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Suppose I have a field $k$ and two extensions $K/k$ and $L/k$ which are both abelian Galois extensions of $k$. Then (assuming $K$ and $L$ are both contained in some bigger field) is the compositum $KL$ an abelian Galois extension of $k$?

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up vote 10 down vote accepted

Let $\psi\colon \text{Gal}(KL/k) \rightarrow \text{Gal}(K/k)\times \text{Gal}(L/k)$ be a map such that $\psi(\sigma) = (\sigma|K, \sigma|L)$. $\psi$ is clearly a group homomorphism. It's easy to see that $\psi$ is injective. Hence $\text{Gal}(KL/k)$ is abelian.

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+1 as you give a complete answer and bit me for several seconda. The OP can also take a look at theorem 1.14 in chapter VI in Lang's "Algebra" –  DonAntonio Sep 6 '12 at 13:05
5  
@Donantonio: he bit you for several seconds??? I think I'll come back to this site only after everyone has had lunch:-) I have upvoted Makoto nevertheless... –  Georges Elencwajg Sep 6 '12 at 13:32

Yes, find the Fixed fields in $\mathrm{Gal}(KL/k)$. You can proceed

  1. If $K$ is splitting field of separable polynomials $\{f_i\}$ in $F$, is $KL$ also a splitting field of those polynomials over $L$?

  2. What is the image of the homomorphism $\mathrm{Gal}(KL/L)$ to $\mathrm{Gal}(K/F)$ by fixing automorphisms of $K$.

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