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I was trying to factor $x^{24}-1$ in $\mathbb{F}_7[x]$, however I ran into some trouble while trying to find certain coefficients of the irreducible polynomials.

My effort:

I computed the cyclotomic cosets modulo $24$ when $q=7$, and got the list below:

Cyclotomic cosets modulo 24

Thus, $\{0,1,2,3,4,5,6,8,9,10,12,13,16,17,20\}$ is a list of representatives. Since $24 |7^2-1$, we take a primitive element $\alpha$ of $\mathbb{F}_{49}$, and the factorization will be of the form

\begin{equation*} x^n-1=\prod_{i=1}^{15} M^{\left(2 s_i \right)}(x), \end{equation*}

where $s_i$ is the $i$-th representative and $M$ is the respective minimal polynomial of $\alpha^{2s_i}$ with respect to $\mathbb{F}_7$. For example, \begin{equation*} M^{(0)}(x)=x-\alpha^0=x-1. \end{equation*}

and

\begin{equation*} M^{(2)}(x)=(x-\alpha^2)(x-\alpha^{14})=x-(\alpha^2+\alpha^{14})+\alpha^{16}. \end{equation*}

How would I then be able to simplify $\alpha^2+\alpha^{14}$ and $\alpha^{16}$? No need for a full answer (I'd like to solve it myself), only an idea on how to proceed.

ImHackingXD
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    Obviously, the value of $\alpha^2+\alpha^{14}$ depends on which particular $\alpha$ you choose. The only thing independent of $\alpha$ will be the product of all the $M^{(k)}$ of same degree. – Ewan Delanoy Nov 25 '22 at 14:56
  • See here and the other threads linked in there. No exact duplicates, I think, but the methodology becomes a bit boring after a few rounds :-) – Jyrki Lahtonen Nov 25 '22 at 14:58

1 Answers1

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To find an element $\alpha$ we first need to make the field $\Bbb{F}_{49}$ a bit more explicit. A simple way of doing that is to observe that as $7\equiv3\pmod4$, $-1$ is a quadratic non-residue modulo $7$. Therefore we can use a zero of $x^2+1$ as an element to be adjoined. Let's call that element $i$, so $$ \Bbb{F}_{49}=\Bbb{F}_7(i)\simeq\Bbb{F}_7[x]/\langle x^2+1\rangle. $$

There are then many ways of finding the required root of unity. We observe that $\omega=2$ can assume the duties of a third root of unity as $2^3=8\equiv1$. From complex numbers we recall that $(1+i)/\sqrt2$ is an eighth root of unity. Here $3^2=9\equiv2$, so $3$ can play the role of $\sqrt2$, and we arrive at $$ u=\frac{1+i}{\sqrt2}=\frac{1+i}3=5+5i $$ as an eighth root of unity. As $3$ and $8$ are coprime, the product $$ \alpha:=\omega u=10+10i=3+3i $$ has order $24$.

Raising to the seventh power is a field automorphism $F$ of $\Bbb{F}_7$. The automorphism can only match "the complex conjugation". In other words, for all $a,b\in\Bbb{F}_7$ we have $$(a+bi)^7=F(a+bi)=a-bi.$$

This tells us that the first quadratic factor $$ \begin{aligned} f_1(x):=(x-\alpha)(x-\alpha^7)&=(x-[3+3i])(x-[3-3i])\\ &=([x-3]-3i)([x-3]+3i)\\ &=(x-3)^2-(3i)^2\\ &=x^2-6x+9+9\\ &=x^2+x+4. \end{aligned} $$ The other factors can be found like this. A few short cuts are available. We can calculate that $\alpha^2=9+18i-9=18i=-3i$ and hence $\alpha^4=(-3i)^2=-9=5$. This is a root of unity of order six (= a primitive root modulo seven), so $\alpha^5=5\alpha=\alpha/3$. Therefore $\alpha^5$ is a root of $$ f_5(x):=3^{-2}f_1(3x)=\frac12(2x^2+3x+4)=x^2+5x+2. $$ We can apply this repeatedly. Leaving it to you to do the rest.

It is worth noting that we can factor polynomials over finite fields in other ways as well. This method does have the advantage that while we produce the factors we also list their zeros. That may be useful when doing more arithmetic in $\Bbb{F}_7(i)$.

The other observation I want to make is that the choice of a primitive root of unity of order $24$ simply permutes the factors of $x^{12}+1$.

Jyrki Lahtonen
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