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I was reading a book on quaternion algebras and found the following exercise and was unable to do it and it has been really frustrating me, Let $(a,b)_F$ be a quaternion algebra with $i^2=a$ , $j^2=b$ and $ij=-ji$

Prove that $(a,b)_F$ , $(b,a)_F$ and $(ac^2,b)_F$ are isomorphic and prove that $(a^2,b)_F$ is not a division algebra.

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    What is $c$? any nonzero real number? – 2017-01-09
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    For the first two, think about reordering the generators. For the third, what happens if you rescale one of the generators, e.g., what relations does $ri$ satisfy for some $r \in F$? – 2017-01-09
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    @rschwieb I imagine $c \in F$, where $F$ is the base field. – 2017-01-09
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    @SpamIAm Yeah, I should have guessed any base field, but it probably can't be zero, can it? – 2017-01-09
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    @rschwieb Ah yes, $c \neq 0$ I'm sure. Good to point it out. – 2017-01-09

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Let's show that $(a,b)_F$ and $(b,a)_F$ are isomorphic. Let's say $(a,b)_F$ is generated by $i,j$ with

$$ i^2=a, \quad j^2=b, \quad ij=-ji $$

and $(b,a)_F$ is generated by $k$ and $\ell$ satisfying

$$ k^2=b, \quad \ell^2=a, \quad k\ell=-\ell k. $$

To define an isomorphism $\phi:(b,a)_F\to(a,b)_F$ it suffices to specify where $\phi$ sends the two generators $k$ and $\ell$. We must send $k$ and $\ell$ to two elements of $(a,b)_F$ where $\phi(k)$ squares to $b$, $\phi(\ell)$ squares to $a$, and $\phi(k)$ and $\phi(\ell)$ anticommute. Can you see two such elements of $(a,b)_F$ to send $k$ and $\ell$ to with $\phi$?

For showing $(a^2,b)_F$ is not a division algebra, try and find a nonzero element that is not invertible; one way this may occur is with zero divisors. Look at the new relation $i^2=a^2$. What can you do to this equation to exhibit zero divisors?

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    So $\phi(k) \mapsto b$ and $\phi(l) \mapsto a$. Couldn't you do this will all quaternion algebras though? I can't see how two quaternion algebras would not be isomorphic using this method but obviously not all quaternion algebras are isomorphic to each other. I can use the fact that quaternion algebras are isomorphic if they are isometric as quadratic spaces, would that make the problem simpler? – 2017-01-13
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    Since $k$ squares to $b$, you need to choose $\phi(k)$ to be an element of $(a,b)_F$ that squares to $b$. (This is because applying $\phi$ to the equation $k^2=b$ yields $\phi(k^2)=\phi(b)$ which is equivalent to $\phi(k)^2=b$. Note that $\phi(b)=b$ because $\phi$ is assumed to be an isomorphism of $F$-algebras, so fixes scalars.) You said $\phi(k)=b$ but $b$ is *not* an element that squares to $b$. There is a *reason* I wrote out the defining relations for $i$ and $j$ in the algebra $(a,b)_F$, it's so it would be obvious what $\phi(k)$ and $\phi(\ell)$ should be. – 2017-01-13
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    Oh wow, just re-read what I put, that was really stupid haha, what I meant to put was $\phi(k) \mapsto j$ and $\phi(l) \mapsto i$, can you answer the questions in my original comment if possible, your help is really appreciated – 2017-01-13
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    What makes you think you could do this with all quaternion algebras? I can't explain why your reasoning is wrong if you don't tell me your reasoning. In any case, I already mentioned how to show $(a^2,b)_F$ will not be a division algebra, so it cannot be isomorphic to quaternion algebras which are division algebras. – 2017-01-14
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    For example, using the notation in your original post, but replacing $(b,a)_F$ with $(a^2,b)_F$ you could map $\phi(i) \mapsto k$ and $\phi(j) \mapsto l$ so $\phi(i)^2=a^2$ and $\phi(j)^2=b$ Obviously I know this is wrong I just can't see a fault. – 2017-01-14
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    Ah actually writing it out has helped, is it the face that $\phi(i)=a$ but also $\phi(i)=k$ which can't be true since $a \in \mathbb{R}$ – 2017-01-14
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    Okay, say we talk about $(b,a)_F$ generated by anticommuting $i,j$ with $i^2=b$ and $j^2=a$, and $(a^2,b)_F$ generated by $k,\ell$ with $k^2=a^2$ and $\ell^2=b$. Then $\phi(i)$ must be something that squares to $b$, so you can choose $\phi(i)=\ell$. No problems there. And $\phi(j)$ must be something that squares to $a$. Can you find any combination of $1,k,\ell,k\ell$ within $(a^2,b)_F$ that squares to $a$? (Note that $k$ squares to $a^2$, not $a$, so setting $\phi(j)=k$ will not work.) – 2017-01-14
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    (In even greater generality, suppose we had $(a,b)_F$ generated by anticommuting $i,j$ with $i^2=a,j^2=b$ and $(c,d)_F$ generated by anticommuting $k,\ell$ with $k^2=c,\ell^2=d$. Without knowing anything about $a,b,c,d$ what do you think the isomorphism could be?) – 2017-01-14