Cyclotomic polynomial over a finite prime field [duplicate]

Solution 1:

We denote by $|S|$ the number of elements of a finite set $S$.

Let $F = \mathbb{Z}/p\mathbb{Z}$. Let $\Omega$ be the algebraic closure of $F$. Let $\omega \neq 1$ be a root of $X^l - 1$ in $\Omega$. Let $K$ be the unique subfield of $\Omega$ such that $|K| = p^f$. $K$ is the set of all the roots of $X^{p^f} - X$ in $\Omega$. $K$ is a finite extension of $F$ and $[K : F] = f$. Let $K^* = K -$ {$0$} be the multiplicative group of $K$. It is well known that $K^*$ is a cyclic group. Since $|K^*| = p^f - 1$ and $l|p^f - 1$, $K^*$ has a unique cyclic subgroup of order $l$. Hence $\omega \in K^*$.

Let $L$ be a proper subfiled of $K$. Let $[L : F] = p^r$. Suppose $\omega \in L$. Then $l|p^r - 1$, i.e. $p^r \equiv 1$ (mod $l$). Since $r < f$, this is a contradiction. Hence $K = F(\omega)$. Hence the minimal polynomial of $\omega$ over $F$ has degree $f$. This completes the proof.