I'm trying to determine if the ring $(\mathbb{A}, +, *)$ is an integral domain, where $\mathbb{A}$ is the set of arithmetic functions and $*$ is the Dirichlet convolution. To do so, I'm trying to either find a zero divisor or prove that there are none. So far I have found that given, $f,g\in \mathbb{A}\setminus \{0\}$, and supposing that $f*g = 0$ we may show that $$0=(f*g)(1) = f(1)g(1)\implies f(1) = 0 \text{ or } g(1)=0$$ From here, we may arbitrarily choose $f(1)$ or $g(1)$ to be zero, as the following is the same regardless of the choice (by commutativity of the Dirichlet convolution). Therefore, we will simply claim $f(1) = 0$. Then, letting $p$ be prime, we see $$0=(f*g)(p)=f(1)g(p)+f(p)g(1) = f(p)g(1)\implies f(p)=0\text{ or } g(1) = 0$$ Now, if we let $f(p)=0$, what ends up happening (this is provable by induction) is that with each next evaluation of a natural number of the form $n = p_1^{e_1}p_2^{e_2}\dots p_n^{e_n}$, we see either that $f(n) = 0$ or that $g(1) = 0$. Choosing the former everytime would simply result in $f = 0$, thus neither $f$ nor $g$ would be zero divisors. Thus, if we seek a zero divisor we must assume that $f(1) = g(1) = 0$.
However, this is where I'm struggling. Now that we've made this assumption, we can try $(f*g)(pq)$ for unique primes $p$ and $q$. This proceeds $$0 = (f*g)(pq) = f(1)g(pq)+f(p)g(q)+f(q)g(p)+f(pq)g(1)\\ = f(p)g(q)+f(q)g(p)$$ But I'm unsure what to do from here. I've tried assigning arbitrary values to $f(p)$ for prime $p$, but there's either some constraint on what these values may be, or there is no such zero divisor, however from here I don't know what to do to find this constraint or prove that $f$ or $g$ must equal $0$. What is the next step I need to take?