# Counting primitive elements in a finite field extension

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 $$\mathbb{F}_{4}$$).

• 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. Jun 14, 2022 at 11:18
• 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 $\phi(3^5-1)$. Jun 14, 2022 at 11:25
• 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. Jun 14, 2022 at 11:27
• 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. Jun 14, 2022 at 11:28

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 $$n$$.

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.