Let $R$ be a commutative ring with $1$ and let $M = \{ f : \Bbb{N} \to R \}$ be the $R$-module of arithmetic functions into $R$.
A basis for $M$ is $(d \mid \cdot) : d \in \Bbb{N}$ where $(d\mid n) = 1 \iff d \mid n$ and is $0$ otherwise.
I'm wondering if you take $M_n = \{ f \in M : f(x) = \sum_{d \in \Bbb{N} \\ \Omega(d) = n}f_d\cdot(d \mid x), \ f_d \in R\}$.
And define $\partial_i : M_i \to M_{i-1}$,
$$ \partial_i (n \mid \cdot) = \partial(q_0 \cdots q_i \mid \cdot) = \sum_{j = 0}^i(-1)^j q_j(\frac{d}{q_j}\mid \cdot) $$
and extend linearly to all of $M_i$. These form differential chain maps. Check by comoposing twice i.e. $\partial^2(pqr\mid \cdot)$ and seeing that everything cancels to $0$.
Also check that $\partial_1(f) = \sum_{q \in \Bbb{P}} f_q q$. So that the condition $f \in \ker \partial_1 \iff \sum_{q} f_q q = 0$. Let's consider $M$ such that we only look at functions that have finitely many non-zero coefficients.
Then $f \in Z_1 = \ker \partial_1$ iff and only if $f(x) = \sum_{q \in \Bbb{P}} f_q(q\mid x)$ and $\sum_{q} f_q q = 0$.
I'm wondering in English what the homology here at $M_1$ counts and say $R = \Bbb{Z}$.
$$ q(p\mid\cdot ) - p(q\mid\cdot) = \partial_2(pq\mid \cdot), \\ \forall p \leq q, \ p,q \in \Bbb{P} $$ is in the boundary submodule $B_1 \leqslant Z_1$. Then ?
Attempt. Let $R = \Bbb{Z}$.
$$ 5 - 3 - 2 = 0 \implies (5\mid \cdot) - (3\mid \cdot) - (2\mid \cdot) \in Z_1 \\ $$
Each $p(q\mid \cdot) - q(p\mid \cdot) \in B_1$, so that for all $2 \leq q$ we have that $2(q\mid \cdot) = q(2\mid \cdot)$ modulo $B_1$. Which means:
$$ f(x) = \sum_{q}f_q(q\mid x) = \sum_{q \neq 2} (f_q)_{(2)}\cdot(q\mid\cdot) + \left((f_2)_{(2)} +\sum_{q} q\left\lfloor\frac{f_q}{2}\right\rfloor\right) (2\mid \cdot) \pmod {B_1} $$
So we can say that the $H_1 = Z_1/B_1 \approx \bigoplus_{q \in \Bbb{P}, \\ q \neq 2} \Bbb{Z}_2 (q \mid x) \oplus \Bbb{Z}(2\mid \cdot)$