The Stacks project

Proof. Follows immediately from Lemmas 15.22.12 and 15.22.7. $\square$

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. $\square$

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 10.23.1. $\square$

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 10.71.9. We take duals to get a sequence

Dualizing again we obtain a commutative diagram

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

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

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

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

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). $\square$

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. $\square$

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

with $N' = \mathop{\mathrm{Im}}(N^{\oplus n} \to N^{\oplus m})$ torsion free. We conclude by Lemma 15.23.5. $\square$

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

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 10.72.6 we conclude $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ has depth $\geq 2$. $\square$

Proof. Since localizing at primes commutes with taking $\mathop{\mathrm{Hom}}\nolimits _ R$ for finite $R$-modules (Algebra, Lemma 10.71.9) parts (1) and (2) follow immediately from Lemma 15.23.10. Part (3) follows from (2) and Lemma 15.22.12. $\square$

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 whose annihilator is $\mathfrak pR_\mathfrak p$ (Algebra, Lemma 10.63.15). 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 10.63.7. $\square$

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 10.63.18. Consider the short exact sequence

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 10.72.6. This implies that $Q$ does not have any associated primes, hence $Q = 0$ by Algebra, Lemma 10.63.7. $\square$

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

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

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. $\square$

Proof. By Algebra, Lemma 10.157.4 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

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. $\square$

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. $\square$

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 10.38.6 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 10.157.6 (or the lemma already cited as $A$ is a reflexive module over itself). $\square$