# Is it true that every element of $\mathbb{F}_p$ has an $n$-th root in $\mathbb{F}_{p^n}$?

It is not hard to prove that every element of $\mathbb{F}_p$ has a square root in $\mathbb{F}_{p^2}$: take any $a \in \mathbb{F}_p$ and consider the polynomial $f = X^2 – a$. If $f$ has a root in $\mathbb{F}_p$, then we are done. Otherwise $f$ is irreducible over $\mathbb{F}_p$, let $\beta \in \overline{\mathbb{F}_p}$ be a root of $f$, then the extension $\mathbb{F}_p(\beta)$ has degree $2$ over $\mathbb{F}_p$, and hence by the uniqueness of finite fields we have $\mathbb{F}_p(\beta) = \mathbb{F}_{p^2}$, so $\beta \in \mathbb{F}_{p^2}$. With exactly the same proof we can see that every element of $\mathbb{F}_p$ has a cube root in $\mathbb{F}_{p^3}$.

But I don’t know if a similar statement holds for an arbitrary $n$. If $n>3$, the fact that $X^n – a$ has no root in $\mathbb{F}_p$ is no longer equivalent to being irreducible. So the thing that we would need to show in this case is that $X^n – a$ has an irreducible factor whose degree is a divisor of $n$. I don’t see how to do that.

The problem would be solved if we could prove that $\mathbb{F}_{p^n}$ contains an element $\beta$ whose multiplicative order is $n(p-1)$, since in that case $\beta^n$ is a primitive root mod $p$. But, from the fact that $\mathbb{F}_{p^n}^{\times}$ is a cyclic group, if follows that the previous condition is equivalent to $n \mid p^{n-1} + p^{n-2} + \ldots + p + 1$, which does not necessarily hold (it does in some special cases, such as $p \equiv 1 \pmod{n}$).

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I think that the following extension to prof. Lubin’s argument settles the question in the affirmative.

First let’s write $n=n_1n_2$, where all the prime factors of $n_1$ are also factors of $p-1$, and $\gcd(n_2,p-1)=1$. For all $a\in \mathbb{F}_p$ the equation $x^{n_2}-a$ has a root $y$ in the prime field, so it suffices to show that $y$ has an $n_1$th root in $\mathbb{F}_{p^{n_1}}\subseteq\mathbb{F}_{p^n}$.

Lemma. Assume that $q$ is a prime, and that the finite field $K$ contains a primitive $q^{th}$ root of unity $\zeta$. Let $\alpha\in K$ be arbitrary. Then the polynomial
$$f(x)=x^q-\alpha$$
has a root in the unique degree $q$ extension $L$ of $K$.

Proof. If $f(x)$ has a root in $K$, then that root is also in $L$. If no such root exists in $K$, then such a root $\beta$ exists in $\overline{K}$.
Because the other zeros of $f(x)$ are gotten from $\beta$ by multiplying it with a power of $\zeta$, we see that $K[\beta]$ is the splitting field of $f(x)$. Let $\sigma$ be a generator of the cyclic Galois group $\operatorname{Gal}(K[\beta],K)$. Then $\sigma(\beta)=\beta\zeta^\ell$ for some exponent $\ell$ coprime to $q$. As $q$ is a prime and $\zeta$ is fixed by $\sigma$, we see that $\sigma$ is of order that is a multiple of $q$. Therefore its order is exactly $q$, and we can conclude that $K[\beta]=L$. Q.E.D.

The claim follows easily from the Lemma. If $q$ is any prime factor of $n_1$, then
$q\mid p-1$, so the prime field already contains the necessary roots of unity.
Hence so do all its extensions, and the Lemma bites. More precisely, if $d\mid dq\mid n_1$, where $q\mid p-1$ is a prime, then, by induction hypothesis the equation
$$x^d=a$$
has a solution $x\in\mathbb{F}_{p^d}$. By the Lemma, the equation $y^q=x$
has a solution $y\in\mathbb{F}_{p^{dq}}$, and $y$ is then a solution of
$$y^{dq}=a.$$ Repeating this step enough many times gets us to $n_1$ settling the claim.

Here’s an unsatisfyingly partial answer, but it’s late and I’m not thinking clearly. We’re in good shape if $n$ is a prime, let’s call it $q$ instead.

Three cases: First, $q=p$ is all right, everything is a $p$-th power already. Second case is that $q$ does not divide $p-1$. Then $\gcd(q,p-1)=1$, and again every element of $\mathbb F_p$ is a $q$-th power. Third case, $q|(p-1)$, then all $q$-th roots of unity are in $\mathbb F_p$, so adjoining one root of $X^q-c$ gets you all of them, and if $c$ wasn’t a $q$-th power in $\mathbb F_p$, then the extension is cyclic of degree $q$, equal to $\mathbb F_{p^q}$. The way I was looking at composite $n$ led me into complications, but I was probably missing something easy.