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Suppose $\alpha, \beta \in \mathbb{F}_{q}$ and $x^q - \alpha x - \beta $ is irreducible in $\mathbb{F}_{q}[x]$ then I have to show that $\beta \ne 0, \alpha = 1$ and $q$ is prime. I clearly see the $\beta \ne 0$ implication, but I don't see the other 2. Maybe if $\alpha \ne 1$, then $x^q - \alpha x - \beta $ contains a root. For the third statement, if $q$ not prime, then $q$ is a prime power, and maybe one can derive a contradiction.

principal-ideal-domain
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user100101212
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  • On the Wikipedia page on finite fields it states that $x^q=x$ for all $x\in\mathbb{F}_q$, but I don't know a proof of this. You can use this to show that $\alpha=1$ must hold. – SmileyCraft Oct 02 '19 at 23:03
  • I think the solution is that if alpha is not 1 then $x^q - \alpha x - \beta = -(\alpha -1) x - \beta $ which has a root. – user100101212 Oct 02 '19 at 23:08
  • Indeed it has a root, but what is it? – SmileyCraft Oct 02 '19 at 23:10
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    The root is $x = -\beta (\alpha -1)^{-1}$, do you have an idea for the prime condition? – user100101212 Oct 02 '19 at 23:12
  • Not really, but it seems to me that it might be false. I think $x^q-x-1$ should always be irreducible. Am I wrong? I am going to check for $\mathbb{F}_4$. – SmileyCraft Oct 02 '19 at 23:13
  • Oh of course, $x^q-x-1$ has no roots, but it could still be the product of higher degree polynomials... That makes things a lot more interesting. – SmileyCraft Oct 02 '19 at 23:15
  • If we can show that $q$ must be prime in order for $ x^q−\alpha x−\beta$ to be irreducible, then the previous statement is true. I think this is the tough part of the problem. – user100101212 Oct 02 '19 at 23:16
  • In $\mathbb{F}_4=\mathbb{F}_2[\alpha]/(\alpha^2+\alpha+1)$ we have $(X^2+X+\alpha)(X^2+X+(\alpha+1))=X^4+X+1$. Maybe someone can generalize this somehow... – SmileyCraft Oct 02 '19 at 23:32
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    @SmileyCraft At least I can tell you why $x^q=x$. This is because the order of the unit group is $q-1$. So for every unit you have $x^{q-1}=1$, hence $x^q=x$ (for $x=0$ as well). – principal-ideal-domain Oct 02 '19 at 23:45
  • @principal-ideal-domain Right, because the multiplicative group of any finite field is cyclic! – SmileyCraft Oct 02 '19 at 23:50
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    @SmileyCraft No, you don't need cyclicality here. It's a general fact about finite groups $G$ (Lagrange's theorem) that $g^{|G|}=e$. – principal-ideal-domain Oct 03 '19 at 08:31
  • See here for ways of factoring $x^q-x-1$ and such over $\Bbb{F}_p$ when $q=p^n, n>1$. And here for when it can be irreducible. – Jyrki Lahtonen Oct 07 '19 at 06:14
  • It was a nice observation that $\beta/(1-\alpha)$ is a root when $\alpha\neq1$. It is a manifestation of the fact that the polynomial $x^q-\alpha x$, as a mapping from e.g. an algebraic closure to itself, is linear over $\Bbb{F}_q$. – Jyrki Lahtonen Oct 07 '19 at 06:19
  • The claim that $n=1$ is necessary for irreducibility then follows by applying the Frobenius to a zero $z$ of $x^q-x-\beta$. We see by induction that $F^j(z)=z+j\beta$ for all $j$. Implying that $z$ has only $p$ conjugates. So the minimal polynomial of $z$ has degree at most $p$. – Jyrki Lahtonen Oct 07 '19 at 06:22

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As already explained in the comments it's clear that $\beta\ne 0$ and $\alpha=1$ otherwise the polynomial has a zero.

I just came up with a proof of one direction:

Let $p$ be prime. We want to show that $f(X):=X^p-X-\beta$ is irreducible. Let $\Phi:\mathbb F_p[X]\to\mathbb F_p[X]$ be the automorphism sending $X$ to $X+1$. It preserves the degree of polynomials. We have $\Phi^p=\operatorname{id}$. Since $\Phi(f)=f$ the automorphism $\Phi$ permutes the factors of $f$. The orbit of each factor has either length $p$ or length $1$. Length $p$ implies that there are at least $p$ factors hence $f$ decomposes into a product of linear factors which can't be the case since $f$ has no root in $\mathbb F_p$. So every factor of $f$ is a fixpoint of $\Phi$. Being a fixpoint $g$ of $\Phi$ means that as function $g:\mathbb F_p\to \mathbb F_p$ is constant. Hence $\deg(g)=0$ or $\deg(g) \ge p$. So $f$ has only one factor, itself, and hence is irreducible.

principal-ideal-domain
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  • This is correct. And would serve in this older thread. You do need to assume that $q=p$ for this to work. Over $\Bbb{F}_q$, $q=p^n$, this generalizes to $x^p-x-\beta$ is irreducible over $\Bbb{F}_q$ if and only if $tr(\beta)\neq0$. – Jyrki Lahtonen Oct 07 '19 at 06:29