The Stacks project

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15.23 Reflexive modules

Here is our definition.

Definition 15.23.1. Let $R$ be a domain. We say an $R$-module $M$ is reflexive if the natural map

\[ j : M \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R) \]

which sends $m \in M$ to the map sending $\varphi \in \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ to $\varphi (m) \in R$ is an isomorphism.

We can make this definition for more general rings, but already the definition above has drawbacks. It would be wise to restrict to Noetherian domains and finite torsion free modules and (perhaps) impose some regularity conditions on $R$ (e.g., $R$ is normal).

Lemma 15.23.2. Let $R$ be a domain and let $M$ be an $R$-module.

  1. If $M$ is reflexive, then $M$ is torsion free.

  2. If $M$ is finite, then $j : M \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ is injective if and only if $M$ is torsion free

Lemma 15.23.3. Let $R$ be a discrete valuation ring and let $M$ be a finite $R$-module. Then the map $j : M \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ is surjective.

Proof. Let $M_{tors} \subset M$ be the torsion submodule. Then we have $\mathop{\mathrm{Hom}}\nolimits _ R(M, R) = \mathop{\mathrm{Hom}}\nolimits _ R(M/M_{tors}, R)$ (holds over any domain). Hence we may assume that $M$ is torsion free. Then $M$ is free by Lemma 15.22.11 and the lemma is clear. $\square$

Lemma 15.23.4. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The following are equivalent:

  1. $M$ is reflexive,

  2. $M_\mathfrak p$ is a reflexive $R_\mathfrak p$-module for all primes $\mathfrak p \subset R$, and

  3. $M_\mathfrak m$ is a reflexive $R_\mathfrak m$-module for all maximal ideals $\mathfrak m$ of $R$.

Proof. The localization of $j : M \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ at a prime $\mathfrak p$ is the corresponding map for the module $M_\mathfrak p$ over the Noetherian local domain $R_\mathfrak p$. See Algebra, Lemma 10.10.2. Thus the lemma holds by Algebra, Lemma 10.22.1. $\square$

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

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

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). We claim $j$ is injective. Assuming the claim a diagram chase shows that the left vertical arrow is an isomorphism, i.e., $M$ is reflexive.

Proof of the claim. Consider the exact sequence $\mathop{\mathrm{Hom}}\nolimits _ R(M',R)\to \mathop{\mathrm{Hom}}\nolimits _ R(M,R)\to Q \to 0$ defining $Q$. One applies Algebra, Lemma 10.10.2 to obtain

\[ \mathop{\mathrm{Hom}}\nolimits _ K(M'\otimes _ R K,K) \to \mathop{\mathrm{Hom}}\nolimits _ K(M\otimes _ R K,K) \to Q\otimes _ R K\to 0 \]

But $M \otimes _ R K \to M' \otimes _ R K$ is an injective map of vector spaces, hence split injective, so $Q \otimes _ R K = 0$, that is, $Q$ is torsion. Then one gets the exact sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(Q,R) \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M,R),R) \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M',R),R) \]

and $\mathop{\mathrm{Hom}}\nolimits _ R(Q,R)=0$ because $Q$ is torsion. $\square$

Lemma 15.23.6. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The following are equivalent

  1. $M$ is reflexive,

  2. there exists a short exact sequence $0 \to M \to F \to N \to 0$ with $F$ finite free and $N$ torsion free.

Proof. Observe that a finite free module is reflexive. By Lemma 15.23.5 we see that (2) implies (1). Assume $M$ is reflexive. Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to \mathop{\mathrm{Hom}}\nolimits _ R(M, R) \to 0$. Dualizing we get an exact sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R) \to R^{\oplus n} \to N \to 0 \]

with $N = \mathop{\mathrm{Im}}(R^{\oplus n} \to R^{\oplus m})$ a torsion free module. As $M = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ we get an exact sequence as in (2). $\square$

Lemma 15.23.7. Let $R \to R'$ be a flat homomorphism of Noetherian domains. If $M$ is a finite reflexive $R$-module, then $M \otimes _ R R'$ is a finite reflexive $R'$-module.

