For a nonzero element $\alpha \in \mathbb F_{p^n}$ (the finite field of cardinality $p^n$) is there a simple criterion to tell whether $\alpha$ is a generator of the cyclic group $\mathbb F_{p^n}^\times$ by looking at the minimal polynomial of $\alpha$ over $\mathbb F_p$?

Here is what I know so far: Let $n\ge 1$ be an integer, $p$ be a prime and $\mathbb F_p = \mathbb Z/p\mathbb Z$ be the prime field of characteristic $p$. It is known that there is a unique (upto isomorphism) field $\mathbb F_{p^n}$ of cardinality $p^n$ which is the splitting field of $g (x) = x^{p^n} - x$.

If $f (x) \in \mathbb F_p [x]$ is monic irreducible of degree $n$ then $f (x)$ is separable and $\mathbb F_{p^n} \cong \mathbb F_p [t]/(f (t))$ is also the splitting field of $f (x)$. Further if $\alpha$ is any root of $f (x)$ in $\mathbb F_{p^n}$ then

  • $\mathbb F_{p^n} = \mathbb F_p (\alpha)$ (so $\alpha$ is a primitive element of the field extension $\mathbb F_{p^n}/\mathbb F_p$)
  • $m_{\alpha, \mathbb F_p} (x) = f (x)$ (the minimal polynomial of $\alpha$ over $\mathbb F_p$)
  • $f (x) = \prod\limits_{i = 0}^{n - 1} (x - \alpha^{p^i})$

Also, there are exactly

  • $M (n, p)/n$ distinct degree $n$ monic irreducible polynomials in $\mathbb F_p [x]$
  • $M (n, p) = \sum\limits_{d \mid n} p^{n/d} \mu (d)$ distinct primitive elements $\beta$ of $\mathbb F_{p^n}/\mathbb F_p$
  • $p^n - 1$ elements in the cyclic group $\mathbb F_{p^n}^\times$
  • $\phi (p^n - 1)$ generators of the cyclic group $\mathbb F_{p^n}^\times$

It is clear that each generator $\beta$ of the cyclic group $\mathbb F_{p^n}^\times$ is an element of degree $n$ over $\mathbb F_{p}$, i.e. $\deg m_{\beta, \mathbb F_p} (x) = n$, and hence is a primitive element of the field extension. It is also clear that all the other roots $\beta^{p^i}$ of the minimal polynomial $m_{\beta, \mathbb F_p} (x)$ are also generators of the cyclic group since $\gcd (p^i, p^n - 1) = 1$.

However in general the number of cyclic generators $\phi (p^n - 1)$ is much less than the number of primitive elements of the extension $M (n, p)$ (for example, for $n = 2$, $\phi (p^2 - 1)$ will be much less than $M (2, p) = p^2 - p$.

So now the question is:

  1. Characterise all the monic irreducible polynomials $f (x) \in \mathbb F_p [x]$ whose roots are cyclic generators of $\mathbb F_{p^n}^\times$.
  2. If $\beta$ is a cyclic generator of $\mathbb F_{p^n}^\times$ what are the indices $0 \le k < p^n$ for which the degree of $\beta^k$ over $\mathbb F_p$ is $n$ (or in general any $d \mid n$)?

P.S. I don't know any Galois theory, so if you use any of that, then I would appreciate if a reference is given.

  • $\begingroup$ See also math.stackexchange.com/questions/2373613/… $\endgroup$ – Chaitanya Tappu Nov 30 '18 at 2:33
  • $\begingroup$ For your information: in the context of finite fields an element is called primitive if and only if it is a generator of the multiplicative group of the extension field. See here for a more verbose explanation. Your tallies of polynomials are correct, but the terminology is off to this extent. $\endgroup$ – Jyrki Lahtonen Nov 30 '18 at 4:01

I only have a partial answer to your first question: Let $ p = 2 $ or $ p \equiv 1 \pmod 4 $. Then, if $ \alpha $ is a cyclic generator of $ \mathbb{F}_{p^n}^{\times} $ with minimal polynomial $ f(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_0 \in \mathbb{F}_{p}[X] $, the last coefficient $ a_0 $ of $ f $ must be a generator of $ \mathbb{F}_{p}^{\times} $.

Proof: The case $ p=2 $ is trivial since the constant coefficient must be non-zero, so let $ p \equiv 1 \pmod 4 $. The roots of $ f $ are $ \alpha, \alpha^p, \cdots, \alpha^{p^{n-1}} $ and hence $ a_0 = (-1)^n \alpha^{\frac{p^n-1}{p-1}} $. Let $ d $ be the order of $ a_0 $. Then we get $$ 1 = a_0^d = (-1)^{nd} \alpha^{d \cdot \frac{p^n-1}{p-1}} \implies \alpha^{d \cdot \frac{p^n-1}{p-1}} = (-1)^{nd} \implies \alpha^{2d \cdot \frac{p^n-1}{p-1}} = 1 $$ It follows that $ p^n - 1 $ divides $ 2d \left( \frac{p^n-1}{p-1} \right) $. Then $ p-1 $ divides $ 2d $ but $ 2d \le 2(p-1) $ and so $ 2d = p-1 $ or $ 2d = 2(p-1) $. If the former case holds, then substituting $ d $ in the above gives $$ 1 = a_0^{\frac{p-1}{2}} = ((-1)^{n \cdot \frac{p-1}{2}}) (\alpha^{\frac{p^n-1}{2}}) = 1 \cdot (-1) = -1 $$ a contradiction. So the latter case holds and $ d = p-1 $ as desired.

This condition is not sufficient as the following example will show: Take $ p =5 $, $ n=2 $ and let $ \alpha $ be a root of the irreducible polynomial $ f(X)=X^2 + 2 \in \mathbb{F}_5[X] $. $ f $ has constant coefficient $ a_0=2 $ which is a generator of $ \mathbb{F}_5^{\times} $ but $ \alpha^2 = 3 \implies \alpha^8 = 3^4 = 1 $, so $ \alpha $ is not a cyclic generator of $ \mathbb{F}_{25}^{\times} $.


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