Suppose $\mathbb{F}_{p^k}$ is the finite field of size $p^k$, where $p$ is prime and $k$ is a positive integer. Also $m$ is a positive integer such that $p \nmid m$, with $K$ being the splitting field of the polynomial $x^{m} − 1$ over the finite field $\mathbb{F}_p$. I want to prove that $K$ is a subfield of $\mathbb{F}_{p^k}$ iff $p^{k} \equiv 1 \mod m$.
I have a hunch that it would useful to know $F = \mathbb{F}_{p^n}$ where $n$ is the smallest such integer such that $p^n \equiv 1 \mod m$, but I'm unsure of how to prove this, or how I would use it in the overall proof.
Best Answer
Let $F = \mathbb{F}_{p^k}, L = \mathbb{F}_{p^m}$. Then $K \subset L$ if and only if $F \subset L$ and $L^{\times}$ contains an element of order $m$.
$L^{\times}$ is a cyclic group of order $p^m - 1$ and thus contains an element of order $m$ if and only if $m$ divides $p^m - 1$.
Recall that there is a unique extension of $\mathbb{F}_p$ of any given degree, and if $E/\mathbb{F}_p$ is an extension of degree $n$, then for any $d$ dividing $n$ there is a subextension $E'$ of degree $d$ over $\mathbb{F}_p$. Therefore we have $F \subset L$ if and only if $k = [F : \mathbb{F}_p]$ divides $m = [L : \mathbb{F}_p]$
So we have that $K$ is a subfield of $L$ if and only if $p^m \equiv 1 \pmod{m}$ and $k \mid m$.
Note that your claimed condition implies that $F^{\times}$ contains an element of order $m$, which would imply that $K = F$.