Let $K$ a field and $\sigma \in Aut(K)$ (the set of automorphism $K\to K$). Let denote $K^\sigma =\{x\in K\mid \sigma (x)=x\}$ the set of fix point.
1) Show that $K^\sigma $ is a subfield of $K$.
2) Let $K=\mathbb R(t)$ and $\sigma :K\to K$ defined by $\sigma (t)=-t$. Show that $\sigma \in Aut(K)$ and find $K^\sigma $.
3) Let $F$ a normale extension of $K$. Show that an automorphism of $K$ cannot always be extend as an automorphism of $F$.
4) Show that if $F$ is a normale extension of $K^\sigma $ then all automorphism of $K$ can be extend of an automorphism on $F$.
What I did:
1) The fact that $K^\sigma $ is a ring is obvious. Let $x\in K\backslash \{0\}$ and $a$ s.t. $$ax=1$$ ($a$ exist since $K$ is a field). Then $$1=\sigma (1)=\sigma (ax)=\sigma (a)\sigma (x)=\sigma (a)x$$ and thus $\sigma (a)$ is the inverse of $x$. By unicity of the inverse, $\sigma (a)=a$ and thus $a\in K^\sigma $.
Is it correct ?
2) Is $\sigma $ suppose to be an homomorphism ? I don't know how to proceed. For $K^\sigma $, let $f(t)=\frac{a_0+...+a_nt^n}{b_0+...+b_mt^m}\in \mathbb R(t)$. Then $$\sigma (f(t))=f(t)\iff(a_0+...+a_n\sigma (t)^n)\sigma \left(\frac{1}{b_0+...+b_mt^m}\right)=\frac{a_0+...+a_nt^n}{b_0+...+b_mt^m}$$
how can I continue.
3) and 4) I don't know.
