Proposition 2

*Let $\mathbf{A}$ be a ring, $\mathbf{F}$ a Cayley $\mathbf{A}$-algebra, e its unit element, $s : x \mapsto \overline{x}$ its conjugation and $\mathbf{N} : \mathbf{F} \to \mathbf{A}$ its Cayley norm (§ 3, no. 4).

(i) For $\mathbf{F}$ to be alternative, it is necessary and sufficient that, for every ordered pair of elements $x, y$ of $\mathbf{F}$, $x^2y = x(xy)$.*

(ii) If F is alternative, then N(xy) = N(x)N(y) for all x, y in F. (iii) Suppose that F is alternative. For an element x ∈ F to be invertible, it is necessary and sufficient that N(x) be invertible in Ae; the inverse of x is then unique and equal to N(x)^{-1}\overline{x}; denoting this by x^{-1},

or all y ∈ F.

The condition $x^2y = x(xy)$ is obviously necessary for F to be alternative (no. 1, Proposition 1). Conversely, if it holds for every ordered pair of elements of F, applying it to $\overline{x}$ and $\overline{y}$, it gives $\overline{x}^2\overline{y} = \overline{x}(\overline{xy})$; applying the conjugation s to this relation, we obtain $yx^2 = (yx)x$, so that conditions (c) of Proposition 1 of no. 1 are satisfied.

Obviously $a(e, x, y) = 0$ for all $x, y$ in F. If F is alternative, we therefore deduce from Proposition 1 (no. 1) that the subalgebra G of F generated by e, x and y is associative. As $\overline{x} = -x + T(x) \in -x + Ae, \overline{x} \in G$ and similarly $\overline{y} \in G$. Then $N(xy) = (xy)(\overline{xy}) = xy.\overline{y}.\overline{x} = N(y)x\overline{x} = N(y)N(x)$, using the fact that $N(y) \in Ae$. This proves (ii).

Finally we prove (iii). If $N(x)$ is invertible in Ae and we write $x' = N(x)^{-1}\overline{x}$, then $xx' = x'x = e$, for $N(x) = x\overline{x} = \overline{x}x$. Conversely if x admits a left inverse $x''$, then $N(x'')N(x) = N(e) = e$ by (ii) and $N(x)$ is invertible in Ae; further, as $x' = N(x)^{-1}\overline{x}$ is in the subalgebra generated by x and e, the elements $x, x', x''$ belong to the associative subalgebra generated by x, $x''$ and e and hence $x'' = x''(xx') = (x''x)x' = x'$, whence the uniqueness assertion. The formulae $x^{-1}(xy) = x(x^{-1}y) = y$ follow from the fact that $x^{-1}, x$ and y are elements of the subalgebra generated by x, y and e, which is associative.