Let $\mathbb K$ be a field of $\text{Char}$($\mathbb K$)$\neq 2$.

Set $Q=Q(a,b\mid \mathbb K)=(a,b)_{\mathbb K}$ be the Quaternion algebra for $a,b \in \mathbb K$, with $\mathbb K$ basis $1, i, j$ and $k$, such that \begin{cases} i^2=a,\\ j^2=b,\\ k^2=-ab,\\ ij=-ji=k \end{cases} And let $Q_0=Q_0(a,b\mid \mathbb K)=\langle \mathbb K i+\mathbb Kj+\mathbb Kk \rangle $ be the set of pure Quaternions.

Then the question is as follows:

Let $[,]:Q \times Q \to Q$ be the commutator, i.e. $[x:y] = xy - yx$ for all $x,y \in Q.$ Show that $[Q_0,Q_0]\subseteq Q_0$ with equality if $a\neq 0 \neq b.$

$\textbf{Some attempt:}$

let $q_1=ix_{11}+jx_{12}+kx_{13}$ and $q_2=ix_{21}+jx_{22}+kx_{23}$ be in $Q_0$. Then \begin{align} q_1 q_2&= bi(x_{13}x_{22} - x_{12}x_{23} +aj(x_{11}x_{23} -x_{13}x_{21})+k(x_{11}x_{22}-x_{12}x_{21})\\&+ax_{11}x_{21}+bx_{12}x_{22}-abx_{13}x_{23} \end{align} and \begin{align} q_2q_1&=-aj(x_{23}x_{11}-x_{21}x_{13})-bi(x_{22}x_{13}-x_{23}x_{12})-k(x_{22}x_{11}-x_{21}x_{12})\\&+ax_{21}x_{11}+bx_{22}x_{12}-abx_{23}x_{13} \end{align}

and so we have $q_1 q_2-q_2q_1=2bi(x_{13}x_{22} - x_{12}x_{23} +2aj(x_{11}x_{23} -x_{13}x_{21})+2k(x_{11}x_{22}-x_{12}x_{21})$ which is an element of $Q_0$ if $a\neq 0\neq b$ and equality holds if $ab\neq 0$.

Can someone let me know if my solution and logic in proving is not correct?


  • $\begingroup$ That is supposeed to be an element of $Q_0$ even if one of $a$ or $b$ happens to be zero. But, what about the equality if $ab\neq0$ part? As in: every element of $Q_0$ is a linear combination of commutators? For that to work you need to assume that the characteristic is not two, but that was probably assumed... $\endgroup$ – Jyrki Lahtonen Apr 16 '18 at 19:10
  • $\begingroup$ @JyrkiLahtonen Many thanks for your comment! Yes, sorry I forgot to add that restriction to $\mathbb K$. I added now. Can you please post your answer? Thanks! $\endgroup$ – Nikita Apr 16 '18 at 19:15
  • $\begingroup$ @JyrkiLahtonen Many thanks for your comment! Sorry I just understood what you said! Yes you are right and I edited it. There is also equality. So I think now it is complete? I think my answer is correct? $\endgroup$ – Nikita Apr 16 '18 at 23:11

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