4

Now if we assume that $x^n - a \in \Bbb{F}_p[x]$, where $n$ is a positive integer and $a \in \Bbb{F}_p \setminus \{0\}$ then my questions are:

  1. For what condition on $n$, we can say that $x^n - a$ is irreducible over $\Bbb{F}_{p}$?
  2. If reducible then for what condition on $n$ we can get two polynomial $g(x), h(x) \in \Bbb{F}_p[x]$ such that $x^n - a = g(x)h(x)$, where $\gcd(g(x),h(x)) = 1$?

Observation 1: If $\gcd(n, p-1) = 1$, then $x^n - a$ has a root in $F_p$, where $a \in F_p$, because $\sigma(x) = x^n$ is an automorphism over $\Bbb{F}_p^*$. Hence $\Bbb{F}_p^n = \{x^n : x \in \Bbb{F}_p\} = \Bbb{F}_p$, hence there exists an $b \in \Bbb{F}_p$ such that $b^n = a$. Hence $x^n - a = x^n - b^n$, which is divisible by $(x - b)$, so not irreducible.

It is reducible but unsure if we can get two relatively prime factorizations of it. If we try the derivative test to find the repeated root test, then $(x^n - a)' = nx^{n-1}$. Now if $p \nmid n$, then as $b \neq 0$, so $nb^{n-1} \neq 0$, so $x -b $ is not a repeated factor of $x^n - a$, hence $\gcd((x - b), x^n - a /(x - b)) = 1.$

So finally from this observation, if $\gcd(n,p-1) = 1$ and $p \nmid n$ then $x^n - a$ is reducible and factorizes into two relatively prime factorizations over $\Bbb{F}_p[x]$.

Obswevation 2: If $p \mid n$, then let $n = pk$, for some $k \in \Bbb{N}$, then $x^n - a = (x^k)^p - a = (x^k)^p - a^p$(As $a^p = a ~\forall~ a \in \Bbb{F}_p$) $= (x^k - a)^p$.

Here we are getting that it is reducible, but the problem is can we say that we can get an answer for question 2, because if $n = p^k$, then $x^n - a = (x - a)^{p^k}$.

I found some Lemmas and a Theorem from a book: Topics in Field Theory, Chapter Name: Radical Extensions, by Gregory Karpilovsky. In $\Bbb{F}_p$, it looks like

Theorem : $x^n - a\ $ is irreducible over $\Bbb{F}_p \iff a \not\in \Bbb{F}_{p}^q,$ for all primes $q \mid n,$ and $\ a \notin -4\Bbb{F}_p^4$ when $\, 4\mid n.$

But I am thinking that can we simplify this theorem over $\Bbb{F}_p$? Like if $q = p$, then $\Bbb{F}_p^q = \Bbb{F}_p$.

In this book, I also found a lemma in the same chapter which stated:

Lemma: Let $F$ be a field, a be an element in $F$, and $m,n$ relatively prime positive integers. Then $X^{mn} - a$ is irreducible over $F \iff x^m - a$ and $x^n - a$ are irreducible over $F$.

Now applying this Lemma over $F = \Bbb{F}_p$, let $v_q(n) = s$, where $q$ is a prime divisor of $n$. Then we can write $n = q^s \cdot m$, where $\gcd(q^s,m) = 1$, then $x^{n} - a = x^{q^sm} - a$ is irreducible iff $x^{q^s} - a$ and $x^m - a$ are irreducible over $\Bbb{F}_p$.

For which values on $n, a$ we can surely say that, either $x^n - a$ will be irreducible or reducible into two relatively prime factors in $\Bbb{F}_p[x]$?

Afntu
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    Discussing the case $p\nmid n$ because otherwise $x^n-a=(x^{n/p}-a)^p$, and the question is reduced to the same one for $n/p$ instead of $n$. A point I want to make is a generalization of your first observation. If $n$ has a prime factor $q$ such that $q\nmid p-1$, then you saw that $f(x)=x^q-a$ factors, i.e. $f(x)=g(x)h(x)$ non-trivially. Consequently also $$x^n-a=f(x^{n/q})=g(x^{n/q})h(x^{n/q})$$ factors. – Jyrki Lahtonen Oct 05 '24 at 13:41
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    So the interesting case is when all the prime factors of $n$ are also factors of $p-1$. In that case the choice of $a$ usually matters. The key parameter is the multiplicative order $m$ of $a$. By Lagrange, $m\mid p-1$. Again, if $q$ is a prime that is not a factor $m$, then raising to $q$th power permutes the elements of the subgroup $\simeq C_m$ generated by $a$, and $x^q-a$ factors. At this point we see that the remaining case is that when all the prime factors of $n$ are also factors of $m$. – Jyrki Lahtonen Oct 05 '24 at 13:46
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    I think I have posted several examples about how that exact situation is resolved. The key is that under those circumstances any zero $\alpha$ of $x^n-a$ in some extension field $K$ of $\Bbb{F}p$ has multiplicative order exactly $mn$. By the basic properties of finite fields we know that $\alpha$ belongs to the field $\Bbb{F}{p^\ell}$, where $\ell$ is the smallest positive integer such that $p^\ell\equiv1\pmod{mn}$. At long last we arrive at the definitive conclusion: Under all these conditions $x^n-a$ is irreducible if and only if $\ell=n$. – Jyrki Lahtonen Oct 05 '24 at 13:51
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    Let me collect a few threads where I have used this technique. 1, 2, 3, may be also 4. Usually the question has been about the irreducibility of a more complicated polynomial when we can still figure out the multiplicative order of the roots. The mother lode. – Jyrki Lahtonen Oct 05 '24 at 14:02
  • Thanks @JyrkiLahtonen for this useful comments. This is really helpful for my problem. I will check all of the links you sheared. – Afntu Oct 05 '24 at 14:11
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    This simple case may be more illuminating. – Jyrki Lahtonen Oct 05 '24 at 14:11
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    Of course, if $p$ is uncomfortably large, finding the value of $m$ may be a tall order :-) – Jyrki Lahtonen Oct 05 '24 at 14:12

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