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Let $n,m$ be coprime. I want to find a formulae for $\Phi_{n\cdot m, \mathbb Q}$. I conjecture that because $$d \mid nm \implies d \mid n \lor d \mid m,$$ that $$ \Phi_{n\cdot m, \mathbb Q} = \Phi_{n, \mathbb Q} \cdot \Phi_{m, \mathbb Q} = \prod_{d|n,~(d,n) = 1} \Phi_{d, \mathbb Q} \cdot \prod_{d|m,~(d,m) = 1} \Phi_{d,\mathbb Q}. $$ Am I right? I didn't find this formula anywhere.

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No, that's not correct. Consider the example of $n=2$, $m=3$, which satisfy $\gcd(n,m)=1$. Then $$\Phi_6=x^2-x+1\neq(x+1)(x^2+x+1)=\Phi_2\Phi_3.$$ By the way, what you've written has other problems. For instance: for any integer $n$, the only $d$ such that $\gcd(d,n)=1$ and $d\mid n$ is $d=1$. Also is not true that $$\Phi_{n} = \prod_{d|n} \Phi_{d}$$ nor is it true that $$\Phi_n=\prod_{\gcd(d,n)=1}\Phi_d.$$ Perhaps what you were thinking of was $$\Phi_n=\prod_{d\mid n}(x^d-1)^{\mu(n/d)}$$ or $$\Phi_n=\prod_{\substack{1\leq k\leq n\\\gcd(k,n)=1}}(x-e^{2\pi ik/n})$$ both of which are true.

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Think I got the right formulae now, its: $\Phi_{nm, \mathbb Q} = \frac{x^{nm} - 1}{(x^n - 1)(x^{m-1} + \ldots + x + 1)}$. –  Stefan Jun 17 '13 at 12:55

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