Theorem 1

Let $M$* be an A-module with a basis $(e_\lambda)_{\lambda \in L}$. Let $L$* be given the structure of a totally ordered set (Set Theory, III, § 2, no. 3, Theorem 1)* and for every finite subset $J$* of $L$* we write

where $(\lambda_k)_{1 \leq k \leq n}$ is the sequence of elements of J arranged in increasing order (Set Theory, III, § 5, no. 3, Proposition 6); we write $e_\varnothing = 1$, the unit element of A. Then the $e_J$, where J runs through the set $\mathscr{F}(L)$ of finite subsets of L, form a basis of the exterior algebra $\bigwedge(M)$.

Since the $e_\lambda$ generate the A-module M, every element of $\bigwedge(M)$ is a linear combination of a finite number of products of the elements $e_\lambda$ and hence (taking account of no. 3, Proposition 5) is a linear combination of a finite number of elements $e_J$ for $J \in \mathscr{F}(L)$. It reduces to proving that the $e_J$ are linearly independent over A. Otherwise, there would exist between these elements a linear relation with coefficients not all zero; the union of the subsets J which correspond to the $e_J$ whose coefficients in this relation are $\neq 0$ is a finite subset K of L (since there is only a finite number of coefficients $\neq 0$). Let N be the submodule of M generated by the $e_\lambda$ such that $\lambda \in K$; N is a direct factor in M, hence (no. 2) $\bigwedge(N)$ is identified with a subalgebra of $\bigwedge(M)$ and, if we show that the $e_J$ with $J \subset K$ form a basis of $\bigwedge(N)$, we shall obtain the desired contradiction.

It therefore reduces to providing Theorem 1 when the basis of M is finite; we may therefore suppose that $L = \{1, m\} \subset N$. For each $i \in L$, let $M_i$ be the free submodule $Ae_i$ of M; M is the direct sum of the $M_i$ and $\bigwedge(M_i)$ is the direct sum of $\bigwedge^0(M_i) = A$ and $\bigwedge^1(M_i) = M_i$ (no. 3, Proposition 6). Let $\bigwedge(M)$ be canonically identified with the A-module the tensor product of the $\bigwedge(M_i)$ (no. 7, Proposition 10); the latter has as basis the tensor product of the bases $(1, e_i)$ of the $\bigwedge(M_i)$ (II, § 3, no. 7, Corollary 2 to Proposition 7); thus we obtain all the elements

where either $u_i = 1$ or $u_i = e_i$; if J is the set of indices i such that $u_i = e_i$, $u_1 \otimes u_2 \otimes \cdots \otimes u_m$ is identical with $e_J$, which completes the proof.