Why is $\mu \star E =e $ , where $\star$ denotes the Dirichlet Convolution operator?

Question

Let $$ E(n) = 1 \qquad \forall n \in \mathbb{Z} $$ be the constant function, and let $\mu$ be the Möbius function. Based on the following definition of the latter function, where $\mu(n) = 1$ for $n=1$, $\mu(n) = 0$ if $a^2 | n$ for some integer $a >1$, and $\mu(n)= (-1)^{r} $ if $n$ has, in total, $r$ different prime factors: why is $$ \mu \star E = e \quad \text{?}$$ (Where $\star$ is the Dirichlet Convolution operator and $e(n) = 1$ for $n=1$ and $e(n) = 0$ for $n \neq 1$ .)

I guess it's quite a simple question for those who had a course in Analytic Number Theory, but I just can't figure it out... Thanks a lot in advance though!

Answer 1

By the definition of the Dirichlet convolution,

$$(\mu \star E)(n) = \sum_{d\mid n} \mu(d) E\left(\tfrac{n}{d}\right)\tag{1}$$

for all $n\in\mathbb{N}\setminus \{0\}$. Since $E(m) = 1$ for all $m$, $(1)$ simplifies to

$$(\mu \star E)(n) = \sum_{d\mid n} \mu(d),\tag{2}$$

and $\mu\star E = e$ is therefore equivalent to

$$\sum_{d\mid n} \mu(d) = \begin{cases} 1 &, n = 1 \\ 0 &, n > 1. \end{cases}\tag{3}$$

For $n = 1$, this is evident, it just says $\mu(1) = 1$. For $n > 1$, we can write $n = m\cdot p^k$ with a prime $p$ and $k > 0$, and an integer $m$ that is not a multiple of $p$. The divisors of $n$ then can be grouped according to the power of $p$ occurring in them,

$$\sum_{d\mid n} \mu(d) = \sum_{\kappa = 0}^k \sum_{f \mid m} \mu(p^\kappa\cdot f).$$

For $\kappa > 1$, the divisors $p^\kappa\cdot f$ are never squarefree, so $\mu(p^\kappa\cdot f) = 0$ then, and we are left with

$$\sum_{d\mid n} \mu(d) = \sum_{f\mid m} \left(\mu(f) + \mu(p\cdot f)\right).$$

Since for every $f$ we have $\mu(p\cdot f) = -\mu(f)$ (because $p\nmid f$), the sum is seen to be zero.

On a more abstract level, $(3)$ follows from the multiplicativity of $\mu$. If $f$ is a multiplicative function, then so is

$$n\mapsto \sum_{d\mid n} f(d),$$

and it suffices to see that

$$\sum_{d\mid n} \mu(d) = 0$$

for prime powers $n = p^k$. But then

$$\sum_{d\mid p^k} \mu(d) = \sum_{\kappa = 0}^k \mu(p^\kappa) = \mu(1) + \mu(p) + \sum_{\kappa = 2}^k \mu(p^\kappa) = 1 + (-1) + 0 = 0$$

is easily seen.

Answer 2

We know that if $m>0$, $(1-1)^m=0$, i.e. that $\displaystyle\sum_{k=0}^m(-1)^k\binom mk=0$ for $m>0$. Now suppose that $n=p_1^{k_1}\cdots p_m^{k_m}$. Then in $\displaystyle \sum_{d\mid n}\mu (d)$ the term $\mu(d)$ will be nonzero precisely when we look at a divisor of the form $p_{i_1}\cdots p_{i_k}$. For each $k$; we can choose $\displaystyle\binom mk$ many of these divisors ($i_k=0$ is considered as the divisor $1$), and they will appear as $(-1)^k=\mu(p_{i_1}\cdots p_{i_k})$. It follows that $$\sum_{d\mid n}\mu(d)=\sum_{k=0}^m \binom mk(-1)^k=0$$ as desired. If $n=1$, $\displaystyle\sum_{d\mid n}\mu(d)=\mu(1)=1$.