Define the trace by $$Tr \quad : \quad \mathbb{F}_{p^n} \ \longrightarrow \mathbb{F}_{p^n} \quad : \quad x \ \longmapsto \ x+x^p+\cdots +x^{p^{n-1}}$$ Now define yet another mapping: $$ L \quad : \quad \mathbb{F}_{p^n} \ \longrightarrow \mathbb{F}_{p^n} \quad : \quad x \ \longmapsto \ x^p-x$$
I know that this is a linear mapping. My syllabus stated the following I could't prove:
- The kernel of the map is $\mathbb{F}_p$.
- If the equation $y^p-y=x$ has a solution then $Tr(x)=0$ (I made it to prove the converse implication)
Effort to prove the first statement
If $x \in \mathbb{F}_p, L(x) = x^p-x = x-x =0$. This establishes the inclusion "$\mathbb{F}_p \subseteq \ker(L)$". Now assume that $x \in \ker(L)$. Then $x^p=x$. How does this imply that $x \in \mathbb{F}_p$?
Effort to prove the second statement
It's sufficient to prove that $\ker(Tr) \subset L(\mathbb{F}_q)$. I failed to prove this though.
I hope you can instruct me how to prove the statements. Thank you for your effort.
The first claim follows from the fact that the Galois group of $\Bbb F_{p^n}/\Bbb F_p$ is generated by the Frobenius map $x \mapsto x^p$. Thus, from Galois theory, the subfield of $\Bbb F_{p^n}$ fixed by $G(\Bbb F_{p^n}/\Bbb F_p)$ is exactly $\Bbb F_p$. For your second statement, recall that $\operatorname{Tr}(y^p) = \operatorname{Tr}(y)$, now apply $\operatorname{Tr}$ on both sides of the equation.