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We have the Trace map defined by:

$$ \mathrm{Tr}\colon \mathbb{F}_q\rightarrow\mathbb{F}_q\colon x\mapsto x+x^p+x^{p^2}+\cdots+x^{p^{n-1}}, $$ where $q=p^n$. Now I have to prove that if $\mathrm{Tr}(y)=0$ then there exists a $x\in\mathbb{F}_q$ such that $x^p-x=y$.

I don't know how to tackle this problem. I need a hint to start with. Thanks.

Jyrki Lahtonen
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Badshah
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    Every finite extension of $\mathbb{F}{p}$ is a Galois extension whose Galois group over $\mathbb{F}{p}$ is generated by the Frobenius map $ \sigma$, (the map that takes $x$ to $x^{p}$). Observe that $\sigma^{0}, \sigma, ..., \sigma^{n-1}$ are linearly independent so that their sum is never zero, but then Tr is surjective. Consider $(\mathrm{id} - \sigma)$. What is its kernel and its image? Show that $\mathrm{Im}(\mathrm{id} - \sigma) = \ker ( \mathrm{Tr})$ – Daniel Akech Thiong Aug 23 '14 at 16:34
  • possible duplicate of Roots of Artin-Schreier equation – Jyrki Lahtonen Aug 23 '14 at 16:38

2 Answers2

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What you are trying to prove is sometimes called the additive version of Hilbert's Theorem 90. Here is a hint:

Since the trace map is nonzero (why is this?), choose some $z \in \mathbb F_q$ with $\operatorname{Tr}(z) \neq 0$, and consider what happens to the sum $$ y z^p + (y+y^p)z^{p^2} + \cdots + (y+y^p + \cdots + y^{p^{n-2}}) z^{p^{n-1}} $$ when it gets hit with the automorphism $x \mapsto x^p$.

Cardboard Box
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We define $\phi(x)=x^p-x$. Then $\mathrm{img} \phi\subset\ker\mathrm{Tr}$ because $\mathrm{Tr}(x^p)=\mathrm{Tr}(x)$. Now if $x\in\ker\phi$, $x\in \mathbb{F}_p$ since this is the definition. Thus $\ker \phi=\mathbb{F}_p$ since it has at most $p$ solutions. Then $\mathrm{img}\phi$ has at least $p^{n}/p=p^{n-1}$ elements. But $\mathrm{Tr}$ is a polynomial with degree $p^{n-1}$, thus at most $p^{n-1}$ solutions. We deduce that $\mathrm{img} \phi=\ker\mathrm{Tr}$.

Golbez
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