Hensel Lemma and cyclotomic polynomial

Suppose that $\omega$ is a non-trivial cube root of unity in $\Bbb Z_p$. If $p\equiv2\pmod3$ then $\omega\equiv1\pmod p$. But this is impossible. For some $n$, $\omega\equiv1+rp^n\pmod{p^{n+1}}$ with $p\nmid r$. Then $\omega^3\equiv 1+3rp^n\pmod{p^{n+1}}$ and so $\omega^3\ne1$.

It is more fun to use some arithmetic to show that the polynomial $X^3 -1$ splits completely in $\mathbf F_p$ iff $p \equiv 1$ mod $3$. Oviously $X^3 - 1$ can be replace by $X^2 + X + 1$, whose discriminant is $-3$, so the problem is equivalent to the computation of the quadratic residue symbol $(\frac {-3}p)$ . The case $p=2$ is obvious. For $p$ odd, the quadratic reciprocity law implies : $(\frac {-3}p)=(\frac {-1}p)(\frac {3}p)=(-1)^{\frac {2(p-1)}4}(\frac {-1}p)(\frac p 3)$ . In the extreme RHS, the first two symbols are the same by Euler's criterion, so finally, $-3$ is a square in $\mathbf F_p$ iff $(\frac p 3)=1$, i.e. $p \equiv 1$ mod $3$ (again by Euler's criterion).