# Artin–Schreier polynomial

Suppose we have the finite field $K=\Bbb{F}_{p^n}$ ($p$ prime and $n>0$) and an Artin–Schreier polynomial $f=x^p-x+\gamma \in K[x]$.

Suppose that $f$ is irreducible. How do we prove that $tr_K(\gamma)=\gamma+\gamma^p+\cdots +\gamma^{p^{n-1}} \neq 0$ ?

I think it helps to see that $tr_K(\gamma)=\sum_{\sigma \in Gal(K/\Bbb{F}_p)} \gamma^\sigma$. But I don't really see the relation between $f$ and the galois group.

## 1 Answer

This follows from Hilbert's Satz 90 for finite cyclic extensions, i.e if $L/K$ is a finite cyclic extension with Galois group generated by $\sigma$, then for an $x \in L$, we have that $\textrm{Tr}_{L/K}(x) = 0$ if and only if $x = \sigma(y) - y$ for some $y \in L$. The extension $\mathbb F_{p^n}/\mathbb F_p$ has cyclic Galois group generated by the Frobenius automorphism $X \to X^p$. Then, we have that $\textrm{Tr}(\gamma) = \textrm{Tr}(-\gamma) = 0$ if and only if $-\gamma = y^p - y$ for some $y \in K$. But then, $y$ is a root of $X^p - X + \gamma$ in $K$, which therefore cannot be irreducible in $K[X]$.

• I see that Satz 90 implies that if $x$ has a norm of $1$, then $x=\frac{\sigma(y)}{y}$ for some $y$. Is it equivalent with what you said ? May 2, 2017 at 22:21
• @JannesBraet There is an additive version of Satz 90 and a multiplicative version. The version you quoted is the multiplicative statement; the version I use in my answer is the additive statement. Their proofs are quite similar. May 2, 2017 at 22:56