Proposition 6

Let $A$ be a ring which is not necessarily commutative and $m$ its Jacobson radical; suppose that $A/m$ is a field. Let $M$ and $N$ be two finitely generated free $A$-modules and $u: M \to N$ a homomorphism. The following properties are equivalent:

(a) $u$ is an isomorphism of $M$ onto a direct factor of $N$; (b) $1 \otimes u: (A/m) \otimes_A M \to (A/m) \otimes_A N$ is injective; (c) $u$ is injective and $\operatorname{Coker}(u)$ is a free $A$-module; (d) the transpose homomorphism ${}^t u: N^* \to M^*$ is surjective.

We know (Algebra, Chapter II, § 1, no. 11, Proposition 21) that, if $N/u(M)$ is free, $u(M)$ is a direct factor of $N$, hence (c) implies (a); conversely, (a) implies that $\operatorname{Coker}(u)$, isomorphic to a complement of $u(M)$ in $N$, is a finitely generated projective $A$-module and a fortiori finitely presented (Chapter I, § 2, no. 8, Lemma 8); hence this module is free by Corollary 2 to Proposition 5 and (a) implies (c). On the other hand, (a) obviously implies (b). For simplicity we write $M' = (A/m) \otimes_A M$, $N' = (A/m) \otimes_A N$; as $M$ and $N$ are finitely generated, the duals ${M'}^*$ and ${N'}^*$ of the $(A/m)$-modules $M'$ and $N'$ are canonically identified with $M^* \otimes_A (A/m)$ and $N^* \otimes_A (A/m)$ and ${}^t(1 \otimes u)$ with $({}^t u) \otimes 1$ (Algebra, Chapter II, § 5, no. 4, Proposition 8); as $M'$ and $N'$ are vector spaces over the field $A/m$, the hypothesis that $1 \otimes u$ is injective implies that ${}^t(1 \otimes u)$ is surjective (Algebra, Chapter 11, § 7, no. 5, Proposition 10); Corollary 1 to Proposition 4 then shows that ${}^t u$ is surjective and we have thus proved that (b) implies (d). Finally we show that (d) implies (a). Suppose that ${}^t u$ is surjective; as $M^*$ is free, there exists a homomorphism $f$ from $M^*$ to $N^*$ such that $1_{M^*} = {}^t u \circ f$ (Algebra, Chapter II, § 1, no. 11, Proposition 21); as $M$ and $N$ are free and finitely generated, there exists a homomorphism $g$ from $N$ to $M$ such that $f = {}^t g$; hence ${}^t 1_M = 1_{M^*} = {}^t u \circ {}^t g = {}^t(g \circ u)$, whence $1_M = g \circ u$; this proves that $u$ is an isomorphism of $M$ onto a submodule which is a direct factor of $N$ (Algebra, Chapter 11, § 1, no. 9, Corollary 2 to Proposition 15).