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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.2012-05-09
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    Coefficients are in fact in $\mathbb{Z} \subseteq \mathbb{Q}$.2012-05-09
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    @JyrkiLahtonen But the Moebius inversion formula instructs to divide polynomials. This explains why it is in $\mathbb{Q}$. But why are coefficients integer ?2012-05-09
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    @JyrkiLahtonen - I did not study complex analysis yet2012-05-09
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    @Sasha - this is proved in my notebook based on the assumption that this polynimial is over $\mathbb{Q}$2012-05-09
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    @Belgi, that application of the Möbius function does not involve complex analysis at all.2012-05-09
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    @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.2012-05-09
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    @JyrkiLahtonen Yes, I am convinced now :) Thanks!2012-05-09
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    @ThomasAndrews - But I do not undertand why they are in $\mathbb{Q}$!2012-05-09
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    Ah, I misread Sasha's comment as being from you, @Belgi. Sorry. I'll delete my comment.2012-05-09
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    http://math.stackexchange.com/questions/130175/cyclotomic-polynomials-in-kx-are-defined-over-the-prime-subfield-of-k/130201#1302012012-05-09

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It follows from the following formula and induction on n.

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

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

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    How does it follow ?2012-05-09
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    @Belgi Then you use Jyrki's comment on division of monic polynomials with integer coefficients (and induction)2012-05-09