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Do we know general formula for factorisation of $X^n-a$ in $\mathbb{F}_p$ for any $a\in\mathbb{F}_p$? At least do we know the degree of the factors?

Some reflexions:

  • if $a=1$ it is the "cyclotomic factorisation" with factorisation of the cyclomic polynomial $\phi_i$ in factor of degree the order of $p$ in $\mathbb{Z}/a\mathbb{Z}$.
  • if $a=-1$ we can get the factorisation of $X^n+1$ from the factorisation of $X^{2n}-1$
  • if $a$ is an $n$-power in $\mathbb{F}_p$ we can get a factorisation with $$ X^n-a=X^n-b^n=b^n\left(\left(\frac{X}{b}\right)^n-1\right) $$
  • if $-a$ is an $n$-power in $\mathbb{F}_p$ we can do the same thing with $X^n+1$
  • if $n=p^k$ then $a$ is a $n$-power (and then Frobenius make the job)
  • I have seen in Lang Algebra (Theo 9.1) that $X^n-a$ will be irreducible $a$ is never a $q$-power for all $q|n$.

Any Idea?

Gabriel Soranzo
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    Things will be easier if you know the (multiplicative) order $d$ of $a$. The details depend on $\gcd(d,n)$. If all the prime factors of $d$ are also factors of $n$, then all the roots of $X^n-a$ will have order $nd$. Implying that the factors will all have the same degree $\ell$ that is the order of $p$ modulo $nd$ (assume $p\nmid n$). In other words $p^\ell\equiv1\pmod{nd}$. Otherwise the orders of the zeros will vary, and it is a bit more complicated. – Jyrki Lahtonen Mar 27 '22 at 20:36
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    See 1, 2, 3,4,5, 6. – Jyrki Lahtonen Mar 27 '22 at 20:47
  • Ok, I will study your referencies! – Gabriel Soranzo Mar 27 '22 at 20:53
  • Have fun! I'm not sure all the aspects have been covered in those. Definitely some overlap there. I have simply used the ideas a few times here. – Jyrki Lahtonen Mar 27 '22 at 21:01
  • Ok the key fact is that $a$ is a root of unity so that if the order of $a$ is $d$, $X^n-a$ will divide $X^{nd}-1$ (as $X^n+1$ divides $X^{2n}-1$). The factors are so of type $\phi_i$ with $i|nd$. The question is what $\phi_i$ exactly? – Gabriel Soranzo Mar 28 '22 at 15:35
  • In the case of $X^n+1$ it was clear: it was the $\phi_i$ with $i|2n$ and not divide $n$... – Gabriel Soranzo Mar 28 '22 at 15:40
  • The cyclotomic polynomials factor modulo $p$ (undoubtedly you knew about this). Sorry to blow my own trumpet, but here is a description of what happens to the cyclotomics modulo $p$. Are you familiar with Galois theory of finite fields? Once you know the order of a root of unity, it is straight forward to enumerate the zeros of its minimal polynomial over the prime field. – Jyrki Lahtonen Mar 28 '22 at 16:45
  • I understand now why all the roots have order $nd$ if $d$ and $n$ have same prime factors (recurrence from your fact 2 of your reference 2). Unfortunately I don't think that I can control the order of $a$... my $a$ is $\left(\frac{l^{q_1-1}}{r^{q_1-1}q_1}\right)^{rq_2}$ and $n=q_1q_2$ (and $q_2$ can be 1). – Gabriel Soranzo Mar 28 '22 at 19:00

2 Answers2

1

I dont' consider it really as a complet answer but it's a begin. Let take an example: $X^6-3$ modulo 5.

We have $3^4=1$ so that if $\alpha$ is a root of $X^6-3$ ie $\alpha^6=6$ then $\alpha^{24}=1$ i.e. $\alpha$ is a root of $X^{24}-1$ and hence $X^6-3$ divides $X^{24}-1$. As a result $X^6-3$ is a product of cycltomic polynomial $\phi_{i,j}$ for $i|24$ (the index $j$ is because the $\phi_i$ are not irreducible in $\mathbb{F}_p$)

We can formalise that by denoting $E_{5,6,3}$ the subset of divisor of $24=6\times\text{ord}_5(3)$ (weighted with the number of subdivisor of $\phi_i$ for each $i$) with the property that $$ X^6-3=\prod\limits_{i\in E_{24,3},j} \phi_{i,j}(X) $$

So the question is: how to find $E_{5,6,3}$?

One way to proceed, which do not give direct (I mean explicit) answer:

\begin{align*} X^{24}-1 &= (X^6)^4-4^4\\ &= 3^4\left[ \left( \frac{X^6}{3} \right)^4-1 \right]\\ &= 3^4\prod\limits_{i|4}\phi_i\left( \frac{X^6}{3} \right)\\ &= \prod\limits_{i|4}3^{\varphi(i)}\phi_i\left( \frac{X^6}{3} \right)\\ &= (X^6-3)(X^6+3)(X^{12}+9) \end{align*}

As a result, $$ X^6-3=\frac{X^{24}-1}{(X^6+3)(X^{12}+9)} $$

And as we know the irreducible factorisation of $X^{24}-1$ it should be not to difficult (here is maybe a problem) to get this one of $X^6-3$:

Here $X^{12}+9=X^{12}-1$ so that

$$ X^6-3=\frac{X^{12}+1}{X^6+3} $$ But $X^{12}+1=\phi_{24}\phi_8$, these two factor decompose in $\mathbb{F}_5$ in factors of degree 2 because 2 is the order of 5 in $\mathbb{Z}/24\mathbb{Z}$ and $\mathbb{Z}/8\mathbb{Z}$ so that $$ X^6-3=\frac{\phi_{24,1}\phi_{24,2}\phi_{24,3}\phi_{24,4}\phi_{8,1}\phi_{8,2}}{X^6+3} $$ so that for "symetric reasons" (hum...) $$ X^6-3=\phi_{24,1}\phi_{24,2}\phi_{8,1} $$

So here $E_{5,6,3}=\{24_2,8_1\}$.

As I said (if the above is correct) it is a little "craft" and it doesn't give directly $E_{p,n,a}$.

Gabriel Soranzo
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0

I think I answered to my question in the two following personal paper: paper 1 and paper 2.

Gabriel Soranzo
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