# So-called Artin-Schreier Extension

Let $F$ be a field of characteristic $p$. Let $K$ be a cyclic extension of $F$ of degree $p$. Prove that $K=F(\alpha)$ where $\alpha$ is a root of the polynomial $p(x) = x^{p} - x - a$ for $a \in \mathbb{F}$.

I've seen, and attempted, a lot of problems that look similar. But I'm not really sure about this one.

• See also this question. My answer gives a standard proof for this - assuming that you are familiar with the linear independce of characters -lemma. – Jyrki Lahtonen Jul 8 '14 at 18:44

Let $\sigma$ be a generator of $G_{K/F}$. The equation $x^p-x=a$ can be written in two different ways:
$$\begin{cases}x(\sigma x)\cdots(\sigma^{p-1}x) & =a \\ x(x+1)\cdots(x+p-1) & =a\end{cases}$$
The first follows because $a$ is the opposite of the constant term of $x$'s minimal polynomial thus giving its norm (as either the polynomial is odd-degree or $1=-1$) and the second follows from simply factoring the polynomial $x^p-x$ in characteristic $p$.
Wouldn't it be nice if the stars aligned and $\sigma x=x+1$, thus guaranteeing that not only were the products equal (both being $a$) but each of their factors were equal in the listed order?
Try to show that given $K/F$ is cyclic $C_p$ there exists an $x\in K$ such that $(\sigma -1)x=1$. Necessarily this would mean that $x\not\in F$ hence $K=F(x)$ and would imply $x^p-x\in F$. (My preferred method would be to construct $x$ as an appropriate polynomial in $\sigma$ applied to a normal basis generator.)