Is the $F$-rank of the multiplicity group of central simple algebra $M_m(D)$ over $F$ equal to $m$? Suppose $F$ is a field, and $A=M_m(D)$ is a central simple algebra over $F$, where $m$ is a natural number and $D$ is a central division algebra over $F$ of rank $d$. Let $G=GL(m,D)$ the multiplicity group of $A$.
Let $S$ be subgroup of $G$ of diagonal matrices with entries in $F^\times\subset D^\times$. I want to prove that $S$ is the maximal split $F-$ subtorus of $G$.
I can make it in case that $F$ is a local field (by writing down $D$ explicitly), but I have no idea other case.
Any one can help me? (at least the global field case)
 A: Claim: The group
$$P=\left\{ \begin{pmatrix}
D^\times & * & \cdots & * \\
0 & D^\times & \ddots & \vdots \\
0 & 0 & \ddots &\ast \\
0 & 0 & 0 & D^\times 
\end{pmatrix}\right\}$$
is a parabolic subgroup of $\mathrm{GL}_n(D)$.
Proof(sketch) Note that there is a natural embedding $\mathrm{GL}_n(D)=\mathrm{GL}_D(D^n)\hookrightarrow \mathrm{GL}_F(D^n)$. Show then that $P=B\cap \mathrm{GL}_D(D^n)$, where $B$ is the $d\times d$ block diagonal matrices $\mathrm{GL}_F(D^n)$. This implies that $P$ is parabolic since then $\mathrm{GL}_n(D)/P$ admits a closed embedding into $\mathrm{GL}_F(D^n)/B$, which is a proper over $\mathrm{Spec}(F)$, and so therefore $\mathrm{GL}_n(D)/P$ must also be proper over $\mathrm{Spec}(F)$. $\blacksquare$
NB: I'm using the standard definition of parabolic subgroup over an arbitrary field as in [Milne, Definition 17.15].
Now, recall (e.g. see [Borel, Proposition 20.6]) that every maximally split torus of $\mathrm{GL}_n(D)$ must be contained in a minimal parabolic, and so, in particular, must be contained in every parabolic. So, $P$ must contain a maximally split torus of $\mathrm{GL}_n(D)$. But, notice that the Levi $L$ of $P$ is precisely $(D^\times)^n$. So, any maximally split torus contained in $P$ must embed into $(D^\times)^n$. But, it's clear then that the maximal such split torus is the diagonal embedding $\mathbb{G}_{m,F}^n\hookrightarrow (D^\times)^n$.
Incidentally, since the center of the identity component of $L$ is exactly our diagonally embedded $\mathbb{G}_{m,F}^n$ we can conclude from loc. cit. that $P$ is actually a minimal parabolic subgroup of $\mathrm{GL}_n(D)$.
EDIT: To see that $\mathbb{G}_m\subseteq D^\times$ is a maximal split torus we may proceed as follows. Suppose not, then there is some maximal split torus $S\subseteq D^\times$ larger than $\mathbb{G}_m$ which, in particular, is non-central (since $\mathbb{G}_m$ is the center of $D^\times$). But by loc. cit. $L=Z_{D^\times}(S)$ is a proper Levi subgroup of $D^\times$. Let $P$ be a parabolic with levi component $L$. Then, $R_u(P)$ is non-trivial, else
$$P=L\rtimes R_u(P)=L\rtimes \{e\}=L$$
but the only parabolic subgroup that is reductive is $G$, which contradicts that $L$ is proper. So, in particular, choosing any non-trivial element $u$ in $R_u(P)(F)$ produces a non-trivial unipotent element of $D^\times$. But, this is impossible. Indeed, note that one has a faithful representation of $F$-algebras
$$\rho\colon D\to \mathrm{End}_F(D)$$
given by sending $d$ to its left-multiplication operator on $D$ (which is $F$-linear). This map also induces a faithful representation
$$\rho\colon D^\times\to\mathrm{GL}_F(D).$$
In particular, $\rho(u)$ is unipotent. This means that $(\rho(u)-1)^n=0$ for some $n\geqslant 1$. Since $\rho$ is an algebra homomorphism this implies that $\rho((u-1)^n)=0$ and since $\rho$ is injective this implies that $(u-1)^n=0$ in $D$. Since $D$ is a division algebra this implies that $u=1$ which is a contradiction.
References:
[Borel] Borel, A., 2012. Linear algebraic groups (Vol. 126). Springer Science & Business Media.
[Milne] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.
