primitive element $a$ of $\mathbb F_{p^n}/\mathbb F_p$ such that $a^n\in\mathbb F_p$

Question

Is it true that for every $n\in \mathbb N$ there exists a prime $p$ such that the extension $\mathbb F_{p^n}/\mathbb F_p$ has a primitive element $a\in \mathbb F_{p^n}$ and $a^n\in\mathbb F_p$?
I tried finding a counter example but didn't mange to think of one. on the other hand I tried to prove it. From Artin's primitive element theorem, for every prime $p$ there exists an element $a\in \mathbb F_{p^n}$ such that \begin{equation*}\mathbb F_{p^n}=\mathbb F_p (a)\end{equation*} and\begin{equation*}[\mathbb F_{p^n}:\mathbb F_p]=n\end{equation*}implies that \begin{equation*}\textrm{deg}(m_{a,\mathbb F_p })=n\end{equation*}Now I dont know how to choose $p$ so \begin{equation*}m_{a,\mathbb F_p }=x^n+a^n\end{equation*}i.e every coefficient of $x^k$ for $k=1,...,n-1$ is $0$

Answer 1

Yes, this is true.

Given a positive integer $n>1$ let $p$ be any prime such that $p\equiv 1\pmod n$. In other words we want $p=1+an$ for some integer $a>0$. By Dirichlet's theorem on infinity of primes in an arithmetic progression there are infinitely many such primes $p$.

Let us fix such a prime $p$. Let $c$ be a primitive root modulo $p$. I claim that $f(x)=x^n-c$ is an irreducible polynomial in the ring $\Bbb{F}_p[x]$.

Lemma. With $n,p$ as above and $k$ any positive integer we have $n(p-1)\mid p^k-1$ if and only if $n\mid k$.

Proof. Clearly $n(p-1)$ is a factor of $p^k-1$, iff $n$ is a factor of $1+p+p^2+\cdots+p^{k-1}$. But here $p^i\equiv1\pmod n$ for all $i$. The claim follows.

We can then prove the irreducibility of $f(x)$ as follows. Assume that $f(x)$ has an irreducible factor of degree $k$. Let $\alpha$ be a zero of such a factor. Let $r$ be the order of $\alpha$, i.e. the smallest positive exponent with the property $\alpha^r=1.$ Here $\alpha^{n(p-1)}=c^{p-1}=1$, so $r\mid n(p-1)$. I claim that we actually have $r=n(p-1)$. Because $c$, an element of the cyclic group generated by $\alpha$, is of order $p-1$ we necessarily have $(p-1)\mid r$. Therefore it suffices to show that for all prime factors $q$ of $n$ we have $\alpha^{n(p-1)/q}\neq1$. Because $n\mid p-1$ we have that $q\mid p-1$. Hence $$ \alpha^{n(p-1)/q}=(\alpha^n)^{(p-1)/q}=c^{(p-1)/q}\neq1. $$

The rest is easy. Because $\alpha\in\Bbb{F}_{p^k}$ we must have $r\mid p^k-1$. The Lemma says that $n\mid k$, so $f(x)$ has no factors of degree $<n$ and must be irreducible.

Answer 2

The Frobenius automorphism $\varphi:\mathbb{F}_{p^n}\to \mathbb{F}_{p^n}$ is a bijection that fixes the prime subfield $\mathbb{F}_p$, so $\varphi(a)\in\mathbb{F}_p$ if and only if $a\in\mathbb{F}_p$, and in fact for all $k$, we have $\varphi^k(a)\in\mathbb{F}_p$ if and only if $a\in\mathbb{F}_{p}$.

Thus, whenever $n$ is a power of $p$, we have $a^n\in\mathbb{F}_p$ if and only if $a\in\mathbb{F}_p$, and such an $a$ certainly cannot be a primitive element for $\mathbb{F}_{p^n}$.