Proposition 7

Let A be a commutative ring and M and N two A-modules. If with every A-linear mapping $g : \bigwedge^n(M) \to N$ ($n \geq 2$) is associated the n-linear mapping

a bijective A-linear mapping is obtained of the A-module $\operatorname{Hom}_A(\bigwedge^n(M), N)$ onto the A-module of alternating n-linear mappings of $M^n$ into N.

We consider the canonical bijection of the A-module $\operatorname{Hom}_A(T^n(M), N)$ onto the A-module $\mathscr{L}_n(M, \ldots, M; N)$ of all n-linear mappings of $M^n$ into N, obtained by associating with every A-linear mapping $f : T^n(M) \to N$ the n-linear mapping

(II, § 3, no. 9). On the other hand, the A-linear mappings $g : \bigwedge^n(M) \to N$ are in one-to-one correspondence with the A-linear mappings $f : T^n(M) \to N$ such that f is zero on $\mathfrak{g}_n''$, by associating with g the mapping $f = g \circ p_n$, where

is the canonical homomorphism (II, § 2, no. 1, Theorem 1). But as $\mathfrak{J}_n''$ is a linear combination of elements of the form

$(x, u_i, v_j$ in $M$), to say that $f$ is of the form $g \circ p_n$ means that the corresponding $n$-linear function $\bar{f}$ satisfies (8), in other words it is alternating.

The A-module $\bigwedge^n(M)$ is called the n-th exterior power of M. For every A-module homomorphism $u : M \to N$, the mapping

which coincides with $\bigwedge(u)$ on $\bigwedge^n(M)$ is called the n-th exterior power of u.