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.