Proposition 10

Let $A, B$ be two rings, $E$ a right $A$-module, $F$ a right $B$-module and $G$ a $(B, A)$-bimodule. Suppose that $F$ is flat. Then, if $E$ is finitely generated (resp. finitely presented), the canonical homomorphism (8) is injective (resp. bijective).

Consider $A, B, F, G$ as fixed and for each right $A$-module $E$ set

and denote the homomorphism (8) by $\nu_E$; for every right $A$-module homomorphism $v : E \to E'$, set $T(v) = 1, \otimes \operatorname{Hom}(v, 1_G)$ and $T'(v) = \operatorname{Hom}(v, 1, \otimes 1_G)$. Let $L, \xrightarrow{v} L_0 \xrightarrow{w} E \to 0$ be a presentation of $E$; we suppose the free module $L_0$ (resp. the free modules $L_0$ and $L_1$) to be finitely generated. The diagram

is commutative and the second row is exact (Algebra, Chapter II, § 2, no. 1, Theorem 1); moreover, the sequence

is exact (loc. cit.), and as $F$ is flat, the first row of (9) is also an exact sequence (no. 3, Proposition 1). Then we know that $\nu_{L_0}$ (resp. $\nu_{L_0}$ and $\nu_{L_1}$) is bijective (resp. are bijective) (Algebra, Chapter II, § 4, no. 2, Proposition 2). If we assume only that $\nu_{L_0}$ is bijective, it follows from (9) that

is injective and hence so is $\nu_E$. If we assume that both $\nu_{L_0}$ and $\nu_{L_1}$ are bijective, it follows from § 1, no. 4, Corollary 2, (ii) to Proposition 2 that $\nu_E$ is surjective, and as we have just seen that $\nu_E$ is injective, it is bijective.