 15 '20 at 13:28
If $R$ is an integral domain and $S=R\setminus{0}$, then $S^{-1}R$ is a field. – user26857 Mar 19 '21 at 20:16

Prove that a ring of fractions of $\mathbb Q[x]$ is a PID

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