I have been shown how to find the number of monic irreducible factors of $X^{(2^8-1)}-1$ in $\mathbb{F}_{2}$ and I'm struggling to apply the method to find the number of monic irreducible factors of $X^{(5^6-1)}-1$ in $\mathbb{F}_5$.
The method for the first question is as follows:
$255 = 2^8-1$ is the order of the multiplicative cyclic group $\mathbb{F}_{256}^*$. For $\alpha \in \mathbb{F}_{256}$, all its conjugates are the roots of the same minimal polynomial over $\mathbb{F}_2$ and this minimal polynomial appears as an irreducible factor for $X^{255}-1$.
The subfields of $\mathbb{F}_{256}$ are $\mathbb{F}_{2}$, $\mathbb{F}_{4}$, $\mathbb{F}_{16}$ and $\mathbb{F}_{256}$.
There is $1$ element $\alpha \in \mathbb{F}_2^*$ is such that $\mathbb{F}_2(\alpha) = \mathbb{F}_4$, which has $1$ conjugate (itself)
There are $4-2 = 2$ elements $\alpha \in \mathbb{F}_4 - \mathbb{F}_2$ such that $\mathbb{F}_2(\alpha) = \mathbb{F}_{4}$, which have $2$ conjugates (they are conjugates of each other)
There are $16-4 = 12$ elements $\alpha \in \mathbb{F}_{16} - \mathbb{F}_4$ such that $\mathbb{F}_2(\alpha) = \mathbb{F}_{16}$, which have $4$ conjugates
There are $256-16 = 240$ elements $\alpha \in \mathbb{F}_{256} - \mathbb{F}_{16}$ such that $\mathbb{F}_2(\alpha) = \mathbb{F}_{256}$, which have $8$ conjugates
Therefore the solution is $1/1 + 2/2 + 12/4 + 240/8 = 35$ monic irreducible polynomials.
Here is my attempt on the second question:
$5^6 - 1$ is the order of the multiplicative group $\mathbb{F}_{5^6}^*$.
The subfields of $\mathbb{F}_{5^6}$ are $\mathbb{F}_{5}$, $\mathbb{F}_{5^2}$, $\mathbb{F}_{5^3}$ and $\mathbb{F}_{5^6}$.
There are $4$ elements $\alpha \in \mathbb{F}_5^*$ is such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5}$, which have $1$ conjugate (themselves)
There are $25-5 = 20$ elements $\alpha \in \mathbb{F}_{5^2} - \mathbb{F}_5$ such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5^2}$, which have $2$ conjugates
There are $125-25 = 100$ elements $\alpha \in \mathbb{F}_{5^3} - \mathbb{F}_{5^2}$ such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5^3}$, which have $3$ conjugates
There are $5^6-5^3 = 15500$ elements $\alpha \in \mathbb{F}_{5^6} - \mathbb{F}_{5^3}$ such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5^6}$, which have $6$ conjugates
This gives $4/1 + 20/2 + 100/3 + 15500/6$ polynomials.
The problem
Clearly I have misunderstood the method because $3$ does not divide $100$. What I think is likely is that I have not understood why these elements have the number of conjugates they have or that I have not understood how to properly count the number of such elements.
If anyone could offer insight into what I've done wrong, or to demonstrate how the method would work on the second question, I would be very grateful.
Thank you to Jyrki Lahtonen for help with this. My error was in the counting of the elements and this answer is my attempt to correct that. I hope I have understood it correctly.
$5^6 - 1$ is the order of the multiplicative group $\mathbb{F}_{5^6}^*$.
The subfields of $\mathbb{F}_{5^6}$ are $\mathbb{F}_{5}$, $\mathbb{F}_{5^2}$, $\mathbb{F}_{5^3}$ and $\mathbb{F}_{5^6}$.
There are $4$ elements $\alpha \in \mathbb{F}_5^*$ is such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5}$, which have $1$ conjugate (themselves)
There are $25-5 = 20$ elements $\alpha \in \mathbb{F}_{5^2} - \mathbb{F}_5$ such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5^2}$, which have $2$ conjugates
There are $125-5 = 120$ elements $\alpha \in \mathbb{F}_{5^3} - \mathbb{F}_{5}$ such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5^3}$, which have $3$ conjugates
There are $5^6-(5^3+5^2-5) = 15480$ elements $\alpha \in \mathbb{F}_{5^6} - \mathbb{F}_{5^3} - \mathbb{F}_{5^2}$ such that $\mathbb{F}_5(\alpha) = \mathbb{F}_{5^6}$ which have 6 conjugates. The number is derived from that the element needs to be in $\mathbb{F_6}$ but not in either of $\mathbb{F_2}$ or $\mathbb{F_3}$. As both $\mathbb{F}_{5^3}$ and $\mathbb{F}_{5^2}$ contains $\mathbb{F}_5$, we must remove $5$ elements as not to double count.
This gives $4/1 + 20/2 + 120/3 + 15480/6 = 2634$ polynomials.