When we consider an unique factorization domain, we get a factorization with finitely many factors. Is it possible to generalize an unique factorization domain by still requiring an unique factorization but allowing for infinitely many factors?
Has this been studied, or is it impossible/not interesting anyway?
EDIT: An example: (I'm not sure yet, whether this is a valid example, but it perhaps helps to see what I'm aiming for.)
Consider the ring $R = \mathbb Z \times \mathbb Z \times \mathbb Z \times \ldots$ of integer sequences with pointwise addition and multiplication, that is for $A=\{a_i\}_i, B=\{b_i\}_i \in R$ consider the addition $A+B = \{a_i+b_i\}_i$ and the multiplication $A\times B = \{a_i \cdot b_i\}_i$
If I'm correct, in this ring $p=(2,1,1,\ldots)$ is a prime, but so is $q = (2,0,0,\ldots)$. That means the factorization is not unique. But we can still factorize some elements into infinitely many prime factors: E.g. $r = (2^1,2^2,2^3,2^4,\ldots)= (2,1,1,1\ldots) \times (1,2,1,1\ldots)^2 \times (1,1,2,1, \ldots)^3 \times \ldots$.
Again if I'm correct the prime numbers here have the form $(u_1, \ldots, u_k, p, u_{k+1}, \ldots )$ where $u_k \in \{0, \pm 1\}$ and $p \in \mathbb Z$ is prime.
We can almost make it unique by factorizing elmenets into primes of the form $(1,1, \ldots,1,p,1,\ldots)$, an unit $\{u_i\}_i$ with $u_i \in \{\pm1\}$ and a element $z=\{z_i\}$ where $z_i \in \{0,1\}$ for zeroing some entries. If we "ignore" the $z$ part, the factorization is unique.