# Transforming numbers of irreducible polynomials

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials of degree $n$ in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?

• I think you mean $M(n)(q)$ to be the number of irreducible polynomials of degree $n$ in $\mathbb F_q[X]$. Oct 22 '16 at 10:30
• The sum is actually $-q$ for $n=1$. Oct 22 '16 at 10:46
• Thanks for the comments - my original question has been corrected accordingly. Oct 22 '16 at 11:14

Yes, it is true. Your expression is the coefficient of $x^n$ in the following product: $$\prod_{P\text{ monic irreducible}} (1-x^{\deg P}) = \prod_{n} (1-x^n)^{M(n)}.$$ The Zeta function of $\mathbb{F}_q[X]$ is $$\sum_{f \text{ monic}} x^{\deg f} = \sum_{n \ge 0} q^n x^n = \frac{1}{1-qx}.$$ The Euler product identity tells us that $$\frac{1}{1-qx} = \prod_{P\text{ monic irreducible}} (1-x^{\deg P})^{-1} =\prod_{n} (1-x^n)^{-M(n)}.$$ Taking its reciprocal, we find that $$1-qx = \prod_{P\text{ monic irreducible}} (1-x^{\deg P}).$$ Now it is just a matter of comparing coefficients on both sides.
Interpretation: Let $\mu: \mathbb{F}_q[X] \to \mathbb{C}$ be the polynomial Möbius function, defined by $$\mu(f) = \begin{cases} 0 & f \text{ not squarefree} \\ (-1)^k & f=p_1 \cdots p_k (p_i \text{ distinct irreducibles}) \end{cases}.$$ The term $\prod_{i=1}^{s} \binom{M(n_i)}{e_i}$ counts the number of monic, squarefree polynomials of degree $n$ whose factorization consists of $e_i$ irreducibles of degree $n_i$. The polynomial Möbius function $\mu(\bullet)$ assumes the value $(-1)^{\sum e_i}$ for each such polynomial. In other words, your sum may be rewritten as $$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$ Since $\mu(f)=0$ for $f$ which is not squarefree, your claim is the same as $$n>1 \implies \sum_{f \text{ monic of degree }n} \mu(f) = 0.$$ This is classical, and due to L. Carlitz, whose proof was exactly as above. It should be compared with the following equivalent formulation of the Riemann Hypothesis: $$\sum_{n \le x } \mu(n) = O(x^{\frac{1}{2}+\varepsilon}),$$ where this time $\mu$ is the integer Möbius function.
It is interesting to ask whether $\sum_{f\text { monic of degree }n} \mu(f)$ may be evaluated without the use of formal power series and the Euler product identity. I will present such a way. Let $M_q$ denote the set of monics in $\mathbb{F}_q[X]$. Given two functions $\alpha, \beta : M_q \to \mathbb{C}$, one defines their "Dirichlet convolution" as the following function: $$(\alpha*\beta) (f) = \sum_{d_1 \cdot d_2 = f} \alpha(d_1) \beta(d_2).$$ Let $\zeta$ denote the constant function $1$ on $M_q$. Let $\delta_{1}$ denote the indicator function of the constant polynomial $1$. The "defining" property of $\mu$ is $$\mu * \zeta = \delta_1.$$ By summing the above over all monic polynomials of degree $n$ (>0) and changing the order of summation, we get that $$\sum_{\deg d \le n} \mu(d) q^{n-\deg d} = 0,$$ or equivalently $$\sum_{\deg d \le n} \frac{\mu(d)}{q^{\deg d}} =0.$$ Plugging $n=m,m+1$ in the above and subtracting, we get that $$m>0 \implies \sum_{\deg d = m+1} \mu(d) = 0.$$
Here is a reformulation of Ofir Gorodetsky's excellent answer to my question: Let $(a(n))_{n \geq 1}$ be a sequence of rational numbers. Define the transformed sequence $T(a)$ of $a$ to have $n$th element \begin{equation*} T(a)(n) = \sum_{n_1^{e_1} \cdots n_s^{e_s} \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{A(n_i)}{e_i}, \qquad n \geq 1 \end{equation*} Then \begin{equation*} 1+ \sum_{n \geq 1} T(a)(n)x^n = \prod_{n \geq 1} (1-x^n)^{a(n)} \end{equation*} We now apply this to the sequence $M(n)(q)$. From the classical formula \begin{equation*} \frac{1}{1-qx} = \prod_{n \geq 1} (1-x^n)^{-M(n)(q)} \end{equation*} we get that \begin{equation*} 1+ \sum_{n \geq 1} T(M(n)(q))x^n = \prod_{n \geq 1} (1-x^n)^{M(n)(q)} = 1-qx \end{equation*} This shows that $T(M)(1)=-q$ and $T(M)(n)=0$ for all $n>1$.