I'm stumbled upon a trivial detail of the proof of Fixed Field Theorem in Artin Algebra. Let $H$ be a finite group of automorphisms of a field $K$, and let $F = K^H$ be its fixed field. Then $K$ is a finite extension of $F$, and its degree $[K:F]$ is equal to the order $|H|$ of the group.
Proof: Let $F = K^H$ and let $n$ be the order of $H$. Theorem 16.5.2 shows that the extension $K / F$ is algebraic, and that the degree over $F$ of any element of $K$ divides $n$. Therefore the degree $[K:F]$ is finite (16.5.3). Let $\gamma$ be a primitive element for this extension. Every element $\sigma$ of $H$ is the identity on $F$, so if $\sigma$ also fixes $\gamma$, it will be the identity map – the identity element of $H$. Therefore the stabilizer of $\gamma$ is the trivial subgroup {l} of $H$ , and the orbit of $\gamma$ has order $n$. Theorem 16.5.2 shows that $\gamma$ has degree $n$ over $F$. Since $K = F(\gamma)$, the degree $[K:F]$ is equal to $n$ too.
The result is based on the assumption that "$\sigma$ fixes $\gamma$". However, we know $\gamma \in K$, and it's possible $\gamma \notin K^H$, which means it's possible $\sigma$ does not fix $\gamma$. So how can we assume "$\sigma$ fixes $\gamma$"
Best Answer
You are mis-reading the proof.
$\gamma$ is a primitive element, so it is not fixed by any $\sigma$, unless $\sigma$ fixes every element of $K$. But then $\sigma$ is the identity.
This is exactly what the proof tells you:
The rest of the proof then follows.