Exercise 5, § 2

Over a commutative field $K$ in which $2 = 0$, every quaternion algebra $E$ of type $(\alpha, 0, \gamma)$ is commutative and the square of every $x \in E$ belongs to $K$; the subalgebra $K(x)$ of $E$ generated by an element $x \in K$ is therefore a quadratic algebra of type $(\lambda, 0)$ over $K$ and $E$ is a quadratic algebra over $K(x)$.

Show that the set $C$ of $x \in E$ whose square is equal to the square of an element of $K$ is a vector subspace of $E$ whose dimension is equal to 1, 2 or 4. If $C$ is of dimension 1 (in which case $C = K$), $E$ is a field. If $C$ is of dimension 2, there exists a quadratic algebra $K(x)$ contained in $E$ which is a field and $E$ has a basis over $K(x)$ consisting of two elements 1 and $u$ with $u^2 = 0$; the set of $y \in E$ such that $y^2 = 0$ is an ideal $a$ of dimension 2 over $K$ and $E/a$ is isomorphic to $K(x)$. Finally, if $C$ is of dimension 4 (and hence equal to $E$), the set $a$ of $y \in E$ such that $y^2 = 0$ is an ideal of dimension 3 and $E/a$ is isomorphic to $K$; there exists in $E$ a basis $(1, e_1, e_2, e_3)$ such that $e_1^2 = e_2^2 = e_3^2 = 0$, $e_1 e_2 = e_3$, $e_1 e_3 = e_2 e_3 = 0$; $Ke_3 = b$ is the only ideal of dimension 1 in $E$, $b$ is the annihilator of $a$ and $a/b$ is the direct sum of two ideals of $E/b$, of dimension 1, which annihilate one another.