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) ?


This follows by adapting the proof of 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. $\square$


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