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This question is based on Atiyah Macdonald, Ex. 1, in Ch. 10.

Let $A = \mathbb{Z}$ be the ring of integers with $p$-adic topology. The topological module $\mathbb{Z}/p\mathbb{Z}$ is then discrete. Construct $M = \oplus_{n=1}^{\infty} \mathbb{Z}/p\mathbb{Z}$, then $M$ is a topological module (direct sum with product topology).

I want to show that scalars can be extended to the completion $\mathbb{Z}_p$ such that $M$ is a topological $\mathbb{Z}_p$-module, for that I have to verify that $M$ is complete.

However, consider the following sequence $f_n:\mathbb{Z}^+\to \mathbb{Z}/p\mathbb{Z}$, where $f_n(i) = 1$ if $i<n$ and $0$ otherwise. The sequence $f_n\in M$ is Cauchy since it projects to a Cauchy sequence in $\mathbb{Z}/p\mathbb{Z}$. But it does not have a limit in $M$. Did Atiyah-MacDonald make a mistake with this exercise?

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  • $\begingroup$ I would say $M$ is discrete, so complete. I don't really understand what you are doing with your $f_n$. $\endgroup$ Apr 14, 2016 at 21:23
  • $\begingroup$ @CaptainLama I thought that $M$ was discrete initially, by my counter-example shows it is not so. The $f_n$ are elements in $M$. Think of them as $(1,1,1,...1,0,0,0,0,0,0,0...)$. $\endgroup$ Apr 14, 2016 at 21:25
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    $\begingroup$ What I don't understand is why it would be a Cauchy sequence. What is the projection here ? The sum of the components ? What does it have to do with being Cauchy ? $\endgroup$ Apr 14, 2016 at 21:27
  • $\begingroup$ @CaptainLama The sequence $f_n \in M$, I claim it is Cauchy. Use the projection map $M\to \mathbb{Z}/p\mathbb{Z}$, it will send that sequence into a Cauchy sequence in $\mathbb{Z}/p\mathbb{Z}$. $\endgroup$ Apr 14, 2016 at 21:34
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    $\begingroup$ @Stahl I am using all projection maps, with respect to each projection map the sequence is Cauchy in $\mathbb{Z}/p\mathbb{Z}$. To answer your question why the sequence must be Cauchy in the original space. Let $G_n$ be (abelian) topological groups. Let $G = \prod_n G_n$, be the direct product group, with the direct product topology. This way $G$ is a topological group. Now $x_n \in G$ is Cauchy if and only if $\pi_i(x_n)\in G_i$ is Cauchy. $\endgroup$ Apr 14, 2016 at 21:45

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You managed to convince me that $M$ is not complete for the product topology : it is not closed in the full product which is compact, whereas any complete subspace in a Hausdorff space must be closed.

This being said, I don't see why it couldn't be a topological $\mathbb{Z}_p$-module. The map $M\to \varprojlim M/p^nM$ is an isomorphism of topological groups, and you get the action of $\mathbb{Z}_p$ on $M$ by $x\cdot (m_1,m_2,\dots) = (\overline{x}m_1,\overline{x}m_2,\dots)$ where $\overline{x}\in \mathbb{Z}/p\mathbb{Z}$ is the natural projection of $x$.

Then $\mathbb{Z}_p\times M\to M$ defined that way is continuous since each $\mathbb{Z}/p^n\mathbb{Z}\times M/p^nM\to M/p^nM$ is continuous.


If we have the same edition of Atiyah-Macdonald then you are mistaken with the meaning of the exercise : the goal is to show that $M$ is complete for the $p$-adic topology. The product topology is nowhere mentioned.

And the fact that the canonical map $M\to \varprojlim M/p^nM$ is an isomorphism exactly means that $M$ is $p$-adically complete.

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  • $\begingroup$ I guess this means that Atiyah-Macdonald did made a mistake in their exercise, unless I am misunderstanding it. If you have the book you can look it up. $\endgroup$ Apr 14, 2016 at 23:07
  • $\begingroup$ After checking in the book you misunderstoof the exercise, I edited to explain a bit. $\endgroup$ Apr 15, 2016 at 11:02
  • $\begingroup$ The main point of the exercise is to find an exact sequence $ M \to N \to 0$ of $\mathbb{Z}$-modules such that $M_p \to N_p \to 0$ is not an exact sequence of $\mathbb{Z}_p$-modules. What exactly is $M$ and $N$ in the exercise of Atiyah-Macdonald? Maybe if I can understand this part, then I will understand the whole exercise. $\endgroup$ Apr 15, 2016 at 17:14

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