Proposition 3

Tag 069I, permanent. Cite it as [Bourbaki, Tag 069I].

Algebra, chapter III, Appendix 0. Alternative algebras. Octonions

Let E be a Cayley A-algebra, γ an element of A and F the Cayley extension of E defined by γ and the conjugation of E (§ 2, no. 5, Proposition 5). For F to be alternative, it is necessary and sufficient that E be associative.

Let $u = (x, y), v = (x', y')$ be two elements of F (where x, y, $x', y'$ are in E). Then (§ 2, no. 5, formula (27))

\begin{cases} u^2v = ((x^2 + \gamma \overline{y}y)x' + \gamma \overline{y}'(y\overline{x} + yx), (y\overline{x} + yx)\overline{x}' + y'(x^2 + \gamma \overline{y}y)) \\ u(uv) = (x(xx' + \gamma \overline{y}'y) + \gamma (x'\overline{y} + \overline{x}.\overline{y}')y, y(\overline{x}'\overline{x} + \gamma \overline{y}y') + (y\overline{x}' + y'x)x). \end{cases}

Using the fact that $\overline{y}y$ and $\overline{x} + x$ are in Ae, examining these formulae shows that the associativity of E implies $u^2v = u(uv)$ and hence the fact that F is alternative (Proposition 2). Conversely, if F is alternative, the equation $u^2v = u(uv)$ applied when $y' = 0$ gives

(y\overline{x} + yx)\overline{x}' = y(\overline{x}'\overline{x}) + (y\overline{x}')x.

Now the left hand side is equal to

(yT(x))\overline{x}' = y(\overline{x}'T(x)) = y(\overline{x}'x + \overline{x}'\overline{x});

comparing with the right hand side, we obtain $(y \bar{x}') x = y (\bar{x}' x)$ , which proves the associativity of E, since $x, y$ and $\bar{x}'$ are arbitrary elements of E.

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

Proposition 3

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