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\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 10.81.13. Let $\varphi : A \to B$ be a ring map. Let $S \subset A$ and $S' \subset B$ be multiplicative subsets such that $\varphi (S) \subset S'$. Let $M \to M'$ be a map of $B$-modules.

  1. If $M \to M'$ is universally injective as a map of $A$-modules, then $(S')^{-1}M \to (S')^{-1}M'$ is universally injective as a map of $A$-modules and as a map of $S^{-1}A$-modules.

  2. If $M$ and $M'$ are $(S')^{-1}B$-modules, then $M \to M'$ is universally injective as a map of $A$-modules if and only if it is universally injective as a map of $S^{-1}A$-modules.

Proof. You can prove this using Lemma 10.81.12 but you can also prove it directly as follows. Assume $M \to M'$ is $A$-universally injective. Let $Q$ be an $A$-module. Then $Q \otimes _ A M \to Q \otimes _ A M'$ is injective. Since localization is exact we see that $(S')^{-1}(Q \otimes _ A M) \to (S')^{-1}(Q \otimes _ A M')$ is injective. As $(S')^{-1}(Q \otimes _ A M) = Q \otimes _ A (S')^{-1}M$ and similarly for $M'$ we see that $Q \otimes _ A (S')^{-1}M \to Q \otimes _ A (S')^{-1}M'$ is injective, hence $(S')^{-1}M \to (S')^{-1}M'$ is universally injective as a map of $A$-modules. This proves the first part of (1). To see (2) we can use the following two facts: (a) if $Q$ is an $S^{-1}A$-module, then $Q \otimes _ A S^{-1}A = Q$, i.e., tensoring with $Q$ over $A$ is the same thing as tensoring with $Q$ over $S^{-1}A$, (b) if $M$ is any $A$-module on which the elements of $S$ are invertible, then $M \otimes _ A Q = M \otimes _{S^{-1}A} S^{-1}Q$. Part (2) follows from this immediately. $\square$


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