Alternative algebras. Octonions

The verification follows immediately from definition (1).

Clearly (a) implies (c). We show that (c) implies (b): by definition (§ 7, no. 3) to prove (b), it suffices to verify that $a(x, x, y) = 0$ and $a(x, y, y) = 0$, which is precisely (c).

To prove that (b) implies (a), we use the following 4 lemmas:

The set $U'$ of $x \in E$ such that $xU \subset U$ and $Ux \subset U$ is obviously a sub-A-module of $E$, which contains $S$ by hypothesis. On the other hand, for $x, y$ in $U'$ and $u \in U$, by hypothesis

on passing to the opposite algebra, we have similarly $u(xy) \in U$. Hence $U'$ is a subalgebra of $E$ and, since it contains $S$, $U' = E$. Hence $EU \subset U$ and a fortiori $UU \subset U$, which proves that $U$ is a subalgebra of $E$; as it contains $S$, $U = E$, which proves the lemma.

A subset $H$ of $F$ is called strongly associative if $a(u, v, w) = 0$ when at least two of the elements $u, v, w$ belong to $H$.

As the set of strongly associative subsets of $F$ is inductive, it suffices to prove that if $H$ is a maximal strongly associative subset of $F$, $H$ is then a subalgebra of $F$. As $H$ is obviously a sub-A-module of $F$, it suffices to verify that for any two elements $u, v$ of $H$, $H \cup \{uv\}$ is also strongly associative, for by virtue of the definition of $H$, this will imply $uv \in H$. Now, for all $z \in H$ and all $t \in F$, by (2)

since $\mathbf{H}$ is strongly associative; as $u, v, z$ are in $\mathbf{H}$, also

whence $a(uv, t, z) = 0$. Using the fact that $a$ is alternating, this shows that $a(p, q, r) = 0$ whenever at least two of the elements $p, q, r$ belong to $\mathbf{H} \cup \{uv\}$ whence the lemma.

$a(u, v, w) = 0$ whenever two of the three elements $u, v, w$ are equal to $x$ and it suffices to apply Lemma 3.

Let $Z$ be the set of $z \in \mathbf{E}$ such that $a(u, v, z) = 0$ for all $u \in \mathbf{X}$ and $v \in \mathbf{Y}$, this is obviously a sub-A-module containing $\mathbf{X}$ and $\mathbf{Y}$ since $\mathbf{X}$ and $\mathbf{Y}$ are strongly associative; by Lemma 2, it will suffice to verify that, for $u \in \mathbf{X}$ and $v \in \mathbf{Y}$, $uZ \subset Z, vZ \subset Z, Zu \subset Z$ and $Zv \subset Z$. Now, for $u, u'$ in $\mathbf{X}$, $v \in \mathbf{Y}$ and $z \in Z$, by (2)

by virtue of the fact that $\mathbf{X}$ is strongly associative and the definition of $Z$. But as $\mathbf{X}$ is strongly associative, $a(u', u, zv) = 0$ and since $u'u \in \mathbf{X}$, $a(u'u, z, v) = 0$ by definition of $Z$. Hence $a(u', uz, v) = 0$, which shows that $uZ \subset Z$. Applying (2) now with $(p, q, r, s) = (v, z, u, u')$, we obtain similarly $Zu \subset Z$. Interchanging the roles of $\mathbf{X}$ and $\mathbf{Y}$ and using the fact that $a$ is alternating, we obtain $vZ \subset Z$ and $Zv \subset Z$; whence the lemma.

It now suffices, to prove that (b) implies (a) in Proposition 1, to take $\mathbf{X} = \{x\}$ and $\mathbf{Y} = \{y\}$, using Lemma 4.

(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.

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))

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

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.

We preserve the notation at the beginning of this no. It follows from formulae (33) of § 2, no. 5 that the set $S$ consisting of the elements $\pm e_0, \pm e_1, \pm e_2, \pm e_3$ is stable under multiplication. Moreover, for $x, y, y'$ in $E$, by formula (22) of § 2, no. 5,

so that the set $T$ consisting of the elements $\pm e_i$ ($0 \leq i \leq 7$) is stable under multiplication; moreover, its multiplication table is independent of the ring $\mathbf{A}$.

In particular, let $\mathbf{A}''$ be the field $\mathbf{Z}/2\mathbf{Z}$ with two elements and let $E''$ be the quaternion algebra of type $(1, 0, 1)$ over $\mathbf{A}''$ and $F''$ the algebra of octonions of type $(1, 0, 1, 1)$ over $\mathbf{A}''$; let $(e''_i)_{0 \leq i \leq 7}$ be the basis of $F''$ formed as described above. Since $-e''_i = e''_i$, the set $T''$ of $e''_i$ has 8 elements and is stable under multiplication; moreover, it follows immediately from the above that the mapping $\theta : T \to T''$ such that $\theta(e_i) = \theta(-e_i) = e''_i$ for $0 \leq i \leq 7$ is a homomorphism for multiplication. Moreover the quaternion algebra $E''$ is in this case commutative and hence $F''$ is associative (§ 2, no. 5, Proposition 5); further, conjugation in $F''$ is in this case the identity. Hence $T''$ is a group and formulae (8) show that it is commutative; these formulae and formulae (33) of § 2, no. 5 show that the square of every element of $T''$ is the unit. If $V$ is used to denote the group $T''$ written additively, $V$ can be given a unique vector space structure over $\mathbf{Z}/2\mathbf{Z}$, necessarily of dimension 3 since

For all $\lambda \in V$, let $e'_\lambda$ then denote the element of $(e_i)_{0 \leq i \leq 7}$ such that $\theta(e'_\lambda) = \lambda$; then $e'_0 = e_0$; moreover, as $\theta$ is a homomorphism and the relation $\theta(x) = \theta(y)$ is equivalent to $x = \pm y$, $e'_\lambda e'_\mu = \pm e'_{\lambda + \mu}$. If $\lambda, \mu, \nu$ are linearly independent over $\mathbf{Z}/2\mathbf{Z}$, they form a basis of $V$ and hence all the elements $e_i$ ($0 \leq i \leq 7$) would belong to the subalgebra generated by $e'_\lambda, e'_\mu$ and $e'_\nu$; when $2 \neq 0$ in $\mathbf{A}$, $e'_\lambda (e'_\mu e'_\nu) = (e'_\lambda e'_\mu) e'_\nu$ is therefore impossible, for $F$ would be associative and hence $E$ would be commutative (§ 2, no. 5, Proposition 5), which contradicts relations (33) of § 2, no. 5. On the other hand, if $\lambda, \mu, \nu$ are linearly dependent in V, the three elements $e'_\lambda, e'_\mu, e'_v$ belong to a subalgebra with 2 generators of F, which is therefore associative (no. 1, Proposition 1); whence the conclusion.