Proposition 1

(i) If $u$ and $v$ are two endomorphisms of a finite-dimensional free $A$-module $M$, then

(2)

(ii) $\det(1_M) = 1$; for every automorphism $u$ of $M$, $\det u$ is invertible in $A$ and

(3)

If $n$ is the dimension of $M$, this follows immediately from the relation $\bigwedge^n(u \circ v) = (\bigwedge^n(u)) \circ (\bigwedge^n(v))$ § 7, no. 2, formula (3)).

Let $M$ be a free $A$-module with a finite basis $(e_i)_{1 \leq i \leq n}$; given a sequence $(x_i)_{1 \leq i \leq n}$ of $n$ elements of $M$, the determinant of this sequence with respect to the given basis $(e_i)$, denoted by $\det(x_1, x_2, \ldots, x_n)$ when no confusion can arise over the basis, is the determinant of the endomorphism $u$ of $M$ such that $u(e_i) = x_i$ for $1 \leq i \leq n$. Then by formula (1)

and this relation characterizes the mapping $(x_i) \mapsto \det(x_1, x_2, \ldots, x_n)$ of $M^n$ into $A$. It shows that this mapping is an alternating n-linear form. As, by virtue of § 7, no. 4, Proposition 7, the $A$-module of alternating $n$-linear forms is canonically isomorphic to the dual of $\bigwedge^n(M)$ and $\bigwedge^n(M)$ is isomorphic to $A$, it is seen that every alternating n-linear form on $M^n$ can be written

for some $\alpha \in A$.