Localization commutes with quotient

Question

Suppose $R$ is a commutative ring and $D$ a multiplicatively closed subset. I'd like to show via universal properties that if $\mathfrak{a} \triangleleft R$ is an ideal, then $D^{-1}R/D^{-1}\mathfrak{a} \cong \bar{D}^{-1}(R/\mathfrak{a})$, where $\bar{D}$ is the image of $D$ in $R/\mathfrak{a}$. This is pretty close to using exactness on the sequence $0 \to \mathfrak{a} \to R \to R/\mathfrak{a} \to 0$, except now we're localizing $R/\mathfrak{a}$ with respect to $\bar{D}$, not just tensoring it with $D^{-1}R$. Essentially I'm trying to show that localization commutes with passing to the quotient by $\mathfrak{a}$.

I could probably brute force this proof, but I'd like to see it in a way that is easy to understand and to remember. I get the feeling it's an easy property in disguise.

Answer 1

Let us define a ring-homomorphism $u:R\to T$ by the universal property that a ring-homomorphism $f:R\to S$ factors (uniquely) through $u$ iff $f$ maps every element of $D$ to a unit and every element of $\mathfrak{a}$ to $0$. From the universal property of $D^{-1}R$, we see that any $f$ which factors through $u$ must factor through the localization map $i:R\to D^{-1}R$ via a unique map $g:D^{-1}R\to R$. Moreover, a map $f$ that factors through $i$ will factor through $u$ iff additionally $f(a)=g(i(a))=0$ for all $a\in \mathfrak{a}$. This happens iff $g$ maps all of $i(\mathfrak{a})$ to $0$. Equivalently, $g$ must factor through the quotient of $D^{-1}R$ by the ideal in $D^{-1}R$ generated by $i(\mathfrak{a})$, which is exactly $D^{-1}\mathfrak{a}$. This shows that maps factoring through $u$ are naturally in bijection with maps factoring through the composition $R\to D^{-1}R\to D^{-1}R/D^{-1}\mathfrak{a}$, so $T\cong D^{-1}R/D^{-1}\mathfrak{a}$.

On the other hand, we can do the same thing but in the reverse order to get that $T\cong \bar{D}^{-1}R/\mathfrak{a}$ as well. By the universal property of $R/\mathfrak{a}$, we see that if $f$ factors through $u$ then it factors through the quotient map $p:R\to R/\mathfrak{a}$ via a unique map $h:R/\mathfrak{a}\to S$. And a map $f$ which factors through $h$ will factor through $u$ iff additionally $f(d)=h(p(d))$ is a unit for all $d\in D$. This is equivalent to saying that $h$ factors through the localization of $R/\mathfrak{a}$ with respect to $p(D)=\bar{D}$. So $f$ factors though $u$ iff it factors through the composition $R\to R/\mathfrak{a}\to \bar{D}^{-1}R/\mathfrak{a}$.

Answer 2

I brute-forced this just for the fun of it, in a slightly different flavour than the above answer. First, the composition $R\to R/\mathfrak{a} \to \overline{D}^{-1}(R/\mathfrak{a})$ sends $D$ to invertible elements and so we get a map $\varphi: D^{-1}R\to \overline{D}^{-1}(R/\mathfrak{a})$ via $\frac{r}{d}\mapsto \frac{[r]}{[d]}$, where the brackets denote mod $\mathfrak{a}$. This map sends $D^{-1}\mathfrak{a}$ to $0$ and so we get a map

$$ \psi: \frac{D^{-1}R}{D^{-1}\mathfrak{a}} \rightarrow \overline{D}^{-1}\left(\frac{R}{\mathfrak{a}}\right) \hspace{.5cm} \text{via} \hspace{.5cm} \left[\frac{r}{d}\right] \mapsto \frac{[r]}{[d]}.$$

Similarly, the composition $R\to D^{-1}R\to \frac{D^{-1}R}{D^{-1}\mathfrak{a}}$ sends $\mathfrak{a}$ to $0$, and so we get a map $\theta: R/\mathfrak{a} \to \frac{D^{-1}R}{D^{-1}\mathfrak{a}}$ via $[r]\mapsto [r/1]$. This map sends $\overline{D}$ to units because $[d/1]\cdot [1/d]=[1/1]$, and so we get a map

$$ \eta: \overline{D}^{-1}\left(\frac{R}{\mathfrak{a}}\right) \rightarrow \frac{D^{-1}R}{D^{-1}\mathfrak{a}} \hspace{.5cm} \text{via} \hspace{.5cm} \frac{[r]}{[d]} \mapsto \left[\frac{r}{d}\right].$$

Thus $\psi$ and $\eta$ are inverses, QED.

Answer 3

Here is an approach that appears to be a little different to the ones already posted.

Theorem. If $A$ is a ring and $S\subseteq A$ is multiplicatively closed, then there is a canonical isomorphism $\frac{S^{-1}A}{S^{-1}I}\cong\overline{S}^{-1}(A/I)$, where $\overline{S}$ is the image of $S$ in $A/I$.

Proof. Let $f$ be the canonical map $A/I\to S^{-1}A/S^{-1}I$. It suffices to show that the pair $(S^{-1}A/S^{-1}I,f)$ satisfies the universal property of localization, i.e. it is initial among $A/I$-algebras $(B,\rho:A/I\to B)$ such that $\rho(x)$ is a unit for all $x\in\overline{S}$.

It is easy to see that $f(x)$ is a unit for all $x\in\overline{S}$. Now suppose $g:A/I\to B$ is a map such that $g(x)$ is a unit for all $x\in\overline{S}$. We must show that there is a unique map $\phi:S^{-1}A/S^{-1}I\to B$ such that $\phi(\overline{a/1})=g(\overline{a})$ for all $a\in A$. Uniqueness is obvious, so it suffices to establish existence.

Let $\pi$ be the quotient map $A\to A/I$. Evidently ($g\circ\pi)(s)$ is a unit for all $s\in S$, so that the map $h:=g\circ\pi:A\to B$ factors as $$ A\longrightarrow S^{-1}A\stackrel{\widetilde{h}}{\longrightarrow} B $$ Moreover, if $x\in S^{-1}I$, say $x=a/s$ where $a\in I$ and $s\in S$, then $$ \widetilde{h}(a/s)=h(a)h(s)^{-1}=0 \, . $$ Hence, $\widetilde{h}$ induces a map $\phi:S^{-1}A/S^{-1}I\to B$ such that for all $a\in A$ and $s\in S$ $$ \phi(\overline{a/s})=\widetilde{h}(a/s)=h(a)h(s)^{-1}=g(\overline{a})g(\overline{s})^{-1} $$ By taking $s=1$, we see that $\phi$ has the desired property. QED.