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I was wondering why do we always get an integer coefficient for each variable and constant everytime we compute $\phi_n$ ?

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Do you mean the coefficients of the cyclotomic pol. over the rationals (in fact, over the integers), or about its values? If it is the first then that's a very nice and non-trivial theorem, and if it is the second then it isn't true... –  DonAntonio Oct 4 '12 at 3:34
    

3 Answers 3

up vote 1 down vote accepted

Primitive $n$-th roots of unity are algebraic integers. Hence the coefficients of $\Phi_n$ are algebraic integers. Since they are rational numbers, they are rational integers because $\mathbb{Z}$ is integrally closed in $\mathbb{Q}$

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It follows from the following formula and induction on n.

$\prod_{d\mid n}\Phi_d(X) = X^n - 1$

Let $g(X) = \prod_{d\mid n, d < n}\Phi_d(X)$. By the induction hypothesis $g(X) ∈ \mathbb{Z}[X]$ and it is monic. Hence $\Phi_n(X) = (X^n - 1)/g(X) ∈ \mathbb{Z}[X]$

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There are several ways we can define the Cyclotomic polynomials, for example there are already three in the introduction of this wikipedia page.

Of those three, two make the integer coefficients part of the definition. For the other definition they list: $$\Phi_n(X) =\prod_\stackrel{1\le k\le n-1}{\gcd(k,n)=1}(X-e^{2i\pi\frac{k}{n}})$$

two options I can see are to either prove the formula Makoto used in his answer, or to show that the primitive roots of unity are a Galois conjugate family in the appropriate field extension, and thus the product on the right hand side is their minimal polynomial.

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