proof of Galois group of the compositum of two Galois extensions


Consider the diagram


(1): Let p(x)K[x] with a root αEF. Then since E (resp. F) is Galois over K, all the roots of p lie in E (resp. F) and thus in EF. The result follows.

(2): We first show that EF is Galois over K. Choose separable polynomialsMathworldPlanetmath p(x),q(x)K[x] so that E (resp. F) is a splitting fieldMathworldPlanetmath for p (resp. q). Then EF is a splitting field for the squarefree part of pq, which is separable since it is squarefreeMathworldPlanetmath and since p(x),q(x) are separable.

Now, define


This map is a group homomorphismMathworldPlanetmath; its kernel is precisely those elements that leave both E and F fixed. Any such element must thus leave EF fixed, so that θ is injective. The image obviously lies in


by construction: (σ|E)|EF=σ|EF=(σ|F)|EF. We will show that H is precisely the image of θ by showing that the order of H is the same as the index of the field extension [EF:K].

For each σGal(E/K), there are precisely |Gal(F/EF)| elements of Gal(F/K) whose restrictions to EF are σ|EF. Thus directly from the definition of H,


By the corollary to the theorem regarding the compositum of a Galois extensionMathworldPlanetmath and another extension (, we have


so that



  • 1 Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004.
Title proof of Galois groupMathworldPlanetmath of the compositum of two Galois extensions
Canonical name ProofOfGaloisGroupOfTheCompositumOfTwoGaloisExtensions
Date of creation 2013-03-22 18:42:01
Last modified on 2013-03-22 18:42:01
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 6
Author rm50 (10146)
Entry type Proof
Classification msc 11R32
Classification msc 12F99