## Existence of Galois splitting fields

Posted: October 1, 2011 in Division Rings, Noncommutative Ring Theory Notes
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Introduction. Let $E/F$ be an algebraic extension. Recall that $G = \text{Gal}(E/F),$ the Galois group of $E/F,$ is the set of all automorphisms $\sigma: E \longrightarrow E$ such that $\sigma(a)=a$ for all $a \in F.$ The fixed field $E^G$ of $G$ is defined by $E^G= \{a \in E : \ \sigma(a)=a, \ \forall \ \sigma \in G \}.$ Clearly $E^G$ is a subfield of $E$ and $F \subseteq E^G.$ If $E^G = F,$ then $E/F$ is called Galois. If $[E:F] < \infty,$ then we have the following two facts from Galois theory.

Fact 1. $E/F$ is Galois $\Longleftrightarrow |G|=[E:F] \Longleftrightarrow E/F$ is separable and $E$ is the splitting field of some polynomial $f(x) \in F[x].$

Fact 2. If $E/F$ is separable, then there exists a field extension $L/E$ such that $L/F$ is Galois and $[L:F] < \infty.$

Notation. For the rest of this post, $D$ is a finite dimensional central division $k$-algebra.

Note. By the theorem in this post, $D$ has a maximal subfield which is separable over $k.$ Notice that such maximal subfields need not be Galois over $k.$ However, as we will see in the following theorem, there exists a finite Galois extension $L/k$ that splits $D.$

Theorem. There exists a finite Galois extension $L/k$ such that $L$ is a splitting field of $D.$

Proof. By the theorem in this post, there exists a maximal subfield $K$ of $D$ such that $K/k$ is separable. Thus, by Fact 2, there exists a field extension $L/K$ such that $L/k$ is Galois and $[L:k]<\infty.$ So the only thing left is to prove that $L$ is a splitting field of $D,$ i.e. $D \otimes_k L \cong M_n(L)$ for some integer $n \geq 1.$ By Corollary 1, $K$ is a splitting subfield of $D$ and so $D \otimes_k K \cong M_n(K)$ for some integer $n \geq 1.$ Thus

$D \otimes_k L \cong D \otimes_k (K \otimes_K L) \cong (D \otimes_k K) \otimes_K L \cong M_n(K) \otimes_K L \cong$ $M_n(K \otimes_K L) \cong M_n(L). \ \Box$

I finish this post with a definition. We will look into it in the future.

Definition. If $D$ has a maximal subfield which is Galois over $k,$ then we say that $D$ is a crossed product.