Theorem 2

*Let $M$ be an $A$-module with a finite basis $(e_i)_{1 \leq i \leq n}$. For a sequence $(x_i)_{1 \leq i \leq n}$ of $n$ elements of $M$ to form a basis of $M$, it is necessary and sufficient that the element $\lambda \in A$ such that

be invertible in $A$.*

Recall that $e_1 \wedge e_2 \wedge \cdots \wedge e_n$ is the unique element of a basis of $\Lambda^n(M)$ (no. 8, Corollary 1 to Theorem 1) so that the element $\lambda \in A$ satisfying (23) is determined uniquely. If $(x_i)_{1 \leq i \leq n}$ is a basis of $M$, $x_1 \wedge x_2 \wedge \cdots \wedge x_n$ is the unique element in a basis of $\Lambda^n(M)$ (no. 8), then $\lambda$ is invertible. Conversely, suppose $\lambda$ is invertible; then the alternating $n$-linear form $f$ corresponding to the linear mapping $g: \Lambda^n(M) \to A$ such that

is such that $f(x_1, x_2, \ldots, x_n) = 1$. For all $x \in M$, obviously

(no. 3, Proposition 6); applying no. 8, Corollary 3 to Theorem 1, we obtain

As $f(x_1, \ldots, x_n) = 1$, this shows that every $x \in M$ is a linear combination of $x_1, x_2, \ldots, x_n$ and, as the latter are linearly independent (since

they form a basis of $M$.