Irreducible polynomials in $\mathbb F_q[T]$

Question

Let $q$ be a power of a prime $p$. Is there an infinite set $S$ of $\mathbb N$ such that for every $l\in S$, the polynomial $T^{q^l}-T-1$ is irreducible in $\mathbb F_q[T]$. It looks like Artin-Schreier theory can not handle that.

Answer 1

No, the set of such natural numbers $\ell$ is always finite. In fact, it always has at most two elements, and that only in the case $q=\ell=2$. Furthermore, unless $q$ is a prime, the set $S$ is empty.

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The key to arriving at such conclusions is the simple Galois theory of finite fields, much like when studying Artin-Schreier polynomials. Let $q=p^n$, $p$ a prime number. Let us fix an algebraic closure $k$ of $\Bbb{F}_q$. Let $\alpha\in k$ be a zero of the polynomial $$ f_\ell(T):=T^{q^\ell}-T-1. $$ Let $F$ be the Frobenius automorphism $F(z)=z^q$ for all $z\in k$ that topologically generates the Galois group of $k/\Bbb{F}_q$. Galois theory tells us that the zeros of the minimal polynomial $m(T)$ of $\alpha$ over $\Bbb{F}_q$ are the iterated images $\alpha, F(\alpha), F(F(\alpha)),\ldots$. If $F^n$ stands for the $n$th iterate, we have $$F^{\ell}(\alpha)=\alpha^{q^\ell}=\alpha^{q^\ell}-f_{\ell}(\alpha)=\alpha+1.$$ By induction on $j$ this implies $F^{j\ell}(\alpha)=\alpha+j\cdot1$ for all natural numbers $j$. Because we are in characteristic $p$, it follows that $$ F^{p\ell}(\alpha)=\alpha. $$ We have just shown that the number of Galois conjugates of $\alpha$ is a factor of $p\ell$ (and also that it is not a factor of $\ell$). Therefore $$\deg m(T)\le p\ell.$$

It is easy to see that $p\ell$ is strictly less that $\deg f_\ell(T)=q^\ell$ except when

• either $q=p,\ell=1$, or
• $q=p=\ell=2$.

In all the other cases $m(T)$ is a proper factor of $f_{\ell}(T)$, and hence the latter cannot be irreducible.

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This reasoning (albeit in the insignificantly simpler context of the prime field) has appeared earlier here. The techniques I described here will say more about the factors of $f_\ell(T)$ — nominally only over the prime field and over the extension field $\Bbb{F}_{q^\ell}$, but it should not be difficult to describe the factors over the intermediate field $\Bbb{F}_q$ either.