proving that a quaternion algebra with anisotropic norm is a division algebra Let $H=\left(\frac{a,b}{K}\right)$, the quaternion algebra over a field $K$. I want to prove that $H$ is a division algebra if and only if the norm form $N(q)=\bar{q}q$ is anisotropic, i.e. $N(q)=0\implies q=0$. The forward direction is literally trivial; naturally, if you have zero divisors, you don't have a division algebra. However, I'm a little stuck on the backward direction. I tried showing the reverse direction by contradiction, but that didn't seem to work out too well. Can I have some direction as to how to prove this?

You can in fact explicitly construct the inverses of elements if $N$ is anisotropic. Let $q \
eq 0$ be an element of $H$. Then $N(q) = q\bar{q}$ is a non-zero element of $K$. Note that this implies that $$\frac{q\bar{q}}{N(q)} = 1,$$ in other words, the inverse of $q$ is $\bar{q}/N(q)$. Thus, $H$ is a division algebra.