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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.

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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.

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

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