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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:

Show that $x \in Q_0$ if and only if $x\notin \mathbb K$ and $x^2 \in \mathbb K$ and conclude that if $\varphi: Q \to Q'$ is a $\mathbb K$-linear isomorphism of such quaternion algebras, then $\varphi (Q_0)=Q_{0}^{'}$.

$\textbf{Some attempt:}$

we have that for $q\in Q$, $q=x_0+q_0$ for $x_0 \in \mathbb K$ and $q_0 \in Q_0$. Then we claim that $q^2\in \mathbb K \Leftrightarrow q\in \mathbb K ~\text{or} ~q ~ \text{is pure}$.

We see that $q^2=(x_0+q_0)^2=x_{0}^{2}+2x_0q_0+q_{0}^{2}$ for $x_0\in \mathbb K$ and $q_0\in Q_0$. And as we know $q_{0}^{2}=(x_1i+x_2j+x_3k)^2=ax_{1}^{2}+bx_{2}^{2}-abx_{3}^{2} \in \mathbb K$. So for to prove that $q^2\in \mathbb K$, we need to show that $x_0q_0 \in \mathbb K$. Which means that $q_0 \in \mathbb K$ or $x_0=0$. And this means that $q\in \mathbb K$ or $q$ is pure. And we can conclude that the pure quaternions in $Q$ are precisely the $q$ satisfying $q^2 \in \mathbb K$ with $q \notin \mathbb K$ along with 0.

And $\varphi: Q \to Q'$ is a $\mathbb K$-linear isomorphism of such quaternion algebras if it takes center to center and we know that $\mathbb K$ is in its center and if for $q\in Z(Q)$, $q\in \mathbb K$ then $q^2\in \mathbb K$ and then by above we can conclude that $q$ is pure or an scalar. So we have $\varphi (Q_0)=Q_{0}^{'}$.

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

Thanks!

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  • $\begingroup$ $(x_1i+x_2j+x_3k)^2=ax_1^2+bx_2^2-abx_3^2$ $\endgroup$ – ArtW Apr 16 '18 at 17:55
  • $\begingroup$ your solution is completely fine, maybe apart from mentioning at the end that also $\mathbb{K}$ is fixed under $\phi$, so that pure quaternions cannot map to scalars or vice versa $\endgroup$ – ArtW Apr 16 '18 at 17:58
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    $\begingroup$ Compare also with this question. $\endgroup$ – Dietrich Burde Apr 16 '18 at 18:02
  • $\begingroup$ @ArtW Many thanks for your comment! Now I will correct it. Can you please edit the last part that says why $\varphi(Q_0)\simeq \varphi(Q_{0}^{'}) ?$ Thanks! $\endgroup$ – Nikita Apr 16 '18 at 18:12
  • $\begingroup$ @DietrichBurde Many thanks for your comment! That one is also me! $\endgroup$ – Nikita Apr 16 '18 at 18:12
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(a) Good, except you need to use $a$ and $b$ when you square $(x_1i+x_2j+x_3k)^2$.

(b) You went from "if $q\in\mathbb{K}$" to "$q$ is pure or a scalar," which is both trivial and fails to show what you're supposed to show (you didn't even use $\varphi$).

Instead, you need to assume $q\in Q_0$, and deduce $\varphi(q)\in {Q_0}'$. Hint: apply $\varphi$ to $q^2$. (Note you must also use the fact that everything in $Z(Q')$ came from something in $Z(Q)$.)

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