Localization of a module

Let $R$ be a commutative ring , let $S\subseteq R$ be a multiplicative set , and let $M$ be an $R$-module. The localization of $M$ at $S$ is an $S^{-1}R$-module, denoted $S^{-1}M$, constructed so that every $s\in S$ acts invertibly on $S^{-1}M$.

Here $S^{-1}R$ is the localized ring .

Construction (fractions in a module)

Define $S^{-1}M$ as equivalence classes of pairs $(m,s)\in M\times S$ under

Write the class of $(m,s)$ as $\frac{m}{s}$. Addition is

and the scalar action of $S^{-1}R$ is given by

The map $\iota_M:M\to S^{-1}M$ given by $\iota_M(m)=\frac{m}{1}$ is $R$-linear.

Universal property

Let $N$ be an $S^{-1}R$-module. Viewing $N$ as an $R$-module via the canonical map $R\to S^{-1}R$, every $s\in S$ acts by an automorphism on $N$. The localization $S^{-1}M$ is characterized by:

For every $R$-linear map $f:M\to N$, there exists a unique $S^{-1}R$-linear map $\widetilde f:S^{-1}M\to N$ with $\widetilde f\circ\iota_M=f$.

In particular, localizing at a prime $\mathfrak p$ means taking $S=R\setminus\mathfrak p$ and writing

in parallel with localization of rings at a prime .

Localization interacts well with exact sequences: it is an exact functor on modules (see exactness of localization and compare with the general notion of an exact sequence ).

Finally, localization can be expressed as a base change: via extension of scalars there is a natural isomorphism

Examples

1. Localizing a quotient. If $I\subseteq R$ is an ideal, then

   $$S^{-1}(R/I)\ \cong\ (S^{-1}R)/(S^{-1}I),$$

   where $S^{-1}I$ denotes the image of $I$ in $S^{-1}R$.

2. Torsion killed by localization. Take $R=\mathbb Z$, $M=\mathbb Z/n\mathbb Z$, and localize at $S=\mathbb Z\setminus (p)$ (so $S^{-1}\mathbb Z=\mathbb Z_{(p)}$).

   • If $p\nmid n$, then $n\in S$ becomes a unit, so $S^{-1}M=0$.
   • If $p\mid n$, then $S^{-1}M\cong \mathbb Z_{(p)}/n\mathbb Z_{(p)}$, which is generally nonzero.

3. Making an element invertible forces a module to vanish. Let $R=k[x]$, $M=R/(x)$, and $S=\{1,x,x^2,\dots\}$. In $S^{-1}R$ the element $x$ is a unit, but $x$ annihilates $M$, so $S^{-1}M=0$.