I state the following theorem from Serge Lang's Book- Algebra(3rd edition).

Theorem: Let $k$ be a field and $n$ an integer $\geq$ 2. Let $a\in k, a\neq 0$. Assume that for all primes $p$ such that $p\mid n$, we have $a\notin k^p$, and if $4\mid n$, then $a\notin -4k^4$. Then $X^n-a$ is irreducible in $k[x]$.

So, for example: If $k=\mathbb{F}_{q}$ is the finite field with $q$ elements, and Let $p$ be a odd prime. Then by this theorem if $a\in \mathbb{F}_{q}^{*}\setminus (\mathbb{F}_{q}^{*})^{p}$, then $x^p-a$ is irreducible in $\mathbb{F}_{q}[x]$.

Now, if we take $m=p^2$, then by the same theorem, if $a\in \mathbb{F}_{q}^{*}\setminus (\mathbb{F}_{q}^{*})^{p}$, then $x^m-a$ is irreducible in $\mathbb{F}_{q}[x]$. Now, in this case $(\mathbb{F}_{q}^{*})^{m} \subseteq (\mathbb{F}_{q}^{*})^{p}$, hence if $a\in (\mathbb{F}_{q}^{*})^{p}\setminus (\mathbb{F}_{q}^{*})^{m}$, then what can be said about the polynomial $x^m-a$?

So, more generally, my question how much is known about the nature of polynomials of the form $x^n-a$, over finite fields, whether there are theorems stronger than the above one, and things like that!

Appreciate your help. Please provide some reference if possible. Thank you!

  • $\begingroup$ In the $m=p^2$ case, it looks like you have this reversed. The theorem says that if $a\in\mathbb{F}_q^*\backslash(\mathbb{F}_q^*)^p$, then $x^m-a$ is irreducible. $\endgroup$ Apr 5, 2019 at 15:27
  • $\begingroup$ sorry , I will edit that! $\endgroup$
    – Riju
    Apr 5, 2019 at 15:28
  • $\begingroup$ The theorem in Lang is nearly an "if and only if". In particular, over a field $k$, $x^p-a$ is irreducible over $k$ if and only if $a$ is not a $p$-th power in $k$. $\endgroup$
    – KConrad
    Apr 5, 2019 at 19:00
  • $\begingroup$ Is there any result if we don't take $n$ to be prime, maybe take $n$ to prime power and things like that! $\endgroup$
    – Riju
    Apr 5, 2019 at 22:30


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