TITLE: Fixed field and automorphisms. Galois Theory.
QUESTION [2 upvotes]: Let $k$ be a field of characteristic $p>0$, and let $a \in k$. Let $f(x) = x^{p} - a^{p-1}x$. Show that $f$ is fixed by the automorphism $\varphi$ of $k(x)$ defined by $\varphi(f(x)/g(x)) = f(x+a)/g(x+a)$ for any $f(x),g(x) \in k[x]$. Show that $k(f)$ is the fixed field of $\varphi$.

Notation. If $S$ is a subset of Aut$(K)$, the fixed field of $S$ is denoted by $\mathcal{F}(S)$.

For the first part:

$$\varphi(f(x))=f(x+a)/1=\underbrace{(x+a)^{p}}_{char(k)=p}-a^{p-1}(x+a)=x^{p}+a^{p}-a^{p-1}x-a^p=x^{p}-a^{p-1}x = f(x).$$

For the second part:

If $c \in k$, $c=c(x)$ is a constant polynomial, so $\varphi(c(x)/1) = c(x+a)=c$. Then $\varphi$ fixed $k$ and $f$. Therefore, $k(f) \subset \mathcal{F}(\varphi)$.
Now, how I show that $\boldsymbol{\mathcal{F}(\varphi) \subset k(f)}$? Thanks for any hint!

REPLY [0 votes]: Observe that $k(f)\le \mathcal{F}(\varphi)\le k(x)$. Moreover, we can show that the minimal polynomial of $x$ over $k(f)$ is 
$$h(y) = y^p - a^{p-1}y -(x^p - a^{p-1}x).$$
(Note that $h(y)=f(y)-f(x)$. $h$ is irreducible over $k(f(x))(y)$ because it is isomorphic to $k(z)(y)$ and $f(y)-z$ is irreducible over $k(z)(y)$.)
Therefore $[k(x):k(f)]=p$ as $k(x) = k(f)(x)$. Hence either $[k(x):\mathcal{F}(\varphi)]=1$ or $[\mathcal{F}(\varphi):k(f)]=1$. If $[k(x):\mathcal{F}(\varphi)]=1$ then every polynomial $g\in k(x)$ must be fixed by $\varphi$, however we know this is not the case. Hence $[\mathcal{F}(\varphi):k(f)]=1$ and we have $\mathcal{F}(\varphi)=k(f)$.
{"set_name": "stack_exchange", "score": 2, "question_id": 2764302}