Let $f$ be irreducible over $\mathbb F_{p^k}$. I want to show that $f$ is irreducible over $\mathbb F_{p^{km}}$ if, and only if, $\gcd(\deg f,m)=1.$
I've been racking my brain on this for some time now, but can't get anything started. Any pushes in the right direction would be appreciated.
Let $d$ be the degree of $f$ and $q=p^k$. Let $\alpha$ be a zero of $f$ is the algebraic closure of $\Bbb F_q$. Then $\Bbb F_q(\alpha)=\Bbb F_{q^{d}}$. Now $f$ is irreducible over $\Bbb F_{q^m}$ iff $\Bbb F_{q^m}(\alpha)=\Bbb F_{q^{md}}$. But $\Bbb F_{q^m}(\alpha)$ is the compositum of the extensions $\Bbb F_{q^d}$ and $\Bbb F_{q^m}$ of $\Bbb F_q$. As $\Bbb F_q$ has precisely one extension of a given degree, then the compositum of $\Bbb F_{q^d}$ and $\Bbb F_{q^m}$ is $\Bbb F_{q^{ \text{lcm}(d,m)}}$. So for $f$ to be irreducible over $\Bbb F_{q^m}$ it's necessary and sufficient that $\text{lcm}(d,m)=dm$, equivalently $\gcd(d,m)=1$.