# Square roots of $-1$ in quaternion ring

In this Wikipedia page it is said that the square roots of -1 in the quaternion ring are the elements of the imaginary sphere. I don't understand why this is so. I don't understand the system that's written there. If I do

$$(a+bi+cj+dk)^2=a^2-b^2-c^2-d^2+2(ab+cd)i+2(ac-bd)j+2(ad+bc)k=-1$$

the system we get is

$$\begin{cases} a^2-b^2-c^2-d^2=-1 \\ ab+cd=0 \\ ac-bd=0 \\ ad+bc=0 \end{cases}$$

I suppose this could be solved to get the same solution, but it is quite tiresome and I'd like to understand how to get to the prettier system the Wikipedia gives.

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The norm of a quaternion is multiplicative i.e. $|q_1q_2|=|q_1||q_2|$, so if the square is $-1$, the norm squared is $a^2+b^2+c^2+d^2=1$. So your first equation tells you that the only way this can be true is if $a^2=0$, i.e. $b^2+c^2+d^2=1$. Oh, and your last three equations are wrong, it's not $ab+cd=0$, it's $ab=0$. This is because when you multiply-out your first line $jk=-kj$, so the two terms with $cd$ in them cancel. – Ryan Budney Oct 1 '11 at 0:48

Add your first equation to $a^2+b^2+c^2+d^2=1$ and you get $a=0$, whence $b^2+c^2+d^2=1$.

Then observe

$$(b\mathbf{i}+c\mathbf{j}+d\mathbf{k})^2=-(b^2+c^2+d^2)+(cd-dc)\mathbf{i}+(db-bd)\mathbf{j}+(bc-cb)\mathbf{k}$$ $$=-1+0\mathbf{i}+0\mathbf{j}+0\mathbf{k}=-1.$$ Hence any quaternion $\mathbf{q}=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ with $a=0,b^2+c^2+d^2=1$ is a square root of $-1$. The error in your calculations is that you didn't take into account anticommutativity, i.e. $$(c\mathbf{j})(d\mathbf{k})=cd\;\mathbf{i}\quad\text{but}\quad (d\mathbf{k})(c\mathbf{j})=-dc\;\mathbf{i}.$$

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They have it right. You have not been careful enough in $\pm$ signs. The coefficient of $i$ does begin with $2ab$ as you say, but then we get $$c j d k + d k c j = c d i - c d i=0.$$ And so forth.

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Ouch, I found my mistake. I must remember that the binomial theorem only works in commutative rings. Thank you! – lentic catachresis Oct 1 '11 at 1:12