## Splitting fields of central simple algebras

Posted: October 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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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 $a \in A. \ \Box$