Return to Answer

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 $\mathbb{F}_q[T]$ 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 $\mu: \mathbb{F}_q[T] \to \mathbb{C}$ 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.

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 $$m>0 \implies \sum_{\deg d = m+1} \mu(d) = 0.$$

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 Euler product identity (for the Zeta function of $\mathbb{F}_q[T]$) 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: 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 $$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$ Note that $\mu(f)=0$ for $f$ which is not squarefree. In other words, 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.

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 $\mathbb{F}_q[T]$ 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 $\mu: \mathbb{F}_q[T] \to \mathbb{C}$ 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.

Yes, it is true.

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 Möbius function $\mu(\bullet)$ assumes the value $(-1)^{\sum e_i}$ for each such polynomial. In other words, your sum may be rewritten as $$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$ Note that $\mu(f)=0$ for $f$ which is not squarefree. In other words, 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. It is easy to prove. We use the Euler product identity: $$\frac{1}{1-qx} = \prod_{P\text{ monic irreducible}} (1-x^{\deg P})^{-1} =\prod_{n} (1-x^n)^{-M(n)}.$$ Taking its reciprocal, we find that $$1-qx = \prod_{P\text{ monic irreducible}} (1-x^{\deg P}) = \sum_{f \text{ monic}}\mu(f) x^{\deg f}.$$ Now just compare coefficient.

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 Euler product identity (for the Zeta function of $\mathbb{F}_q[T]$) 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: 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 $$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$ Note that $\mu(f)=0$ for $f$ which is not squarefree. In other words, 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.