The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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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