In finite field why $\overline c = -c \ne 0$ for $\overline c = c^{p^f}$.

Question

Let $GF(p^{2f})$ be a finite field of order $p^{2f}$. Consider the map $\overline x := x^{p^f}$ for $x \in GF(p^{2f})$. Let $b \in GF(p^{2f}) - GF(p^f)$ and set $c := b - \overline b$.

Why do we have $\overline c = -c$ and $c \ne 0$?

I see that $\overline{\overline{c}} = c$. But is this enough to conlude both statements?

Answer 1

As the subfield $GF(p^f)$ consists exactly of the roots of $X^n - X$ for $n = p^f$, we have $c \ne 0$ for $b \notin GF(p^f)$. Further as $(x+y)^p = x^p + y^p$ in finite fields we have $\overline{x + y} = \overline x + \overline y$. This gives $\overline c = \overline{b - \overline b} = \overline b - b = -(b - \overline b) = -c$.