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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.

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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] $.

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  • $\begingroup$ 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 ? $\endgroup$ May 2, 2017 at 22:21
  • $\begingroup$ @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. $\endgroup$
    – Ege Erdil
    May 2, 2017 at 22:56

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