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Let for some prime power $q$ $\mathbb{F}_q$ be a finite field and consider $f \in \mathbb{F}_q\left[X\right]$.
I want to show the implication mentioned in the title, i.e. $$ \deg\left(\,f\right) = \min\left(\left\{d \in \mathbb{N}^\times;\; f\,\vert\, X^{q^d} - X\right\}\right) \implies f \;\textit{is irreducible} $$

I could really use some help proving this.
If you want to see my current approach (which I couldn't yet turn into a proof), read on.

I already know that for $d \in \mathbb{N}^\times$, $P_d:=X^{q^d} - X$ is the squarefree product of all irreducible polynomials $r \in \mathbb{F}_q[X]$ such that $\deg\left(r\right)\,\mid\,d$.
Another fact that might be useful is that if $r$ is an irreducible factor of $P_d$, then $\deg\left(r\right) \,\mid\, d$.

My current proof idea is as follows:
Assume that $\deg\left(f\right)$ has the minimality property mentioned above and that $f$ is not irreducible. We hence have, for $2 \leq m \in \mathbb{N}^\times$ and prime factors $f_i \in \mathbb{F}_q\left[X\right]$ $$ f = \prod_{i=1}^m f_i $$

I'm not sure if this helps, but on a side note, all the $f_i$ are coprime ($f$ is squarefree, since $P_d$ is).

Anyways, consider $\mu:=\text{lcm}\left(\left\{\deg\left(f_i\right);\; 1 \leq i \leq m\right\}\right)$.
Using one of the facts mentioned above, note that for all $i$, since $f_i$ is an irreducible factor of $f$ which divides $P_d$, we have $\deg\left(f_i\right)\,\mid\,deg\left(f\right)$. By definition of $\mu$, this implies $\mu \,\mid\, \deg\left(f\right)$.
It thus would suffice to show that $\mu \lt \deg\left(f\right)$ to arrive at a contradiction to the minimality property of $\deg\left(f\right)$.

Can somebody help me out here?

UPDATE: The answer here leads me to believe my approach might be wrong. Not sure though, just found this.

Batominovski
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polynomial_donut
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1 Answers1

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This is false. For instance, if $q=2$ and $f(x)=x(x^2+x+1)(x^3+x+1)$ then $f$ is not irreducible but the smallest $d>0$ such that $f$ divides $x^{2^d}-x$ is $6$, the degree of $f$. Indeed, the irreducible factors of $f$ have degree $1,2,$ and $3$, and so $f$ divides $x^{2^d}-x$ iff $d$ is divisible by $1,2,$ and $3$. The least such $d$ is $6$.

In general, by the same reasoning, if $f$ is squarefree then the least such $d$ will be the least common multiple of the degrees of the irreducible factors of $f$.

Eric Wofsey
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