TITLE: Splitting $\Phi_{15}$ in irreducible factors over $\mathbb{F}_7$
QUESTION [6 upvotes]: I have to split $\Phi_{15}$ in irreducible factors over the field $\mathbb{F}_7$. It has been a while that I did this kind of stuff, and to be the honest, I've never really understood this matter. I'd really be grateful if someone could show me how this works. I don't ask you to do this whole job. It would be better if I would get some subtle advice instead.

I use the following definition:
$$
\Phi_n(X) \quad = \quad \prod_{ord(\zeta) = n} (X - \zeta)
$$

REPLY [2 votes]: +1 to the question and both answers. 
Koenraad, a different way of reaching the conclusion in Jack's answer is to use Galois theory. You hopefully remember that the Galois groups of finite fields are generated by the Frobenius map - in this case $F:a\mapsto a^7$. So if $\zeta$ is a fixed primitive fifteenth root of unity in whatever extension field $L$ of $K=\Bbb{F}_7$ it resides, its conjugates are
$$
\zeta^7,\quad \zeta^{49}=\zeta^{3\cdot15+4}=\zeta^4,\quad\zeta^{4\cdot7}=\zeta^{15+13}=\zeta^{13},\quad\zeta^{7\cdot13}=\zeta^{6\cdot15+1}=\zeta.
$$
So by iterating the Frobenius map we generated the listed four conjugates. Therefore the
polynomial
$$
p(x)=(x-\zeta)(x-\zeta^7)(x-\zeta^4)(x-\zeta^{13})
$$
is the minimal polynomial of $\zeta$ over $K$. The nice trick in Thomas' answer is based on the observation that we can write these conjugates in the form
$$
\begin{aligned}
\zeta&=\zeta^{10}\zeta^6\\
\zeta^7&=\zeta^{10}\zeta^{12}\\
\zeta^4&=\zeta^{10}\zeta^9\\
\zeta^{13}&=\zeta^{10}\zeta^3,
\end{aligned}
$$
where the recurring factor $\zeta^{10}$ on the right hand side is of order three (and can be identified as an element of $K$), and the other factors are exactly the roots of unity of order five.

Don't know if this helps. Ask, if you need more details.