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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 $nge 1$ be an integer, $p$ be a prime and $mathbb F_p = mathbb Z/pmathbb Z$ 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 $mathbb F_p$)


• $f (x) = prodlimits_{i = 0}^{n - 1} (x - alpha^{p^i})$

Also, there are exactly

• 
  $M (n, p)/n$ distinct degree $n$ monic irreducible polynomials in $mathbb F_p [x]$
  


• 
  $M (n, p) = sumlimits_{d mid n} p^{n/d} mu (d)$ 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.

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} $.