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!