# How to find a non-surjective and non-injective tensor products of the formal completion?

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 $\lim_{n\to\infty} a_n=0$.
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.