Exercise 6, § 2

Let $K$ be a commutative field in which $2 \neq 0$ and $E$ a quaternion algebra over $K$ of type $(0, 0, \gamma)$. (a) If $\gamma$ is not a square in $K$, there exists no left (resp. right) ideal in $E$ of dimension 1 over $K$; the set $a$ of $x \in E$ such that $x^2 = 0$ is a two-sided ideal of dimension 2 over $K$ and $E/a$ is a field, a quadratic algebra over $K$. (b) If $\gamma$ is a square $\neq 0$ in $K$, there exists in $E$ a basis $(e_i)_{1 \leq i \leq 4}$ with the following multiplication table:

<table> <tr> <th></th> <th>e_1</th> <th>e_2</th> <th>e_3</th> <th>e_4</th> </tr> <tr> <th>e_1</th> <td>e_1</td> <td>0</td> <td>e_3</td> <td>0</td> </tr> <tr> <th>e_2</th> <td>0</td> <td>e_2</td> <td>0</td> <td>e_4</td> </tr> <tr> <th>e_3</th> <td>0</td> <td>e_3</td> <td>0</td> <td>0</td> </tr> <tr> <th>e_4</th> <td>e_4</td> <td>0</td> <td>0</td> <td>0</td> </tr> </table>

The set $a$ of $x \in E$ such that $x^2 = 0$ is a two-sided ideal of dimension 2, the direct sum of the two two-sided ideals $Ke_3$ and $Ke_4$ which annihilate one another; the latter are the only (left or right) ideals of dimension 1 in $E$. The quotient algebra $E/a$ is isomorphic to the product of two fields isomorphic to $K$.

(c) If $\gamma = 0$, the set $a$ of $x \in E$ such that $x^2 = 0$ is a two-sided ideal of dimension 3; $Kk = b$ is the only (left or right) ideal of dimension 1 in $E$; it is a two-sided ideal, the left and right annihilator of $a$. In the quotient algebra $E/b$, $a/b$ is the direct sum of two two-sided ideals of dimension 1 which annihilate one another; finally $E/a$ is a field isomorphic to $K$.