# Does quotient commute with localization?

Let $R$ be a commutative ring, and $I \subset R$ an ideal. If we choose an element $x \in R$ we can consider $(R/I)_x$ and $R_x/I_x$. In general, does localization commute with quotient? i.e. $(R/I)_x \simeq R_x/I_x$? If not... are there hypotheses on $x$, $R$ or $I$ under which localization commutes with quotient?

In general, if $$0\to M\to N\to P\to0$$ is a short exact sequence of $R$-modules and $S$ is a multiplicative set in $R$, the localized sequence $$0\to M_S\to N_S\to P_S\to0$$ is also exact.

If $I\subseteq R$ is an ideal, we have a short exact sequence $$0\to I\to R\to R/I\to0$$ of $R$-modules, and therefore for $x\in R$ we get that $$0\to I_x\to R_x\to (R/I)_x\to0$$ is also exact. This means, among other things, that $(R/I)_x$ is isomorphic to $R_x/I_x$.

This isomorphism is as a $R$-module, and you probably want it to be as rings: that's a little extra work: the above map, which we have just observed to be a bijection, is actually a map of rings —checking this is simply a matter of writing down what it means.

• In particular my dubt is the follow: $R=\mathbb{C}[x,y,z]/(x^3+y^3+z^3)$, and $I=(y,bz-x)$, with $b^3=-1$, then is it true that $(R_x)/I \simeq \mathbb{C}[x]_x$? Sep 23, 2013 at 13:54
• Because I used that localization commute with the quotient, and maybe this is a general result. Sep 23, 2013 at 13:59
• What I wrote is the general result. Sep 23, 2013 at 13:59
• But my quotient is a quotient of rings... However I think that is correct to say that $(R_x)/I \simeq \mathbb{C}[x]_x$ Sep 23, 2013 at 14:02
• Well... I suggest you read a bit more about what the ideals in localizations look like. I am not getting anywhere here, so I will now stop. Sep 23, 2013 at 15:40

Let me add some details to Mariano's answer. The exact sequence gives an isomorphism of $$R_x$$‑modules $$\varphi:R_x/I_x\to (R/I)_x$$ where explicitely we have $$\varphi(\overline{a/s})=\overline{a}/s$$.

Now we want to show that this is actually an isomorphism of rings. Every isomorphism of groups $$\psi$$ between a ring and a group $$G$$ gives a ring structure on $$G$$ by defining multiplication between two elements of $$G$$ using $$\psi$$; explicitely $$g\times g'=\psi(\psi^{-1}(g)\psi^{-1}(g'))$$. Moreover, with $$G$$ under this additional structure, $$\psi$$ gives an isomorphism of rings.

In our case, notice that $$R_x/I_x$$ is a ring because the scalar multiplication structure we get when localizing a ring as a module over itself is exactly the same, by definition, as the multiplication structure we get when localizing a ring as a ring. Therefore we can apply the previous paragraph to $$\varphi$$ and obtain a ring structure on the $$R_x$$‑module $$(R/I)_x$$, plus $$\varphi$$ is now an isomorphism of rings. The multiplication in $$(R/I)_x$$ is given by $$(\overline{a}/s)\times (\overline{b}/s')=\overline{ab}/ss'.$$

Consider the ring $$(R/I)_\overline{x}$$; be careful, there's a bar over the $$x$$, we're localizing the ring $$R/I$$ at its element $$\overline{x}$$. Now consider $$i:R/I\to (R/I)_\overline{x}$$ and $$j:R/I\to (R/I)_x$$ the canonical maps of rings. For every $$\overline{s}$$ in the multiplicative set $$S$$ generated by $$\overline{x}$$ in $$R/I$$, $$j(\overline{s})$$ is a unit with inverse $$\overline{1}/s$$. Therefore by the universal property of localization of a ring there exists a unique morphism of rings $$\vartheta:(R/I)_\overline{x}\to(R/I)_x$$ such that $$\vartheta\circ i=j$$. The explicit expression of this morphism is $$\vartheta(\overline{a}/\overline{s})=\overline{a}/s$$.

Let's show that $$\vartheta$$ is actually an isomorphism of rings. It is clearly surjective. Now to prove that it is injective: suppose $$\vartheta(\overline{a}/\overline{s})=\overline{a}/s=0$$. This means there exists some $$u\in S$$ such that $$u\overline{a}=0$$ in the $$A$$‑module $$A/I$$. Thus $$\overline{u}\cdot\overline{a}=0$$, which implies $$\overline{a}=0$$ in $$(R/I)_\overline{x}$$. This shows that the kernel of $$\vartheta$$ is trivial.

To conclude, we have proven that $$R_x/I_x$$ and $$(R/I)_\overline{x}$$ are isomorphic (as rings). Notice that we could have replaced the localization at $$x$$ with localization at an arbitrary muliplicative set $$S$$ and the argument would still work. So in general $$R_S/I_S$$ and $$(R/I)_S$$ are isomorphic as rings.