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I want to find the no. of elements $\alpha \in \mathbb{F}_{3^5}$ so that $\mathbb{F}_{3}(\alpha) = \mathbb{F}_{3^5}$(minimal polynomial of $\alpha$ is of degree 5). I know such things do exist but how to count them?

I need to basically know how man irreducible factors of $x^{3^5}-x$ are of degree $5$ (over $\mathbb{F}_{3}$).

By ad hoc method I am getting for $\mathbb{F}_{2^4}$ the answer is $12$ since the minimal polynomial can only have irreducible polynomials whose degree divides $4$ and so there are 3 degree $4$ and $1$ degree 2(other combinations dont work due to size of $\mathbb{F}_{4}$).

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    $\begingroup$ Sorry, I think I have now answered the question. Basicall the key is note that any subextensions will have degrees which divide that of the original one. So, for $3^5$ if the only irreducible factors are of degree 1 or 5 and so one easily gets 240. $\endgroup$
    – Subham Jaiswal
    Jun 14, 2022 at 11:18
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    $\begingroup$ 240 is the correct answer to the number of of elements $\alpha$ such that $\Bbb{F}_{243}=\Bbb{F}_3(\alpha)$. Do observe that in the context of finite fields the term primitive element more commonly refers to generators of the multiplicative group (generalizing the notiion of a primitive root modulo a prime number). The answer to that question is gotten by calculating the Euler $\phi$-function $\phi(3^5-1)$. $\endgroup$
    – Jyrki Lahtonen
    Jun 14, 2022 at 11:25
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    $\begingroup$ I explain the differences between various uses of primitive in the context of extensions of fields in general and finite fields in particular here. I also share my impressions as to the origin of the difference, but I'm not a historian, so that is partly speculation. $\endgroup$
    – Jyrki Lahtonen
    Jun 14, 2022 at 11:27
  • $\begingroup$ But that's somewhat marginal. Feel free to post your argument as an answer. That way you can get some feedback on it. I do suspect that we covered this question in an earlier thread, but I don't have the time to look for one right now. $\endgroup$
    – Jyrki Lahtonen
    Jun 14, 2022 at 11:28

2 Answers 2

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Given any element $a\in\mathbb F_{p^n}\setminus \mathbb F_p$ whose degree over $\mathbb F_p$ is $d>1$, we have $\mathbb F_{p^n}$ is a vector space over $\mathbb F[a]$, hence $|\mathbb F_{p^n}|=|\mathbb F[a]|^m=p^{dm}\Rightarrow n=dm$ for some natural number $m$.

When $n$ is a prime number (such as $5$ in your case), there must be $d=n$ as $d|n$ and $d>1$. That is, there is no intermediate field between $\mathbb F_p$ and $\mathbb F_{p^n}$. In particular, the number of elements with degree $n$ is exactly $p^n-p$.

When $n$ is not a prime number, we need to use a more complicated analysis related to the factorization of $n$.

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I am posting a general answer as suggested by Jyrki Lahtonen,

The facts we need:

  1. All elements of $\mathbb{F}_{p^n}$ are roots of $x^{p^n} - x$ in some algebraic closure over $\mathbb{F}_{p}$.

  2. $\mathbb{F_{q}}$ is a subfield of $\mathbb{F}_{p^n}$ IFF $q= p^m$ for some $m$ dividing $n$ and their is a unique such subfield. One can proof the uniqueness using the fact that these are also cyclic groups(under multiplication omitting the zero).

So, the irreducible factors of $x^{p^n}-x$ will only have degree $k$ dividing $n$.

Every element leads to an extension of the possible degree mentioned and by uniqueness we get (denoting the answer to be $f(m)$ for $\mathbb{F}_{p^m}$):

$\sum _{d | n} f(d) = p^{n}$.

One can apply Mobius inversion to get a more explicit formula.

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