Lemma 15.23.15. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The following are equivalent
$M$ is reflexive,
for every prime $\mathfrak p$ of $R$ one of the following happens
$M_\mathfrak p$ is a reflexive $R_\mathfrak p$-module, or
$\text{depth}(M_\mathfrak p) \geq 2$.
Proof. If (1) is true, then $M_\mathfrak p$ is a reflexive module for all primes of $\mathfrak p$ by Lemma 15.23.4. Thus (1) $\Rightarrow $ (2). Assume (2). Set $N = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ so that
for every prime $\mathfrak p$ of $R$. See Algebra, Lemma 10.10.2. We apply Lemma 15.23.14 to the map $j : M \to N$. This is allowed because $M$ is finite and $N$ is torsion free by Lemma 15.22.12. In case (2)(a) the map $M_\mathfrak p \to N_\mathfrak p$ is an isomorphism and in case (2)(b) we have $\text{depth}(M_\mathfrak p) \geq 2$. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: