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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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    What do you mean by the $p$-adic norm on $A$?2012-07-30
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    @QiaochuYuan Added it.2012-07-31
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    Okay, then I don't see how your sequence $s_n$ is Cauchy. It seems to me that by your definition $|s_n - s_{n+1}| = 1$. (More generally $|x|_p$ can only take the values $0$ or $1$, so a sequence is Cauchy iff it's eventually constant.)2012-07-31
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    @QiaochuYuan Oh no, I thought of it as one over the norm : / Ok, so perhaps the group is already complete. But I still don't understand how to compute the completion as done in Dylan's answer in the linked post : /2012-07-31
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    Argh, confused myself twice : ( The highest power that divides one is $p$ to the zero. So yes, the distance is always one.2012-07-31
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    But thanks, you have answered my question. Perhaps I'll make a different follow up post to help me understand Dylan's answer.2012-07-31
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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. What I still don't understand is why the transition maps are forced to be the identity maps $\pi_{ij}$ if our inverse system is $(X_i = A, \pi_{ij})$. I understand that $A$ satisfies the universal property of the direct limit of this system if $\pi_{ij} = id_A$ though.2012-07-31
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    Dear @MattN., the transition don't have to be the identity but in this case, they are. By construction, the maps $A/p^{n+1}A \to A/p^nA$ of the inverse system are obtained from the "reduction mod $p^n$" maps, which can be canonically identified with the identity maps in this case (the projection map $A \to A/(0)$ can be canonically identified with the identity map of $A$).2012-07-31
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    Dear @Bruno, thank you very much. I think I'm starting to understand. In this example, where we have $A = \bigoplus_n \mathbb Z / p \mathbb Z$, the sequence of groups in our inverse system is $G_n = A/0 = A$. Our morphisms are $f_n: A/p^nA \to A/p^{n-1}A$, $\bar{x} \mapsto x \mod p^{n-1}$. But $\bar{x}$ was already "mod $p$" because $x \in A = \bigoplus_n \mathbb Z / p \mathbb Z$ so that applying $f_n$ to $x$ does nothing to $x$ and $f_n = id$.2012-07-31
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    Dear @Bruno, now I'm trying to understand why the inverse limit of an inverse system $(G_n = G, f_n = id_G)$ is $G$.2012-07-31
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    @MattN: Dear Matt, Great! You're almost there. For this last issue, try using the characterization of the inverse limit via the universal property. Regards,2012-07-31
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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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    Ok, so if our inverse system is $(A, id)$, i.e. our sequence of groups is $G_n = A$ and our transition maps are $t_n = id_A$, then an element in the inverse limit looks like a sequence of elements in $A$ such that the $n$-th index of the sequence looks like the $n-1$-th term which means that an element in $\varprojlim A$ looks like $(a,a,a, \dots)$ which is hence isomorphic to $A$.2012-07-31
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    Does that make sense?2012-07-31
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    @MattN: Dear Matt, Yes, that makes sense (although, as my comment will have indicated, I prefer to think in terms of the universal property rather than via the explicit construction). Regards,2012-07-31
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    Dear @MattE, thank you for checking my work. I am working on using the universal property rather than the explicit construction but I only just started to learn about inverse limits some days ago and for now I'm quite happy that I could develop my understanding as far as I have. I will now develop it further, I have not ignored your hint in your answer, telling me that it's better to think of it in terms of universal properties. (And I will post a complete answer to my previous question later today.)2012-07-31
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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