Bourbaki

Proposition 4

Tag 05Z1, permanent. Cite it as [Bourbaki, Tag 05Z1].

Algebra, chapter III, § 2. Examples of algebras

Let E be an A-algebra and s and s' antiautomorphisms of E such that (E, s) and (E, s') are Cayley algebras. If E admits a basis containing the unit element e, then s' = s.

Clearly s′(u)=s(u)=us'(u) = s(u) = u for all u∈Aeu \in Ae. If T, N (resp. T', N') are the trace and norm functions for (E, s) (resp. (E, s')), it follows from (13) that

(T(u)−T′(u)).u−(N(u)−N′(u))=0. (T(u) - T'(u)).u - (N(u) - N'(u)) = 0.

Let B be a basis of E containing e and u an element of B distinct from e; then T(u)−T′(u)=0T(u) - T'(u) = 0, whence s(u)=s′(u)s(u) = s'(u). As s and s' coincide on B, they are equal.

In what follows, we shall write uˉ=s(u)\bar{u} = s(u), so that

{u+uˉ=T(u),uuˉ=uˉu=N(u),uˉ=uu+v=uˉ+vˉ,αu=αuˉ,uv=vˉ.uˉ \left\{\begin{array}{lll} u + \bar{u} = T(u), & u\bar{u} = \bar{u}u = N(u), & \bar{u} = u \\ u + v = \bar{u} + \bar{v}, & \alpha u = \alpha \bar{u}, & uv = \bar{v}.\bar{u} \end{array}\right.

for u,vu, v in E, α∈A\alpha \in A; moreover

T(e)=2e,N(e)=e. T(e) = 2e, \quad N(e) = e.

From the formula

T(uv)=uv+uv‾=uv+vˉ.uˉ=uv+(T(v)−v)(T(u)−u), T(uv) = uv + \overline{uv} = uv + \bar{v}.\bar{u} = uv + (T(v) - v)(T(u) - u),

we deduce that

uv+vu=T(u)v+T(v)u+(T(uv)−T(u)T(v)) uv + vu = T(u)v + T(v)u + (T(uv) - T(u)T(v))

whence, exchanging uu and vv,

T(vu)=T(uv). T(vu) = T(uv).

On the other hand, N(u+v)=(u+v)(uˉ+vˉ)=N(u)+N(v)+T(uvˉ)N(u + v) = (u + v)(\bar{u} + \bar{v}) = N(u) + N(v) + T(u\bar{v}), whence

(18)

T(vuˉ)=T(uvˉ)=N(u+v)−N(u)−N(v). T(v\bar{u}) = T(u\bar{v}) = N(u + v) - N(u) - N(v).

Now, (16) applied with uu replaced by uˉ\bar{u} gives

T(uvˉ)=T(u)T(v)+uˉv+vuˉ−T(u)v−T(v)uˉ=T(u)T(v)−uv−vˉ.uˉ; T(u\bar{v}) = T(u)T(v) + \bar{u}v + v\bar{u} - T(u)v - T(v)\bar{u} = T(u)T(v) - uv - \bar{v}.\bar{u};
whence

(19)

T(vuˉ)=T(uvˉ)=N(u+v)−N(u)−N(v)=T(u)T(v)−T(uv). T(v\bar{u}) = T(u\bar{v}) = N(u + v) - N(u) - N(v) = T(u)T(v) - T(uv).

Finally, clearly for all α∈A\alpha \in A,

(20)

N(αu)=α2N(u); N(\alpha u) = \alpha^2 N(u);
in particular N(2u)=4N(u)N(2u) = 4N(u), so that formula (19) gives

(21)

(T(u))2−T(u2)=2N(u). (T(u))^2 - T(u^2) = 2N(u).

Clearly TT is a linear form on the (Ae)(Ae)-module EE. As (u,v)↦T(vuˉ)(u, v) \mapsto T(v\bar{u}) is a bilinear form on this module, it follows from (18) and (20) that NN is a quadratic form (cf. IX, § 3, no. 4).

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

  • IX, § 3, no. 4 not yet on this site: A IX
Transcribed from the printed edition, 1998, Springer. Not translated.