proof that the compositum of a Galois extension and another extension is Galois


The diagram of the situation of the theorem is:


To see that EF/F is Galois, note that since E/K is Galois, E is a splitting fieldMathworldPlanetmath of a set of polynomialsMathworldPlanetmathPlanetmathPlanetmath over K; clearly EF is a splitting field of the same set of polynomials over F. Also, if fK[x] is separablePlanetmathPlanetmath over K, then also f is separable over F. Thus EF is normal and separable over F, so is Galois. E is obviously Galois over EF since EFK.

Let r be the restrictionPlanetmathPlanetmathPlanetmath map


r is clearly a group homomorphismMathworldPlanetmath, and since E is normal over K, r is well-defined.

Claim r is injectivePlanetmathPlanetmath. For suppose σGal(EF/F) and σ|E is the identityPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. Then σ is fixed on F (since it is in Gal(EF/F) and on E (since its restriction to E is the identity), so is fixed on EF and thus is itself the identity.

Now, the image of r is a subgroupMathworldPlanetmathPlanetmath of Gal(E/K) with fixed field L, and thus the image of r is Gal(E/L). Claim EF=L. is obvious: any element xEF is fixed by σ|E for each σH since σ fixes F. Thus EFL. To see the reverse inclusion, choose xL; then x is fixed by each r(σ) for σH. But xLE, so that (as an element of E), x is fixed by each σH. Thus xF so that xEF.

Thus L=EF, and r is then an isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath Gal(EF/F)Gal(E/EF). ∎


Title proof that the compositum of a Galois extensionMathworldPlanetmath and another extensionPlanetmathPlanetmathPlanetmathPlanetmath is Galois
Canonical name ProofThatTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois
Date of creation 2013-03-22 18:41:58
Last modified on 2013-03-22 18:41:58
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 6
Author rm50 (10146)
Entry type Proof
Classification msc 11R32
Classification msc 12F99