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Suppose $A$ is a finite dimensional central simple algebra over a field $F$, $B$ is a simple subalgebra of $A$. $C$ is the centralizer of $B$ in $A$.

From Wedderburd-Artin theorem and the double centralizer theorem We know

  1. $A\cong M_n(\Delta_A)$,where $\Delta_A$ is a central division algebra over $F$
  2. $B\cong M_m(\Delta_B)$ ,where $\Delta_B$ is a division algebra over $F$
  3. $C\cong M_k(\Delta_C)$ ,where $\Delta_C$ is a division algebra over $F$

so what is the relationship between $\Delta_A$, $\Delta_B$ and $\Delta_C$, and $m,n,k$?

if we write their schur index, the dimension of a division algebra over its maximal subfield, as $Ind_A$, $Ind_B$, $Ind_C$, what is the relationship between these three numbers?

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The double centralizer theorem (cf. Knapp, Anthony W. (2007), Advanced algebra, Cornerstones p 115 Theorem 2.43) implies that $$\textrm{dim}_F B \cdot \textrm{dim}_F C = \textrm{dim}_F A.$$ Thus we should have $$m^2\textrm{dim}_F \Delta_B \cdot k^2\textrm{dim}_F \Delta_C = n^2\textrm{dim}_F \Delta_A.$$ Furthermore, $C$ is simple, and $B$ is the centralizer of $C$ in $A$.

As a corollary of the double centralizer theorem, If $\Delta$ is a central finite dimensional division algebra over a field $F$, and if $K$ is the maximal subfield of $\Delta$, then $$\textrm{dim}_F\Delta=(\textrm{dim}_F K)^2.$$ So, for $\Delta_A$, we have $\textrm{dim}_F \Delta_A = (Ind_A)^2$. For $\Delta_B$ and $\Delta_C$, note that $Z(B)=B\cap C = Z(C)$. Thus, $$\textrm{dim}_F\Delta_B=\textrm{dim}_{B\cap C}\Delta_B\cdot\textrm{dim}_F(B\cap C)=(Ind_B)^2\cdot\textrm{dim}_F(B\cap C)$$ $$\textrm{dim}_F\Delta_C=\textrm{dim}_{B\cap C}\Delta_C\cdot\textrm{dim}_F(B\cap C)=(Ind_C)^2\cdot\textrm{dim}_F(B\cap C).$$

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I don't know if we may expect a "simple"(pun unintended) answer. You can take $B={\mathbb H}$ the hamiltonian quaternions, embed it in $M_4({\mathbb R})$ and the latter can be embedded in $M_n({\mathbb R})$ any old how. (Then $C$ may be a bit more complicated).

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  • $\begingroup$ I don't understand you would further embed $M_4(\mathbb{R})$ in $M_n(\mathbb{R})$. Based on OP, the field $\mathbb{F}=\mathbb{R}$, and your example is $B=\mathbb{H}$, $A=M_4(\mathbb{R})$, $C$ is the centralizer of $\mathbb{H}$ in $M_4(\mathbb{R}$. By direct computation, $C$ is isomorphic to $\mathbb{H}$ with different matrices corresponding to $i$, $j$, and $k$. $\endgroup$ – Sungjin Kim Feb 26 '13 at 21:27

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