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This is a question related to this.

Let $G= \mathbb Z / p \mathbb Z$ for some prime $p$. Let $A = \bigoplus_{n\in \mathbb N} G$, that is, all sequences in $G$ with all but finitely many terms zero. Now I'm interested in the $p$-adic completion of this group. The answer is supposed to be that the completion of $A$ is $A$ itself. But I don't see how that is true. The completion is the same as the inverse limit, which is a subset of the product $\prod_{n \in \mathbb N} A$. So it's sequences of sequences. But consider the sequence $s_n$ where the first $n$ terms are one. Then clearly this is a Cauchy sequence in the $p$-adic norm but its limit is not in $A$. So how can $A$ be complete? Thanks for your help.

Edit

$|x|_p = \frac{1}{\text{highest power of } p \text{ that divides } x}$, $|0|_p = 0$

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    Dear Matt, Certainly any abelian group that is killed by some fixed power of $p$ (like $A$) is already $p$-adically complete. Regards,2012-07-31

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First, when thinking about these kind of completion questions, it is probably better not to use Cauchy sequences, etc., but rather to use the inverse limit point of view.

Secondly, the inverse limit is characterized by its universal property. One way to construct it is as a certain subset of the product, but it is not normally helpful to think in terms of this construction.

So, if $A$ is any abelian group, its $p$-adic completion is the inverse limit of the modules $A/p^n A$, with the transition map $A/p^{n+1} A \to A/p^n$ being the obvious projection.

Suppose now that $A$ is annihilated by $p$. Then $A/p^n A = A$ for all $n \geq 1$, and the transition map is just the identity $A \to A$. Thus the $p$-adic completion of $A$ is the inverse limit of copies of $A$ with transition maps being the identity. Clearly this inverse limit is canonically isomorphic to $A$ itself. (Think in terms of the universal property.)

[I belive that Dylan explained this to you in a comment on your earlier question.]

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    Dear MattE, thank you, I think I did that already, see the comment to Bruno's answer. : )2012-07-31
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The real question is, what do you mean by the $p$-adic completion?

If you mean

$\varprojlim A/p^nA$

then this is indeed $A$ since $pA=0$ (so each term in the inverse limit is $A$, with the identity maps).

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    Dear @MattE, it took a bit longer than expected but I finally posted a complete answer to my other question.2012-08-02