# How are these Quaternion Algebras isomorphic? $\Big(\dfrac{\mathit{a,b}}{\mathit{F}}\Big) \simeq \Big(\dfrac{\mathit{b,a}}{\mathit{F}}\Big)$

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?

• $j^2=a$ in the second quaternion algebra. – jgon Dec 11 '18 at 13:16
• 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 – Kimball Dec 12 '18 at 17:13