The Stacks project

Lemma 31.12.4. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module.

  1. If $\mathcal{F}$ is reflexive, then $\mathcal{F}$ is torsion free.

  2. The map $j : \mathcal{F} \longrightarrow \mathcal{F}^{**}$ is injective if and only if $\mathcal{F}$ is torsion free.

Proof. Omitted. See More on Algebra, Lemma 15.23.2. $\square$


Comments (2)

Comment #6273 by Matthieu Romagny on

Add a point at the end of condition (2).

There are also:

  • 2 comment(s) on Section 31.12: Reflexive modules

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Your email address will not be published.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 0AY2. Beware of the difference between the letter 'O' and the digit '0'.