# Nature of polynomials of the form $x^n-a$ over finite fields

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!

• 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. Apr 5, 2019 at 15:27
• 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 $k$. Apr 5, 2019 at 19:00
• Is there any result if we don't take $n$ to be prime, maybe take $n$ to prime power and things like that!