Consider the finite field extension $\mathbb{F}_{{q}^{d}}$ over $\mathbb{F}_{q}$, where $q=p^a$ for some prime $p$. We assume $d\geq 2$. Let, $$ S=\{ \alpha \in \mathbb{F}_{q^d}\hspace{0.1 cm} | \hspace{0.1 cm} \mathbb{F}_{q}(\alpha)=\mathbb{F}_{q^d} \}$$

In other words, $S$ consists of all those elements in $\mathbb{F}_{q^d}$, whose minimal polynomial over $\mathbb{F}_{q}$ has degree $d$, or to say another way $S$ consists of all field generators of $\mathbb{F}_{q^d}$ Over $\mathbb{F}_{q}$. Let, $S^m= \{ s^m | s\in S\} $, where $m$ is a positive integer $\geq 2$. Then, $$ |S\cap S^m|=?$$

More, precisely, $S\cap S^m$ is the set of all field generators in $\mathbb{F}_{q^d}$ (over $\mathbb{F}_q$), which are $m^{th}$ powers in $\mathbb{F}_{q^d}$.

I calculated, this for $m=2$. The answer depends on whether $d$ is odd or even. We have,

$$ |S\cap S^2|= \begin{cases} \frac{|S|}{2} & if \hspace{0.2 cm} d \text{ is odd}\\ \frac{1}{2}[|S|-\frac{(q^{d/2}-1)}{d}] & if \hspace{0.2 cm} d \text{ is even} \end{cases} $$

And, $|S|=dM(d,q)$, where $M(d,q)$ denote the number of irreducible polyomials of degree $d$ over $\mathbb{F}_{q}$.

I couldn't generalize my method to $m>2$. My, idea is that this problem seems to have been well studied in the literature of finite fields. So, I am hoping for some kind of help or suitable references in this case.

Thank you!


1 Answer 1


$$| S \cap S^m | = \sum_{n|d} \mu(n) \frac{ \gcd ( m (q^{d/n}-1), q^{d}-1 )}{ \gcd (q^{d} - 1, m) } $$

First note that $|S \cap S^m | = |S \cap \mathbb F_{q^d}^m |$ because every $m$th power that generates is an $m$th power of a generator.

We can count elements of $S$ by an inclusion-exclusion argument, subtracting and adding the number of elements in subfields. This gives the term $ \mu(n) q^{d/n}$, or $ \mu(n)( q^{d/n}-1)$ if we only count nonzero elements. To count elements of $S$ that are $m$th powers, we use inclusion-exclusion to count the number of $m$th powers in subfields.

To count the number of elements of $\mathbb F_{q^{d/n}}^\times$ that are $m$th powers in $\mathbb F_{q^d}^\times$, we observe that their $m$th roots are both $ m (q^{d/n}-1)$st roots of unity and $(q^d-1)$st roots of unity, hence are $\gcd ( m (q^{d/n}-1),(q^d-1))$th roots of unity, and each of them has $\gcd(q^d-1, m)$ $m$th roots in $\mathbb F_{q^d}^\times$, so the total number of them is $\frac{ \gcd ( m (q^{d/n}-1), q^{d}-1 )}{ \gcd (q^{d} - 1, m) } $.

Inclusion-exclusion gives our formula.

  • $\begingroup$ Thank you Will for the expression. I was thinking along the same lines, by doing exclusion inclusion, involving the sub field value sets, but somehow I couldn’t tackle the inclusion- exclusion business, using Mobius function. Anyway I will like to use this formula for a certain computation in my ongoing research work, and hence I ask your permission for the same, of course with acknowledgment. Thank you very much again. $\endgroup$
    – Riju
    Jul 19, 2019 at 21:32
  • $\begingroup$ @Riju you’re welcome to use it. $\endgroup$
    – Will Sawin
    Jul 20, 2019 at 6:50

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