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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 $0$

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  • $\begingroup$ what's a primitive element ? $\endgroup$ – mercio Jul 16 '15 at 9:40
  • $\begingroup$ @mercio Given a finite separable field extension $K/F$ a primitive element for $K/F$ is some $a \in K$ such that $K=F(a)$. See en.wikipedia.org/wiki/Primitive_element_theorem $\endgroup$ – Crostul Jul 16 '15 at 9:46
  • $\begingroup$ I feel the need to add that while clearly the OP (and Artin's result) specify that a primitive element is one that generates the field extension in the context of finite fields only generators of the multiplicative group of the said field are called primitive. This is in line with the concept of a primitive root modulo $p$ - a residue class that gives all the other residue classes modulo $p$ as its powers. Anyway, here the usual notion of primitive was specified (and is the only one that makes sense), so I based my answer on that meaning of primitive. $\endgroup$ – Jyrki Lahtonen Jul 16 '15 at 10:31
  • $\begingroup$ A related question. $\endgroup$ – Jyrki Lahtonen Jul 16 '15 at 10:32
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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.

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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 $\mathbb{F}_{p^n}$.

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    $\begingroup$ okay but $p$ can be chosen not to divide $n$. $\endgroup$ – Nathan Sikora Jul 16 '15 at 9:54

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