I've been trying to solve the following problem, but I am having troubles with the backward implication.
If $|K| = q$ and $f \in K[x]$ is irreducible, then $f$ divides $x^{q^n} – x$ if and only if $\text{deg}(f)$ divides $n$.
My idea: For the direct implication, we take $L$ to be the extension of $K$ of degree $n$. That extension is unique up to $\mathbb{Z}_p$-isomorphism, and it is precisely the splitting field of $x^{q^n} – x$. Since $f$ divides $x^{q^n} – x$, all the roots of $f$ are in $L$, which means that $K(u_1)$ is an intermediate extension of $L/K$, where $u_1$ is a root of $f$. Clearly, $[K(u_1):K] = \text{deg}f$, and the direct implication follows.
For the backward implication I am super confused because any finite extension of a finite field is the splitting field of a polynomial of the form $x^{p^n} – x$, but in this problem we are trying to prove that $f$ which is an arbitrary irreducible polynomial divides an expression like the one before. That led me to think that $f$ is generated by the multiplication of terms of the form $(x – \zeta)$ where $\zeta$ is a root of unity. Anyway, any ideas for the backward implication and my philosophical question?
Best Answer
One route. Do tell me if a step needs something you have not covered.
So we assume that $f(x)$ is irreducible over $K$, and $m=\deg f(x)$ is a factor of $n$. Then
This is probably overkill in the number of steps. Depending on how familiar you are with making deductions like these about divisibility of polynomials you may be able to do several steps at once. I tried to break it into tiny steps. Don't know if that is pedagogically optimal.