Theorem 3
Tag 0DNT, permanent. Cite it as [Bourbaki, Tag 0DNT].
Algebra, chapter IX, § 9. Algèbres de Clifford
Suppose that the dimension of is an odd number (and hence that is of characteristic ).
a) The algebra is central simple. If Q is of maximum index , is isomorphic to the endomorphism algebra of a vector space of dimension over .
b) The algebra is separable. Its center is of dimension 2, and is isomorphic to , and hence is simple or a direct sum of two simple subalgebras.
Let be a non-isotropic vector of , and let be the orthogonal of ; denote by the quadratic form on ; it is clear that is non-degenerate. Since (for ), we have , and consequently the mapping of into extends to a homomorphism of into (No. 1, Prop. 1). Now is simple (Th. 2) and has the same dimension as ; this implies, since , that is an isomorphism. Moreover, if is of index , can be chosen in such a way that is also of index (§ 4, No. 2, Prop. 3), which proves .
Soit now an orthogonal basis of ; let . One verifies immediately that commutes with for , and hence belongs to the center of . Let be the subspace of generated by and ; it is a subalgebra of the center of and a quadratic extension of , since is odd and is equal to the scalar . Consider the homomorphism of into defined by . Since and is invertible, the mapping is an isomorphism of the module onto , which entails that contains and , and hence coincides with . Since and have the same dimension , is an isomorphism; this proves , 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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