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I'm trying to solve the following exercise:

Let $\mathbb{F}_q$ be a finite field with $q$ elements and characteristic $p$. Show that, for every $n > 1$ coprime with $p$, and for every $a \in \mathbb{F}_q$, the polynomial $x^{q^{n}} - x + a$ is not irreducible over $\mathbb{F}_q[x]$. What can we say (or conjecture) in the case where $p$ divides $n$?

Could anyone give me any tips? I imagine that the exercise has to do with the fact that the extension $\mathbb{F}_{q^{n}}$ over $\mathbb{F}_q$ can be seen as the spliting field of the polynomial $x^{{q}^{n}} - x$ over $\mathbb{F}_q$ but I was unable to use this information to my advantage. I also tried unsuccessfully to find some explicit factorization. Thanks for the help.

Monteiro_C
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  • Do you meant the polynomial $x^{q^n}-x+a$ instead of $x^n-x+a$? If that is the case try to have a look at the roots of the polynomial. To be more specific, show that if $\alpha$ is a root so is $\alpha+a$ for all $a\in\mathbb{F}_{q^n}$. – schiepy Nov 20 '24 at 23:03
  • This follows directly from Rabin's Irreducibility Test – masky Nov 20 '24 at 23:28
  • @schiepy thank you for help! – Monteiro_C Nov 21 '24 at 16:33
  • @ CosmicOscillator thank you for help! – Monteiro_C Nov 21 '24 at 16:33
  • @schiepy I developed an argument based on what you said and it seems to be right and prove the statement, but he was unable to explicitly factor this polynomial, do you have any idea how to do this? – Monteiro_C Nov 23 '24 at 00:03
  • It's a bit overkill for the purposes of this question, but we also have this on how to fully factor this in either $\Bbb{F}_q[x]$ or $\Bbb{F}_p[x]$. There we have $a$ in the prime field $\Bbb{F}_p$, so the answer does not work as such. The method still gives you a lot of information about the factors. – Jyrki Lahtonen Nov 23 '24 at 05:31
  • Explicitly factoring it seems hard. You can try to factor it over $\mathbb{F}_{q^n}$ with the help of what Jyrki linked or Theorem 3.80 in Lidl and Niederreiter's book "Finite Fields" and then finding arguments to go down to $\mathbb{F}_q$. However, i tried it and could only find an implicit factorization in terms of what the roots should be and in what extension they can be found. – schiepy Nov 23 '24 at 11:34
  • Actually the argument from the linked question can be recycled to cover this case as well, I think. Let $\alpha$ be a zero of that polynomial (in some extension field). Let's try and bound the number of $\Bbb{F}_q$-conjugates it has. We obviously have $$\alpha^{q^n}=\alpha-a.$$ As $a\in\Bbb{F}_q$ raising this to power $q^n$ gives $$\alpha^{q^{2n}}=\alpha^{q^n}-a^{q^n}=\alpha-2a.$$ Continuing this way, we arrive at $$\alpha^{q^{pn}}=\alpha-pa=\alpha.$$ This means that $\alpha$ belongs to the fiel of $q^{pn}$ elements. – Jyrki Lahtonen Dec 02 '24 at 18:50
  • Therefore the minimal polynomial of $\alpha$ over $\Bbb{F}_q$ has degree $pn$ at most. It is easy to see that $pn<q^n$ unless $p=n=q=2$, so that is the only possible exception. – Jyrki Lahtonen Dec 02 '24 at 18:51
  • So at this time I'm inclined to call this a duplicate of the linked thread and also this oldie. – Jyrki Lahtonen Dec 02 '24 at 18:52

1 Answers1

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Recycling the argument from an earlier answer, as explaining a minor modification in a comment is apparently now against the rules.

Let $\Omega=\overline{\Bbb{F}_q}$ be an algebraic closure, and $F:\Omega\to\Omega, z\mapsto z^q$ the Frobenius automorphism. If $\alpha\in\Omega$ is a zero of $L(x):=x^{q^n}-x+a$, then iterating $F$ shows that $$F^n(\alpha)=\alpha^{q^n}=\alpha-a.$$ As $F(a)=a$, an obvious induction shows that for any natural number $j$ we get $$F^{jn}(\alpha)=\alpha-j\cdot a.$$ Because we are in characteristic $p$, this implies that $$F^{pn}(\alpha)=\alpha.$$ By the well known Galois theory of finite fields, this means that the degree of the minimal polynomial of $\alpha$ over $\Bbb{F}_q$ is a factor of $pn$.

But, like here $pn$ is strictly less than $q^n$ except in the two cases A) $q=p, n=1$, and B) $q=p=2, n=2$. In all the other cases the minimal polynomial of $\alpha$ is a non-trivial factor of $L(x)$.

Jyrki Lahtonen
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