What is the minimal polynomial of $x$ over $k(x^p - x)$? Let $k=\Bbb F_p$, and let $k(x)$ be the rational function field in one variable over $k$. Define $\phi:k(x)\to k(x)$ by $\phi(x)=x+1$. I know that $\phi$ has order $p$ in $\operatorname{Gal}(k(x)/k)$ and $k(x^p-x)$ is the fixed field of $\phi$, but I'm confused about the minimal polynomial of $x$ over $k(x^p-x)$. (This is problem 17 of Fields and Galois Theory by Patrick Morandi, page 27.)
Please explain about that minimal polynomial or calculate it!
 A: The hardest part, IMO, of this problem is not confusing yourself.
Define $y = x^p - x$. So you are looking at the field $K = k(y)$. (or maybe use $\alpha$ instead of $y$ if you prefer)
$x$ is algebraic over $K$. Your goal is to find a polynomial over $K$ -- an element $f \in K[t]$ -- such that $f(x) = 0$.
So can you find a polynomial expression in $x$ with coefficients in $K$ that vanishes? If so, you can then write down an $f(t)$, that makes for a good starting point to begin solving the problem.
A: If you have absolutely no idea to get the minimal polynomial, and you know that there is an isomorphism $\Bbb F_p \to Gal(k(x)/k(x^p-x))$ given by $t \mapsto (x \mapsto x+t)$, then you know that the conjugates of $x$ are the $x+t$, and then you can simply compute the product $\prod_{t \in \Bbb F_p}(T-(x+t))$. This product is invariant by $x \mapsto x+1$ so its coefficients will be in $k(x^p-x)$.
You will find $\prod_{t \in \Bbb F_p}(T-(x+t)) = \prod_{t \in \Bbb F_p}((T-x)-t) = (T-x)^p - (T-x) = (T^p-T)-(x^p-x)$
This is a polynomial $P(T)$ with coefficients in $k(x^p-x)$, and $P(x)=0$.
More generally, if you consider the extension $k(f(x)) \subset k(x)$ with $f(x) = P(x)/Q(x)$, then $Q(T)f(x)-P(T)$ is an obvious polynomial over $k(f(x))$ annihilating $x$
