The following statement seems to be quite easy but somehow I can't manage to prove it:
Let $n = l_1^{v_1}...l_n^{v_n}$ be a prime decomposition of the natural number n. Let $\xi$ be a primitive n-th root of unity. Then $\xi_i = \xi^{n/{l_i^{v_i}}}$ is a primite $l_i^{v_i}$-th root of unity and one has $\mathbb{Q}(\xi) = \mathbb{Q}(\xi_1)...\mathbb{Q}(\xi_n)$ and (the part I don't understand):
$$ \mathbb{Q}(\xi_1)...\mathbb{Q}(\xi_{i-1}) \cap \mathbb{Q}(\xi_i) = \mathbb{Q} $$
To me it it kind of makes sense that this holds, because any product involving only powers of $l_1,...l_{i-1}$ is coprime to $l_i$ but I can't manage to prove it nicely.
Not only $l_1,...,l{i-1}$ is coprime to $l_i$, but it happens that you can not construct a chord of a polygon $l_i$ from the other examples. Since the cyclotomic set $Q(i)$ corresponds to the span of chords of {i}, and this by the root $\exp \pi/l$.
The trick here is that any of the roots of $Q(l_i)$ will solve the equation $x^{l_i}+1=0$. This has several factors including $x^1+1=0$ (which leads to $Q$), but nothing in the set of numbers coprime to $l_1$ can solve $x^{l_1}+1=0$.
The solution here is not dissimilar to that of $10^{l_i}-1$ shares with ${10^n}-1$, only the common factor $10-1$. It is because both cyclotomic numbers and bases call down to a common structure, and the proof lies therein.