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 .

Definition

The localization of $M$ at $S$, written $S^{-1}M$, can be defined in either of two equivalent ways:

1. Fractions: elements are symbols $\frac{m}{s}$ with $m\in M$, $s\in S$, modulo the relation

   $$\frac{m}{s}=\frac{m'}{s'} \iff \exists\, t\in S\text{ such that } t(s'm-sm')=0 \text{ in } M.$$

   Operations are

   $$\frac{m}{s}+\frac{m'}{s'}=\frac{s'm+sm'}{ss'},\qquad \frac{r}{u}\cdot \frac{m}{s}=\frac{rm}{us}.$$

   (Here $\frac{r}{u}\in S^{-1}R$ acts on $\frac{m}{s}$.)

2. Tensor product: there is a canonical isomorphism

   $$S^{-1}M \;\cong\; M\otimes_R S^{-1}R,$$

   where $\otimes$ is the tensor product and $S^{-1}R$ is the localized ring .

This makes $S^{-1}M$ naturally into an $S^{-1}R$-module.

Examples

1. Localizing the ring itself.
   Taking $M=R$, one recovers the ring localization:

   $$S^{-1}R \cong S^{-1}M.$$

2. Invert 2 on an abelian group.
   With $R=\mathbb Z$, $M=\mathbb Z$, and $S=\{1,2,2^2,\dots\}$,

   $$S^{-1}M \cong \mathbb Z\left[\tfrac12\right]$$

   as a module over $S^{-1}\mathbb Z$.

3. A localization that kills a module.
   Let $R=k[x]$, $M=R/(x)$, and $S=\{1,x,x^2,\dots\}$. Since $x$ acts as $0$ on $M$ but becomes invertible after localization, one gets

   $$S^{-1}M = 0.$$