For the definition of splitting fields of central simple algebras see here. Throughout, D is a division algebra with the center k and \dim_k D < \infty. For a subalgebra B of D, we will denote the centralizer of B in D by C_D(B).

Theorem. Let L be a subfield of D and suppose that k \subseteq L. Then D \otimes_k L \cong M_n(C_D(L)), where n = \dim_{C_D(L)} D = \dim_k L.

Proof. Let R = D \otimes_k L. As we saw in here, D has a structure of a (right) R-module and

End_R(D) \cong C_D(L). \ \ \ \ \ \ \ (1)

Note that R is a finite dimensional central simple L-algebra. In particular, it is primitive. We are now going to show that D is a faithful simple R-module. First, D is faithful because the annihilator of D in R is an ideal of R and, since R is a simple algebra, the annihilator is zero. Now, let I be a non-zero R-submodule of D. Let x \in D and y \in I. Then, since I is an R-module, we have yx = y(x \otimes_k 1) \in I. Thus I is a right ideal of D and so I=D. Hence D is a simple R-module. So, by (1) and the structure theorem for primitive rings, we have R \cong M_n(C_D(L)), where n = \dim_{C_D(L)} D. To complete the proof of the theorem, we only need to show that \dim_k L=n. Well, we have

\dim_k D = \dim_L R = \dim_L M_n(C_D(L))=n^2 \dim_L C_D(L)

and thus

(\dim_k L)^2(\dim_k D)=n^2 (\dim_k C_D(L))(\dim_k L)=n^2 \dim_k D. \ \ \ \ \ \ \ (2)

The last equality in (2) is true by the lemma in this post. Therefore \dim_k L = n. \ \Box

Corollary 1. If L is a maximal subfield of D, then L is a splitting subfield of D. In fact, D \otimes_k L \cong M_n(L), where n = \deg D = \dim_L D=\dim_k L.

Proof. If L is a maximal subfield, then k \subset L and C_D(L)=L. \ \Box

Corollary 2. If L is a splitting subfield of D, then L is a maximal subfield of D.

Proof. We only need to show that C_D(L)=L. By Theorem 2, D \otimes_k L \cong M_n(C_D(L)), where n = \dim_k L. Since L is a splitting subfield of D, we also have D \otimes_k L \cong M_m(L), where m = \deg D. Thus M_n(C_D(L)) \cong M_m(L). So m = n and C_D(L) \cong L. Thus \dim_k C_D(L)=\dim_k L and hence C_D(L)=L because L \subseteq C_D(L). \ \Box

Corollary 3. Let L be a subfield of D. Then the following statements are equivalent.

1) L is a maximal subfield of D.

2) L is a splitting subfield of D.

3) k \subset L and \dim_k L = \deg D.

4) C_D(L)=L.

Proof. Straightforward! \Box

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