Let $A$ be a commutative ring with unit endowed with $I$-adic topology where $I$ is the ideal of $A$. Let $\hat A$ be the formal completion of $A$ for the $I$-adic topology, and $M$ an $A$-module. Let $\hat M$ be the formal completion of $A$ for the $I$-adic topology. I know that if $A$ is Noetherian ring and $M$ a finitely generated $A$-module then Artin–Rees lemma gives that $M\otimes_A\hat{A}\to \hat{M}$ is an isomorphism.

But how can I choose $A$ and $M$ such that $M\otimes_A\hat{A}\to \hat{M}$ is not surjective? And similarly, how can I choose $A$ and $M$ such that $M\otimes_A\hat{A}\to \hat{M}$ is not injective?

Qing Liu, Algebraic Geometry and Arithmetic Curves exercise 1.3.4.


A counterexample for surjectivity is the following: $A=\mathbb Z_p$ the ring of $p$-adic integers, and $M=A^{(\mathbb N)}$ a countable direct sum of copies of $A$. (The $p$-adic topology is considered for $A$ and for $M$ as well.) Then $\hat A\otimes_A M=M$ since $\hat A=A$. On the other side, $\hat M$ is the submodule of $A^{\mathbb N}$ (a countable direct product of copies of $A$) consisting of sequences $(a_n)$ with $\lim_{n\to\infty} a_n=0$.

(Let me remark that surjectivity holds for finitely generated modules over any commutative ring.)

A counterexample for injectivity can be found in Bourbaki, but I think one could find simpler examples.


A little tardy to the party but here's my idea for a non-injective map: take $A=\mathbb{Z}$, $M=\mathbb{Q}$ and complete them with respect to $I=p\mathbb{Z}$ with $p$ prime. On one hand you get the $p$-adics $\mathbb{Z}_p$ and the other you get the null ring since $\forall n, p^n\mathbb{Q}=\mathbb{Q}$: $$\hat{\mathbb Q}=0$$ However we get: $$\mathbb{Q}\otimes_{\mathbb Z}\mathbb{Z}_p = \mathbb Q_p $$ since $\mathbb Q$ is the localization of $\mathbb Z$ with respect to $S=\mathbb Z-\{0\}$, so the right hand side must be $S^{-1}\mathbb Z_p=\mathbb Q_p$.

So clearly $M\otimes_A \hat{A}\rightarrow \hat M$ can't be injective.


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