Example

(1) If we take $\mathbf{E} = \mathbf{A}$ (and hence $s = 1_A$), the algebra $\mathbf{F}$ is a quadratic A-algebra with basis $(e, j)$ where $j^2 = \gamma e$.

(2) Take E to be a quadratic algebra of type $(\alpha, \beta)$, so that the underlying module of E is $\mathbf{A}^2$, with multiplication table (2) (no. 3) for the canonical basis. Take s to be conjugation of E (no. 3, Proposition 2). Then, for all $\gamma \in \mathbf{A}$, the Cayley extension F of $(\mathbf{E}, s)$ defined by $\gamma$ is called the quaternion algebra of type $(\alpha, \beta, \gamma)$, which is associative by no. 3, Proposition 1 and Proposition 5 above; its underlying module is $\mathbf{A}^4$ and, if $(e, i, j, k)$ denotes the canonical basis of $\mathbf{A}^4$, the corresponding multiplication table is given by

(30)

Further, for $u = \rho e + \xi i + \eta j + \zeta k$ (with $\rho, \xi, \eta, \zeta$ in $\mathbf{A}$), we have (writing $\bar{u}$ instead of $t(u)$ and identifying $\mathbf{A}$ with $\mathbf{Ae}$):

Formulae (30) follow from (28) and (29) and formulae (31) from (23) and (24), taking account of the formulae for the quadratic algebra E.

Then, for $u, v$ in F,

for $N_F(uv) = uv.\overline{uv} = uv(\overline{v}.\overline{u}) = u(v\overline{v})\overline{u} = (u\overline{u})(v\overline{v})$ by virtue of the associativity and the fact that $N_F(u)$ belongs to the centre of F.

An A-algebra isomorphic to a quaternion algebra is also called a quaternion algebra; if a basis of such an algebra has multiplication table (30), it is called a basis of type $(\alpha, \beta, \gamma)$. By an abuse of language, a quaternion algebra is said to be of type $(\alpha, \beta, \gamma)$ when it has a basis of type $(\alpha, \beta, \gamma)$.

When $\beta = 0$, formulae (30) and (31) simplify to

and

Then $(\alpha, \beta, \gamma)$ is replaced throughout by $(\alpha, \gamma)$ in the above expressions. It is immediate that the quaternion algebras of types $(\alpha, \gamma)$ and $(\gamma, \alpha)$ are isomorphic.

Note that formulae (32) show that F is not commutative when $-1 \neq 1$ in $\mathbf{A}$.

*Taking A to be the field $\mathbf{R}$ of real numbers and $\alpha = \gamma = -1, \beta = 0$, the corresponding algebra F is called the algebra of Hamiltonian quaternions and is denoted by $\mathbf{H}$. If $u = \rho e + \xi i + \eta j + \zeta k$ ($\rho, \xi, \eta, \zeta$ in $\mathbf{R}$) is an element $\neq 0$ in $\mathbf{H}$, the formula $u\bar{u} = \bar{u}u = \rho^2 + \xi^2 + \eta^2 + \zeta^2$ (cf. (34)) shows that $N(u) \neq 0$ in $\mathbf{R}$, so that $u$ admits an inverse $u^{-1} = N(u)^{-1}\bar{u}$ in $\mathbf{H}$ and that $\mathbf{H}$ is therefore a non-commutative field.

(3) If E is taken to be a quaternion algebra (cf. Example 2), the Cayley extension of E defined by an element $\delta \in \mathbf{A}$ is in general non-associative

(Proposition 5); it is called an octonion algebra over A (cf. Appendix, no. 3).