$15$-th Cyclotomic polynomial over $\mathbb{Q}$ and $\mathbb{F}_p$ Recall that the $n$-th cyclotomic polynomial $\Phi_{n,K}(x)$ associated with a field $K$ (with characteristic coprime with $n$) is $\prod_{1\leq j\leq n, (j,n) = 1}(x - \omega^j)$ where $\omega$ is a generator of the group $\mu_n(L) = \{\alpha \in L: \alpha^n = 1\}$, with $L$ being a splitting field of $x^n - 1 \in K[x]$. Treat that $\omega$ always exists and that $\Phi_{n,K} \in K[x]$, without proof.
Find all prime numbers $p \neq 3,5$ such that $\Phi_{15,\mathbb{Q}}(x) \pmod p = \Phi_{15,\mathbb{F}_p}(x)$.
I know the Galois structure in $\mathbb{Q}$ for $n = 15$ (the group of units of $\mathbb{Z}/15\mathbb{Z}$ is $C_2 \times C_4$, generated by $2$ and $11$ for example) and also an explicit form for the polynomial in this case, but have no idea if any of this is useful.
I can also independently show that the $\mathbb{Q}$ polynomial is reducible $\pmod p$ for every $p \neq 3,5$.
Any help appreciated!
 A: Try this. Since $\deg(\Phi_{15,\Bbb Q})=8$, you’re looking for $p$ such that $\deg(\Phi_{15,\Bbb F_p})=8$ as well. Now, for the latter degree to be eight, it means that adjoining a fifteenth root of unity to $\Bbb F_p$ would give you an $8$-degree field extension. That means exactly that $15|(p^8-1)$ and (since the biggest proper divisor of $8$ is $4$) at the same time that $15$ does not divide $p^4-1$. In other words, we are seeking $p$ such that $p^8\equiv1$ but $p^4\not\equiv1\pmod{15}$.
But we’ve got a problem here: as you know, every number $m$ prime to $15$ has $m^4\equiv1\pmod{15}$.
Thus it would seem that there are no such $p$.

EDIT:
I finally realize that we were communicating at cross purposes. You want a polynomial that catches all and only the primitive $15$-th roots of unity; I was wanting a minimal polynomial for such a root of unity.
But then the answer to your question is simple: irrespective of characteristic (avoiding $p=3,5$), the polynomial you’re looking for is always
$$
\frac{(X^{15}-1)(X-1)}{(X^5-1)(X^3-1)}\quad=\quad X^8-X^7+X^5-X^4+X^3-X+1\,.
$$
