Proposition 4

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

Algebra, chapter IX, § 4. Sous-espaces totalement isotropes. Théorème de Witt

A non-degenerate quadratic form Q on E such that there exists a vector $x \neq 0$ of E such that $Q(x) = 0$ being given, for every element a of A, there then exists $y \in E$ such that $Q(y) = a$ .

Indeed, according to Cor. 1 of Prop. 2, there exists a subspace $G = F + F'$ ( $F, F'$ : totally singular subspaces of dimension 1) of E, of dimension 2, such that the restriction of Q to G is neutral. If $\{ e, e' \}$ ( $e \in F, e' \in F'$ ) is a basis of G, we have

Q(xe + x'e') = bxx' \quad (x \in A, x' \in A, b \in A, b \neq 0).

It suffices therefore to take for

y

the vector

ae + b^{-1}e'

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

Cited by

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Cites

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