Does non-zero abolute trace of an element $\alpha$ imply the irreducibility of $f(x)=x^p-x-\alpha$

Question

I am currently reading the paper "Fast Contruction of Irreducible Polynomials over Finite Fields" by Couveignes and Lercier.

On page 81, it reads, "... So $1/(1-b)$ is a root of the separable polynomial $x^p-x=a^{-1}$. This polynomial is irreducible over $\mathbb{F}_{p^k}[x]$ because the absolute trace of $a^{-1}$ is non-zero."

I don't understand why that last sentence is true. Let me define what $a$ and $b$ are such that the reader doesn't need to access the paper.

Let $\mathbb{K}=\mathbb{F}_q=\mathbb{F}_{p^w}$ and $\Omega$ be an algebraic closure of $\mathbb{K}$.

Let $l,k$ be positive integers such that $l$ divides $k$. Define the polynomial, $$T_{l,k}(x)=x+x^{p^l}+x^{p^{2l}}+\dots+x^{p^{(k/l-1)l}}.$$

For any $k$, define the set $\mathcal{A}_k\subset\Omega$ by $a\in \mathcal{A}_k$ if the following hold

1. $\mathbb{F}_p(a)=\mathbb{F}_{p^k}$.

2. $T_{1,k}(a)\neq 0$ (This is called the absolute trace of $a$.)

3. $T_{1,k}(a^{-1})\neq 0$

Finally define the rational function, $$I(x)=\frac{x^p-1}{x+x^2+\dots+x^{p-1}}.$$

$b$ is an element such that $I(b)=a$.

My question is thus, does having a non-zero trace of an element make the polynomial above (the Artin-Schreier polynomial) irreducible?

Answer 1

Let $K=\Bbb{F}_{p^k}$. Yes, the irreducibility of $f(x)=x^p-x-\alpha$ in the polynomial ring $K[x]$ is a consequence of the assumption $tr(\alpha)\neq0$.

The argument goes as follows.

Observation. The polynomial $f(x)$ has a zero in the field $K$ if and only if $tr(\alpha)=0$.

Remark. Sometimes this observation is called the additive version of Hilbert's Satz 90.

Proof. If $z\in K$ then a basic property of the trace is that $tr(z^p)=tr(z)$. So if $f(z)=0$, then $tr(\alpha)=tr(z^p)-tr(z)=0$. Conversely, the mapping $\phi(z)=z^p-z$ is an $\Bbb{F}_p$-linear mapping from $K$ to itself. Clearly $\mathrm{Ker}(\phi)$ is equal to the prime field $\Bbb{F}_p$. By rank-nullity the image of $\phi$ is thus a $(k-1)$-dimensional subspace of $K$. It follows that $$\mathrm{Im}(\phi)=\mathrm{Ker}(tr).$$ Basically because $tr(z)=tr(z^p)$ gives the inclusion "$\subseteq$", and both sets have $p^{k-1}$ elements.

The irreducibility of $f(x)$ follows from this observation. I think that most (if not all) the answers to a closely related question generalize to this case, even though that question specifies $k=1$. We can use the other observation that if $z$ is a zero of $f(x)$, the other zeros are $z+j, j\in\Bbb{F}_p$, and the arguments in the linked thread still work.