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Let $Q = (i,j,k)$ be the quaternion ring with multiplication defined by: $$ij = k, \quad jk = i, \quad i^2 = j^2 = k^2 = −1$$ and addition defined as a formal vector space over $\mathbb R$ with basis: $1, i, j, k.$

Show that $x^2 + 1$ has an infinite number of roots in Q.

My approach:

Let $a, b, c$ be real numbers satisfying $a^2 + b^2 + c^2 = 1$ and let $x = ai + bj + ck$. Then $$ x^2 = (ai + bj + ck)(ai + bj + ck) = -(a^2 + b^2 + c^2) = -1. $$

I'm not quite sure this approach is true in $\mathbb Q$, (I saw this approach was used when $x \in \mathbb H$)

Dietrich Burde
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Ilan Aizelman WS
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2 Answers2

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Yes, it is correct, but, of course, you must explain why there are infinitely many elements $(a,b,c)\in\mathbb R^3$ such that $a^2+b^2+c^2=1$. And you should not write $\mathbb Q$ when you mean $Q$. Actually, I suppose that you mean $\mathbb H$ here.

J. W. Tanner
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José Carlos Santos
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Write a quaternion as $w=r+q$, where $r\in\mathbb{R}$ and $q=ai+bj+ck$ (with $a,b,c\in\mathbb{R}$). Then $q^*=-q$ (the asterisk denoting quaternion conjugation).

We have $w^2=(r+q)^2=r^2+2rq+q^2$, due to the fact that $r$ commutes with $q$. Hence $$ w^2=(r+q)^2=r^2+2rq-qq^*=(r^2-qq^*)+2rq $$ If we want $w^2\in\mathbb{R}$, we need either $r=0$ or $q=0$. The latter case yields no solution to $w^2=-1$. With $r=0$ we have $$ w^2=-qq^*=-(a^2+b^2+c^2) $$ which equals $-1$ if and only if $a^2+b^2+c^2=1$. This is a spherical surface in $\mathbb{R}^3$, so there are infinitely many solutions. And we found all solutions to $w^2=-1$.

egreg
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