Processing math: 100%

The Stacks project

Lemma 31.12.15. Let X be a regular scheme of dimension \leq 2. Let \mathcal{F} be a coherent \mathcal{O}_ X-module. The following are equivalent

  1. \mathcal{F} is reflexive,

  2. \mathcal{F} is finite locally free.

Proof. It is clear that a finite locally free module is reflexive. For the converse, we will show that if \mathcal{F} is reflexive, then \mathcal{F}_ x is a free \mathcal{O}_{X, x}-module for all x \in X. This is enough by Algebra, Lemma 10.78.2 and the fact that \mathcal{F} is coherent. If \dim (\mathcal{O}_{X, x}) = 0, then \mathcal{O}_{X, x} is a field and the statement is clear. If \dim (\mathcal{O}_{X, x}) = 1, then \mathcal{O}_{X, x} is a discrete valuation ring (Algebra, Lemma 10.119.7) and \mathcal{F}_ x is torsion free. Hence \mathcal{F}_ x is free by More on Algebra, Lemma 15.22.11. If \dim (\mathcal{O}_{X, x}) = 2, then \mathcal{O}_{X, x} is a regular local ring of dimension 2. By More on Algebra, Lemma 15.23.18 we see that \mathcal{F}_ x has depth \geq 2. Hence \mathcal{F} is free by Algebra, Lemma 10.106.6. \square


Comments (1)

Comment #9900 by Vihaan Dheer on

Small typo at the end: we've shown is free, not .

There are also:

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

Your email address will not be published.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.