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 15.23.5. Let $R$ be a Noetherian domain. Let $0 \to M \to M' \to M''$ an exact sequence of finite $R$-modules. If $M'$ is reflexive and $M''$ is torsion free, then $M$ is reflexive.

Proof. We will use without further mention that $\mathop{\mathrm{Hom}}\nolimits _ R(N, N')$ is a finite $R$-module for any finite $R$-modules $N$ and $N'$, see Algebra, Lemma 10.70.9. We take duals to get a sequence

\[ \mathop{\mathrm{Hom}}\nolimits _ R(M, R) \leftarrow \mathop{\mathrm{Hom}}\nolimits _ R(M', R) \leftarrow \mathop{\mathrm{Hom}}\nolimits _ R(M'', R) \]

Dualizing again we obtain a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M', R), R) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M'', R), R) \\ M \ar[u] \ar[r] & M' \ar[u] \ar[r] & M'' \ar[u] } \]

We do not know the top row is exact. But, by assumption the middle vertical arrow is an isomorphism and the right vertical arrow is injective (Lemma 15.23.2). Now a diagram chase shows that the left vertical arrow is an isomorphism, i.e., $M$ is reflexive. $\square$


Comments (1)

Comment #3620 by JuanPablo on

It is not clear to me that is injective, which seems to be required for the diagram chase. This is because in the exact sequence one applies Lemma 0583 to obtain . But is split injective so , that is, is torsion. Then one gets the exact sequence and because is torsion.


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 0EB8. Beware of the difference between the letter 'O' and the digit '0'.