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I would like a verification of a proof for the following statement. Let $S$ be a multiplicatively closed subset of a ring $R$. If $R$ is a PID, then $S^{-1}R$ is a PID.

Let $I = \left<r_1/s_1, r_2/s_2\ldots\right>$ be an ideal of $S^{-1}R$. Since $1/s_i$ is a unit in $S^{-1}R$, we have $\left<r_i/s_i\right> = \left<r_i\right>$. But then $I = \left<r_1/1,r_2/1,\ldots\right>$ and so we may think of $I$ as an ideal in $R$. Since $R$ is a PID, we have $I = \left<x\right>$ for some $x \in R$. So $I = \left<x/1\right>$, thus $S^{-1}R$ is a PID.

Thanks!

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  • $\begingroup$ The proof is not valid as written. $I$ is a subset of $S^{-1}R$, so you need to justify "thinking of it" as a subset of $R$. The ideal with the same generators is principal in $R$, but you need to prove that this implies that $I$ is principal, not just state it. $\endgroup$
    – Slade
    Dec 14 '15 at 3:18
  • $\begingroup$ Ah, ok. But what I have written does establish that an ideal of $S^{-1}R$ is a of the form $JS^{-1}R$, where $J$ is an ideal of $R$, correct? $\endgroup$ Dec 14 '15 at 3:22
  • $\begingroup$ I would say that it comes close. Too much is implicit here for my tastes. $\endgroup$
    – Slade
    Dec 14 '15 at 3:29
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Let me give a proof to

If $A$ is a PID, and $\mathfrak{p}$ a prime ideal in $A$, then the localization $A_{\mathfrak{p}}$ is also a PID.

Proof. Suppose $\mathfrak{a} \subset A_{\mathfrak{p}}$ be an ideal. We need to show that $\mathfrak{a}$ is principal, i.e. $\mathfrak{a}= (f)$ for some $f\in A_{\mathfrak{p}} $. Recall (see Prop. 3.11 in Atiyah-Macdonald) that every ideal in the localization is an extended ideal. Hence there exists an ideal $\mathfrak{b} =(g)\subset A$ such that $(g)^e = \mathfrak{a}$. Now given the cannonical map $\varphi:A\to A_{\mathfrak{p}}, \ x \mapsto x/1$, $$ \mathfrak{a} = \mathfrak{b}^e :=\langle \varphi(g) \rangle = (g/1). $$ Thus $A_{\mathfrak{p}}$ is a PID.

This is just a remark that above result gives a nice way of describing the Zariski-open sets in the scheme $\mathrm{Spec}A_{\mathfrak{p}}$ if $A$ is a PID and $\mathfrak{p} \subset A$ is a prime ideal. For a PID $A$, one shows that open sets are same as basic open sets $D(f)$. Indeed, if $U\subset \mathrm{Spec}A$ is open, then $U=V(\mathfrak{a})^c=V(f)^c=:D(f)$ for some $f\in A$ that generates the ideal $\mathfrak{a}.$ Because of the above result in the grey box, we know $A_{\mathfrak{p}}$ is a PID, and hence every open set in $\mathrm{Spec}A_{\mathfrak{p}}$ is of the form $D(g)$ for some $g\in A_{\mathfrak{p}}.$

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the idea behind your method is sound, but it is a little confusing, perhaps, to argue merely that we may think of I as an ideal in R.

the difficulty that this phraseology avoids is that we cannot a priori rule out the presence of a non-finitely generated ideal in $S^{-1}R$. to rule this out we may seek to use the result that the localization of a Dedekind domain at a multiplicative set is again a Dedekind domain. thus any ideal of the localized domain $S^{-1}R$ is a product of prime ideals. this finiteness condition allows us to slide the ideal back up into $R$ through multiplication by a suitably chosen element of $S$.

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Since every ideal of a localization is an extended ideal, so the result is true.

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