Let $\mu(m)$ be the Möbius function on monic polynomials in $\mathbb{F}_q[x]$ ($q$ is power of prime) where $\mu(m) = 0$ if $m$ is not square-free and $\mu(m) = (-1)^k$ if $m$ is square-free and can be factored as product of $k$ irreducible monic polynomials. Consider the sum $\sum_{\deg(m) = n} \mu(m)$ over monic polynomials of degree $n$. Show the value of this sum is $0$ if $n > 1$.
I am unsure how to approach this problem but I believe it may involve considering a function field analog of the Mertens function $M(x) = \sum_{n \leq x} \mu(n)$. Any help is greatly appreciated.
We define a zeta function for $\mathbb{F}_q[x]$ in the following way: $$ \zeta_q(s)=\sum_{P: \mathrm{monic}}\frac1{(q^{\mathrm{deg}P})^s}. $$ Since the number of monic polynomials of degree $n$ equals $q^n$, we have $$ \zeta_q(s)=\sum_{n\geq 0} \frac{q^n}{q^{ns}} = \frac1{1-q^{1-s}}, \ \Re(s)>1. $$ By the factorization of monic polynomials into a product of irreducible polynomials, we have $$ \zeta_q(s)= \prod_{P: \mathrm{irreducible \ monic}} \left(1-\frac1{(q^{\mathrm{deg}P})^s}\right)^{-1}. $$ Then we have $$ \zeta_q(s)^{-1} = \prod_{P: \mathrm{irreducible \ monic}} \left(1-\frac1{(q^{\mathrm{deg}P})^s}\right) = \sum_{P: \mathrm{monic}} \frac{\mu(P)}{(q^{\mathrm{deg}P})^s}$$
$$=\sum_{n\geq 0} \frac{ \sum_{\mathrm{deg}P=n} \mu(P) }{q^{ns}}=1-q^{1-s}. $$
Comparing coefficients of Dirichlet series in the last identity, we have $$ \sum_{\mathrm{deg}P=n} \mu(P)=\begin{cases}- q &\mbox{ if } n=1,\\ 0 &\mbox{ if } n>1. \end{cases} $$