# Completion and extension by scalars

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. $$\square$$