Localization - Key Proof

We prove that localization is exact, one of its most important properties, and demonstrate the local-global principle.

TheoremExactness of Localization

Let $S \subseteq R$ be a multiplicative set. If $M' \xrightarrow{f} M \xrightarrow{g} M''$ is an exact sequence of $R$ -modules, then: $S^{-1}M' \xrightarrow{S^{-1}f} S^{-1}M \xrightarrow{S^{-1}g} S^{-1}M''$ is exact.

ProofProof of Exactness

We need to show $\text{Im}(S^{-1}f) = \ker(S^{-1}g)$ .

( $\subseteq$ ): If $m'/s \in \text{Im}(S^{-1}f)$ , then $m'/s = (S^{-1}f)(m/t)$ for some $m/t \in S^{-1}M'$ . Applying $S^{-1}g$ : $(S^{-1}g)(m'/s) = (S^{-1}g)((S^{-1}f)(m/t)) = (g \circ f)(m)/st = 0/st = 0$ since $g \circ f = 0$ by exactness of the original sequence.

( $\supseteq$ ): Suppose $(S^{-1}g)(m/s) = 0$ in $S^{-1}M''$ . Then $g(m)/s = 0$ , so there exists $t \in S$ with $t \cdot g(m) = g(tm) = 0$ in $M''$ .

By exactness, $tm \in \ker(g) = \text{Im}(f)$ , so $tm = f(m')$ for some $m' \in M'$ . Therefore: $\frac{m}{s} = \frac{tm}{ts} = \frac{f(m')}{ts} = (S^{-1}f)\left(\frac{m'}{ts}\right)$

Thus $m/s \in \text{Im}(S^{-1}f)$ , completing the proof.

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Remark

The proof relies crucially on the ability to "clear denominators" using elements of $S$ . This technique is fundamental throughout localization theory.

TheoremLocal-Global Principle (Detailed)

An $R$ -module $M$ is zero if and only if $M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$ .

ProofProof of Local-Global Principle

( $\Rightarrow$ ): Clear, since localizing the zero module gives zero.

( $\Leftarrow$ ): Suppose $M \neq 0$ and let $m \in M$ be non-zero. Consider the annihilator: $I = \text{ann}(m) = \{r \in R : rm = 0\}$

This is a proper ideal since $1 \notin I$ (as $m \neq 0$ ). By the existence of maximal ideals, there exists a maximal ideal $\mathfrak{m}$ containing $I$ .

In $M_\mathfrak{m}$ , the element $m/1 \neq 0$ because if $m/1 = 0$ , then there exists $s \in R \setminus \mathfrak{m}$ with $sm = 0$ , contradicting $s \notin \mathfrak{m} \supseteq I = \text{ann}(m)$ .

Therefore, if $M_\mathfrak{m} = 0$ for all maximal $\mathfrak{m}$ , we must have $M = 0$ .

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ExampleApplication to Injectivity

To show $f: M \to N$ is injective, it suffices to show $f_\mathfrak{m}: M_\mathfrak{m} \to N_\mathfrak{m}$ is injective for all maximal $\mathfrak{m}$ .

This follows by applying the local-global principle to $\ker(f)$ : if $\ker(f)_\mathfrak{m} = \ker(f_\mathfrak{m}) = 0$ for all maximal $\mathfrak{m}$ , then $\ker(f) = 0$ , so $f$ is injective.

ProofProof of Universal Property

Given $\phi: R \to T$ with $\phi(S) \subseteq T^\times$ , define $\tilde{\phi}: S^{-1}R \to T$ by $\tilde{\phi}(r/s) = \phi(r)\phi(s)^{-1}$ .

Well-defined: If $r/s = r'/s'$ , then $t(rs' - r's) = 0$ for some $t \in S$ . Applying $\phi$ : $\phi(t)(\phi(r)\phi(s') - \phi(r')\phi(s)) = 0$

Since $\phi(t)$ is a unit, $\phi(r)\phi(s') = \phi(r')\phi(s)$ . Multiplying by $\phi(s)^{-1}\phi(s')^{-1}$ gives $\phi(r)\phi(s)^{-1} = \phi(r')\phi(s')^{-1}$ .

Homomorphism: Direct verification shows $\tilde{\phi}$ respects addition and multiplication.

Uniqueness: Any homomorphism $S^{-1}R \to T$ extending $\phi$ must send $r/s$ to $\phi(r)\phi(s)^{-1}$ since $(r/s) \cdot (s/1) = r/1$ in $S^{-1}R$ .

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These proofs establish the foundational properties of localization through elementary but careful manipulation of fractions and ideals.