Element of $\mathbb{F}_{p^2}$ of order $p^2$ raised to $p+1$th power is element of $\mathbb{F}_p$?

Question

For any prime number $p$ and any element $a$ of the finite field $\mathbb{F}_{p^2}$ of order $p^2$, do we have$$a^{p+1} \in \mathbb{F}_p \subset \mathbb{F}_{p^2}?$$

Answer 1

Since no one else bites, I elevate my comment to an answer.

The Frobenius map $F:\Bbb{F}_{p^2}\to\Bbb{F}_{p^2}, F(x)=x^p$ is an automorphism of the field $L=\Bbb{F}_{p^2}$. Its fixed field is the prime field $K=\Bbb{F}_p$. Because $L$ consists of the zeros of $x^{p^2}-x$ we know that $F$ is of order two in $\operatorname{Gal}(L/K)$.

But whenever $\sigma$ is an order two automorphism of a field $F$ we have that $z\sigma(z)$ is in the fixed field of $\sigma$. It is, after all, a product of images under all distinct powers of $\sigma$ so applying $\sigma$ to it just permutes the factors. The example a beginner is most likely familiar with is the usual complex conjugation that has the subfield of real numbers as its fixed field. In that case this fact implies that $z\overline{z}\in\Bbb{R}$ for all complex numbers $\Bbb{C}$.

Anyway, for all $a\in L$ we have $$ a^{p+1}=a\cdot a^p=a F(a), $$ so this is an element of the fixed field $K$.

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The map $a\mapsto a F(a)$ is also the norm map $N_K^L$. The result generalizes in that form to all cyclic field extensions. If $L/K$ is any finite Galois extension such that $\operatorname{Gal}(L/K)$ is generated by a single automorphism $\sigma$ of order $n$, then for all $z\in L$ the norm $$ N^L_K(z)=\prod_{i=0}^{n-1}\sigma^i(z)\in K. $$ Any extension of finite fields is cyclic, so this analogy applies to all of them.