Existence of Galois splitting fields

Posted: October 1, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , ,

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.

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