Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $n \in \mathbb{N}^{\ast}$ and $\Phi_{n}(X)$ be the $n$-th cyclotomic polynomial defined by :

$$ \Phi_{n}(X) = \prod \limits_{\substack{1 \leq k \leq n-1 \\ \gcd(k,n)=1}} \Big( X - \exp \big( \frac{2ik\pi}{n} \big) \Big) $$

I know that, for all $n \in \mathbb{N}^{\ast}$,

$$ X^{n}-1 = \prod \limits_{d \vert n} \Phi_{d}(X) $$

where $d$ runs over all positive divisors of $n$. Let $n=p_{1}^{\alpha_{1}}\ldots p_{k}^{\alpha_{k}}$ where $p_{1},\ldots,p_{k}$ are prime numbers and $\alpha_{1},\ldots,\alpha_{k}$ are positive integers. I read in a book the following statement, which I do not understand :

$$ \prod \limits_{\substack{d \vert n \\ 1 < d < n}} \Phi_{d}(1) = \prod \limits_{j=1}^{k} \prod \limits_{t=1}^{\alpha_{j}} \Phi_{p_{j}^{t}} (1) \tag{$\star$} $$

I do not understand this equality because I think that $n$ has more strict divisors than $p_{j}^{t}$ where $1 \leq t \leq \alpha_{j}$. A strict divisor of $n$ would be of the form :

$$ p_{1}^{t_{1}} \ldots p_{k}^{t_{k}} $$

where $(t_{1},\ldots,t_{k}) \notin \left\{ (0,\ldots,0) , (\alpha_{1},\ldots,\alpha_{k}) \right\}$. What am I missing ?

Edit : I found the answer to my question ! The formula $(\star)$ is incomplete. In fact, we have :

$$ \prod \limits_{\substack{d \vert n \\ 1 < d < n}} \Phi_{d}(1) = \prod \limits_{d \in E} \Phi_{d}(1) \prod \limits_{j=1}^{k} \prod \limits_{t=1}^{\alpha_{j}} \Phi_{p_{j}^{t}} (1) $$

where $E$ denotes the set of the divisors $d$ of $n$ which satisfy $1 < d < n$ and are not a power of a prime number.

share|cite|improve this question
This question answers itself. – Greg Martin Nov 15 '13 at 22:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.