## 31.12 Reflexive modules

This section is the analogue of More on Algebra, Section 15.23 for coherent modules on locally Noetherian schemes. The reason for working with coherent modules is that $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is coherent for every pair of coherent $\mathcal{O}_ X$-modules $\mathcal{F}, \mathcal{G}$, see Modules, Lemma 17.22.5.

Definition 31.12.1. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The reflexive hull of $\mathcal{F}$ is the $\mathcal{O}_ X$-module

$\mathcal{F}^{**} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}( \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X), \mathcal{O}_ X)$

We say $\mathcal{F}$ is reflexive if the natural map $j : \mathcal{F} \longrightarrow \mathcal{F}^{**}$ is an isomorphism.

It follows from Lemma 31.12.8 that the reflexive hull is a reflexive $\mathcal{O}_ X$-module. You can use the same definition to define reflexive modules in more general situations, but this does not seem to be very useful. Here is the obligatory lemma comparing this to the usual algebraic notion.

Lemma 31.12.2. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The following are equivalent

1. $\mathcal{F}$ is reflexive,

2. for $U \subset X$ affine open $\mathcal{F}(U)$ is a reflexive $\mathcal{O}(U)$-module.

Proof. Omitted. $\square$

Remark 31.12.3. If $X$ is a scheme of finite type over a field, then sometimes a different notion of reflexive modules is used (see for example [bottom of page 5 and Definition 1.1.9, HL]). This other notion uses $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits$ into a dualizing complex $\omega _ X^\bullet$ instead of into $\mathcal{O}_ X$ and should probably have a different name because it can be different when $X$ is not Gorenstein. For example, if $X = \mathop{\mathrm{Spec}}(k[t^3, t^4, t^5])$, then a computation shows the dualizing sheaf $\omega _ X$ is not reflexive in our sense, but it is reflexive in the other sense as $\omega _ X \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\omega _ X, \omega _ X), \omega _ X)$ is an isomorphism.

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. $\square$

Lemma 31.12.5. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The following are equivalent

1. $\mathcal{F}$ is reflexive,

2. $\mathcal{F}_ x$ is a reflexive $\mathcal{O}_{X, x}$-module for all $x \in X$,

3. $\mathcal{F}_ x$ is a reflexive $\mathcal{O}_{X, x}$-module for all closed points $x \in X$.

Proof. By Modules, Lemma 17.22.3 we see that (1) and (2) are equivalent. Since every point of $X$ specializes to a closed point (Properties, Lemma 28.5.9) we see that (2) and (3) are equivalent. $\square$

Lemma 31.12.6. Let $f : X \to Y$ be a flat morphism of integral locally Noetherian schemes. Let $\mathcal{G}$ be a coherent reflexive $\mathcal{O}_ Y$-module. Then $f^*\mathcal{G}$ is a coherent reflexive $\mathcal{O}_ X$-module.

Proof. Omitted. See More on Algebra, Lemma 15.22.4 for the algebraic analogue. $\square$

Lemma 31.12.7. Let $X$ be an integral locally Noetherian scheme. Let $0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{F}''$ an exact sequence of coherent $\mathcal{O}_ X$-modules. If $\mathcal{F}'$ is reflexive and $\mathcal{F}''$ is torsion free, then $\mathcal{F}$ is reflexive.

Proof. Omitted. See More on Algebra, Lemma 15.23.5. $\square$

Lemma 31.12.8. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$, $\mathcal{G}$ be coherent $\mathcal{O}_ X$-modules. If $\mathcal{G}$ is reflexive, then $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is reflexive.

