Exercise. Determine all the arithmetic multiplicative functions that are idempontent to the convolution product, i.e., determine all the functions $f$ such that, for every $a \in \Bbb N$, we have:
$$ (f * f)(n) = \sum_{d|n} f(d)f\left(\frac{n}{d}\right) = f(n) $$
My attempt (not complete resolution).
Assuming that $n$ is square-free, we can guarantee that $\left(d,\frac{n}{d}\right) = 1$ and thus we can use the fact that $f$ is multiplicative. In this cases, $ f(d)f\left(\frac{n}{d}\right) = f(n) $ and thus the sum above is no more than $f(n) \tau(n)$, where $\tau(n)$ is the function that represents the number of divisors of $n$. Thus,
$$ f(n)\tau(n) = f(n) \Leftrightarrow f(n)(\tau(n)-1) = 0 \Leftrightarrow \tau(n) = 1 \vee f(n) = 0. $$ The only $n$ for which $\tau(n) = 1$ is $n=1$ and thus the only function that verifies this is the following:
$$ f(n) = \begin{cases} 1, \text{ if } n = 1 \\ 0, \text{ if } n>1 \end{cases} = \epsilon(n) $$ Now we have to deal with the case where $n$ isn't square-free, i.e., $n = p_1^{a_1}\dots p_k^{a_k}$ where each of the exponents $a_i$ can be superior than two. I understand that since $f(n)$ is multiplicative, we only have to study the case where $n = p^a$ and extend this result in the end. But how would one do so?
Thanks for any help in advance.
The ring of arithmetic functions under Dirichlet convolution (with values in, say, $\mathbb{C}$, or more generally any integral domain) is an integral domain, so if $f^2 = f$ then $f(1 - f) = 0$ which gives $f = 0, 1$. We don't need to assume that $f$ is multiplicative.
This is easiest to see by passing to formal Dirichlet series, sending $f$ to
$$\sum_{n \ge 1} \frac{f(n)}{n^s}$$
which shows, once we write $n^{-s}$ as a product $\prod p_i^{-e_i s}$ using the prime factorization of $n$, that the ring of arithmetic functions under Dirichlet convolution, or equivalently the ring of formal Dirichlet series, is a formal power series ring in countably many variables $p^{-s}$, one for each prime; this is an integral domain.