Finite field as splitting fields It is very well-know that for any finite extension $\mathbb{F}_q/\mathbb{F}_p$. Assume $q=p^n$. One can view, $\mathbb{F}_q$ as the splitting field of the polynomial $X^{p^n}-X\in\mathbb{F}_p[X]$.
What about the case where we start with $\mathbb{F}_r/\mathbb{F}_q$ where $r=q^m=p^{mn}$. Surely $\mathbb{F}_r$ can be viewed as the splitting field of $X^{q^m}-X=X^{p^{mn}}-X\in\mathbb{F}_p[X]\subseteq\mathbb{F}_q[X]$. But is it correct to view it as the splitting field of $X^{p^m}-X\in\mathbb{F}_q[X]$? I was thinking $X^{p^{mn}}-X$ has already too many roots which are already in $\mathbb{F}_q$. Then probably we just need a polynomial of degree $p^m$ as in the case when the ground field is $\mathbb{F}_p$.
 A: First of all, in the $\Bbb F_q/\Bbb F_p$ case, the polynomial of degree $p^n$ is simple, but it is by no means the smallest degree possible. Indeed, there exist irreducible polynomials over $\Bbb F_p$ of degree $n$, and the splitting field of any one of those over $\Bbb F_p$ is $\Bbb F_q$. So in the $\Bbb F_r/\Bbb F_q$ situation, a polynomial of degree $p^m$ (instead of $p^{mn}$) might be somewhat more desirable, but really there exist irreducible polynomials of degree $m$ over $\Bbb F_q$, and the splitting field of any one of those over $\Bbb F_q$ is $\Bbb F_r$.
As for your specific question, the answer is no (although you're not too far off). The problem is that $m$ and $n$ might have factors in common. In the most extreme case, when $m=n$, then $X^{p^m}-X$ already splits over $\Bbb F_q$, and so we will be far from getting $\Bbb F_r$ that way. In general, the splitting field of $X^{p^m}-X$ over $\Bbb F_q$ is $\Bbb F_{p^{\mathop{\rm lcm}[m,n]}}$. (This is the same as the splitting field of $(X^{p^m}-X)(X^{p^n}-X)$ over $\Bbb F_p$, by your first paragraph, and both $X^{p^m}-X$ and $X^{p^n}-X$ divide $X^{p^{\mathop{\rm lcm}[m,n]}}-X$.) So it works when $\gcd(m,n)=1$, but not in general.
