Characterising the irreducible polynomials in positive characteristic whose roots generate the (cyclic) group of units of the splitting field

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.

• – Chaitanya Tappu Nov 30 '18 at 2:33
• 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. – 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 $$\mathbb{F}_{25}^{\times}$$.