Proof. Choose a short exact sequence $0 \to M \to F \to N \to 0$ with $F$ finite free and $N$ torsion free, see Lemma 15.23.6. Since $R \to R'$ is flat we obtain a short exact sequence $0 \to M \otimes _ R R' \to F \otimes _ R R' \to N \otimes _ R R' \to 0$ with $F \otimes _ R R'$ finite free and $N \otimes _ R R'$ torsion free (Lemma 15.22.4). Thus $M \otimes _ R R'$ is reflexive by Lemma 15.23.6. $\square$

Lemma 15.23.8. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. Let $N$ be a finite reflexive $R$-module. Then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ is reflexive.

Proof. Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M \to 0$. Then we obtain

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to N^{\oplus n} \to N' \to 0 \]

with $N' = \mathop{\mathrm{Im}}(N^{\oplus n} \to N^{\oplus m})$ torsion free. Choose a sequence $0 \to N \to F \to N'' \to 0$ with $N''$ torsion free as in Lemma 15.23.6. We obtain an injective map $\delta : \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to F^{\oplus n}$. A snake lemma argument shows there is a short exact sequence

\[ 0 \to N' \to \mathop{\mathrm{Coker}}(\delta ) \to (N'')^{\oplus n} \to 0 \]

Thus $\mathop{\mathrm{Coker}}(\delta )$ is an extension of torsion free modules, hence torsion free (Lemma 15.22.5). $\square$

Definition 15.23.9. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The module $M^{**} = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ is called the reflexive hull of $M$.

This makes sense because the reflexive hull is reflexive by Lemma 15.23.8. The assignment $M \mapsto M^{**}$ is a functor. If $\varphi : M \to N$ is an $R$-module map into a reflexive $R$-module $N$, then $\varphi $ factors $M \to M^{**} \to N$ through the reflexive hull of $M$. Another way to say this is that taking the reflexive hull is the left adjoint to the inclusion functor

\[ \text{finite reflexive modules} \subset \text{finite modules} \]

over a Noetherian domain $R$.

Lemma 15.23.10. Let $R$ be a Noetherian local ring. Let $M$, $N$ be finite $R$-modules.

  1. If $N$ has depth $\geq 1$, then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ has depth $\geq 1$.

  2. If $N$ has depth $\geq 2$, then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ has depth $\geq 2$.

Proof. Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M \to 0$. Dualizing we get an exact sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to N^{\oplus n} \to N' \to 0 \]

with $N' = \mathop{\mathrm{Im}}(N^{\oplus n} \to N^{\oplus m})$. A submodule of a module with depth $\geq 1$ has depth $\geq 1$; this follows immediately from the definition. Thus part (1) is clear. For (2) note that here the assumption and the previous remark implies $N'$ has depth $\geq 1$. The module $N^{\oplus n}$ has depth $\geq 2$. From Algebra, Lemma 10.71.6 we conclude $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ has depth $\geq 2$. $\square$

Lemma 15.23.11. Let $R$ be a Noetherian ring. Let $M$, $N$ be finite $R$-modules.

  1. If $N$ has property $(S_1)$, then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ has property $(S_1)$.

  2. If $N$ has property $(S_2)$, then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ has property $(S_2)$.

  3. If $R$ is a domain, $N$ is torsion free and $(S_2)$, then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ is torsion free and has property $(S_2)$.

Proof. Since localizing at primes commutes with taking $\mathop{\mathrm{Hom}}\nolimits _ R$ for finite $R$-modules (Algebra, Lemma 10.70.9) parts (1) and (2) follow immediately from Lemma 15.23.10. Part (3) follows from (2) and Lemma 15.22.12. $\square$

Lemma 15.23.12. Let $R$ be a Noetherian ring. Let $\varphi : M \to N$ be a map of $R$-modules. Assume that for every prime $\mathfrak p$ of $R$ at least one of the following happens

  1. $M_\mathfrak p \to N_\mathfrak p$ is injective, or

  2. $\mathfrak p \not\in \text{Ass}(M)$.

Then $\varphi $ is injective.

Proof. Let $\mathfrak p$ be an associated prime of $\mathop{\mathrm{Ker}}(\varphi )$. Then there exists an element $x \in M_\mathfrak p$ which is in the kernel of $M_\mathfrak p \to N_\mathfrak p$ and is annihilated by $\mathfrak pR_\mathfrak p$ (Algebra, Lemma 10.62.15). This is impossible in both cases. Hence $\text{Ass}(\mathop{\mathrm{Ker}}(\varphi )) = \emptyset $ and we conclude $\mathop{\mathrm{Ker}}(\varphi ) = 0$ by Algebra, Lemma 10.62.7. $\square$

Lemma 15.23.13. Let $R$ be a Noetherian ring. Let $\varphi : M \to N$ be a map of $R$-modules. Assume $M$ is finite and that for every prime $\mathfrak p$ of $R$ one of the following happens

  1. $M_\mathfrak p \to N_\mathfrak p$ is an isomorphism, or

  2. $\text{depth}(M_\mathfrak p) \geq 2$ and $\mathfrak p \not\in \text{Ass}(N)$.

