Localization is Exact

Statement

Let $R$ be a commutative ring and $S\subseteq R$ a multiplicative set . For any $R$-module $M$, its localization is

Theorem (Exactness of localization).
If

is a short exact sequence of $R$-modules (see short exact sequence ), then

is a short exact sequence of $S^{-1}R$-modules.

Equivalently,

(using tensor product ), so localization is exact because $S^{-1}R$ is a flat $R$-module.

See also: localization of rings .

Examples

1. Killing torsion away from a prime.
   The sequence

   $$0 \to \mathbb Z \xrightarrow{\cdot n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0$$

   is exact. Localize at the prime $p$ (i.e. take $S=\mathbb Z\setminus p\mathbb Z$, giving $\mathbb Z_{(p)}$).
   If $p\nmid n$, then $n$ becomes a unit in $\mathbb Z_{(p)}$, so multiplication by $n$ is an isomorphism after localization. Exactness forces

   $$(\mathbb Z/n\mathbb Z)_{(p)} = 0.$$

2. Localizing at an element makes it invertible.
   In $R=k[x]$, the sequence

   $$0 \to R \xrightarrow{\cdot x} R \to R/(x) \to 0$$

   is exact. Localize at $S=\{1,x,x^2,\dots\}$, obtaining $R_x$.
   Since $x$ is a unit in $R_x$, the map $\cdot x: R_x\to R_x$ is an isomorphism, so exactness gives

   $$(R/(x))_x = 0.$$

3. Injectivity can be checked after localization (common use).
   If $f:M\to N$ is a map of $R$-modules and $S^{-1}f$ is injective, then $\ker(f)$ localizes to $0$.
   Exactness identifies $\ker(S^{-1}f)=S^{-1}\ker(f)$, so injectivity after localization is equivalent to $\ker(f)$ being “$S$-torsion” (every element killed by some $s\in S$).