Proposition 8

Let $R$ and $S$ be two commutative rings, $p : R \to S$ a ring homomorphism and $M$ a finitely generated $R$-module. Suppose that $R$ is Noetherian and that $S$ is a flat $R$-module. Then, if $M$ is reflexive, so is the $S$-module $M_{(S)} = M \otimes_R S$. We know (Chapter I, § 2, no. 10, Proposition 11) that there exists a canonical isomorphism $\omega_M : (M^*)_{(S)} \to (M_{(S)})^*$, such that

€or x \in M, x^* \in M^*. As M is a quotient of a finitely generated free R-module L, M* is isomorphic to a sub-R-module of the dual L* and L* is free and finitely generated; since $\mathbf{R}$ is Noetherian, M* is therefore also a finitely generated R-module, whence an isomorphism $\omega_{M^*}: (M^{**})_{(S)} \to ((M^*)_{(S)})^*$ such that

for $x^* \in M$ and $x^{**} \in M^{**}$. On the other hand, there is an isomorphism $^t\omega_M: (M_{(S)})^{**} \to ((M^*)_{(S)})^*$, whence by composition a canonical isomorphism:

such that, in the above notation:

(1)

We consider now the canonical homomorphism $c_M: M \to M^{**}$ and show that the composite homomorphism :

(2)

is just the canonical homomorphism $c_{M_{(S)}}$. This follows immediately from (1) which gives the relations :

and from the fact that the elements $\omega_M(x^* \otimes 1)$ generate $(M_{(S)})^*$. This being so, the hypothesis that M is reflexive means that $c_M$ is bijective, hence so is $c_M \otimes 1$ and therefore $\psi = c_{M_{(S)}}$ is bijective, which shows the proposition.