The uniqueness of $\bar{f}$ follows from the fact that $C(Q)$ is generated by $\rho_Q(E)$. Let $h$ be the unique homomorphism of $T(E)$ into D which extends $f$ ($h$ is defined by $h(x_1 \otimes \cdots \otimes x_n) = f(x_1) \ldots f(x_n)$). We have

and consequently $h$ vanishes on $I(Q)$ and defines by passing to the quotient the homomorphism $\bar{f}$ in question.

Prop. 1 expresses that $C(Q)$ is the solution of a universal mapping problem (Sets, Chap. IV, § 3, No. 1).

Take in particular for D the opposite algebra of $C(Q)$ and for $f$ the mapping $\rho$; prop. 1 entails that there exists one and only one anti-automorphism $\beta$ of $C(Q)$ whose restriction to $\rho(E)$ is the identity; it is called the principal antiautomorphism of $C(Q)$. It is clear that $\beta^2 = 1$.

D’autre part let $Q'$ be a quadratic form on an A-module $E'$ and $f$ a linear mapping of $E$ into $E'$ such that $Q' \circ f = Q$. We have $\rho_Q(f(x))^2 = Q'(f(x)).1 = Q(x).1$, and consequently there exists one and only one homomorphism $C(f)$ of $C(Q)$ into $C(Q')$ such that $C(f) \circ \rho_Q = \rho_{Q'} \circ f$. If $f$ is the identity, $C(f)$ is the identity; if $Q''$ is a quadratic form on an A-module $E''$ and $g$ a linear mapping of $E'$ into $E''$ such that $Q'' \circ g = Q'$, we have $C(g \circ f) = C(g) \circ C(f)$. When $E'$ is a submodule of $E$ and $f$ the canonical injection of $E'$ into $E$ (so that $Q'$ is the restriction of $Q$ to $E'$), we say that $C(f)$ is the canonical homomorphism of $C(Q')$ into $C(Q)$.

Let us take in particular $Q' = Q$ and for $f$ the mapping $x \to -x$; we see that there exists one and only one automorphism $\alpha$ of $C(Q)$ such that $\alpha \circ \rho = -\rho$; it is called the principal automorphism of $C(Q)$. It is clear that $\alpha^2 = 1$, and that the restriction of $\alpha$ to $C^+$ (resp. $C^-$) is the identity (resp. the mapping $u \to -u$).

It suffices to prove that the algebra $C' = A' \otimes C(Q)$ and the mapping $1 \otimes \rho_Q$ of $E'$ into $C'$ form a solution of the same problem of universal mapping as $C(Q')$ and $\rho_{Q'}$. Now, let D’ be an algebra over $A'$ and let $f'$ be an $A'$-linear mapping of $E'$ into D’ such that $f'(x')^2 = Q'(x').1$ for every $x' \in E'$. The mapping $g : x \to f' (1 \otimes x)$ of E into D’ (considered as an A-module by means of the homomorphism $\varphi$) is A-linear and we have $g(x)^2 = Q' (1 \otimes x).1 = Q(x).1$ for every $x \in E$. There therefore exists one and only one A-homomorphism $\overline{g}$ of $C(Q)$ into D’ such that $\overline{g}(\rho_Q(x)) = f'(1 \otimes x)$. Consequently there exists one and only one $A'$-homomorphism $\overline{f}'$ of $C'$ into D’ such that $\overline{f}'(1 \otimes \rho_Q(x)) = f'(1 \otimes x)$ for every $x \in E$; by linearity it follows that $\overline{f}'((1 \otimes \rho_Q)(x')) = f'(x')$ for every $x' \in E'$. QED.

T(E) generated by the kernel of f. The ideal I(Q) is stable under $i_f$; passing to the quotient, $i_f$ therefore defines a linear mapping (still denoted by $i_f$) of $C(Q)$ into itself.

Indeed formula (5) can be written

As (4) completely determines $i_f$ on $T^0$, and since (6) determines $i_f$ on $T^n$ if its values on $T^{n-1}$ are known, the uniqueness of $i_f$ is proved. On the other hand, for $x \in E$ and $u \in T^{n-1}$, the right-hand side of (6) is bilinear on $E \times T^{n-1}$; this proves the existence of $i_f$ by induction on $n$ (Chap. III, § 1, No. 2) and also proves, by induction on $n$, that $i_f(T^n) \subset T^{n-1}$. If $f = ag + bh$ (a, b in A, g, h in E*) it is clear that $ai_g + bi_h$ satisfies (4) and (5), hence is equal to $i_f$. We have

and, since $(i_f)^2(1) = 0$, it follows by induction on $n$ that $(i_f)^2$ is zero on $T^n$. Replacing $f$ by $f + g$, the relation $(i_f)^2 = 0$ gives $i_f \circ i_g + i_g \circ i_f = 0$. An analogous reasoning by induction to the preceding ones shows that $i_f$ is zero on the subalgebra generated by the kernel of $f$. Finally (6) entails that the set of elements $u$ of $I(Q)$ such that $i_f(u) \in I(Q)$ is a left ideal of $T(E)$; moreover, if $u = (x \otimes x - Q(x) \cdot 1) \otimes \nu$ ($x \in E,\ \nu \in T(E)$), we have

therefore $I(Q)$ is stable under $i_f$, and the last assertion follows. QED.

Let F be a bilinear form on E; in the remainder of this paragraph we shall denote by $i_x^F$ ($x \in E$) the mapping $i_f$ corresponding to the linear form $f : y \to F(x, y)$ on E.

For any $f \in E^*$, we have

Indeed, formula (8) is equivalent to

Comme (7) completely determines $\lambda_F$ on $T^0$, and since (10) determines $\lambda_F$ on $T^n$ if its values on $T^{n-1}$ are known, the uniqueness of $\lambda_F$ is demonstrated. On the other hand, for $x \in E$ and $u \in T^{n-1}$, the right-hand side of (10) is bilinear on $E \times T^{n-1}$; this demonstrates the existence of $\lambda_F$ by induction on $n$. It remains to prove (9), which we shall do by induction; both members of (9) are zero on $T^0$; suppose (9) to be true on $\sum_{h=0}^{n-1} T^h$; then, for $x \in E$ and $u \in T^{n-1}$, we have:

whence our last assertion.

Indeed, $\lambda_F \circ \lambda_G$ has the characteristic properties (7) and (8) of $\lambda_{F+G}$: we have $(\lambda_F \circ \lambda_G)(1) = 1$, and

On the other hand, if $F = 0$, we have $i_x^F = 0$ for every $x \in E$, and consequently $\lambda_F = I$, which entails that, for every F, we have $\lambda_F \circ \lambda_{-F} = \lambda_{-F} \circ \lambda_F = I$.

Since $\lambda_F$ is a bijection whose inverse bijection is $\lambda_{-F}$ (Lemma 3), it suffices to prove the inclusion $\lambda_F(I(Q')) \subset I(Q)$. Since $I(Q)$ is a left ideal stable under $i_x^F$ (Lemma 1), (8) shows that the set of $u \in T(E)$ such that $\lambda_F(u) \in I(Q)$ is a left ideal. It therefore suffices to prove that, whatever $u \in T(E)$ and $x \in E$, we have $\lambda_F(x \otimes x \otimes u - Q'(x)u) \in I(Q)$. Now, according to (8) and Lemma 1, we have

whence

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:

It suffices indeed to prove that $p_2(a_2)p_1(b_1) = \varepsilon p_1(b_1)p_2(a_2)$, and for this purpose we may suppose that $p_2(a_2)$ (resp. $p_1(b_1)$) is a product $x_1 \ldots x_h$ (resp. $y_1 \ldots y_k$) of elements of $\rho_Q(E_2)$ (resp. $\rho_Q(E_1)$). Since $E_1$ and $E_2$ are orthogonal, we have

whence

The conclusions of cor. 3 and 4 remain true if we omit the hypothesis that $E_1$ and $E_2$ are free modules (cf. exerc. 1).

Indeed, since Q is neutral, E can be decomposed into the direct sum of two totally singular subspaces N and P of dimensions r (§ 4, no. 2, cor. 1 of prop. 2). The restriction of Q to N being zero, the subalgebra S of C(Q) generated by N is identified with the exterior algebra of N (no. 3, lemma 4). For $n \in \mathbf{N}$, we shall denote by $e'_n$ the mapping $t \to nt$ of S into itself.

