Quaternion algebra and norm

Question

Let $a \in \mathbb{Q}$ be a nonzero rational number and set $(5,a)$ and (for the associated division algebras over $\mathbb{Q}$). Let us suppose that $b$ is the norm of some element of $\mathbb{Q}[\sqrt{5}]$. How can I write down an explicit isomorphism between $(5,a)$ and $(5,ba)$?

Answer 1

Let $b=N(x+y\sqrt5)=x^2-5y^2$, $\ x,y\in\Bbb Q$.

Let $\mathfrak A:=(5,a)=\Bbb Q[i,j,k]/_{\displaystyle{(i^2=5,\,\ j^2=a,\,\ ij=-ji=k)}}$
and let $\mathfrak B:=(5,ab)$, this has $j^2=ab$.

Now consider the linear mapping $\psi:\mathfrak B\to\mathfrak A$ which sends $$\matrix{1\mapsto 1 && j\mapsto j(x+yi) \\ i\mapsto i && k\mapsto k(x+yi)}\,,$$ verify that it indeed will give an algebra homomorphism.

For the inverse $\varphi:\mathfrak A\to\mathfrak B$, do it similarly using $\displaystyle\frac{1}{x+yi}=\frac{x-yi}b$: $$\varphi:=\quad\matrix{1\mapsto 1 && j\mapsto j(x-yi)/b \\ i\mapsto i && k\mapsto k(x-yi)/b}\,.$$