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I'm reading Qing Liu's Algebraic geometry book, and here's a problem that I can't find an answer:

Find a ring $A$ and a $A$-module $M$ such that $M\otimes_{A} \hat{A} \to \hat{M}$ is not surjective, where $\hat{A} = \lim A/I^{n}$ and $\hat{M} =\lim M/I^{n}M$ are $I$-adic completions for some ideal $I\subset A$.

I know that the map is surjective if $M$ is finitely generated over $A$, so $M$ should be infinitely generated over $A$ to be a counterexample. I tried $A = \mathbb{Z}, M = \mathbb{Q}$ and $I =p\mathbb{Z}$, which gives an example of non-injective case, not surjective. Can anyone give a hint to find such example? Thanks in advance.

Seewoo Lee
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    I think the following works. Set $A:=K[x]$ ($K$ field, $x$ indeterminate), $I:=(x)$, $M:=A^{(\mathbb N)}$, $e_n=(0,...,1,0,...)\in M$ ($1=n$-th coordinate). Then $s:=\sum x^ne_n\in\hat M$, but $s$ is not in the image of $\hat A\otimes_AM\to\hat M$. – Pierre-Yves Gaillard Oct 23 '19 at 14:44
  • @Pierre-YvesGaillard What is $\hat{M}$ in this case? Since $M\simeq k[x, y]$, I thought $\hat{M} \simeq k[[x]][y]$, but it seems that you are saying $M \simeq k[[x, y]]$. – Seewoo Lee Oct 25 '19 at 13:05
  • Define the sequence $s$ in $K[x,y]$ by $s_n:=\sum_{i=0}^nx^iy^i$. Do you agree that $s$ is Cauchy with respect to the $(x)$-adic topology of $K[x,y]$? – Pierre-Yves Gaillard Oct 25 '19 at 13:29
  • @Pierre-YvesGaillard Yes I agree with that. – Seewoo Lee Oct 25 '19 at 14:44

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