because $ Tr (a) = Tr (a ^ p) $ where $ a \in \mathbb {F} _q$ (finite field) and p is characteristic of this field

Question

Can anyone explain to me why this relationship? Where Tr is the absolute trace function $ Tr: \mathbb{F}_q \to \mathbb{F}_p $

Answer 1

Assume $q=p^n$

$$\begin{aligned}Tr(a^p)&=(a^p)+(a^p)^p+\cdots+(a^p)^{p^{n-2}}+(a^p)^{p^{n-1}}\\ &=a^p+a^{p^2}+\cdots+a^{p^{n-1}}+a^{p^n}\\ &=a^p+a^{p^2}+\cdots+a^{p^{n-1}}+a\\ &=Tr(a). \end{aligned}$$