Fixedpoints of involution of field extensions Suppose we have a field extension $L | K$ and a nontrivial $K$-automorphism $\phi: L \rightarrow L$, ie. an automorphism on $L$ which fixes the elements of $K$, which is an involution, ie. selfinverse. What conditions do we have to impose on $L$ or $\phi$ to have that $\phi(x)=x \Rightarrow x \in K$?
 A: Claim. The extension $L/L^{\langle \phi\rangle}$ has degree $2$. 
Since $L^{\langle \phi\rangle}$ contains $K$, we deduce immediately that $L^{\langle \phi\rangle}=K$ if and only if $L/K$ has degree $2$.
Proof of the claim.
This follows directly from Artin's lemma, which says that  $L/L^{\langle \phi\rangle}$ is in fact Galois of degree $2$, but we do not need the galois property, and there is a direct way to compute the degree, which I give now. 
Let $E=L^{\langle \phi\rangle}$.
Let $\alpha\in L^\times$ such that $\phi(\alpha)\neq \alpha$. I claim that $1,\alpha$ is an $E$-basis of $L$.
First, they are $E$-linearly independent, since otherwise $\alpha\in E$.
Now let $x\in L$, and set $b=(\phi(x)-x)(\phi(\alpha)-\alpha)^{-1}$, and $a=x-b\alpha$.
Then $\phi(b)=b$ (this uses that $\phi$ is an automorphism satisying $\phi\circ\phi=\phi$). Moreover, $\phi(a)=\phi(x)-b\phi(\alpha)$, so $\phi(a)-a=\phi(x)-x-b(\phi(\alpha)-\alpha)=0$ by definition of $b$. Hence $a,b,\in E$. Now ,by definition of $a$, $x=a+b\alpha$, and we are done.
