# The quaternion division ring contains an infinite number of elements $u$ satisfying $u^2=-1$

Show that the quaternion division ring contains an infinite number of elements $$u$$ satisfying $$u^2=-1$$

I was trying to solve the above exercise on Page 133, Basic Algebra, Jacobson. Maybe it is convenient to consider $$\mathbb H=\left\{ \left[\begin{matrix} \alpha & \beta\\-\bar{\beta} & \bar{\alpha} \end{matrix} \right]: \alpha, \beta\in\mathbb C \right\}.$$

Of course, $$i= \begin{bmatrix} \sqrt{-1} & 0\\ 0 & -\sqrt{-1} \end{bmatrix},\quad j=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},\quad k=\begin{bmatrix} 0 & \sqrt{-1}\\ \sqrt{-1} & 0 \end{bmatrix}$$ satisfy $$u^2=-1$$. But how to show that there are infinitely many elements in $$\mathbb H$$ satisfying the identity?

• "Maybe it is convenient....". It isn't. – Lord Shark the Unknown Jan 13 at 8:49

## 2 Answers

The proof of this statement can be found on Wikipedia. Let the square root of $$-1$$ be $$x=a+bi+cj+dk$$, then $$x^2=-1$$ is equivalent to $$a^2-b^2-c^2-d^2=-1\qquad2ab=2ac=2ad=0$$ If $$a\ne0$$, $$b=c=d=0$$ and the first equation becomes $$a^2=-1$$, impossible since $$a\in\mathbb R$$. Hence $$a=0$$, at which point the second set of equations is automatically satisfied and the first equation becomes $$b^2+c^2+d^2=1$$. The quaternionic square roots of $$-1$$ thus form the unit sphere in $$\mathbb R^3$$, so there are infinitely many solutions.

Hint: Forget about the matrices for a moment. What is $$(ai+bj)^2$$?