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Let $A$ be a commutative ring, $S \subset A $ a multiplicatively closed set and $M$ an $A$-module.

For every $s \in S$ we denote by $M_{s}$ the localization of $M$ with respect to $\{ 1, s, s^2, ...\}$. Then we have a direct system $\{ M_{s}\}_{s \in S}$ with respect to inclusion. Infact for every $s, s' \in S$ we have $$M_{s} \hookrightarrow M_{ss'}$$ $$M_{s'} \hookrightarrow M_{ss'}$$ I have to prove that $$S^{-1}M = \underrightarrow{\lim}(M_{s})$$

Obviously $S^{-1}M$ with inclusions $i_{s} : M_{s} \to S^{-1}M$ is a cocone, but why it has the colimit universal property ?

WLOG
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1 Answers1

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$\DeclareMathOperator{\h}{Hom}$ $S^{-1}M$ is characterized by the universal property $\h_{S^{-1}A}(S^{-1}M,T) \cong \h_A(M,T_{|A})$. In order to show that $L := \lim_{s \in S} M_s$ satisfies this property there are two things to do:

  1. Each $s \in S$ acts as an automorphism on $L$ (so $L$ carries a natural $S^{-1}A$-module structure).
  2. For each $A$-Module $T$ with the same property and each homomorphism $f : M \to T$ there is exactly one homomorphism $f^* : L \to T$ extending $f$.

The first point follows from the fact that each $s \in S$ acts as an automorphism on $M_{ss'}$ for all $s' \in S$, and every element of $L$ lies in the image of some $\iota_s : M_s \to L$. Regarding the second point, note that $f$ extends to morphisms $f_s : M_s \to T$ for each $s \in S$ (by the universal property of localization) which leads to a cocone with apex $T$ giving the desired unique morphism $L \to T$.

Dune
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