# Localization commutes with quotient

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.

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 $R\to R/\mathfrak{a}\to \bar{D}^{-1}R/\mathfrak{a}$.

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.

• This is the most natural proof! Jan 23, 2020 at 16:33