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! $\Box$