For each $x \in GF(p^{2n})$ it is $1 - x^{p^n + 1} \in GF(p^n)$

Question

Let $GF(p^{2n})$ be a finite field of order $p^{2n}$. Then $GF(p^n) \subseteq GF(p^{2n})$. Why do we have $1 - x^{p^n+1} \in GF(p^n)$?

I know that $x^{p^n} - x = 0$ holds iff $x \in GF(p^n)$, but why is $1 - x^{p^n + 1} \in GF(p^n)$?

Answer 1

Because $x \mapsto x^{p^{n}+1}$ is the trace map from $\mathbb{F}_{p^{2n}} \to \mathbb{F}_{p^{n}}$, so is always in $\mathbb{F}_{p^{n}}$.

Notice that $x \mapsto x^{p^{n}}$ is an automorphism of $\mathbb{F}_{p^{2n}}$ that fixes precisely the elements of $\mathbb{F}_{p^{n}}$. Since $$(x^{p^{n}+1})^{p^{n}} = x^{p^{2n}+p^{n}} = x^{p^{2n}}x^{p^{n}} = x x^{p^{n}} = x^{p^{n}+1}$$ this element is fixed by this map, and therefore must be in $\mathbb{F}_{p^{n}}$.