
Lemma 10.11.15. Let $M$ be an $R$-module. Then the $S^{-1}R$-modules $S^{-1}M$ and $S^{-1}R \otimes _ R M$ are canonically isomorphic, and the canonical isomorphism $f : S^{-1}R \otimes _ R M \to S^{-1}M$ is given by

$f((a/s) \otimes m) = am/s, \forall a \in R, m \in M, s \in S$

Proof. Obviously, the map $f' : S^{-1}R \times M \to S^{-1}M$ given by $f((a/s, m)) = am/s$ is bilinear, and thus by the universal property, this map induces a unique $S^{-1}R$-module homomorphism $f : S^{-1}R \otimes _ R M \to S^{-1}M$ as in the statement of the lemma. Actually every element in $S^{-1}M$ is of the form $m/s$, $m\in M, s\in S$ and every element in $S^{-1}R \otimes _ R M$ is of the form $1/s \otimes m$. To see the latter fact, write an element in $S^{-1}R \otimes _ R M$ as

$\sum _ k \frac{a_ k}{s_ k} \otimes m_ k = \sum _ k \frac{a_ k t_ k}{s} \otimes m_ k = \frac{1}{s} \otimes \sum _ k {a_ k t_ k}m_ k = \frac{1}{s} \otimes m$

Where $m = \sum _ k {a_ k t_ k}m_ k$. Then it is obvious that $f$ is surjective, and if $f(\frac{1}{s} \otimes m) = m/s = 0$ then there exists $t'\in S$ with $tm = 0$ in $M$. Then we have

$\frac{1}{s} \otimes m = \frac{1}{st} \otimes tm = \frac{1}{st} \otimes 0 = 0$

Therefore $f$ is injective. $\square$

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