# Why is the cyclotomic polynomial over $\mathbb{Q}$?

As defined in Wikipedia (and this is the same definition I was given in class), it is not clear to me why the cyclotomic polynomial is over $\mathbb{Q}$.

It is over $\mathbb{C}$, but I don't see a reason for the coefficient to be in $\mathbb{Q}$.

Can anyone help with this one?

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It follows from for example the Möbius inversion formula on that Wikipedia page. – Jyrki Lahtonen May 9 '12 at 20:25
Coefficients are in fact in $\mathbb{Z} \subseteq \mathbb{Q}$. – Sasha May 9 '12 at 20:25
@Belgi, that application of the Möbius function does not involve complex analysis at all. – Jyrki Lahtonen May 9 '12 at 20:30
@Sasha, when you divide two monic polynomials with integer coefficients, the result will also have integer coefficients. Think about what happens when you do long division. – Jyrki Lahtonen May 9 '12 at 20:32

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