I have a rather complicated inverse limit which I am trying to compute. I will try to distil the question to its barest form, but if more details are necessary I can supply them.
Let $R$ be an integral domain and equip $\mathbb N$ with the usual ordering. Consider an inverse system $(A_i, f_{ij})_{\mathbb N}$ in $\text{Mod}_R$ defined by $$f_{i,i+1}: A_{i+1} \to A_i, \quad a \mapsto r_{i+1} a, \qquad r_i \neq 0,1.$$ I would like to determine $\varprojlim A_i$.
Things seem to be as nice as possible: we're in a bicomplete category so we know the limit should exist, the directed system is totally ordered, etc. However, in trying to construct an element of this limit, it seems like every component must consist of an infinite product of the $r_i$'s. Does this mean that the only element of this limit is zero, and hence that $\varprojlim A_i = 0$?
Edit: Very little may be assumed about the $r_i$. For example, we do not know if they are units, and even if they are, we do not know whether their inverses appear in any morphism.
The inverse limit is not zero. You have a functor $A: \mathbb{N}^{op} \to Mod_R$, and you want a module that satisfies the universal property. The module:
$$ \{ x \in \prod_{n \in \mathbb{N} } A_n | x_n = f_{n,m}(x_m) \ for \ all \ n \leq m \} $$
has the universal property of the inverse limit, so this is isomorphic to the inverse limit.
The elements are not infinite products, but infinite vectors indexed by the natural numbers.
Also, this construction does not depend of the index set neither the ring (you can take any ring not necessarily commutative, with unit).