# Roots of $x^{p^{n-1}}+\ldots+x^p+x$ in $\mathbb{F}_{p^n}$

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?

• 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 ____? – Jyrki Lahtonen May 17 at 7:37

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 $$\Bbb F_p$$.

• I combined your answer with Jyrki's comment to fill in all the details. Thanks. – Michael Morrow May 17 at 20:31

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.)

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 $$a^q-a=f(b)=0$$.