1) Let p(x) be the 13th cyclotomic polynomial $x^{12} + x^{11} + ... +x+ 1$ and suppose p is a prime congruent to 1 modulo 13. I want to show p(x) splits completely in the finite field of p elements.
2) This time let p be congruent to 2 modulo 13, I need to show that p(x) is irreducible in the finite field of p elements
For the first question I tried to use the properties of cyclotomic polynomials i.e $x^{13} - 1 = (x-1)\times(x^{12} + ... + x + 1)$ then $p(x) = \frac{x^{13}-1}{x-1}$, if we worked in $F_{13}$ then it splits, but not sure how to go from there. On the second question I have no clue. Any hint/help is appreciated, thanks.
For the first one, let $u$ be a generator for the group of units. (This means that $u^{p - 1} = 1$ and $u^n \ne 1$ if $1 \le n < p - 1$.) Now, since $13 \mid p - 1$, the element $v = u^{(p - 1)/13}$ exists. What can you say about $v$?
For the second one, use the fact that $2$ is a primitive root mod $13$. Look at the extension $\mathbf{F}_p[\zeta_{13}] = \mathbf{F}_{p^d}$ where $\zeta_{13}$ is a primitive $13$-th root of unity and $d = (\mathbf{F_p}[\zeta_{13}] : \mathbf{F_p})$. Since $\zeta_{13}$ has order $13$, it follows that $13$ divides the order of $\mathbf{F}_{p^d}^\times$, which is $p^d - 1$. Now use the fact that $p \equiv 2 \pmod {13}$ and has order $12$.