Unimodular groups Let $F$ be a non-archimedean local fields of characteristic $p$ (for $p$ any given prime number): the field of formal Laurent series $F_q((T))$ over a finite field $F_q$ (where $q$ is a power of $p$).
Is $GL(n,F)$ unimodular?
More generally, are all reductive algebraic groups over local ﬁelds unimordular?
 A: Yes. Let $\mathbf{G}$ be a connected reductive group over a local field $k$; we assume the base field $k$ is local that $G=\mathbf{G}(k)$ is a locally compact group and therefore admits a Haar measure. One can show that the modulus character of a Haar measure on $G$ (left or right, depending on your favorite convention) is the absolute value of the homomorphism $G\to k^\times$ induced by an algebraic character $\delta_\mathbf{G}:\mathbf{G}\to\mathbf{G}_m$. We can actually show that $\delta_\mathbf{G}$ is trivial. Since $\mathbf{G}$ is smooth (really geometrically reduced is what's important, but this is equivalent to smoothness in the setting of $k$-groups locally of finite type) and $\mathbf{G}_m$ is $k$-separated, it is enough to prove that the induced homomorphism $\delta_{\mathbf{G},\overline{k}}:\mathbf{G}(\overline{k})\to\overline{k}^\times$ is trivial, where $\overline{k}$ is a choice of algebraic closure of $k$. But because $\mathbf{G}$ is connected reductive, its radical is equal to its center $Z_\mathbf{G}$, a torus, and the $k$-homomorphism $\mathscr{D}\mathbf{G}\to\mathbf{G}/Z_\mathbf{G}$ is an isogeny. In particular, the induced map on $\overline{k}$-points is surjective. Concretely, since groups of $\overline{k}$-points of the scheme-theoretic derived subgroup $\mathscr{D}\mathbf{G}$ and the scheme-theoretic center $Z_\mathbf{G}$ are equal to the derived group $\mathscr{D}(\mathbf{G}(\overline{k}))$ and the center $Z_{\mathbf{G}(\overline{k})}$ of $\mathbf{G}(\overline{k})$, respectively, this says that $\mathbf{G}(\overline{k})$ is generated by its center together with its derived subgroup (I'm using here that $(\mathbf{G}/Z_{\mathbf{G}})(\overline{k})=\mathbf{G}(\overline{k})/Z_{\mathbf{G}}(\overline{k})$, for which it is important that we take points in an algebraically closed field, although I guess maybe one can get around this using an argument with Zariski density of $k$-points, but I haven't thought that through). Any homomorphism from $\mathbf{G}(\overline{k})$ to a commutative group kills its derived subgroup. On the other hand, the description of the algebraic homomorphism $\delta_\mathbf{G}$ in terms of pullback of a choice of left-invariant differential form (see $\S$ 4.2 of the book Néron Models, specifically Corollary 3 and Proposition 4) makes it clear that this homomorphism kills the scheme-theoretic center, in particular that $\delta_{\mathbf{G},\overline{k}}$ kills $Z_{\mathbf{G}}(\overline{k})=Z_{\mathbf{G}(\overline{k})}$, so $\delta_{\mathbf{G},\overline{k}}$ is trivial, which is what we wanted.
I'm not sure if there is a more elementary argument. I don't know one, and it seems to me that one has to use facts about reductive groups since the question is specific to $k$-points of reductive groups.
