# non commuative division ring with a discrete valuation

I do not manage to exhib a non commutative ring of division with a discrete valuation. Can anyone show me one? Examples with quaternions would be a plus !!

Thanks in advance.

## 1 Answer

I’m sure this quaternion algebra is well known, and I apologize for presenting it in a probably nonstandard way.

Consider $$\Bbb Q_p$$, the $$p$$-adic integers, and its (unique) unramified quadratic extension $$U$$, definable as the field of $$(p^2-1)$$-th roots of unity over $$\Bbb Q_p$$. Consider the two-dimensional $$U$$-vector space $$E$$ with basis $$\{1,\pi\}$$, where $$\pi^2=p$$. Now consider the non-commuting relation $$\pi u=u^\sigma\pi$$ for any $$u\in U$$, where $$\sigma$$ is the Frobenius automorphism of $$U$$, sending the $$(p^2-1)$$-th root of unity $$\zeta$$ to $$\zeta^p$$. Now you show that $$E$$ is a four-dimensional vector space over $$\Bbb Q_p$$, it’s the central division algebra of rank four and of “invariant” $$\frac12$$ over $$\Bbb Q_2$$. I do leave it to you to show that it’s a division algebra, and that the expected valuation $$\upsilon_p$$ is discrete.

The “expected valuation” is, for $$z\in U$$, $$\upsilon_E(z)=\frac12\upsilon_p\mathbf N(z)$$, where $$\mathbf N$$ is the reduced norm from $$E$$ to $$\Bbb Q_p$$. If you don’t know about the reduced norm, use the “unreduced norm” of $$z$$, $$\mathbf N'(z)$$, which is the determinant of $$z$$ in the regular representation of $$E$$ as a four-dimensional $$\Bbb Q_p$$-space, and then we have $$\upsilon_E(z)=\frac14\upsilon_p\mathbf N'(z)$$.

By the way, I’m pretty sure you can extend the $$2$$-adic valuation to the familiar rational quaternion algebra $$\Bbb Q(i,j,k)$$ in just this way as well.

• Thank you. I have to think to your comment. I will try to prove all your assertions. – joaopa Dec 25 '18 at 9:47