Does $I \varprojlim M_n = \varprojlim I M_n$?

Question

Let $I$ be an ideal of a Noetherian ring $A$ and $(M_n)_{n \geq 0}$ an inverse system of finitely-presented $A$-modules whose inverse limit can also be finitely presented. Is it then the case that $$I\varprojlim M_n = \varprojlim IM_n ?$$ (In the hoped-for application, $A$ is a power series ring in finitely many indeterminates over a subring of $\mathbb Q$.)

Answer 1

I don't know if you have sorted out so far (in case you have, feel free to give a comment upon what you have found out). I thought about your question for a while and here is a version of what you're asking (some facts are probably well known and elementary, so if you already know them skip them out!)

I misread previously what you had written so this is another proof:

I will assume moreover that the above inverse system $(M_n)_{n \geq 0}$, consists of finite length, flat modules such that the inverse limit is flat and that the ideal $I$ is finitely generated. It's true that the inverse limit of flats need not be flat (in fact in our case finitely presented $+$ flat implies projective, and inverse limits of projectives need not be projective even in ideal cases as for abelian groups. A standard counterexample is the inverse system $(\prod_{i=1}^{m}\mathbb{Z}, \pi_m),$ where the maps are projections). In that case we have a characterization for an any commutative ring $A$ and a given $A$-module $M$, of the form $$ M \textit{ is flat iff } \thinspace I \otimes_A M \cong IM \textit{ is an isomorphism for all ideals } \thinspace I \trianglelefteq A . $$

Tensor products although preserve direct limits, they do not necessarily preserve inverse limits. In fact the latter is true whenever the inverse system consists of modules of finite length (that is Noetherian and Artinian modules, and in our case just Artinian suffices), and the tensor product is taken with respect a finitely presented $A$-module. Though any Noetherian ring is coherent, and the latter is equivalent with the fact that finitely generated ideals are finitely presented. So in $A$ our ideal is finitely-presented hence from our assumption that the inverse limit is flat we get the following isomorphism $$ I \otimes_A (\varprojlim_{n\in \mathbb{N}} M_n) \cong \varprojlim_{n\in \mathbb{N}}(I \otimes_A M_n),$$ which implies for the left-hand side $I \otimes_A (\varprojlim_{n\in \mathbb{N}} M_n) \cong I(\varprojlim_{n\in \mathbb{N}} M_n)$, whilst for the right one $\varprojlim_{n\in \mathbb{N}}(I \otimes_A M_n) \cong \varprojlim_{n\in \mathbb{N}} IM_n$. By the latter your answer follows.

I couldn't figure out a proof without this assumption, or a concrete counterexample to prove the opposite. In fact I believe that this is probably not true in the general case. If you have found out something more general or a counterexample please do let me know.