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*}

Proposition 10.88.3. Let $M$ be an $R$-module. The following are equivalent:

  1. $M$ is finitely presented.

  2. For every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, the canonical map $M \otimes _ R \left( \prod _{\alpha } Q_{\alpha } \right) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$ is bijective.

  3. For every $R$-module $Q$ and every set $A$, the canonical map $M \otimes _ R Q^{A} \to (M \otimes _ R Q)^{A}$ is bijective.

  4. For every set $A$, the canonical map $M \otimes _ R R^{A} \to M^{A}$ is bijective.

Proof. First we prove (1) implies (2). Choose a presentation $R^ m \to R^ n \to M$ and consider the commutative diagram

\[ \xymatrix{ R^ m \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d]^{\cong } & R^ m \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d]^{\cong } & M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ \prod _{\alpha } (R^ m \otimes _ R Q_{\alpha }) \ar[r] & \prod _{\alpha } (R^ n \otimes _ R Q_{\alpha }) \ar[r] & \prod _{\alpha } (M \otimes _ R Q_{\alpha }) \ar[r] & 0. } \]

The first two vertical arrows are isomorphisms and the rows are exact. This implies that the map $M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to \prod _{\alpha } ( M \otimes _ R Q_{\alpha })$ is surjective and, by a diagram chase, also injective. Hence (2) holds.

Obviously (2) implies (3) implies (4), so it remains to prove (4) implies (1). From Proposition 10.88.2, if (4) holds we already know that $M$ is finitely generated. So we can choose a surjection $F \to M$ where $F$ is free and finite. Let $K$ be the kernel. We must show $K$ is finitely generated. For any set $A$, we have a commutative diagram

\[ \xymatrix{ & K \otimes _ R R^ A \ar[r] \ar[d]_{f_3} & F \otimes _ R R^ A \ar[r] \ar[d]_{f_2}^{\cong } & M \otimes _ R R^ A \ar[r] \ar[d]_{f_1}^{\cong } & 0 \\ 0 \ar[r] & K^ A \ar[r] & F^ A \ar[r] & M^ A \ar[r] & 0 . } \]

The map $f_1$ is an isomorphism by assumption, the map $f_2$ is a isomorphism since $F$ is free and finite, and the rows are exact. A diagram chase shows that $f_3$ is surjective, hence by Proposition 10.88.2 we get that $K$ is finitely generated. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 059K. Beware of the difference between the letter 'O' and the digit '0'.