Then $\varphi $ is an isomorphism.

Proof. By Lemma 15.23.12 we see that $\varphi $ is injective. Let $N' \subset N$ be an finitely generated $R$-module containing the image of $M$. Then $\text{Ass}(N_\mathfrak p) = \emptyset $ implies $\text{Ass}(N'_\mathfrak p) = \emptyset $. Hence the assumptions of the lemma hold for $M \to N'$. In order to prove that $\varphi $ is an isomorphism, it suffices to prove the same thing for every such $N' \subset N$. Thus we may assume $N$ is a finite $R$-module. In this case, $\mathfrak p \not\in \text{Ass}(N) \Rightarrow \text{depth}(N_\mathfrak p) \geq 1$, see Algebra, Lemma 10.62.18. Consider the short exact sequence

\[ 0 \to M \to N \to Q \to 0 \]

defining $Q$. Looking at the conditions we see that either $Q_\mathfrak p = 0$ in case (1) or $\text{depth}(Q_\mathfrak p) \geq 1$ in case (2) by Algebra, Lemma 10.71.6. This implies that $Q$ does not have any associated primes, hence $Q = 0$ by Algebra, Lemma 10.62.7. $\square$

Lemma 15.23.14. Let $R$ be a Noetherian domain. Let $\varphi : M \to N$ be a map of $R$-modules. Assume $M$ is finite, $N$ is torsion free, and that for every prime $\mathfrak p$ of $R$ one of the following happens

  1. $M_\mathfrak p \to N_\mathfrak p$ is an isomorphism, or

  2. $\text{depth}(M_\mathfrak p) \geq 2$.

Then $\varphi $ is an isomorphism.

Proof. This is a special case of Lemma 15.23.13. $\square$

Lemma 15.23.15. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The following are equivalent

  1. $M$ is reflexive,

  2. for every prime $\mathfrak p$ of $R$ one of the following happens

    1. $M_\mathfrak p$ is a reflexive $R_\mathfrak p$-module, or

    2. $\text{depth}(M_\mathfrak p) \geq 2$.

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

\[ N_\mathfrak p = \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak p}( \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak p}(M_\mathfrak p, R_\mathfrak p), R_\mathfrak p) \]

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 $\text{depth}(M_\mathfrak p) \geq 2$. $\square$

Lemma 15.23.16. Let $R$ be a Noetherian domain. Let $M$ be a finite reflexive $R$-module. Let $\mathfrak p \subset R$ be a prime ideal.

  1. If $\text{depth}(R_\mathfrak p) \geq 2$, then $\text{depth}(M_\mathfrak p) \geq 2$.

  2. If $R$ is $(S_2)$, then $M$ is $(S_2)$.

Proof. Since formation of reflexive hull $\mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ commutes with localization (Algebra, Lemma 10.10.2) part (1) follows from Lemma 15.23.10. Part (2) is immediate from Lemma 15.23.11. $\square$

Example 15.23.17. The results above and below suggest reflexivity is related to the $(S_2)$ condition; here is an example to prevent too optimistic conjectures. Let $k$ be a field. Let $R$ be the $k$-subalgebra of $k[x, y]$ generated by $1, y, x^2, xy, x^3$. Then $R$ is not $(S_2)$. So $R$ as an $R$-module is an example of a reflexive $R$-module which is not $(S_2)$. Let $M = k[x, y]$ viewed as an $R$-module. Then $M$ is a reflexive $R$-module because

\[ \mathop{\mathrm{Hom}}\nolimits _ R(M, R) = \mathfrak m = (y, x^2, xy, x^3) \quad \text{and}\quad \mathop{\mathrm{Hom}}\nolimits _ R(\mathfrak m, R) = M \]

and $M$ is $(S_2)$ as an $R$-module (computations omitted). Thus $R$ is a Noetherian domain possessing a reflexive $(S_2)$ module but $R$ is not $(S_2)$ itself.

