Difference between F-Automorphism and Identity morphism

Question

In reading Shiffrin Abstract Algebra, in the section on Galois Theory, it gives the following definition:

For K a field extension of F, a ring isomorphism $\phi:K\to K$ is an F-Automorphism if

$\phi(a) = a$, $\forall a \in F$

My question is, how is this different from the identity morphism? Isn't (by this definition) the set of all F-automorphisms, or the Galois group, just the identity mapping? I'm slightly confused.

Thank you very much!

Answer 1

The difference is that the restriction of $\phi$ to $F$ is the identity morphism on $F$. The definition doesn't say anything about what $\phi$ should do to the rest of $K$ (other than be a bijective homorphism, of course). It may not be true that $\phi$ fixes the rest of $K$. For example, complex conjugation is an $\mathbb{R}$-automorphism of $\mathbb{C}$.

Answer 2

Even though elements of $F$ are fixed, $F$-automorphisms can send elements of $K$\ $F$ to other such elements of $K$. For example, in $\mathbb{Q}[\sqrt{2}]$, one such $\mathbb{Q}$-automorphism could send $\sqrt{2} \mapsto -\sqrt{2}$. In fact, this is the only nontrivial $\mathbb{Q}$-automorphism of this extension.

P.S. I'm learning from Shifrin's text too. :)