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[parent] Galois group of the compositum of two Galois extensions (Theorem)
Theorem 1   Let $E$ and $F$ be Galois extensions of a field $K$ . Then:
  1. The intersection $E\cap F$ is Galois over $K$ .
  2. The compositum $EF$ is Galois over $K$ . Moreover, the Galois group $\Gal(EF/K)$ is isomorphic to the subgroup $H$ of the direct product $G=\Gal(E/K)\times \Gal(F/K)$ given by: $$H=\{ (\sigma, \psi) : \sigma|_{E\cap F}=\psi|_{E\cap F} \}$$ i. e. $H$ consists of pairs of elements of $G$ whose restrictions to $E\cap F$ are equal.
Corollary 1   Let $E$ and $F$ be Galois extensions of a field $K$ such that $E\cap F=K$ . Then $EF$ is Galois over $K$ and the Galois group is isomorphic to the direct product: $$\Gal(EF/K)\cong \Gal(E/K) \times \Gal(F/K).$$




"Galois group of the compositum of two Galois extensions" is owned by alozano.
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See Also: the compositum of a Galois extension and another extension is Galois, Galois extension

Keywords:  compositum, Galois group

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proof of Galois group of the compositum of two Galois extensions (Proof) by rm50
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Cross-references: restrictions, direct product, subgroup, isomorphic, Galois group, compositum, intersection, field, Galois extensions
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This is version 2 of Galois group of the compositum of two Galois extensions, born on 2005-02-22, modified 2005-03-10.
Object id is 6794, canonical name is GaloisGroupOfTheCompositumOfTwoGaloisExtensions.
Accessed 2110 times total.

Classification:
AMS MSC12F99 (Field theory and polynomials :: Field extensions :: Miscellaneous)
 11R32 (Number theory :: Algebraic number theory: global fields :: Galois theory)

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