Proof. The statement makes sense because $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is coherent by Cohomology of Schemes, Lemma 30.9.4. To see the statement is true, see More on Algebra, Lemma 15.23.8. Some details omitted. $\square$

Remark 31.12.9. Let $X$ be an integral locally Noetherian scheme. Thanks to Lemma 31.12.8 we know that the reflexive hull $\mathcal{F}^{**}$ of a coherent $\mathcal{O}_ X$-module is coherent reflexive. Consider the category $\mathcal{C}$ of coherent reflexive $\mathcal{O}_ X$-modules. Taking reflexive hulls gives a left adjoint to the inclusion functor $\mathcal{C} \to \textit{Coh}(\mathcal{O}_ X)$. Observe that $\mathcal{C}$ is an additive category with kernels and cokernels. Namely, given $\varphi : \mathcal{F} \to \mathcal{G}$ in $\mathcal{C}$, the usual kernel $\mathop{\mathrm{Ker}}(\varphi )$ is reflexive (Lemma 31.12.7) and the reflexive hull $\mathop{\mathrm{Coker}}(\varphi )^{**}$ of the usual cokernel is the cokernel in $\mathcal{C}$. Moreover $\mathcal{C}$ inherits a tensor product

$\mathcal{F} \otimes _\mathcal {C} \mathcal{G} = (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G})^{**}$

which is associative and symmetric. There is an internal Hom in the sense that for any three objects $\mathcal{F}, \mathcal{G}, \mathcal{H}$ of $\mathcal{C}$ we have the identity

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {C}(\mathcal{F} \otimes _\mathcal {C} \mathcal{G}, \mathcal{H}) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {C}(\mathcal{F}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{H}))$

see Modules, Lemma 17.22.1. In $\mathcal{C}$ every object $\mathcal{F}$ has a dual object $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X)$. Without further conditions on $X$ it can happen that

$\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \not\cong \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X) \otimes _\mathcal {C} \mathcal{G} \quad \text{and}\quad \mathcal{F} \otimes _\mathcal {C} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X) \not\cong \mathcal{O}_ X$

for $\mathcal{F}, \mathcal{G}$ of rank $1$ in $\mathcal{C}$. To make an example let $X = \mathop{\mathrm{Spec}}(R)$ where $R$ is as in More on Algebra, Example 15.23.17 and let $\mathcal{F}, \mathcal{G}$ be the modules corresponding to $M$. Computation omitted.

Lemma 31.12.10. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The following are equivalent

1. $\mathcal{F}$ is reflexive,

2. for each $x \in X$ one of the following happens

1. $\mathcal{F}_ x$ is a reflexive $\mathcal{O}_{X, x}$-module, or

2. $\text{depth}(\mathcal{F}_ x) \geq 2$.

Proof. Omitted. See More on Algebra, Lemma 15.23.15. $\square$

Lemma 31.12.11. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$ be a coherent reflexive $\mathcal{O}_ X$-module. Let $x \in X$.

1. If $\text{depth}(\mathcal{O}_{X, x}) \geq 2$, then $\text{depth}(\mathcal{F}_ x) \geq 2$.

2. If $X$ is $(S_2)$, then $\mathcal{F}$ is $(S_2)$.

Proof. Omitted. See More on Algebra, Lemma 15.23.16. $\square$

Lemma 31.12.12. Let $X$ be an integral locally Noetherian scheme. Let $j : U \to X$ be an open subscheme with complement $Z$. Assume $\mathcal{O}_{X, z}$ has depth $\geq 2$ for all $z \in Z$. Then $j^*$ and $j_*$ define an equivalence of categories between the category of coherent reflexive $\mathcal{O}_ X$-modules and the category of coherent reflexive $\mathcal{O}_ U$-modules.

Proof. Let $\mathcal{F}$ be a coherent reflexive $\mathcal{O}_ X$-module. For $z \in Z$ the stalk $\mathcal{F}_ z$ has depth $\geq 2$ by Lemma 31.12.11. Thus $\mathcal{F} \to j_*j^*\mathcal{F}$ is an isomorphism by Lemma 31.5.11. Conversely, let $\mathcal{G}$ be a coherent reflexive $\mathcal{O}_ U$-module. It suffices to show that $j_*\mathcal{G}$ is a coherent reflexive $\mathcal{O}_ X$-module. To prove this we may assume $X$ is affine. By Properties, Lemma 28.22.5 there exists a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with $\mathcal{G} = j^*\mathcal{F}$. After replacing $\mathcal{F}$ by its reflexive hull, we may assume $\mathcal{F}$ is reflexive (see discussion above and in particular Lemma 31.12.8). By the above $j_*\mathcal{G} = j_*j^*\mathcal{F} = \mathcal{F}$ as desired. $\square$

If the scheme is normal, then reflexive is the same thing as torsion free and $(S_2)$.

