Let FFF be a field and AAA a simple subalgebra of a finite dimensional FFF-algebra BBB. Prove that dimF(A)\dim_F\p{A}dimF(A) divides dimF(B)\dim_F\p{B}dimF(B).
This problem is false as written. For example, let F=CF = \CF=C, A=M2(C)A = M_2\p{\C}A=M2(C), and B=M2(C)×M3(C)B = M_2\p{\C} \times M_3\p{\C}B=M2(C)×M3(C). Then AAA is a simple FFF-subalgebra of BBB, but 444 does not divide 131313.