Prove the sum of the Mobius function over monic polynomials of degree $n$ is $0$ if $n > 1$

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.

1 Answer

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 $$\sum_{\mathrm{deg}P=n} \mu(P)=\begin{cases}- q &\mbox{ if } n=1,\\ 0 &\mbox{ if } n>1. \end{cases}$$