# Is $(R_S)_{\mathfrak{p}R_S}$ isomorphic to $R_{\mathfrak{p}}$?

Let $R$ be an integral domain, let $S$ be a multiplicative subset of $R$, not intersecting $\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal of $R$. Hence $\mathfrak{p}R_S$ (the ideal generated by $\mathfrak{p}$ in $R_S$) is a prime ideal of $R_S$, and we can take localization $(R_S)_{\mathfrak{p}R_S}$. I'm asking if these two rings: $(R_S)_{\mathfrak{p}R_S}$ and $R_{\mathfrak{p}}$ are isomorphic or not.

I think there is something canonical here, hence i tried to show the existence of an iso only doing general considerations, not dealing with elements and their (possibly complicated) expressions. I used universal property of localization, in this sense: since $S\subseteq R - \mathfrak{p}$, then $R_S$ is contained in $R_{\mathfrak{p}}$ (both subrings of field of fractions of $R$), and let $i$ be the natural inclusion of $R_S$ in $R_{\mathfrak{p}}$. Clearly, $i$ sends elements of $S$ in invertible elements of $R_{\mathfrak{p}}$. Now, let $\phi$ be the canonical morphism from $R_S$ to its localization $(R_S)_{\mathfrak{p}R_S}$. By universal property, there is exactly one morphism of rings , say $h$, from $(R_S)_{\mathfrak{p}R_S}$ to $R_{\mathfrak{p}}$ such that $h\circ\phi=i$, which is obviously injective.

Now, is $h$ also surjective?

Yes, this is true for arbitrary rings. There is a canonical isomorphism between the two rings. This is Corollary 4 in section 4 of Matsumura's Commutative Ring Theory.

The map $R\rightarrow R_\mathfrak{p}$ factors uniquely through $R_S$ because $S\subseteq R\setminus\mathfrak{p}$, so you get a map $R_S\rightarrow R_\mathfrak{p}$ which visibly sends $(R_S\setminus\mathfrak{p}R_S)$ to units, so you get an induced map $\varphi:(R_S)_{\mathfrak{p}R_S}\rightarrow R_\mathfrak{p}$. For the other direction, consider the composite $R\rightarrow R_S\rightarrow(R_S)_{\mathfrak{p}R_S}$. This sends $R\setminus\mathfrak{p}$ to units, so it factors uniquely through $R_\mathfrak{p}$, giving a map $\psi:R_\mathfrak{p}\rightarrow(R_S)_{\mathfrak{p}R_S}$.

Now you want to verify that these maps are inverses. This amounts to the fact that localizations have no endomorphisms, i.e., to prove that the composite $\varphi\circ\psi:R_\mathfrak{p}\rightarrow R_\mathfrak{p}$ is the identity, it suffices to verify that it is an $R$-algebra map (the universal property of localization then implies it must be the identity). But this is true by construction. The other composite is similar (although maybe a little more fussy).

This gets used a lot, for example in proving the various equivalent characterizations of finite locally free modules. It is also the key to proving that the morphism $\mathrm{Spec}(R_f)\rightarrow\mathrm{Spec}(R)$ is an open immersion for any $f\in R$.

• but, canonical isomorphism is the $h$ in my question? If yes, how can i see surjectivity? – Federica Maggioni Mar 30 '13 at 16:32
• You can also see this as a special case of exercise 3 in Chapter 3 of Atiyah-MacDonald (which is a bit more general). – A.P. Mar 30 '13 at 16:36
• Yes, it is $h$, but I wouldn't recommend trying to show directly that it is surjective. Just show that it has an inverse. Everything follows from the universal properties of the localizations involved. Showing directly surjectivity is just messy (fractions of fractions and so forth). – Keenan Kidwell Mar 30 '13 at 16:37

If $T$ is a commutative ring, then

$$\begin{array}{ccl} \hom((R_S)_{\mathfrak{p} R_S},T) & \cong & \{f \in \hom(R_S,T) : f(R_S \setminus \mathfrak{p} R_S) \subseteq T^*\} \\ & \cong & \{f \in \hom(R,T) : f(S) \subseteq T^*, f(R \setminus \mathfrak{p}) \subseteq T^*\} \\ & = & \{f \in \hom(R,T) : f(R \setminus \mathfrak{p}) \subseteq T^*\} \\ & \cong & \hom(R_{\mathfrak{p}},T)\end{array}$$

The Yoneda Lemma implies $(R_S)_{\mathfrak{p} R_S} \cong R_{\mathfrak{p}}$.

Some more details: The first and last isomorphisms come from the universal property of localizations. The equality comes from the assumption $S \subseteq R \setminus \mathfrak{p}$ (and basically this is already the whole idea behind the proof). The second isomorphism comes from the universal property of the localization $R_S$, as well as the observation $R_S \setminus \mathfrak{p} R_S$ differs only by units from the image of $R \setminus \mathfrak{p}$ in $R_S$.

Bonus exercise: Try to find the proof of the Yoneda Lemma in Keenan's answer.

Alternative proof using schemes: Using directed colimits, we may assume that $S=\{1,f,f^2,\dotsc\}$ for some $f \in R$. Open immersions are isomorphisms on stalks. Applying this to $D(f) \hookrightarrow \mathrm{Spec}(R)$, we get $(R_f)_{\mathfrak{p} R_f} = R_\mathfrak{p}$.