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.


  • $\begingroup$ 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, 2015 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, 2015 at 3:22
  • $\begingroup$ I would say that it comes close. Too much is implicit here for my tastes. $\endgroup$
    – Slade
    Dec 14, 2015 at 3:29

3 Answers 3


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 $g\in A_{\mathfrak{p}}.$


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 $S$.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.