0
$\begingroup$

I want to prove, that the only multiplicative quadratic form $Q$ (so $Q(xy)=Q(x)Q(y) \forall x,y$) on a quaternion algebra $\Big(\dfrac{a,b}{F}\Big)$ is the norm $\mathrm{Nr}$, which is isometric to the form $\langle 1,-a,-b,ab\rangle$ in the orthogonal basis $1,i,j,ij$. One easily sees, that $Q(1)=Q(1^2)=Q(1)^2\Rightarrow Q(1)=1$ and similarly $Q(i)^2=a^2$, $Q(j)^2=b^2$ and therfore $Q(ij)=(ab)^2$. But how do we show $Q(i)=-a$ and $Q(j)=-b$? Or is it even necessary? It seems like I'm missing something obvious.

$\endgroup$
  • $\begingroup$ A couple of things to keep in mind: (1) your argument that $Q(1) = 1$ is only valid if you assume $Q(1) \ne 0$, and (2) in general, a quadratic form is not determined by its values on a basis. $\endgroup$ – Kimball Jan 24 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.