# Galois extension

1. Feb 22, 2010

### math_grl

This stuff is killing me...

Let $$K \leq M \leq L$$ be fields such that L is galois over M and M is galois over K. We can extend $$\phi \in G(M/K)$$ to an automorphism of L to show L is galois over K.

I need help filling in the details in why exactly L is galois over K.

2. Feb 23, 2010

### Tinyboss

Something like this, I think: let $$[M:K]=a$$ and $$[L:M]=b$$. Since $$M/K$$ is Galois, there are $$a$$ automorphisms of $$M$$ that fix $$K$$. For the same reason, there are $$b$$ automorphisms of $$L$$ that fix $$M$$. So there are $$ad$$ automorphisms of $$L$$ that fix $$K$$. Since $$ad$$ is also the degree of $$L/K$$, it's Galois.

If you're using Dummit and Foote, check out theorems 13.8 and 13.27.