Lemma 15.23.18. Let $R$ be a Noetherian normal domain with fraction field $K$. Let $M$ be a finite $R$-module. The following are equivalent

  1. $M$ is reflexive,

  2. $M$ is torsion free and has property $(S_2)$,

  3. $M$ is torsion free and $M = \bigcap _{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}$ where the intersection happens in $M_ K = M \otimes _ R K$.

Proof. By Algebra, Lemma 10.151.4 we see that $R$ satisfies $(R_1)$ and $(S_2)$.

Assume (1). Then $M$ is torsion free by Lemma 15.23.2 and satisfies $(S_2)$ by Lemma 15.23.16. Thus (2) holds.

Assume (2). By definition $M' = \bigcap _{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}$ is the kernel of the map

\[ M_ K \longrightarrow \bigoplus \nolimits _{\text{height}(\mathfrak p) = 1} M_ K/M_\mathfrak p \subset \prod \nolimits _{\text{height}(\mathfrak p) = 1} M_ K/M_\mathfrak p \]

Observe that our map indeed factors through the direct sum as indicated since given $a/b \in K$ there are at most finitely many height $1$ primes $\mathfrak p$ with $b \in \mathfrak p$. Let $\mathfrak p_0$ be a prime of height $1$. Then $(M_ K/M_\mathfrak p)_{\mathfrak p_0} = 0$ unless $\mathfrak p = \mathfrak p_0$ in which case we get $(M_ K/M_\mathfrak p)_{\mathfrak p_0} = M_ K/M_{\mathfrak p_0}$. Thus by exactness of localization and the fact that localization commutes with direct sums, we see that $M'_{\mathfrak p_0} = M_{\mathfrak p_0}$. Since $M$ has depth $\geq 2$ at primes of height $> 1$, we see that $M \to M'$ is an isomorphism by Lemma 15.23.14. Hence (3) holds.

Assume (3). Let $\mathfrak p$ be a prime of height $1$. Then $R_\mathfrak p$ is a discrete valuation ring by $(R_1)$. By Lemma 15.22.11 we see that $M_\mathfrak p$ is finite free, in particular reflexive. Hence the map $M \to M^{**}$ induces an isomorphism at all the primes $\mathfrak p$ of height $1$. Thus the condition $M = \bigcap _{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}$ implies that $M = M^{**}$ and (1) holds. $\square$

Lemma 15.23.19. Let $R$ be a Noetherian normal domain. Let $M$ be a finite $R$-module. Then the reflexive hull of $M$ is the intersection

\[ M^{**} = \bigcap \nolimits _{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}/(M_\mathfrak p)_{tors} = \bigcap \nolimits _{\text{height}(\mathfrak p) = 1} (M/M_{tors})_\mathfrak p \]

taken in $M \otimes _ R K$.

Proof. Let $\mathfrak p$ be a prime of height $1$. The kernel of $M_\mathfrak p \to M \otimes _ R K$ is the torsion submodule $(M_\mathfrak p)_{tors}$ of $M_\mathfrak p$. Moreover, we have $(M/M_{tors})_\mathfrak p = M_\mathfrak p/(M_\mathfrak p)_{tors}$ and this is a finite free module over the discrete valuation ring $R_\mathfrak p$ (Lemma 15.22.11). Then $M_\mathfrak p/(M_\mathfrak p)_{tors} \to (M_\mathfrak p)^{**} = (M^{**})_\mathfrak p$ is an isomorphism, hence the lemma is a consequence of Lemma 15.23.18. $\square$

Lemma 15.23.20. Let $A$ be a Noetherian normal domain with fraction field $K$. Let $L$ be a finite extension of $K$. If the integral closure $B$ of $A$ in $L$ is finite over $A$, then $B$ is reflexive as an $A$-module.

Proof. It suffices to show that $B = \bigcap B_\mathfrak p$ where the intersection is over height $1$ primes $\mathfrak p \subset A$, see Lemma 15.23.18. Let $b \in \bigcap B_\mathfrak p$. Let $x^ d + a_1x^{d - 1} + \ldots + a_ d$ be the minimal polynomial of $b$ over $K$. We want to show $a_ i \in A$. By Algebra, Lemma 10.37.6 we see that $a_ i \in A_\mathfrak p$ for all $i$ and all height one primes $\mathfrak p$. Hence we get what we want from Algebra, Lemma 10.151.6 (or the lemma already cited as $A$ is a reflexive module over itself). $\square$


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