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*}

Example 10.88.1. Take $R = \mathbf{Z}$, $M = \mathbf{Q}$, and consider the family $Q_ n = \mathbf{Z}/n$ for $n \geq 1$. Then $\prod _ n (M \otimes Q_ n) = 0$. However there is an injection $\mathbf{Q} \to M \otimes (\prod _ n Q_ n)$ obtained by tensoring the injection $\mathbf{Z} \to \prod _ n Q_ n$ by $M$, so $M \otimes (\prod _ n Q_ n)$ is nonzero. Thus $M \otimes (\prod _ n Q_ n) \to \prod _ n (M \otimes Q_ n)$ is not injective.

On the other hand, take again $R = \mathbf{Z}$, $M = \mathbf{Q}$, and let $Q_ n = \mathbf{Z}$ for $n \geq 1$. The image of $M \otimes (\prod _ n Q_ n) \to \prod _ n (M \otimes Q_ n) = \prod _ n M$ consists precisely of sequences of the form $(a_ n/m)_{n \geq 1}$ with $a_ n \in \mathbf{Z}$ and $m$ some nonzero integer. Hence the map is not surjective.


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