Proposition 1

A quadratic algebra E is associative and commutative.

The fact that E is commutative follows from the equation $e_1 e_2 = e_2 e_1$ in (2); similarly, to verify associativity, it suffices to see that $x(yz) = (xy)z$ when $x, y, z$ are each equal to $e_1$ or $e_2$. Now, this relation is obvious if at least one of the elements $x, y, z$ is equal to $e_1$; it is also true for $x = y = z = e_2$ since E is commutative; whence the proposition.

Let $e$ denote the unit element of a quadratic algebra E and let $(e, i)$ be a basis of E of type $(\alpha, \beta)$; every other basis of E containing $e$ is therefore of the form $(e, j)$ with $j = \gamma e + \delta i$ (II, § 7, no. 2, Corollary to Proposition 3); moreover, for $(e, j)$ to be a basis of E, it is necessary and sufficient that $\delta$ be invertible in A; the condition is obviously sufficient; conversely, if $\bar{i}$ is the canonical image of $i$ in $E/Ae$, $\bar{i}$ and $\bar{j} = \delta \bar{i}$ must each form a basis of $E/Ae$, whence the necessity of the condition. Then

thus it is seen that E is of type

for all invertible $\delta \in A$ and all $\gamma \in A$. In particular, if $E$ is of type $(\alpha, 2\beta')$, $i$ is also of type $(\alpha + {\beta'}^2, 0)$ as is seen by taking $\gamma = -\beta'$ and $\delta = 1$.