We want to prove that
$$ \Phi_n(x) = \prod_{d|n} \left( x^{\frac{n}{d}} – 1 \right)^{\mu(d)} $$
where $\Phi_n(x)$ in the n-th cyclotomic polynomial and $\mu(d)$ is the Möbius function defined on the natural numbers.
We were instructed to do it by the following stages:
Using induction we assume that the formula is true for $n$ and we want to prove it for $m = n p^k$ where $p$ is a prime number such that $p\not{|}n$.
a) Prove that $$\prod_{\xi \in C_{p^k}}\xi = (-1)^{\phi(p^k)} $$ where $C_{p^k}$ is the set of all primitive $p^k$-th roots of unity, and $\phi$ is the Euler function. I proved that.
b) Using the induction hypothesis show that
$$ \Phi_m(x) = (-1)^{\phi(p^k)} \prod_{d|n} \left[ \prod_{\xi \in C_{p^k}} \left( (\xi^{-1}x)^{\frac{n}{d}} – 1 \right) \right]^{\mu(d)} $$
c) Show that
$$ \prod_{\xi \in C_{p^k}} \left( (\xi^{-1}x)^{\frac{n}{d}} – 1 \right) = (-1)^{\phi(p^k)} \frac{x^{\frac{m}{d}}-1}{x^{\frac{m}{pd}} – 1} $$
d) Use these results to prove the formula by substituting c) into b).
I am stuck in b) and c).
In b) I tried to use the recursion formula $$ x^m – 1 = \prod_{d|m}\Phi_d(x) $$ and
$$ \Phi_m(x) = \frac{x^m-1}{ \prod_{\stackrel{d|m}{d<m}} \Phi_d(x)} . $$
In c) I tried expanding the product by Newton's binom using $\phi(p^k) = p^k ( 1 – 1/p)$. I also tried replacing the product by $\xi \mapsto [ \exp(i2\pi / p^k) ]^j$ and let $j$ run on numbers that don't divide $p^k$. In both way I got stuck.
I would appreciate help here.
Best Answer
For b), you have to prove this by saying that $\Phi_n(x) = \prod_{\xi \in C_n} (x - \xi)$, so you have to relate $C_m$ with $C_n$ and $C_{p^k}$. Can you find a relation between $\phi(m)$, $\phi(n)$, and $\phi(p^k)$ ? If so it should give you an idea for a way to describe $C_m$ in terms of $C_n$ and $C_{p^k}$.
For c), maybe you can compute $\Phi_{p^k}$, and use that result to show that both expressions are equal to $(-1)^{\phi(p^k)} \Phi_{p^k}(x^{n/d})$