Notation and Terminology:
An algebra over a field is always associative and unitary, and subalgebras always possess the unity.
A subfield of an $F$-algebra $A$ is a commutative $F$-subalgebra $B$ of $A$ such that for every $b\in B$ there is $c\in B$ such that $bc=cb=1$.
An algebra $A$ over a field $F$ is said to be a central simple algebra of degree $n$ over $F$ if there is an extension field $K$ of finite dimension over $F$ such that the scalar extension $A_K:=K\otimes_FA$ is isomorphic as $K$-algebra to $M_n(K)$, the set of all square matrices of size $n$ over $K$, and in this case we say that $K$ is a splitting field of $A$ over $F$.
Question:
If $A$ is a central simple algebra of degree $n$ over $F$ and $Z$ is a subfield of $A$ of dimension $m$ over $F$, is the centralizer $C_A(Z)$ a central simple algebra of degree $\frac{n}{m}$ over $Z$?
My Attempt:
I know the following facts:
Fact 1: Let $A\cong M_n(F)$ be an $F$-algebra and $B\cong M_m(F)$ be an $F$-subalgebra of $A$. Then the centralizer $C_A(B)\cong M_{\frac{n}{m}}(F)$ and $A\cong B\otimes_FC_A(B)$ as $F$-algebras.
Fact 2: Let $A$ be a finite dimensional $F$-algebra and $B\cong M_m(F)$ be an $F$-subalgebra of $A$. Then $A\cong B\otimes_FC_A(B)$.
Fact 3: Let $A$ be a finite dimensional $F$-algebra and $B$ be a central simple $F$-subalgebra of $B$ with splitting field $K$, then $A\cong B\otimes_FC_A(B)$ and $C_{A_K}(B_K)=(C_A(B))_K$.
Fact 4: Let $A\cong M_n(F)$ be an $F$-algebra and $B$ be a central simple $F$-subalgebra with splitting field $K$, then $C_A(B)$ is a central simple $F$-subalgebra of $A$ with splitting field $K$.
Fact 5: If $A\cong M_n(F)$ is an $F$-algebra and $Z$ is a subfield of dimension $m$ over $F$, then the centralizer $C_A(Z)\cong M_\frac{n}{m}(Z)$ as an $Z$-algebra.
If $D$ is a central simple algebra of degree $n$ over $F$ and $Z$ is a subfield of $D$ of dimension $m$ over $F$, then, by the regular representation of $D$, $D$ is an $F$-subalgebra of an $F$-algebra $M\cong M_{n^2}(F)$ and, if $D_0=C_M(D)$, then $D_0$ is central simple over $F$ and $M\cong D\otimes_F D_0$, so $D\subseteq C_M(D_0)$, $C_M(D_0)$ is central simple over $F$ and $M\cong C_M(D_0)\otimes_F D_0$, so $C_M(D_0)=D$. Also, if $A=C_M(Z)$, then $A\cong M_{\frac{n^2}{m}}(Z)$ as an $Z$-algebra, $Z\subseteq A$ and $D_0\subseteq A$. Moreover, if $Q=A\cap D$, then $Q=C_D(Z)=C_A(D_0)$, so $A\cong Q\otimes_FD_0$ as $F$-algebra.
I do not know if $D_0$ is a central simple algebra over $Z$, but if is so, then I can conclude that $C_D(Z)=Q$ is central simple over $Z$.
Indeed, $D$ is a subalgebra of an algebra $M\cong M_{n^2}(F)$, so if $D_0=C_M(D)$, then $D_0$ is central simple of degree $n$ over $F$ and also $M=D\otimes_FD_0$, so $(D_0)_Z=Z\otimes_F D_0\subseteq D\otimes_F D_0=M$, so if $A=C_M(Z)$ then $(D_0)_Z=Z\otimes_FD_0\subseteq A$ because $Z\subseteq D$ and $Z$ is commutative, and $A\cong M_{\frac{n^2}{m}}(Z)$, hence, as $D_0$ is central simple of degree $n$ over $F$, then $(D_0)_Z$ is central simple of degree $n$ over $Z$, so if $E=C_M(D_0)$, as $D_0$ is central simple over $F$, then $M=E\otimes_FD_0$, so $\mathrm{dim}(E)=\mathrm{dim}(D)$, but $D\subseteq E$, so $D=E$. Let $Q=C_A(D_0)$, then $Q=A\cap D=C_D(Z)$ and $Q=C_A((D_0)_Z)$ is central simple of degree $\frac{n}{m}$ over $Z$ and $A=Q\otimes_Z(D_0)_Z$, so that $m\mid n$, so $C_D(Z)$ is central simple of degree $\frac{n}{m}$ over $Z$.