Lemma 4

If the quadratic form $Q$ is zero, $C(Q)$ is none other than the exterior algebra of E.

Indeed the exterior algebra of $E$ is none other than the quotient of $T(E)$ by the two-sided ideal $J$ generated by the elements of the form $a \otimes \nu \otimes a$ ($a \in E, \nu \in T(E)$) (Chap. III, § 5, No. 5 and No. 9). It is clear that $I(Q) \subset J$. It is therefore enough to show that $a \otimes \nu \otimes a \in I(Q)$. This is obvious if $\nu \in T^0$. Suppose this assertion proved for $\nu \in \sum_{h=0}^{n-1} T^h$, and let $x \in E$ and $u \in T^{n-1}$; we have

and the four terms of the right-hand side belong to $I(Q)$.

Suppose in particular that the module $E$ admits a basis $(x_i)_{i \in L}$, and totally order the set of indices $L$. We know (Chap. III, § 5, No. 6) that the exterior algebra of $E$ admits as a basis the family formed by the elements $x_H$, where $H$ runs through the set of finite subsets of $L$ and where $x_H$ is the element $x_{h_1} \wedge \ldots \wedge x_{h_q}$, $(h_1, \ldots, h_q)$ denoting the strictly increasing sequence of the elements of $H$.

On the other hand, consider the bilinear form $F$ defined by $F(x_i, x_j) = -\Phi(x_i, x_j)$ if $i > j$, $F(x_i, x_j) = 0$ if $i < j$ and $F(x_i, x_i) = -Q(x_i)$. It is clear that $Q(x) + F(x, x) = 0$; with the notations of Prop. 3, it follows from this proposition and Lemma 4 that $\bar{\lambda}_F$ is an isomorphism of $\wedge E$ onto $C(Q)$, which is therefore a free $A$-module. Now prove, by induction on the number of elements of $H$, that we have

(where $(h_1, \ldots, h_q)$ is the strictly increasing sequence of the elements of $H$). This is obvious if $H$ has 0 or 1 element. Suppose (11) verified for the subsets having at most $q - 1$ elements. Consider a subset $H$ having $q$ elements, denote its smallest element by $j$, and write $H = \{ j \} \cup K$, where $K$ is a subset having $q - 1$ elements. We have, by (8) and the induction hypothesis

putting, for every finite subset $J$ of $L$, $x'_J = \rho(x_{j_1}) \ldots \rho(x_{j_s})$, where $(j_1, \ldots, j_s)$ is the increasing sequence of the elements of $J$. Now, for $i \in K$, $x_i$ belongs to the kernel of the linear form $y \to F(x_j, y)$, hence $i^F_{x_j}(x'_K) = 0$ (Lemma 1). This proves the desired result.

We have therefore proved the following theorem: