Bourbaki

Theorem 3

Tag 0DNT, permanent. Cite it as [Bourbaki, Tag 0DNT].

Algebra, chapter IX, § 9. Algèbres de Clifford

Suppose that the dimension of EE is an odd number m=2r+1m = 2r + 1 (and hence that AA is of characteristic ≠2\neq 2).

a) The algebra C+(Q)C^+(Q) is central simple. If Q is of maximum index rr, C+(Q)C^+(Q) is isomorphic to the endomorphism algebra of a vector space of dimension 2r2^r over AA.

b) The algebra C(Q)C(Q) is separable. Its center ZZ is of dimension 2, and C(Q)C(Q) is isomorphic to Z⊗AC+(Q)Z \otimes_A C^+(Q), and hence is simple or a direct sum of two simple subalgebras.

Let x0x_0 be a non-isotropic vector of EE, and let FF be the orthogonal of x0x_0; denote by Q1Q_1 the quadratic form y→−Q(x0)Q(y)y \to -Q(x_0)Q(y) on FF; it is clear that Q1Q_1 is non-degenerate. Since x0y=−yx0x_0 y = -y x_0 (for y∈Fy \in F), we have (x0y)2=−Q(x0)Q(y)=Q1(y)(x_0 y)^2 = -Q(x_0)Q(y) = Q_1(y), and consequently the mapping y→x0yy \to x_0 y of FF into C+(Q)C^+(Q) extends to a homomorphism hh of C(Q1)C(Q_1) into C+(Q)C^+(Q) (No. 1, Prop. 1). Now C(Q1)C(Q_1) is simple (Th. 2) and has the same dimension 22r2^{2r} as C+(Q)C^+(Q); this implies, since h(1)=1h(1) = 1, that hh is an isomorphism. Moreover, if QQ is of index rr, x0x_0 can be chosen in such a way that Q1Q_1 is also of index rr (§ 4, No. 2, Prop. 3), which proves aa.

Soit now (x1,…,x2r)(x_1, \ldots, x_{2r}) an orthogonal basis of FF; let z=x0x1…x2rz = x_0 x_1 \ldots x_{2r}. One verifies immediately that zz commutes with xjx_j for j=0,…,2rj = 0, \ldots, 2r, and hence belongs to the center of C(Q)C(Q). Let ZZ be the subspace of C(Q)C(Q) generated by 11 and zz; it is a subalgebra of the center of C(Q)C(Q) and a quadratic extension of AA, since zz is odd and z2z^2 is equal to the scalar (−1)rQ(x0)…Q(x2r)(-1)^r Q(x_0) \ldots Q(x_{2r}). Consider the homomorphism θ\theta of Z⊗AC+(Q)Z \otimes_A C^+(Q) into C(Q)C(Q) defined by θ(u⊗v)=uv\theta(u \otimes v) = uv. Since z∈C−z \in C^- and is invertible, the mapping u→zuu \to zu is an isomorphism of the module C+C^+ onto C−C^-, which entails that θ(Z⊗C+)\theta(Z \otimes C^+) contains C+C^+ and C−C^-, and hence coincides with C(Q)C(Q). Since Z⊗C+Z \otimes C^+ and C(Q)C(Q) have the same dimension 22r+12^{2r+1}, θ\theta is an isomorphism; this proves bb, in view of the results of Chap. VIII, § 7.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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