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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.

Flose
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  • An error you made is that $\Bbb{F}{5^2}$ is not a subset of $\Bbb{F}{5^3}$. This is because the extension degree is not a factor of the higher degree, in other words $2\nmid 3$. Therefore $\Bbb{F}_{5^3}$ has $125-5=120$ elements with 3 conjugates (rather than $100$). – Jyrki Lahtonen May 07 '20 at 17:34
  • Also, an element of $\Bbb{F}{5^6}$ has six conjugates unless it is an element of EITHER $\Bbb{F}{5^2}$ OR $\Bbb{F}_{5^3}$. So you have to throw out $5^3+5^2-5$ elements out of the $5^6$ rather than $5^3$ which is what you tried. – Jyrki Lahtonen May 07 '20 at 17:36
  • But, +1 for sharing your thoughts in detail. Makes it super easy to spot the problem! – Jyrki Lahtonen May 07 '20 at 17:37
  • The keys are that $\Bbb{F}{p^n}$ is a subset of $\Bbb{F}{p^m}$ if and only if $n\mid m$. And, that if both $d_1\mid m$ and $d_2\mid m$, then the intersection $\Bbb{F}{p^{d_1}}\cap \Bbb{F}{p^{d_2}}$ is exactly the subfield $\Bbb{F}_{p^\ell}$ with $\ell=\gcd(d_1,d_2)$. And you must not double count the intersection. – Jyrki Lahtonen May 07 '20 at 17:41
  • With a higher $m$ (one with more prime factors) you need to use exclusion-inclusion principle to avoid over/undercounting, and end up using the Möbius function. – Jyrki Lahtonen May 07 '20 at 17:42
  • @Jyrki Lahtonen Thank you so much! That solves it. – Flose May 07 '20 at 17:45
  • For more feedback feel free to post a revised calculation as an answer! Your call, of course. This is a special case of a more general question, so I should not post one (my stance on site politics I don't want to compromise, but that need not apply to you). – Jyrki Lahtonen May 07 '20 at 17:48
  • Will do. Thanks again. – Flose May 07 '20 at 17:50
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    Does this answer your question? Monic irreducible polynomials of degree 6 in $F_{5}[X]$ – Jyrki Lahtonen Jun 02 '20 at 20:43

1 Answers1

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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.

Flose
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