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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 $(5,ba)$?

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  • $\begingroup$ Sorry, what does exactly $(5,a)$ denote, and what are these 'associated division algebras'? And how is it related to quaternions? $\endgroup$ – Berci Apr 16 '13 at 11:57
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    $\begingroup$ Berci: (5,a) is the same as the notation they ahve for (a,b)_F en.wikipedia.org/wiki/Quaternion_algebra here. It is a standard notation. $\endgroup$ – Metes Apr 16 '13 at 12:02
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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 $\displaystyle\frac{1}{x+yi}=\frac{x-yi}b$: $$\varphi:=\quad\matrix{1\mapsto 1 && j\mapsto j(x-yi)/b \\ i\mapsto i && k\mapsto k(x-yi)/b}\,.$$

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  • $\begingroup$ Berci: Is this really an algebra homomorphism? $\endgroup$ – Metes Apr 16 '13 at 13:03
  • $\begingroup$ I think so. Calculate $\psi(j)^2$, $\psi(k)^2$, $\psi(j)\psi(k)$ and $\psi(k)\psi(j)$... $\endgroup$ – Berci Apr 16 '13 at 13:17
  • $\begingroup$ When I calculate $\varphi(j)^2$ I do not seem to get that it is an algebra isomorphism. $\varphi(j^2) = a $ while $\varphi(j)^2 = ab(x-yi)^2/b^2 = a(x-yi)^2/b $. If you are interpreting i as sqrt(5), I still do not see how the supposed equality should hold. $\endgroup$ – Metes Apr 16 '13 at 13:24
  • $\begingroup$ Hey hey!! It's not commutative. $\varphi(j)^2=1/b^2\,j(x-yi)\,j(x-yi)=1/b^2\,j(x-yi)\,(x+yi)j$ as $jx=xj$ and $-ji=+ij$. $\endgroup$ – Berci Apr 16 '13 at 13:29
  • $\begingroup$ OF COURSE! THANK YOU! $\endgroup$ – Metes Apr 16 '13 at 13:29

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