Show $\mathbb{F}_p(b)=\mathbb{F}_{p^{pk}}$ where $1/(1-b)$ is a root of $x^p-x-a^{-1}$

Question

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

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

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

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

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

It can be shown that $1/(1-b)$ is a root of the polynomial $f(x)=x^p-x-a^{-1}$ (see the paper "Fast Contruction of Irreducible Polynomials over Finite Fields" by Couveignes and Lercier page 81)

My question is why is $\mathbb{F}_p(b)=\mathbb{F}_{p^{pk}}$?

We have that $\mathbb{F}_p(a)=\mathbb{F}_{p^k}$ by the definition of $a$, and to add $1/(1-b)$, we need to take the quotient $\mathbb{F}_{p^k}[x]/(f(x))=\mathbb{F}_{p^{pk}}$, but why is this $\mathbb{F}_{p}(b)$?

Answer 1

By definition $\Bbb{F}_p(a)=\Bbb{F}_{p^k}$. If $\omega\in\Omega$ is a root of $f(x):=x^p-x-a^{-1}\in\Bbb{F}_{p^k}[x]$ then clearly $$\Bbb{F}_{p^k}\subset\Bbb{F}_p(\omega)\subset\Bbb{F}_{p^{pk}},$$ because $f(x)$ is either irreducible, or splits into linear factors. Of course $$\Bbb{F}_p(b)=\Bbb{F}_p(1/(1-b))=\Bbb{F}_p(\omega),$$ for some such $\omega$, so it suffices to show that $f$ does not have a root in $\Bbb{F}_{p^k}$. That is to say, it suffices to show that $a^{-1}\neq c^p-c$ for all $c\in\Bbb{F}_{p^k}$. But this follows immediately from the assumption that the absolute trace of $a^{-1}$ is nonzero.