Field Automorphisms Quite a general question: For a subfield $K$ of $\mathbb{C}$, how can we prove that every automorphism of $K$ fixes every rational number, and secondly, that the set of such automorphisms forms a group?
 A: Any field automorphism must map $1$ to $1$ (it must map it to an element that satisfies $x^2=x$, and the only such elements are $0$ and $1$). 
If the field has characteristic $0$, and so contains a copy of $\mathbb{Q}$, then since it maps $1$ to $1$ then it maps $n$ to $n$ for every nonnegative integer $n$ (it must map $0$ to $0$); hence, since $f(-n) = -f(n)$, it maps $a$ to $a$ for every integer $a$. Hence, since $f(x^{-1}) = f(x)^{-1}$, it maps $\frac{1}{a}$ to $\frac{1}{a}$ for every nonzero integer $a$. Hence it maps $\frac{p}{q} = p\left(\frac{1}{q}\right)$ to $\frac{p}{q}$ for every rational number $\frac{p}{q}$.
This works for any field of characteristic $0$, whether it is contained in $\mathbb{C}$ or not.
The same argument, only simpler, shows that if $K$ is a field of characteristic $p$: every such field contains a copy of $\mathbf{F}_p$, generated by the multiplicative identity, and any automorphism will fix that copy of $\mathbf{F}_p$ element-wise.
The fact that automorphisms form a group follows by verifying that:


*

*Composition of functions is associative;

*Composition of automorphisms is an automorphism;

*The identity map is an automorphism; and

*The inverse of an automorphism is an automorphism.

