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I want to show that $X^n-1 = \prod_{d|n}{\Phi_d}$, and I found this proof on internet:

$X^n-1=\prod_{k=1}^{n}{(X-e^{i\frac{2k\pi}{n}})}=\prod_{d|n}{\prod_{k=1,gcd(k;n)=d}^{n}{(X-e^{i\frac{2k\pi}{n}})}}=\prod_{d|n}{\Phi_{n/d}}=\prod_{d|n}{\Phi_d}$

What I don't understand :

1) How to split the product in a double one (right hand side of the second equality) and how to justify it.

2) How to pass from $\Phi_{n/d}$ at $\Phi_d$ at the end? Is it beacause there is a bijection $D(n) -> D(n); d -> n/d$ (where $D(n) = \left\{d|n , d\ge0\right\}$)

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    2) is easily solved by examine the map $d\mapsto n/d$ is one-to-one correspondence over the set of all divisors of $n$. – 2017-02-15
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    The main point here, imo, is to realize that if a number $\;\zeta\;$ is a $\;d\,-$ th root of unity, then $\;\zeta\;$ is an $\;n\,-$ root of unity **for any multiple of** $\;d\;$ , and together with this to realize that *any* $\;n\,-$ th root of unity is a **primitive root of unity of order** $\;d\;$ for one single $\;d\,\mid\,n\;$ . – 2017-02-15

2 Answers 2

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1) We can write the set $\{1,\cdots,n\}$ as the disjoint union of the sets $\{k;\,1\le k\le n\,\mathrm{and}\,gcd(k,n)=d\}$ for all $d$ belonging to the set of (positive) divisors of $n$.

2) Yes, the bijection you mention is the main key (another reason lies in the fact that multiplication of integers is a commutative law).

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Since the $n$-th roots of unity form a cyclic group, $$ X^n-1=\prod_{ord(\omega) \mid n}{X-\omega} =\prod_{d \mid n} \prod_{ord(\omega)=d}{X-\omega} =\prod_{d \mid n} \Phi_d(X) $$ where $ord(\omega)$ is the least positive exponent $k$ such that $\omega^k=1$.