# Extending valuation to division algebra over a non-Archimedean local field?

Let $D$ be a division algebra over a non-Archimedean local field $K$. I would like to extend the discrete valuation on $K$ to $D$.

For any $x \in D$, the subfield $K(x)$ of $D$ has a unique extension of the valuation on $K$, so I can define $v: K \to \mathbb Z \cup \{\infty\}$ by setting $v(x) = v_D(x): = v_{K(x)}(x)$. But why is $v$ defined this way multiplicative? That is, why is $v_{K(xy)} (xy) = v_{K(x)} (x) + v_{K(y)}(y)$?

It's clear in the case that $x$ and $y$ commute: then all three fields $K(xy)$, $K(x)$, and $K(y)$ are all subfields of the field $K(x, y)$, and the valuations on the three subfields agree with the restriction on the valuation on $K(x, y)$. But this should be true for any $x, y \in D$.

(Here's a silly argument I can make. Construct $v$ instead using the reduced norm. Then it's clear that $v$ is multiplicative. Also show that $v$ extends the valuation on $K$ and on intermediate fields, and is non-Archimedean -- this last using the trick that, by multiplicativity, $v(x + y) = v(x) + v(1 + y/x)$, and then $y/x$ and $1$ live in the same subfield. So this $v$ a valuation, and because it agrees on subfields with the first construction, that first construction also gives a valuation, which in particular must be multiplicative. So I guess that works, but I don't like it!)

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Your silly argument is not at all silly. In fact that's the way Reiner does it. But IMHO it does need an extra step explaining why the set of "integers" you get from the reduced norm is closed under addition. Reiner does that step using and argument resembling Eisenstein's criterion. See also this related question. – Jyrki Lahtonen Nov 19 '11 at 9:21
I was hoping for an elementary argument, maybe something like this: Find an element $y'$ in the algebraic closure $\overline{K(x)}$ of $K(x)$ that satisfies the same minimal polynomial over $K$ as $y$; in particular, $v(y) = v(y')$. Since $x$ and $y'$ commute, we have $v(xy') = v(x) + v(y')$. So all we need to show is that $xy$ and $xy'$ have the same valuation... (But there's no need for $y$ and $y'$ to have even the same degree over $K$, let alone the same minimal polynomial: if $D$ is the real quaternions with $x = i$ and $y = j$ so that $y' = \pm i$, then $xy' = \pm 1$ whereas $xy = k$.) – sibilant Nov 19 '11 at 22:05

One construction is via the Haar measure point of view (explained e.g. here), namely: if $x \in D\setminus \{0\}$, you define $|x|$ to be the factor by which multiplication by $x$ scales the additive Haar measure on $D$.