Show there exists an $\alpha\in F_{p^n}$ such that $\alpha +\alpha^p+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}\neq 0$

Question

This is part of a problem, 7.22c in "Ireland and Rosen" (self-study).

In the prior problem it is shown that for prime $p$ and $\alpha\in F_{p^n}$, $f(x)=(x-\alpha)(x-\alpha^p)(x-\alpha^{p^{2}})\cdots(x-\alpha^{p^{n-1}})$ is in $\mathbb{Z}/p{\mathbb{Z}}[x]$.

Show there exists an $\alpha\in F_{p^n}$ such that Tr$(\alpha)=\alpha +\alpha^p+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}\neq 0$ for some $\alpha\in F$.

So far my approach is that with $f(x)\in \mathbb{Z}/p{\mathbb{Z}}[x]$, the coefficients are in $\mathbb{Z}/p{\mathbb{Z}}$. And by the symmetric polynomial functions, the Tr$(\alpha)$ as defined is a coefficient, it follows it is in $\mathbb{Z}/p{\mathbb{Z}}$.

Thus I would think that Tr$(\alpha)\neq 0$ is equivalent to $p\nmid \alpha +\alpha^p+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}$

If this is correct I would please appreciate help as to how to show it.

Otherwise, I would appreciate further help.

Thanks

Answer 1

Hint: Remember that a polynomial of degree $m$ can only have at most $m$ roots in any given field.

Bigger hint:

Apply this to the polynomial $g(x)=x +x^p+x^{p^2}+\cdots+x^{p^{n-1}}$.

Answer 2

The claim follows from the fact that the trace is a non-degenerated bilinear form on a field extension iff it is a separable one, and finite fields are always perfect so any algebraic extension of them is separable.

Answer 3

Suppose that for every $\alpha \in \mathbb{F_{p^n}}$ we have $\text{Tr}(\alpha) = 0$. Then by the additive version of Hilbert 90, we have that $$\{y^p - y | y \in \mathbb{F}_{p^n}\} = \mathbb{F}_{p^n},$$ so the map $\mathbb{F_{p^n}} \rightarrow \mathbb{F_{p^n}}$ given by $y \mapsto y^p-y$ is a bijection. However, for $\beta, \gamma \in \mathbb{F_{p}}$ we have $$\beta^p-\beta=\gamma^p-\gamma=0,$$ a contradiction.