Exactness of localization

Localization is not just a way to invert elements in a ring (localization of rings) ; it is also a well-behaved operation on modules (localization of modules) . One of its most important formal properties is that it preserves exact sequences.

Theorem

Let $R$ be a commutative ring , let $S$ be a multiplicative set in $R$, and let

be an exact sequence of $R$-modules . Then the localized sequence

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

Equivalently, for any $R$-linear map $h\colon M\to N$,

• $S^{-1}(\ker h)=\ker(S^{-1}h)$,
• $S^{-1}(\operatorname{im} h)=\operatorname{im}(S^{-1}h)$, so localization commutes with kernels and images.

One convenient conceptual reformulation is that localization is extension of scalars along the ring map $R\to S^{-1}R$:

and this tensor description is what makes the exactness behave so cleanly.

Examples

1. Localizing a short exact sequence over $\mathbb Z$.
   Consider

   $$0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0.$$

   Localize at $S=\mathbb Z\setminus (p)$ (so $S^{-1}\mathbb Z=\mathbb Z_{(p)}$). Exactness says

   $$0\to \mathbb Z_{(p)} \xrightarrow{\times n} \mathbb Z_{(p)} \to (\mathbb Z/n\mathbb Z)_{(p)} \to 0$$

   is exact. Concretely:

   • if $p\nmid n$, then multiplication by $n$ becomes an isomorphism on $\mathbb Z_{(p)}$, so $(\mathbb Z/n\mathbb Z)_{(p)}=0$;
   • if $n=p^k m$ with $p\nmid m$, then $m$ becomes invertible in $\mathbb Z_{(p)}$, and the cokernel is “the $p$-primary part,” isomorphic to $\mathbb Z/p^k\mathbb Z$ as a $\mathbb Z_{(p)}$-module.

2. Localization can kill torsion.
   Let $R=k[x]$ and consider the exact sequence

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

   Localize at $S=\{1,x,x^2,\dots\}$, so $S^{-1}R=R_x$. Then $x$ becomes a unit in $R_x$, hence multiplication by $x$ on $R_x$ is an isomorphism. Exactness forces

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

   which reflects that $R/(x)$ is “$x$-torsion.”

3. Exactness on kernels and images.
   Let $R=\mathbb Z$, $h\colon \mathbb Z\to\mathbb Z$ be multiplication by $6$, and localize at $S=\{1,2,2^2,\dots\}$. Then $\ker h=0$ and exactness implies $\ker(S^{-1}h)=0$ as well. Meanwhile $\operatorname{im} h=6\mathbb Z$ localizes to $6\mathbb Z[1/2]$, which equals the image of the localized map $\mathbb Z[1/2]\xrightarrow{\times 6}\mathbb Z[1/2]$.

Exactness is also a key input to the prime correspondence under localization; see prime correspondence under localization .