Let $k = \mathbb{F}_p$, and let $k(x)$ be the rational function field in one variable over $k$. Define $\varphi: k(x) \to k(x)$ by $\varphi(x) = x + 1$. Show that the $\varphi$ has finite order in $Gal(k(x)/k)$. Determine this order, find a $u$ so that $k(u)$ is fixed field of $\varphi$, determine the minimal polynomial over $k(u)$ of $x$, and find all the roots of this minimal polynomial.
I couldn't to find the minimal polynomial of $x$ over $k(u)$ (consequently, find all the roots) nor conclude $k(u) = \cal{F}(\varphi)$. I don't know if my idea is correct, but I'm going to write bellow.
Idea. Note that $$(\varphi(x))^p = \varphi^p (x) = x = \rm{id}$$ so, $o(\varphi) = p$. Take $u = x^p - x$, then we have $$\varphi(x^p - x) = \varphi(x)^p - \varphi(x) = \underbrace{(x + 1)^p}_{\mathrm{char}(k) = p} - (x + 1) = x^p - x$$ so, since $\varphi$ fixed $u$ and fixed $k$ (by assumption), $\varphi$ fixed $k$ for all $k \in k(u)$. Then $k(u) \subset \cal{F}(\varphi)$ and $\cal{F}(\varphi) \subset k(x)$... how I prove that $\boldsymbol{k(u)=\cal{F}(\varphi)}$?
Now, $x$ is root of $p(X) = X^p - X - u$... Is this polynomial irreducible?
Thanks for the any hint.
For proving equality: Notice that if $\frac{q(x)}{r(x)}=\varphi(\frac{q(x)}{r(x)})=\frac{q(x+1)}{r(x+1)}$ (where $q(x),r(x)\in k[x]$) then by properties of polynomials, $q(x)=q(x+1)$ and $r(x)=r(x+1)$. So for finding $\cal{F}(\varphi)=k(x)^{<\varphi>}$, we need to only work with elements of $k[x]$. Now, if $q(x)$ is fixed, then for any $a\in k$, $q(a)-q(0)=0$. So $x(x-1)\cdots(x-(p-1))=x^p-x$ divides $q(x)-q(0)$. Can you finish the proof?
To see that the polynomial is irreducible, we notice that it is equal to the polynomial $(X-x)(X-\varphi(x))(X-\varphi^2(x))\cdots(X-\varphi^{p-1}(x))$ which we know is the minimal polynomial of $x$.