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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 $k^{\times}$.

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My recollection may be off, but I think there is a result in Reiner's Maximal Orders to the effect that when $k$ is a $p$-adic field complete w.r.t to a valuation and valuation ring $\mathfrak{o}$, then we get an $\mathfrak{o}$-order of $D$ as the preimage $\nu^{-1}(\mathfrak{o})$. As any order is a f.g. free module over $\mathfrak{o}$, this is compact giving support to your guess. I admit that I know nothing about the more general setting, so cannot help you there. – Jyrki Lahtonen Jul 30 '11 at 18:23
You don't mention that $k$ has any topology until after you ask whether the map is proper. This is a bit confusing: are you actually interested only in locally compact $k$, or what? – Pete L. Clark Jul 30 '11 at 18:38
@Pete L. Clark: Well, this is a completely algebraic situation, so I thought there might be a completely algebraic argument. But yes, I'm actually interested in the case that $k$ is local and non-Archimedean. – Justin Campbell Jul 30 '11 at 19:30
@Jyrki Lahtonen: Why don't you make that an answer? That handles the question in the case I really care about. – Justin Campbell Jul 30 '11 at 19:30
@Justin: Sorry, the general case does not go as easily as the special case of $n=2$ that I was testing the argument with (I deleted that comment). It is best to wait until I go to my office and look it up from Reiner. Meanwhile somebody else can probably give an answer. – Jyrki Lahtonen Jul 30 '11 at 21:12
up vote 4 down vote accepted

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.

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I agree that this is equivalent, but why is it true? – Justin Campbell Jul 31 '11 at 2:30

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.

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