I'm looking for counterexamples to Proposition 10.13 in Atiyah-Macdonald when the hypotheses are not satisfied (throughout, $A$ is a ring, and $\hat{\phantom{M}}$ denotes the $I$-adic completion for some fixed ideal $I \triangleleft A$):
Since we have a natural homomorphism $A \to \hat{A}$, we can regard $\hat{A}$ as an $A$-algebra and so for any $A$-module $M$ we can form an $\hat{A}$-module $\hat{A} \otimes_A M$. [...] Now the $\hat{A}$-module homomorphism $M \to \hat{M}$ defines an $\hat{A}$-module homomorphism $$\hat{A} \otimes_A M \to \hat{A} \otimes_A \hat{M} \to \hat{A} \otimes_{\hat{A}} \hat{M} = \hat{M}.$$ In general, for arbitrary $A$ and $M$, this is neither injective nor surjective, but we do have:
Proposition 10.13. For any ring $A$, if $M$ is finitely-generated, $\hat{A} \otimes_A M \to \hat{M}$ is surjective. If, moreover, $A$ is Noetherian then $\hat{A} \otimes_A M \to \hat{M}$ is an isomorphism.
Namely, I'm looking for:
- Non-finitely generated modules such that the map is not surjective.
- Finitely generated modules over a non-Noetherian ring such that the map is not injective.
- Any module over a non-Noetherian ring such that the map is neither injective nor surjective.
For the first type, is the following example correct?
Take $A = \mathbb{Z}$, $M = \bigoplus_{i = 0}^\infty \mathbb{Z}$, $I = p\mathbb{Z}$. Then the natural map is $\mathbb{Z}_p \otimes M \to \hat{M} = \lim_{\leftarrow,n} M/p^nM$. Let $(e_i)$ be the canonical basis for $M$. Then the element $e_0 + pe_1 + p^2e_2 + \cdots$ is in $\hat{M}$ but does not come from $\hat{A} \otimes M$ since elements in the latter can only use a finite number of distinct $e_i$.
Would I also be correct in saying the following? If in the example above we replace $M = \bigoplus_{i = 1}^\infty \mathbb{Z}$ with $\prod_{i = 1}^\infty \mathbb{Z}$, then even though $M$ is not a finitely generated $\mathbb{Z}$-module, it does not constitute a counterexample to 10.3 because elements such as $e_0 + pe_1 + p^2e_2 + \cdots$ already exist in $M$ without completion.