Simple Proof that a Reductive Group is Unimodular? Let $G$ be a connected, reductive group over a local field $k$ of characteristic zero.  I thought of a simple proof that $G(k)$ is unimodular, but I realize it is almost certainly wrong: $G(k)$ is generated by its derived group and its center, and the modular character is trivial on both of these, Q.E.D.
I am relying on some results from algebraic groups over algebraically closed fields which I am not certain carry over to arbitrary fields.  So what I wanted to know which of these results are false for arbitrary $k$ (all being true in the case $k = \overline{k}$)
-$[G,G](k) = [G(k),G(k)]$
-$Z(G)(k) = Z(G(k))$
-$G(k)$ is generated by $[G,G](k)$ and $Z(G)(k)$
Moreover, can the idea of this proof be modified to correctly show that $G(k)$ is modular?
 A: Since you assume that $k$ has characteristic zero, one can make use of the Lie algebra. If $G$ is an arbitrary Lie group over a locally compact field $k$ of characteristic zero, $G$ is unimodular if and only of the adjoint action of $G$ on its Lie algebra is by elements of determinant of modulus 1. In the case of $G=\mathbb{G}_k$, with $\mathbb{G}$ reductive connected, the Lie algebra has the form $\mathfrak{s}\oplus\mathfrak{z}$ with $\mathfrak{s}$ semisimple and $\mathfrak{z}$ central, and the adjoint $G$-action preserves this decomposition and is trivial on $\mathfrak{z}$. Moreover the automorphism group of $\mathbb{s}$ has a finite index subgroup acting with determinant 1 (because self-derivations of $\mathfrak{s}$ are inner and have trace zero because $\mathfrak{s}$ is perfect). This entails the unimodularity statement. 
A: The modular character $\Delta: G(k)\rightarrow {\mathbb R}_{>0}$ on $G(k)$ is trivial on the commutator subgroup and is trivial on a compact open subgroup and also on the centre. These three groups generate a subgroup of finite index since the abeliansation of $G(k)$ modulo centre is compact. Hence the modular character is trivial on an open subgroup of finite index, and is hence trivial. This works in arbitrary characteristic. 
I am sure that I have seen this argument somewhere in the literature, but old age and failing memory prevent me from recalling the reference.  
