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So let $\Big(\dfrac{\mathit{a,b}}{\mathit{F}}\Big)$ be a quaternion Algebra over a field $F$ with char $\neq 2$, and let $i,j$ be the standard generators for the quat. Algebra, meaning $i^2=a$ and $j^2= b$. Apparently $\Big(\dfrac{\mathit{a,b}}{\mathit{F}}\Big) \simeq \Big(\dfrac{\mathit{b,a}}{\mathit{F}}\Big)$ by interchanging $i$ and $j$, $(i,j)\mapsto(j,i)$.

But a homomorphism $\phi$ of $F$-Algebras restricts to the identity on $F$, meaning $\phi(a)=a$, but if $\phi(i)=j \Rightarrow a = \phi(a) = \phi (i^2) = \phi (i)^2 = j^2 = b$, which in not given.

Am I missing something? Where am I thinking wrong?

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    $\begingroup$ $j^2=a$ in the second quaternion algebra. $\endgroup$ – jgon Dec 11 '18 at 13:16
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    $\begingroup$ yes, to go along with @jgon's comment, what's confusing you is that you're using i, j to mean different things in your 2 quaternion algebras. perhaps if you labelled them $i_1, j_1$ and $i_2, j_2$, this would help $\endgroup$ – Kimball Dec 12 '18 at 17:13

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