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.
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 $\bar p$ ./.