Let $A$ be a coherent ring, complete with respect to the adic topology generated by a finitely generated ideal $I$.
Define the category $Coh(A,I)$ whose objects are inverse systems $\{M_n\}$ of $A$-modules $M_n$ such that:
(1) $I^nM_n = 0$ for all $n$
(2) $M_n$ is a coherent $A$-module (hence coherent $A/I^n$-module)
(3) there are compatible isomorphisms of $A$-modules $M_{n+1}/I^nM_{n+1}\cong M_n$
and morphisms are compatible systems of maps.
Is $Coh(A,I)$ equivalent to the category of coherent $A$-modules?
In the stacks project, this is proved if $A$ is Noetherian, in which case one has many good properties: finitely generated $A$-modules are $I$-adically complete and separated, coherent modules and finitely generated modules are synonyms, etc.
When we drop Noetherianity, I would expect $Coh(A,I)$, even with $A$ coherent, does not have a concrete description in terms of topological modules over $A$, by sending $(M_n)$ to $\varprojlim M_n$. Am I right?
There's a functor $Coh(A)\to Coh(A,I)$ sending $M$ to $M_n := M/I^nM$, but it's not clear that this should be reversible. The candidate inverse functor is $(M_n) \mapsto \varprojlim_n M_n =:M'$, so the question is, is $M'$ a coherent $A$-module? Is it complete?
For $A$ Noetherian, it is enough to check it is finitely generated, still true with same argument. Since $A$ is coherent, it is enough to say whether $M'$ is finitely presented or not. Is it?
I don't see a reason why $M'$ should be finitely presented. Would be great to have an explicit counterexample.
My motivation is an étale cohomology question.
References:
[StacksProject] Coherent rings, here.
[StacksProject] Coherent formal modules, here.
