I have seen that: $$(\omega\star\mu)(n)=\sum_{d\vert n}\mu(d)\omega\left(\frac{n}{d}\right)=\begin{cases}1 & n\ \text{is prime}\\ 0 &\text{otherwise} \end{cases}$$ where $\mu(n)=\delta_{\omega(n)}^{\Omega(n)} \lambda(n)$ is the Mobius function, $\omega(n)$ is the small prime omega (counting prime factors without multiplicity) and $\star$ is their Dirichlet convolution.
I was able to show that if $n$ is a prime then this is equal to 1, and if $n$ is any power $\ge2$ of a prime this is 0. However, I'm kinda confused about what to do with other cases.
I tend to think one should look at 2 other cases:first, when $n$ is the product of more than one prime and $n$ is square free, and the other case is when $n=p_1^{a_1}\cdot..\cdotp_k^{a_k}$ and $a_1,..,a_k\ge2$. However, I suspect I'm wrong and in fact, the entire statement can be proven in one fell swoop.
So I'd really like a proof that this convolution is indeed the indicator function for the primes, and I'll appreciate any comments on my attempt.
Consider the Dirichlet convolution of the prime indicator function, $1_{p}=\begin{cases}1& p\ \text{is prime}\\ 0& \text{otherwise}\end{cases}$ with $1_{\mathbb{N}}$, which is 1 for every $n$, i.e.; $$(1_{p}\star 1_{\mathbb{N}})(n)=\sum_{d\vert n}1_{p}(d)$$ this sum is over all prime divisors of $n$, since they are the only nonzero terms, so we have: $$\sum_{d\vert n}1_{p}(d)=\sum_{p\vert n}1=\omega(n)$$ by definition, since $\omega(n)$ counts the number of prime factors without multiplicity. This shows that: $$(1_{p}\star1_{\mathbb{N}})(n)=\omega(n)$$ and hence by The Mobius inversion theorem: $$(\omega\star\mu)(n)=1_{p}.$$