Proposition 3

Let E be a quadratic A-algebra of type $(\alpha, \beta)$.

(i) If A is a field and contains no element $\zeta$ such that $\zeta^2 = \alpha + \beta \zeta$, E is a (commutative) field (cf. V, § 3).

(ii) If the ring A contains an element $\zeta$ such that $\zeta^2 = \alpha + \beta \zeta$ and $\beta - 2\zeta$ is invertible (resp. zero), E is isomorphic to A $\times$ A resp. is of type $(0, 0))$.

We prove (i). Let $\xi, \eta$ be two elements of A and $u = \xi e + \eta i$. If $\eta \neq 0$ and we write $\theta = -\xi \eta^{-1}$, then $\mathrm{N}(u) = \eta^2 (\theta^2 - \beta \theta - \alpha)$ by (5), whence $\mathrm{N}(u) \neq 0$ by virtue of the hypothesis on A; if $\eta = 0$, then $\mathrm{N}(u) = \xi^2$. In any case, if $u \neq 0$, then $\mathrm{N}(u) \neq 0$, hence $\mathrm{N}(u)$ is invertible in A and therefore $u$ is invertible in E.

We now prove (ii). The canonical basis $(e_1, e_2)$ of the algebra A $\times$ A is of type $(0, 1)$. We have seen (formula (3)) that E is of type

for all $\gamma \in A$ and all $\delta$ invertible in A. If $\beta - 2\zeta$ is invertible, take $\delta = (\beta - 2\zeta)^{-1}$ and $\gamma = -\zeta (\beta - 2\zeta)^{-1}$; then $2\gamma + \beta \delta = 1$ and

thus E is of type $(0, 1)$ and hence isomorphic to A $\times$ A. If $\beta - 2\zeta = 0$, it has already been remarked that E is of type $(\alpha + \zeta^2, 0)$ and hence of type $(0, 0)$ since $\alpha + \zeta^2 = 2\zeta^2 - \beta \zeta = 0$.

A quadratic A-algebra of type $(0, 0)$ is also called an algebra of dual numbers over A.