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)=u for all u∈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.
Let B be a basis of E containing e and u an element of B distinct from e; then T(u)−T′(u)=0, whence s(u)=s′(u). As s and s' coincide on B, they are equal.
In what follows, we shall write uˉ=s(u), so that
{u+uˉ=T(u),u+v=uˉ+vˉ,uuˉ=uˉu=N(u),αu=αuˉ,uˉ=uuv=vˉ.uˉ
for u,v in E, α∈A; moreover
T(e)=2e,N(e)=e.
From the formula
T(uv)=uv+uv=uv+vˉ.uˉ=uv+(T(v)−v)(T(u)−u),
we deduce that
uv+vu=T(u)v+T(v)u+(T(uv)−T(u)T(v))
whence, exchanging u and v,
T(vu)=T(uv).
On the other hand, N(u+v)=(u+v)(uˉ+vˉ)=N(u)+N(v)+T(uvˉ), whence
(18)
T(vuˉ)=T(uvˉ)=N(u+v)−N(u)−N(v).
Now, (16) applied with u replaced by uˉ gives
T(uvˉ)=T(u)T(v)+uˉv+vuˉ−T(u)v−T(v)uˉ=T(u)T(v)−uv−vˉ.uˉ; whence
(19)
T(vuˉ)=T(uvˉ)=N(u+v)−N(u)−N(v)=T(u)T(v)−T(uv).
Finally, clearly for all α∈A,
(20)
N(αu)=α2N(u); in particular
N(2u)=4N(u), so that formula (19) gives
(21)
(T(u))2−T(u2)=2N(u).
Clearly T is a linear form on the (Ae)-module E. As (u,v)↦T(vuˉ) is a bilinear form on this module, it follows from (18) and (20) that N 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.
Nothing.