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Let $\mathbb{F}_q$ denote a field with $q=p^n$ elements, where $p$ is prime. Consider the polynomial $f=x^{p^{n-1}}+\ldots+x^p+x$ and the sets $$ \begin{align*} S&=\{a^p-a:a\in\mathbb{F}_q\},\\ T&=\{b\in\mathbb{F}_q:f(b)=0\}. \end{align*} $$ Show that $S=T$.

My ideas: I can show $S\subset T$ as follows. Let $x\in S$. Then $x=a^p-a$ for some $a\in\mathbb{F}_q$. We have $$ \begin{align*} f(x)&=(a^p-a)^{p^{n-1}}+(a^p-a)^{p^{n-2}}+\ldots+(a^p-a)^p-(a^p-a)\\ &=a^{p^n}+a^{p^{n-1}}+\ldots+a^{p^2}+a^p-a^{p^{n-1}}-a^{p^{n-2}}-\ldots-a^p-a\\ &=a^{p^n}-a\\ &=a-a\\ &=0. \end{align*} $$ Thus $x\in T$. However, I have no idea how to show $T\subset S$. Any advice? Maybe we can make use of the fact that $a\mapsto a^p-a$ is $\mathbb{F}_p$-linear?

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  • $\begingroup$ You can use linearity (together with the rank-nullity theorem). The kernel of $g:a\mapsto a^p-a$ has dimension ___? Therefore the image of $g$ has dimension ____? The degree of $f$ is $p^{n-1}$. Therefore the kernel of $f$ has dimension at most ____? You have shown that the image of $g$ is contained in the kernel of $f$. Therefore ____? $\endgroup$ – Jyrki Lahtonen May 17 at 7:37
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You have proved $S\subseteq T$. As $f$ has degree $p^{n-1}$ it has at most $p^{n-1}$ zeroes: $|T|\le p^{n-1}$.

The polynomial $g(x)=x^p-x$ has degree $p$. For $a\in\Bbb F_q$, $g(x)=a$ has at most $p$ solutions. Therefore the image $S=\{g(b):b\in \Bbb F_q\}$ has $|S|\ge q/p=p^{n-1}$. Therefore $$p^{n-1}\le|S|\le|T|\le p^{n-1}.$$ Consequently $|S|=|T|$, and so $S=T$.

Note that $f$ represents the trace map from $\Bbb F_q$ to $\Bbb F_p$.

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  • $\begingroup$ I combined your answer with Jyrki's comment to fill in all the details. Thanks. $\endgroup$ – Michael Morrow May 17 at 20:31
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Let $b \in T$. Since the degree of $f$ is $p^{n-1} < q$, you can take $c \in \mathbb{F}_q$ such that $f(c) \neq 0$. Replacing $c$ by $f(c)^{-1}c$, you may assume $f(c)=1$.(In this argument, use the fact that $f(c) \in \mathbb{F}_p$ and that $x \mapsto f(x)$ is $\mathbb{F}_p$-linear.)

And put $a = bc^p+(b+b^p)c^{p^2}+ \cdots +(b+b^p+ \cdots +b^{p^{n-2}})c^{p^{n-1}}$.

Then

$b+a^p\\ =b\\ \quad +b^pc^{p^2}+(b^p+b^{p^2})c^{p^3}+ \cdots+(b^p+ \cdots +b^{p^{n-1}})c^{p^n}\\ =b(c+c^p+ \cdots c^{p^{n-1}})\\ \quad +b^pc^{p^2}+(b^p+b^{p^2})c^{p^3}+ \cdots+(b^p+ \cdots +b^{p^{n-1}})c\\ =(b+b^p+ \cdots b^{p^{n-1}})c\\ \quad +bc^p+(b+b^p)c^{p^2}+ \cdots +(b+b^p+ \cdots b^{p^{n-2}})c^{p^{n-1}}\\ =0+a\\ =a$

Here the second equality comes from $c+c^p+ \cdots c^{p^{n-1}}=f(c)=1$ and $c^{p^n}=c$, the third one is sorting with respect to $c$, and the fourth uses $b \in T$ and the definition of $a$.

Now $b=a^p-a$ inplies $b \in S$.

Background

This proposition is a special case of Hilbert's theorem 90. (Cosnsidering a cyclic extention $\mathbb{F}_q / \mathbb{F}_p$, the set S is coboundary and T is cocycle. Hibert 90 states these two sets coincide.)

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Suppose that $f(b)=0$ for $b\in\mathbb{F}_q$. Let $a$ be a root of $x^p-x-b$ in its splitting field, so that $b=a^p-a$. It remains to show that $a\in\mathbb{F}_q$. In fact, $$\begin{split}f(a^p-a)&=(a^p-a)+(a^p-a)^p+\dots+(a^p-a)^{p^{n-1}}\\&=(a^p-a)+(a^{p^2}-a^p)+\dots+(a^{p^n}-a^{p^{n-1}})\\&=a^{p^n}-a.\end{split}$$ It follows that $a^q-a=f(b)=0$.

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