Does an irreducible polynomial over a finite field F divide the splitting fields polynomials for which F is a subfield?

A subfield of $F_{p^n}$ has order $p^d$ where $d\mid n$, and there is one such subfield for each $d$.

Let $q = p^n$

We have that any irreducible polynomial of degree $n$ over $F_p$ is a factor of $X^q - X \in F_p[X]$. For any positive integer $n$

Suppose, we have a subfield as in the quoted statement and an irreducible polynomial of degree $k$ over it, such that $n = dk$.

Is this polynomial also a factor of $X^q - X$ ?

• How's that "any polynomial of the form $X^q - X \in F_p[X]$"? There is only one such polynomial, and it is $X^q - X$. – Marc van Leeuwen Jun 28 '17 at 6:54
• @MarcvanLeeuwen You're right, it's changed – Donno Jun 28 '17 at 7:01

One has $$k \mid n \qquad\text{iff}\qquad p^{k} - 1 \mid p^{n} - 1 \qquad\text{iff}\qquad x^{p^{k}} - x \mid x^{p^{n}} - x.$$
Now $k$ divides $n = d k$, and your irreducible polynomial of degree $k$ divides $x^{p^{k}} - x$, and thus divides $x^{p^{n}} - x$.
Suppose $n = d k$, and let $F$ be a finite field of order $p^{d}$. Suppose the (monic) polynomial $f \in F[x]$ is irreducible in $F[x]$. Does $f$ divide $x^{p^{n}} - x$?
The answer is still yes in this formulation. If $\alpha$ is a root of $f$ in some extension of $F$, then $E = F[\alpha]$ is a finite field of order $(p^{d})^{k} = p^{n}$. If $g$ is the minimal polynomial of $\alpha$ over $F_{p}$, then $f \mid g$. Since $g$ has a root in $E$, $g$ divides $x^{p^{n}} - x$, and so does $f$.