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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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    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 $xy = k$.)2011-11-19

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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 $D$.