Let $K$ be an extension of $F$. When is $ \operatorname{Aut}(K)= \operatorname{Gal}(K/F)$?

Question

Let $K$ be an extension of $F$. The Galois group of $K$ is defined to be the set of all $F$-automorphisms of $K$, i.e., all automorphisms of $K$ which fix $F$.

An extension $K$ over $F$ is Galois if $| \operatorname{Gal}(K/F)|=[K:F]$.

(I am aware some call the set of all $F$-automorphisms of $K$ the Galois group when $K$ is Galois over $F$.)

It is clear $\operatorname{Gal}(K/F) \subset \operatorname{Aut}(K)$.

When is $ \operatorname{Aut}(K)= \operatorname{Gal}(K/F)$?

Answer 1

When $F$ is the prime field of $K$, i.e. the subfield of $K$, which does not contain any proper subfield, then we have equality. This is due to the fact that the prime field is fixed by any automorphism.

You can check this in two steps:

• first prove that the prime field is always isomorphic to either $\mathbb{Q}$ (if $char(K)=0$) or $\mathbb{F}_p$ (if $char(K)=p$)
• then show that any automorphism fixes the prime field (use that $\sigma(1)=1, \sigma(a+b)=\sigma(a)+\sigma(b)$ and $\sigma(a\cdot b)=\sigma(a)\sigma(b)$)