Proposition 3

Let $M$ be a finite-dimensional free $A$-module. For every endomorphism $u$ of $M$, the following conditions are equivalent: (f) $u$ is injective; (g) det $u$ is not a divisor of zero in $A$.

With the same notation as in the proof of Theorem 1, for $u$ to be injective, it is necessary and sufficient that the $x_i$ be linearly independent. By § 7, no. 9, Proposition 12, it is necessary and sufficient for this that the relation $\lambda x_1 \wedge x_2 \wedge \cdots \wedge x_n = 0$ (with $\lambda \in A$) imply $\lambda = 0$. But this is equivalent to $\lambda (\det u) = 0$ since $e_1 \wedge \cdots \wedge e_n$ is a basis of $\Lambda^n(M)$; whence the proposition.