Throughout this post, k is a field and A is a finite dimensional central simple k-algebra.

Definition. A field L is called a splitting field of A if k \subseteq L and A \otimes_k L \cong M_n(L), as L-algebras, for some integer n \geq 1.

Remark. By this lemma, the algebraic closure of k is a splitting field of any finite dimensional central simple k-algebra. Also, if f L is a splitting field of A, then \dim_k A = \dim_L A \otimes_k L = \dim_L M_n(L)=n^2 and so n = \deg A.

Theorem 1. Let A = M_n(D), where D is some finite dimensional central division k-algebra and let L be a field containing k. Then L is a splitting field of A if and only if L is a splitting field of D.

Proof. Let \dim_k D = m^2. Then \dim_k A = (mn)^2 and so \deg A = mn. If L is any field containg k, then

A \otimes_k L \cong M_n(D) \otimes_k L \cong M_n(D \otimes_k L). \ \ \ \ \ \ \ \ \ (*)

Now, suppose that L is a splitting field of A. Then A \otimes_k L \cong M_{mn}(L). Also, since D \otimes_k L is a finite dimensional central simple L-algebra, D \otimes_k L \cong M_r(D') for some division ring D' and some integer r. Therefore, by (*), we have M_{mn}(L) \cong A \otimes_k L \cong M_{rn}(D') and thus m=r and D' \cong L. Hence D \otimes_k L \cong M_m(L). Conversely, if L is a splitting field of D, then D \otimes_k L \cong M_m(L) and (*) gives us A \otimes_k L \cong M_{mn}(L). \ \Box

Corollary. There exists a finite Galois extension L/k such that L is a splitting field of A.

Proof. Trivial by Theorem 1 and the theorem in this post. \Box

Notation. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Note that, since F is a field, \phi is injective. Now, given an integer n \geq 1, we define the map \phi_m: M_n(F) \longrightarrow M_n(E) by \phi_m([u_{ij}]) = [\phi(u_{ij})] \in M_n(E). Clearly \phi_m is a k-algebra injective homomorphisms.

Theorem 2. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Suppose also that F is a spiltting field of A with an F-algebra isomorphism f : A \otimes_k F \longrightarrow M_n(F). Then E is a splitting field of A and there exists an E-algebra isomorphism g : A \otimes_k E \longrightarrow M_n(E) such that g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1)) for all a \in A.

Proof. The map \phi_m: M_n(F) \longrightarrow M_n(\phi(F)) is an isomorphism and so

A \otimes_k \phi(F) \cong A \otimes_k F \cong M_n(F) \cong M_n(\phi(F)).

Therefore

A \otimes_k E \cong A \otimes_k (\phi(F) \otimes_{\phi(F)} E) \cong (A \otimes_k \phi(F)) \otimes_{\phi(F)} E \cong M_n(\phi(F)) \otimes_{\phi(F)} E \cong

M_n(\phi(F) \otimes_{\phi(F)} E) \cong M_n(E).

So we have an isomorphism g: A \otimes_k E \longrightarrow M_n(E). It is easy to see that g(a \otimes_k 1)=\phi_m(f(a \otimes_k 1)) for all a \in A. \ \Box

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