Do you know a generalization of a formula about cyclotomic polynomial?

Question

Let n be a natural number. Then n-th cyclotomic polynomial is defined as follows: $$\Phi_{n}(x)=\prod_{k\in\mathbb{N}_{<k},(n,k)=1}(x-\zeta^k)$$ where $\mathbb{N}_{<k}$ means the set of natural number less than $k$ and $(n,k)$ means the greatest common divisor of $n$ and $k$, and $\zeta=\exp\frac{2\pi i}{n}.$

The following is well known: $$\Phi_{n}(x)=\prod_{d\mid n}(x^d-1)^{\mu(\frac{n}{d})}.$$

This can be written as

$$\Phi_{n}(x)=\prod_{d\mid n}\Phi_1(x^d)^{\mu(\frac{n}{d})}.$$

Then I want to try to generalize above formula and make formula with form:

$$\Phi_{n}(x)=\prod_{d}\Phi_{m}(x^d)^{\mu(\frac{n}{d})}.$$

Is this already given and known to some people or mathematicians? If you know anything, could you teach me?

Answer 1

Theorem. Whenever $n$ is a multiple of $m$ such that $m$ and $\frac nm$ are coprime, the following holds:

$$\Phi_{n}(x)=\prod_{d|\frac nm }\Phi_{m}(x^d)^{\mu(\frac{n}{md})}$$

Proof. do a Möbius inversion to the formula $$\Phi_m(x^n)=\prod_{d|n} \Phi_{dm}(x)\tag{1}$$

by reading it as $G(n)=\prod_{d|n} F(d)$ where $F(d)=\Phi_{dm}(x)$ and $G(n)=\Phi_m(x^n)$.

Formula $(1)$ itself can be derived by considering the roots of the polynomials on the left and on the right (which is where you need for $m$ and $\frac nm$ to be coprime).

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Edit and generalisation (removing the coprimality condition).

Let $m,n$ be any pair of integers such that $m\mid n$. Write $m=ab$ where $a$ is the highest factor of $m$ such that $\gcd(a,\frac n a)=1$. Then:

$$\large\Phi_{n}(x)=\prod_{d|a}\Phi_{\frac nm}(x^{bd})^{\mu(\frac{a}{d})}$$

Proof.

Lemma. If $n\in \Bbb N$ and $p$ is a prime such that $p^2\mid n$, then $$\Phi_n(x)=\Phi_\frac{n}{p}(x^p)$$ Proof. It is easy to see that the polynomials on both sides have the same roots.

Using the lemma repeatedly yields $$\Phi_n(x)=\Phi_\frac{n}{b}(x^b)$$

since by definition all primes that divide $b$ also divide $\frac{n}{b}$.

Now the result follows from the theorem, given that $\gcd(a,\frac{n/b}{a})=1$.