reduced norm is proper If $D$ is a central division algebra of dimension $n^2$ over a field $k$ we can consider the reduced norm $\nu : D \to k$, which satisfies $\nu^n = N_{D/k}$. In particular we get a group homomorphism $\nu : D^{\times} \to k^{\times}$. My question is: is this map proper? I was hoping that in case $k$ is a local field and we choose a multiplicative Haar measure $\mu$ on $D^{\times}$, the pushforward $\nu_*\mu$ is a multiplicative Haar measure on $k^{\times}$.
 A: If $k$ is a local field and $D$ is a division algebra over $k$, then the elements of reduced norm $1$ in $D$ are a compact subgroup (under multiplication).  From this it follows that the reduced norm $D^{\times} \to k^{\times}$ is proper.
(So the pushforward of Haar measure on $D^{\times}$ should give Haar measure on
$k^{\times}$.)
Added at OP's request: As explained in Serre's discussion of local class field theory in Cassels and Frolich, if $D$ is a division algebra over $k$ (a local field), then $D$ has a unique maximal order $\mathcal O_D$ which is compact,
and a uniformizer $\pi$, so that $D^{\times} = \mathcal O_D^{\times} \times
\pi^{\mathbb Z}.$  Thus the elements of norm $1$ are a closed subgroup
of $\mathcal O_D^{\times}$.  On the other hand, $\mathcal O_D^{\times}
= \mathcal O_D \setminus \pi \mathcal O_D$ is closed in the compact set 
$\mathcal O_D$, hence is compact.
A: A bit late, but since I had a chance to take a peek at Reiner's book (chapter 3, section 12), here's an outline of his argument.
Assume that $k$ is the field of fractions of a complete DVR $\mathfrak{o}$. Consider the map $\tilde\nu$ given by the reduced norm composed with the valuation of $k$. An argument involving a suitable variant of Hensel's lemma tells us that all the elements of $D$ with reduced norms in $\mathfrak{o}$ are actually integral over $\mathfrak{o}$. It then follows that the integral elements of $D$ form an $\mathfrak{o}$-order $R$. This follows from the multiplicativity of the reduced norm and the fact that if $b$ is integral, so is $1+b$ (because $1$ and $b$ commute, the argument of the commutative case covers this). This implies that $R$ is a ring. Obviously any element of $D$ can be multiplied with a high enough power of a prime element $\pi$ of $\mathfrak{o}$ such that the result has a reduced norm in $\mathfrak{o}$. Therefore $R\cdot k=D$. It is also easy to see $R$ is contained in a free $\mathfrak{o}$-module of rank $n^2$ is thus one itself. Therefore $R$ is, indeed, an order (in fact it is the unique maximal order of $D$). As a free $\mathfrak{o}$-module of a finite rank the ring $R$ is compact, and it is the inverse image of $\mathfrak{o}$ under the reduced norm map, so in this case your claim follows.
