Let $R\subset S$ be commutative rings, $I\trianglelefteq R$ an ideal and $M$ be an $R$-module. Suppose that


1) $R$ is Noetherian and $I$-adically complete.


2) $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)


3) $S$ is a flat $R$-algebra.


4) $S$ is $I$-adically complete


5) $M/IM$ is free module over $R/I$,



> 
> Is it true that under the above assumptions $S\otimes\_{R}M$ is $I$-adically complete?
> 
> 
> 


I am only able to prove the above under the assumption that


6) $\operatorname{Tor}^{R}\_{i}(R/I^n,M)=0$ for all $i,n>0$
in the following way:


Consider a resolution of $M$ in $R$-$\operatorname{Mod}$
$$\ldots \rightarrow R^{\oplus m\_2}\rightarrow R^{\oplus m\_1}\rightarrow R^{\oplus m\_0}\rightarrow M\rightarrow 0$$
by finite free modules. Applying $-\otimes\_{R}R/I^n$ we obtain an exact sequence 
$$\ldots \rightarrow R^{\oplus m\_1}/I^n R^{\oplus m\_1}\rightarrow R^{\oplus m\_0}/I^n R^{\oplus m\_0} \rightarrow M/I^nM\rightarrow 0$$
using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences
$$\ldots \rightarrow S^{\oplus m\_1}/I^n S^{\oplus m\_1}\rightarrow S^{\oplus m\_0}/I^n S^{\oplus m\_0} \rightarrow S\otimes\_{R}M/I^n (S\otimes\_{R}M)\rightarrow 0$$
by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude 
$\varprojlim\_{m} S\otimes\_{R}M/I^n (S\otimes\_{R}M)$ is the cokernel
of $S^{\oplus m\_1}\rightarrow S^{\oplus m\_0}$ hence it's isomorphic $S\otimes\_{R}M$. 


Can one show this without assumption 6) at least for the case when $I$ is principal, perhaps involving the second half of condition 1) and condition 5) ? 



Completion and extension by scalars
This follows by adapting the proof of Tag [00MA](https://stacks.math.columbia.edu/tag/00MA), even without assumption 5. We also don't need $S$ to be an algebra; a complete $R$-module suffices. Finally, we never use that $R$ is $I$-adically complete!


Indeed, by assumption 2 there exists a short exact sequence $0 \to K \to F \to M \to 0$ with $F$ finite free. By assumption 3, the sequence
$$0 \to K\_S \to F\_S \to M\_S \to 0$$
is exact, where $(-)\_S = (-) \otimes\_R S$. For each $n$, this gives a short exact sequence
$$0 \to K\_S/(I^nF\_S \cap K\_S) \to F\_S/I^nF\_S \to M\_S/I^nM\_S \to 0.$$
Since all systems satisfy the Mittag-Leffler condition, the limit sequence
$$0 \to \lim\_{\substack{\longleftarrow \\ n}} K\_S/(I^nF\_S \cap K\_S) \to (F\_S)^\wedge \to (M\_S)^\wedge \to 0\label{Eq 1}\tag{1}$$
is exact. By the Artin–Rees lemma, there exists $c \geq 0$ such that $I^nK \subseteq I^nF \cap K \subseteq I^{n-c}K$ for all $n \geq c$. By flatness of $S$, for any $R$-module $N$ and any $n \in \mathbf Z\_{\geq 0}$, the surjection
$$N\_S \twoheadrightarrow N\_S/I^nN\_S \cong (N/I^nN)\_S$$
identifies $I^nN\_S$ with $(I^nN)\_S$ as submodules of $N\_S$. Similarly, the map
$$F\_S \to F\_S/I^nF\_S \oplus F\_S/K\_S \cong (F/I^nF \oplus F/K)\_S$$
identifies $I^nF\_S \cap K\_S$ with $(I^nF \cap K)\_S$ as submodules of $F\_S$. Thus, we conclude that
$$I^nK\_S \subseteq I^nF\_S \cap K\_S \subseteq I^{n-c}K\_S$$
for all $n \geq c$. In particular, (\ref{Eq 1}) reads as
$$0 \to (K\_S)^\wedge \to (F\_S)^\wedge \to (M\_S)^\wedge \to 0.$$
We now get a commutative diagram with exact rows
$$\begin{array}{ccccccccc}0 & \to & K\_S & \to & F\_S & \to & M\_S & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & (K\_S)^\wedge & \to & (F\_S)^\wedge & \to & (M\_S)^\wedge & \to & 0.\! \end{array}$$
By assumption 4, the middle vertical arrow is an isomorphism. This immediately implies that the right vertical arrow is surjective, and applying the same reasoning to $K$ gives the same statement for the left vertical arrow. Then the five lemma shows that all vertical maps are isomorphisms. $\square$


