Reference request: multiplicative group of a central simple algebra, their reductivity and parabolic subgroups During my study of the theory of automorphic forms and $L$-functions, I never found any literature dealing with the following:
Suppose $D$ is a central division algebra over a local or global field $F$, and then $G=GL(m,D)$ (which is the multiplicative group of the central simple algebra $M(m,D)$) is an algebraic group defined over $F$ (these groups are the "inner forms" of the group $GL(n)$).
My question is: I didn't find any literature proving that this algebraic group is reductive, nor could I give a direct proof by myself.
And moreover, are the parabolic subgroups of $G=GL(m,D)$ (which are defined over $F$) all conjugate (over $F$) to the "standard" blockwise upper triangular ones (with diagonal elements in $GL(m_i,D)$ and $m_i$'s is a partition of $m$), just as the case $GL(n)$?
I think this must be true, but I didn't find any text or paper containing a clear proof of this seemingly reasonable fact, neither could I obtain a proof by myself. An embedding of $G$ into $GL(mr)$ (where $r$ is the rank of $D$ over $F$) doesn't seem to do any help to this question. So can anybody give me some help or hint? Thanks a lot in advance!
 A: By proposition 19.13 in Milne "Algebraic Groups", it suffices to show that the base change to any field extension of $F$ is reductive. Let $K/F$ be a splitting field for $D$.
Suppose for a moment that $G$ is any affine $F$-group scheme and $K/F$ is any field extension. $G=\mathrm{Spec}(B)$. Then the functor that $G$ represents is given by $A\mapsto\mathrm{Hom}_{F\textrm{-Alg}}(B,A)$. The functor that $G_K$ represents is given by $A\mapsto\mathrm{Hom}_{K\textrm{-Alg}}(B_K,A)$, but by the adjunction of scalar extension and restriction for algebras, this is just $A\mapsto \mathrm{Hom}_{F\textrm{-Alg}}(B,\mathrm{Res}_{K/F}A)$. The takeaway is that the functor the scalar extension $G_K$ represents is just the composition of the functor $G$ represents with the scalar restriction functor $\mathrm{Res}_{K/F}:K\textrm{-Alg}\to F\textrm{-Alg}$.
The functor that $\mathrm{GL}(m,D)$ represents is $A \mapsto (A\otimes_F \mathrm{M}_{m\times m}(D))^\times$. If $A$ is actually a $K$-algebra, then we have $A\cong A\otimes_K K$, so $$A\otimes_F \mathrm{M}_{m\times m}(D)\cong A\otimes_K K \otimes_F \mathrm{M}_{m\times m}(D)\cong A\otimes_K (K\otimes_F \mathrm{M}_{m\times m}(D))$$ Because $K$ is a splitting field for $D$, we have $K\otimes_F D\cong \mathrm{M}_{n\times n}(K)$, where $n=\sqrt{\mathrm{dim}_F(D)}$, hence $$K\otimes_F \mathrm{M}_{m\times m}(D) \cong K\otimes_F D \otimes_F \mathrm{M}_{m\times m}(F)\cong\mathrm{M}_{n\times n}(K) \otimes_F \mathrm{M}_{m\times m}(F) \cong \mathrm{M}_{mn\times mn}(K)$$
Thus $$A\otimes_F \mathrm{M}_{m\times m}(D)\cong \mathrm{M}_{mn\times mn}(A)$$
So we get that $\mathrm{GL}(m,D)_K$ represents the functor $A \mapsto \mathrm{M}_{mn\times mn}(A)^\times=\mathrm{GL}(mn,A)$. Thus we have shown that $\mathrm{GL}(m,D)_K \cong \mathrm{GL}(mn,K)$, so $\mathrm{GL}(m,D)_K$ is reductive and so is $\mathrm{GL}(m,D)$.
