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Define the mth cyclotomic polynomial $\displaystyle \Phi_m = \prod_{i \in U(m)} (X - \xi^i)$, where $ \xi$ is a primitive mth root of unity.

Why is true that $X^m - 1 = \displaystyle \prod_{d|m} \Phi_d$?

I can see that $X^m -1 = \displaystyle \prod_{i \in \mathbb Z / m \mathbb Z} (X - \xi^i)$ where $\xi$ is a primitive mth root of unity, but I can't see why this is equal to $ \displaystyle \prod_{d|m} \Phi_d$. The degrees of these two polynomials are the same, but that's about all I'm managing to notice...

Any help appreciated. Thanks

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up vote 2 down vote accepted

Hint: Think that clearly $x^m-1$ can be factored into linear factors with the $n^{\text{th}}$ roots of unity as cosntants. But, the $n^{\text{th}}$ roots of unity form a cyclic group of order $n$. Now, let me ask you, is it true that $\mathbb{Z}_n$ is the union over $d\mid n$ of $O_d$ where $O_d$ is the set of elements of order $d$ in $\mathbb{Z}_n$?

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