Localization of Modules
Key Definitions
Definition.
Let $R$ be a commutative ring with unity, and let $S \subset R$ be a multiplicatively closed subset.
If $M$ is an $R$–module, the localization of $M$ with respect to $S$ is
\[S^{-1}M = \left\{ \frac{m}{s} : m \in M, \ s \in S \right\} / \sim,\]
where the equivalence relation is given by
\[\frac{m_1}{s_1} \sim \frac{m_2}{s_2} \quad \Longleftrightarrow \quad \exists\, s \in S \ \text{such that } \ s(s_1 m_2 - s_2 m_1) = 0.\]
$S^{-1}M$ is naturally both an $R$–module and an $S^{-1}R$–module.
If $\mathfrak{p}$ is a prime ideal of $R$, we define the localization of $M$ at $\mathfrak{p}$ by
\[M_{\mathfrak{p}} = S_{\mathfrak{p}}^{-1}M, \quad S_{\mathfrak{p}} = R \setminus \mathfrak{p}.\]
Key Theorems
Theorem (Localization is an exact functor).
The localization functor
\[S^{-1} : \text{Category of $R$–modules} \;\longrightarrow\; \text{Category of $S^{-1}R$–modules}\]
is exact. In other words, it preserves kernels and cokernels (more generally, all limits and colimits).
Theorem (Localization as Tensor Product).
For any $R$–module $M$, there is a natural isomorphism
\[S^{-1}R \otimes_R M \;\cong\; S^{-1}M,\]
given by $\big(\tfrac{r}{s}, m\big) \mapsto \tfrac{rm}{s}$.
Theorem.
For an $R$–module $M$, the following are equivalent:
- $M = 0$.
- $M_{\mathfrak{p}} = 0$ for all prime ideals $\mathfrak{p}$ of $R$.
- $M_{\mathfrak{m}} = 0$ for all maximal ideals $\mathfrak{m}$ of $R$.
Remark.
From a geometric perspective, modules correspond to quasi-coherent sheaves on the affine scheme $X = \operatorname{Spec}(R)$, which we may think of as collections of generalized functions on a geometric space.
- Maximal ideals correspond to closed points of $X$.
- Localizing $M$ at a maximal ideal $\mathfrak{m}$ gives the germ of sections near the point $\mathfrak{m}$.
The theorem above can be interpreted as saying: if a function vanishes at every point (i.e., its germ is zero everywhere), then the function is identically zero globally.
Corollary.
Let $f : M \to N$ be an $R$–module homomorphism. Then the following are equivalent
- $f$ is injective/surjective.
- $f_{\mathfrak{p}} : M_{\mathfrak{p}} \to N_{\mathfrak{p}}$ is injective/surjective for all prime ideals $\mathfrak{p}$ of $R$.
- $f_{\mathfrak{m}} : M_{\mathfrak{m}} \to N_{\mathfrak{m}}$ is injective/surjective for all maximal ideals $\mathfrak{m}$ of $R$.
Proof (Sketch).
Since localization is an exact functor, it preserves kernels. In particular,
\[(\ker f)_{\mathfrak{p}} = \ker(f_{\mathfrak{p}}) \quad \text{for all prime ideals } \mathfrak{p} \subset R.\]
Therefore, $\ker f = 0$ if and only if $\ker f_\mathfrak{p} = (\ker f)_\mathfrak{p} = 0$ for all prime ideals $\mathfrak{p}$ of $R$. This translates to
$f$ is injective if and only if $f_\mathfrak{p}$ is injective for all prime ideals $\mathfrak{p}$ of $R$.