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I understand the proof of the Möbius inversion formula by moving the inner summation of

$$ \sum_{d|n}\mu(n/d)\sum_{k|d}g(k) \\ $$

outside, giving:

$$ \sum_{k|n}g(k)\sum_{q|n/k}\mu(n/kq) = g(n) $$

I am trying to prove the equivalent for monic polynomials in $\mathbb{F}[x]$ for some finite field $\mathbb{F}$. In this case $F,G : \{f \in \mathbb{F}[x] : f \text{ is monic}\} \to \mathbb{C}$.

We could also have

$$ \mu(f) = \begin{cases} (-1)^r, & \text{if $f$ is square free and $f=p_1\cdots p_r$} \\ 0, & \text{otherwise} \end{cases} $$

where $p_i$ is irreducible.

I was wondering if the exact same proof can apply or whether there is more complexity to doing the proof on polynomials. I don't really want a solution, just an idea on what to look out for, if anything.

Thanks!

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    What di you mean with "for monic polynomials"? Is $g$ a multiplicative map $\Bbb N\to \Bbb F[x]$? – 2017-02-10
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    @HagenvonEitzen Ah sorry I wrote the question in a rush. $F, G : \{\text{monic } f : f \in \mathbb{F}[x]\} \to \mathbb{C}$. – 2017-02-10
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    Hm. What then should $\mu$ of a polynomial be? – 2017-02-10
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    I was thinking $\mu(f) = (-1)^r$ if $f$ is square free for $f = p_1p_2\cdots p_r$ where $p_i$ is irreducible. And $\mu(f) = 0$ otherwise. – 2017-02-10
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    In $\mathbb{Z}$, note that with $\mu(-n) = - \mu(n)$ then $\sum_{\pm d | n} \mu(d) = 2_{n=1}$. In general, you have a natural Möbius function in any [UFD](https://en.wikipedia.org/wiki/Unique_factorization_domain) with finitely many units. Otherwise, you can look only at the monic polynomials, it is a multiplicatively closed commutative monoid, with unique factorization and only one unit. Yes the proof is exactly the same (every such monoid is in bijection with $\mathbb{Z}_{\ge 1}$) – 2017-02-10

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