Let $p$ be a prime that does not divide $m$ and let $\omega$ be a primitive $m$th root of unity over$\mathbb{F}_p$. Show that $[\Bbb{F}_p(\omega):\Bbb{F}_p]=k$ where $k$ is the order of $\bar{p}=p+m\Bbb{Z}$ in the group $U_m=(\Bbb{Z}/m\Bbb{Z})^\times$. Also, show that the $m$th cyclotomic polynomial $\Phi_m(x)\in\Bbb{Z}[x]$ is irreducible over $\Bbb{F}_n$ if and only if $U_m$ is a cyclic group generated by $\bar{p}$
I'm very lost in this one and don't even know how to start. Any help is appreciated.
 A: The Frobenius automorphism is the main tool in Galois theory of finite fields. We know that any finite extension F$/\mathbf F_p$ is cyclic, with Galois group generated by $\phi:x\to x^p$, hence the degree of the extension equals the order of $\phi$, i.e. the smallest integer $k$ s.t. $\phi^k=Id$. Here F$=\mathbf F_p (\omega)$, where $\omega$ is a primitive $m$-th root of $1$ (in an algebraic closure of $\mathbf F_p$), so $\phi^k=Id$ iff $\omega^{p^k}=\omega$. Since $\omega$ is an $m$-th root of $1$, in the latter condition the exponent $p^k$ can be viewed as the class ${\bar p}^k \in \mathbf Z/m$, and it remains to show that $\omega^{p^k}=\omega$ iff ${\bar p^k}=\bar 1$. But because $\omega$ is a primitive $m$-th root of $1$, $\omega^{p^k}=\omega$ iff $m\mid (p^k -1)$ q.e.d.
Now consider the $m$-th cyclotomic polynomial $\Phi_m(X)\in \mathbf Q[X]$, of degree equal to the order $\phi(m)$ of $(\mathbf Z/m)^*$. Look  at the proof that any primitive $m$-th root of $1$ (in an algebraic closure of $\mathbf Q$) is a root of $\Phi_m(X)$, and you'll see that this is actually a universal property, i.e. it's valid for any field of characteristic not dividing $m$ (this is explicitely stressed e.g. in Lang's "Algebra", chap. VIII,§3 in the proof of the corollary to theorem 6). In particular, here, $\omega$ is a root of $\Phi_m(X)\in \mathbf F_p[X]$, and $\Phi_m(X)$ is irreducible over $\mathbf F_p$ iff the degree $k=[\mathbf F_p(\omega):\mathbf F_p]$ is equal to  $\phi_m(X)$. But we have seen previously that $k$ is the order of the class $\bar p$ in $(\mathbf Z/m)^*$. Equivalently, $\Phi_m(X)$ is irreducible iff $(\mathbf Z/m)^*$ is generated by $\bar p$ ./.