Let $(n_1, \ldots, n_r)$ be a basis of N; we shall denote by $(p_1, \ldots, p_r)$ the basis of P such that $\Phi(n_i, p_j) = \delta_{ij}$ (§ 4, no. 2, prop. 2). For $p \in P$, we shall denote by $p'$ the linear form $n \to \Phi(n, p)$ on N, and by $i_p$ the endomorphism of S obtained by passing to the quotient from the endomorphism $i_{p'}$ of $T(N)$ associated with $p'$ as was said in lemma 1 of no. 2. By (5), we have

Put, for $x = n + p \in E$ (with $n \in \mathbf{N}$ and $p \in \mathbf{P}$), $s(x) = e'_n + i_p$. It is clear that s is a linear mapping of E into $\mathscr{L}(S)$. Since we have

by virtue of (12) and lemma 1 (no. 2), s extends to a homomorphism (which we shall still denote by s) of C(Q) into $\mathscr{L}(S)$ (no. 1, prop. 1). We shall now prove that this homomorphism is surjective, which, since C(Q) and $\mathscr{L}(S)$ are both of dimension $2^{2r}$, will imply that s is an isomorphism and will prove our first assertion.

Let us indeed denote by $I$ the interval $[1, r]$. For any subset H of I, we shall put $H' = I - H$ and denote by $n_H$ (resp. $p_H$) the product of the $n_i$ (resp. $p_i$) for $i \in H$, arranged in increasing order of the indices. Recall that the $n_H$ form a basis of S (No. 3, Theorem 1). Finally, for any two subsets H, K of I, put $x_{H, K} = n_H p_I n_K$. We shall now prove that the elements $s(x_{H, K})$ of $s(C(Q))$ generate $\mathscr{L}(S)$. Now, if $j \notin H$, we have $s(p_j)(n_H) = i_{p_j}(n_H) = 0$ by Lemma 1, since the $n_i$ for $i \in H$ belong to the kernel of the linear form $n \to \Phi(n, p_j)$ on N; on the other hand, we have

(by (12)). Since s is a homomorphism, it follows that, for any two subsets H, K of I, $s(p_K)(n_H) = 0$ if $K \not\subset H$, and that $s(p_K)(n_H) = \pm n_{H-K}$ if $K \subset H$. Since, for $M \subset I$ and $L \subset I$, by definition $s(n_M)(n_L) = n_M n_L$, and $n_M n_L$ is zero if $M \cap L \neq \emptyset$ and is equal to $\pm n_{M \cup L}$ otherwise, it follows from the above that, for any subsets $H, K, L$ of $I$, $s(x_{H,K})(n_L) = s(n_H)s(p_I)s(n_{K'}) (n_L)$ is zero if $K \neq L$ and is equal to $\pm n_H$ if $K = L$. This shows that the $s(x_{H,K})$ generate $\mathscr{L}(S)$ and completes the proof of the first assertion.

To prove the second assertion, put $S^+ = S \cap C^+$ and $S^- = S \cap C^-$; it is clear that $S^+$ (resp. $S^-$) is the subspace of S generated by the $n_H$ such that H has an even (resp. odd) number of elements, that S is the direct sum of $S^+$ and $S^-$, and that $s(C^+)$ leaves $S^+$ and $S^-$ stable. Consequently $s$ maps $C^+$ into a subalgebra of $\mathscr{L}(S)$, isomorphic to the product $\mathscr{L}(S^+) \times \mathscr{L}(S^-)$; the restriction of $s$ to $C^+$ is an isomorphism of $C^+$ onto this subalgebra, since $s$ is injective and $C^+$ and $\mathscr{L}(S^+) \times \mathscr{L}(S^-)$ both have dimension $2^{2r-1}$ (No. 2, Corollary 1 of Theorem 1). QED.

Indeed, let $A'$ be the algebraic closure of A, and $Q'$ the quadratic form on $E' = A' \otimes_A E$ deduced from Q by extension of scalars. We have seen that $C(Q')$ is isomorphic to $A' \otimes_A C(Q)$ (prop. 2), and it is clear that $C^+(Q')$ is isomorphic to $A' \otimes_A C^+(Q)$. Since $Q'$ is neutral ($§ 4$, No. 2, Cor. 2 of prop. 3), the corollary is an immediate consequence of Th. 2 and the permanence theorems of Chap. VIII, $§ 7$.