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?
Asked
Active
Viewed 1.1k times
2 Answers
47
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.
Mariano Suárez-Álvarez
- 135,076
-
4In 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 $(R_x)/I \simeq \mathbb{C}[x]_x$? – ArthurStuart Sep 23 '13 at 13:54
-
If that is your doubt, then why not ask about it in the question?! – Mariano Suárez-Álvarez Sep 23 '13 at 13:55
-
4Because I used that localization commute with the quotient, and maybe this is a general result. – ArthurStuart Sep 23 '13 at 13:59
-
2What I wrote is the general result. – Mariano Suárez-Álvarez Sep 23 '13 at 13:59
-
2But my quotient is a quotient of rings... However I think that is correct to say that $(R_x)/I \simeq \mathbb{C}[x]_x$ – ArthurStuart Sep 23 '13 at 14:02
-
1If you read what I wrote, you will notice that I did say that the isomorphism is of rings... – Mariano Suárez-Álvarez Sep 23 '13 at 14:06
-
2In any case, your statement does not make sense: $I$ is not an ideal in $R_x$. – Mariano Suárez-Álvarez Sep 23 '13 at 14:06
-
2@MarianoSuárez-Alvarez why $I$ is not an ideal in $R_x$? – ArthurStuart Sep 23 '13 at 15:24
-
1Because it isn't! To pick a simpler example: the ideal $I=(x)$ in the ring $\mathbb C[x]$ is not an ideal of the ring $\mathbb C[x]_x$, simply because $I$ is not closed by multiplication by arbitrary elements of $\mathbb C[x]_x$, like $\tfrac1x$. – Mariano Suárez-Álvarez Sep 23 '13 at 15:26
-
2@MarianoSuárez-Alvarez $I=(y,bz-x)$ doesn't contain $x$... – ArthurStuart Sep 23 '13 at 15:38
-
10Well... 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. – Mariano Suárez-Álvarez Sep 23 '13 at 15:40
5
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.
