Artin's lemma in Galois Theory: when is [E:F] < |G|?

[Boorglar]

I am trying to understand Galois Theory and reading through various theorems and lemmas, some of which are still confusing me.

A lemma proved by Artin states that if F is the fixed field of a finite group G of automorphisms in a field E, then the degree [E:F] ≤ |G| = n. The proof relies on setting up a linear dependence relationship for any set of n+1 elements in E, using n equations in n+1 unknowns with coefficients obtained by every permutation in G.

I am trying to get an intuitive understanding of the result. I already know that in the case where G is the Galois group G(E/F) and E is a splitting field of a separable polynomial over F, then F is the fixed field of G and [E:F] = |G|. But what properties of E and G could cause [E:F] < |G|?

I can't find any example where strict inequality arises.

 

[Terandol]

You can't find a counterexample because I believe none exists. In fact, given any set $S$ of $n$ distinct isomorphisms of a field $E_1$ into a field $E_2$ (here they don't even need to form a group just some set of isomorphisms), then $[E:\mathrm{Fix}(S) ] \geq n$.

Whatever source you are using probably split the proof that if $S$ is a group then $[E:\mathrm{Fix}(S) ] =n$ into separate lemmas so I would imagine there is also a proof of the above fact nearby the proof of the other inequality in your book.

 

[Boorglar]

Yes, I think you are right. I found out later in the book, they prove that if F is a fixed field of E under the finite group of automorphisms of E, then E is a finite, normal, separable extension of F (in fact the two properties are equivalent). This implies that [E:F] = |G(E/F)| ≥ |G|, which is the other inequality.

I was scratching my head for a while before I realized this.

An interesting side-effect of [E:F] = n is that it proves the set of all automorphisms of G is linearly independent over F, since from Artin's proof, the nxn system of equations must have only the trivial solution.