Lemma 31.12.13. Let $X$ be an integral locally Noetherian normal scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The following are equivalent

1. $\mathcal{F}$ is reflexive,

2. $\mathcal{F}$ is torsion free and has property $(S_2)$, and

3. there exists an open subscheme $j : U \to X$ such that

1. every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$,

2. $j^*\mathcal{F}$ is finite locally free, and

3. $\mathcal{F} = j_*j^*\mathcal{F}$.

Proof. Using Lemma 31.12.2 the equivalence of (1) and (2) follows from More on Algebra, Lemma 15.23.18. Let $U \subset X$ be as in (3). By Properties, Lemma 28.12.5 we see that $\text{depth}(\mathcal{O}_{X, x}) \geq 2$ for $x \not\in U$. Since a finite locally free module is reflexive, we conclude (3) implies (1) by Lemma 31.12.12.

Assume (1). Let $U \subset X$ be the maximal open subscheme such that $j^*\mathcal{F} = \mathcal{F}|_ U$ is finite locally free. So (3)(b) holds. Let $x \in X$ be a point. If $\mathcal{F}_ x$ is a free $\mathcal{O}_{X, x}$-module, then $x \in U$, see Modules, Lemma 17.11.8. If $\dim (\mathcal{O}_{X, x}) \leq 1$, then $\mathcal{O}_{X, x}$ is either a field or a discrete valuation ring (Properties, Lemma 28.12.5) and hence $\mathcal{F}_ x$ is free (More on Algebra, Lemma 15.22.11). Thus $x \not\in U \Rightarrow \dim (\mathcal{O}_{X, x}) \geq 2$. Then Properties, Lemma 28.10.3 shows (3)(a) holds. By the already used Properties, Lemma 28.12.5 we also see that $\text{depth}(\mathcal{O}_{X, x}) \geq 2$ for $x \not\in U$ and hence (3)(c) follows from Lemma 31.12.12. $\square$

Lemma 31.12.14. Let $X$ be an integral locally Noetherian normal scheme with generic point $\eta$. Let $\mathcal{F}$, $\mathcal{G}$ be coherent $\mathcal{O}_ X$-modules. Let $T : \mathcal{G}_\eta \to \mathcal{F}_\eta$ be a linear map. Then $T$ extends to a map $\mathcal{G} \to \mathcal{F}^{**}$ of $\mathcal{O}_ X$-modules if and only if

• for every $x \in X$ with $\dim (\mathcal{O}_{X, x}) = 1$ we have

$T\left(\mathop{\mathrm{Im}}(\mathcal{G}_ x \to \mathcal{G}_\eta )\right) \subset \mathop{\mathrm{Im}}(\mathcal{F}_ x \to \mathcal{F}_\eta ).$

Proof. Because $\mathcal{F}^{**}$ is torsion free and $\mathcal{F}_\eta = \mathcal{F}^{**}_\eta$ an extension, if it exists, is unique. Thus it suffices to prove the lemma over the members of an open covering of $X$, i.e., we may assume $X$ is affine. In this case we are asking the following algebra question: Let $R$ be a Noetherian normal domain with fraction field $K$, let $M$, $N$ be finite $R$-modules, let $T : M \otimes _ R K \to N \otimes _ R K$ be a $K$-linear map. When does $T$ extend to a map $N \to M^{**}$? By More on Algebra, Lemma 15.23.19 this happens if and only if $N_\mathfrak p$ maps into $(M/M_{tors})_\mathfrak p$ for every height $1$ prime $\mathfrak p$ of $R$. This is exactly condition $(*)$ of the lemma. $\square$

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. $\square$

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