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Below is a proof that the cyclotomic polynomial $\Phi_n(x)=\prod_{d|n}(x^d-1)^{\mu(n/d)}$ using Möbius inversion. However, it requires that we take the log of a polynomial, which (to my knowledge) is not necessarily well defined. Is there any way to fix this, or rigorously define a logarithm of a polynomial?

We have $x^n-1=\displaystyle\prod_{d|n}\Phi_d(x)$. Taking the log of both sides yields: $$\log\left(\prod_{d|n}\Phi_d(x)\right)=\sum_{d|n}\log(\Phi_d(x))=\log(x^n-1).$$ By Möbius inversion we have \begin{align*}\log(\Phi_n(x)) &= \log(x^n-1)*\mu \\ &=\sum_{d|n}\log(x^d-1)\mu\left(\frac{n}{d}\right)\\ &=\sum_{d|n}\log\left[(x^d-1)^{\mu(n/d)}\right] \\ &=\log\left(\prod_{d|n}(x^d-1)^{\mu(n/d)}\right), \end{align*} and exponentiating both sides gives the desired result.

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no need to take log -Moebius inversion works also multiplicatively. It works with any Abelian group, so it doesn't matter whether the operation is denoted $+$ or $\times$. –  user8268 Oct 10 '13 at 6:36

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up vote 4 down vote accepted

You can think of the logarithm as taking the pointwise logarithm, but restricted to values of $x$ such that $\Phi_d(x) > 0$ for all $d$. There are infinitely many of these, and verifying a polynomial identity for infinitely many values of $x$ verifies it identically.

By writing the identity as an identity about $1 - x^n$ you can also consider the logarithm as a formal power series.

But as user8268 says in the comments something more fundamental is going on: the proof of Möbius inversion works multiplicatively as well as additively.

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