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I'm trying to do the following exercise:

Let $p$ be a prime and for $n\geq 1$ let $\alpha_n :\mathbb Z/p \mathbb Z \to \mathbb Z/p^n \mathbb Z$ be the injection of abelian groups given by $1 \mapsto p^{n−1}$. Consider the direct sum $\alpha : A \to B$ of these maps where $A$ is a countable direct sum of copies of $\mathbb Z/p \mathbb Z$ and $B$ is the direct sum of the groups $\mathbb Z/p^n \mathbb Z$. Show that the $p$-adic completion of $A$ is just $A$ but that the completion of $A$ for the topology induced from the $p$-adic topology on $B$ is the direct product of the $\mathbb Z/p \mathbb Z$. Deduce that $p$-adic completion is not a right exact functor on the category of all $\mathbb Z$-modules.

At first I thought $A$ was just the normal integers but it's not since for example for $p=2$, $-1 = 01111\dots$ is not in the space. The direct sum are all things with only finitely many non-zero terms, so for example the sequence $a_0 = 10000\dots a_1 = 110000\dots, a_2 = 111000\dots$ is a sequence in $A$ with a limit not in $A$.

I guess I am confused about what "$p$-adic completion" means: I assumed it meant that I take the equivalence classes of Cauchy sequences (Cauchy with respect to $|\cdot|_p$) where two sequences are equivalent if their difference tends to zero. But if that was what "$p$-adic completion" really meant then the sequence $a_k$ I gave above would be Cauchy and didn't have a limit in $A$ which is a counter example to what the exercise asks me to show.

Would someone explain to me what "$p$-adic completion" means? Thanks.

Edit

I'm bumping this question because the answerer is on holiday and I still have a bunch of questions. Thanks for your help.

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    $\begingroup$ I suppose it means this sort of completion. $\endgroup$
    – Asaf Karagila
    Commented Jul 26, 2012 at 7:04
  • $\begingroup$ @AsafKaragila I think if I take the $I$-adic completion in this case it's just what part (ii) asks me to do: the inverse limit. Which gives me the $p$-adic integers which is not $A$. But part (i) asks me to show that the completion of $A$ is $A$. So it must be something else, or am I missing something? $\endgroup$ Commented Jul 26, 2012 at 7:08
  • $\begingroup$ Could you clarify why you think it must be something else? Also, Atiyah-McDonald, right? BTW, $A$ is clearly not the integers because it is entirely torsion, and I don't see that we're dealing with metrics at all here. $\endgroup$
    – anon
    Commented Jul 26, 2012 at 7:18
  • $\begingroup$ @anon I think the second part of the question asks me to take the inverse limit on $B$ which is the $p$-adic integers. Then the topology induced on $A$ by this inverse limit topology on $B$ is $O$ is open in $A$ if and only if there exists $O^\prime$ open in $B$ such that $O = \alpha^{-1}(O^\prime)$. And I think that this thing is the same as the topology induces by the norm $|\cdot|_p$. Hence the completion of $A$ with respect to it gives me the $p$-adic integers. (we're dealing with metrics since the norm induces a metric which induces the topology, no?) $\endgroup$ Commented Jul 26, 2012 at 7:29
  • $\begingroup$ "And I think that this thing is the same as the topology induced by the norm" - could you expand on this? What is the norm on $B$? I think that the completion of $A$ wrt the induced topology is supposed to be like $\varprojlim \alpha(A)/p^nB\cap\alpha(A)$. $\endgroup$
    – anon
    Commented Jul 26, 2012 at 7:36

2 Answers 2

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I thought it would be a good exercise to post an answer of my own:

(i) We want to know the completion of the topological ring $A = \bigoplus_{n \in \mathbb N} \mathbb Z / p \mathbb Z$ with respect to the $p$-adic topology, i.e., the topology induced by neighbourhoods of zero of the form $A_n = p^n A$ so that $A=A_0 \supset A_1 \supset A_2\supset \dots$.

From chapter 10 in Atiyah-MacDonald we know that since this is a topological Abelian group with a countable neighbourhood basis of zero such that $A_n \supset A_{n+1}$, the completion is isomorphic to the inverse limit of the inverse system $X_n = A/A_n$ and the transition maps $f_n : A/A_n \to A/A_{n-1}$, $(x \mod p^{n}) \mapsto (x \mod p^{n-1})$. But since $pA = 0$ we get $X_n = A$ and $f_n = id_A$. Now the inverse limit are sequences $\vec{a} \in \bigoplus_n A/A_n = \bigoplus_n A$ such that $id(a_n) = a_{n-1}$, that is, constant sequences. Now clearly, $\varprojlim_n A \cong A$ via the map $(a,a,a, \dots ) \mapsto a$.


(ii) Next we would like to compute the completion of $A$ with respect to the topology induced by the $p$-adic topology on $B = \bigoplus_n \mathbb Z / p^n \mathbb Z$. The map $\alpha : A \to B$ is the inclusion map so that a set $U$ is a neighbourhood in $A$ if and only if $\alpha^{-1}(V) = V \cap A = U$ for $V$ open in $B$. Open sets in $B$ are of the form $p^k B$. Since $B = \bigoplus_n \mathbb Z / p^n \mathbb Z$, $p^k B = \bigoplus_{n=0}^{\infty} p^k \mathbb Z / p^n \mathbb Z$ where for $n \leq k$ the component is zero. We compute the inverse image $\alpha^{-1}(p^kB)$ as follows: Let $(A)_n = \mathbb Z / p \mathbb Z$ be the $n$-th component of $A$. For the $n$-th component of $p^k B$ we get

$$ (p^kB)_n = \begin{cases} 0 & n \leq k \\ p^k \mathbb Z / p^n \mathbb Z & n > k\\ \end{cases} $$

We have $\alpha(A)_n = \alpha_n(\mathbb Z / p \mathbb Z) = p^{n-1} \mathbb Z / p^n \mathbb Z$ so that $$ \alpha^{-1}(p^k B) = \begin{cases} 0 & n \leq k \hspace{0.2cm} (\text{since } \alpha \text{ is injective}) \\ \mathbb Z / p \mathbb Z & n > k \hspace{0.2cm} (\text{since } \mathrm{im}(\alpha ) = p^{n-1} \mathbb Z / p^n \mathbb Z \subset p^{k} \mathbb Z / p^n \mathbb Z)\\ \end{cases}$$

Hence open sets in this topology on $A$ look like $O_k = 0 \oplus \dots \oplus 0 \oplus \mathbb Z / p \mathbb Z \oplus \mathbb Z / p \mathbb Z \dots $ where the first $k$ entries are zero.

For our inverse system this gives us $X_n = A / O_k = \mathbb Z / p \mathbb Z \oplus \dots \oplus \mathbb Z / p \mathbb Z \oplus 0 \oplus \dots $ where the first $k$ entries are non-zero. For the transition maps $t_n: X_n \to X_{n-1}$ this means $(x_1, x_2, \dots,x_{k-1}, x_k, 0 , 0 , \dots) \mapsto (x_1, x_2, \dots,x_{k-1}, 0 , 0 , \dots)$.

For the inverse limit of this system this means that it is all sequences with $x_n \in X_n$, which gives us $\varprojlim_n X_n = \prod_n X_n = \prod_n \mathbb Z / p \mathbb Z$.


Now to conclude that $\varprojlim_n$ is not a right-exact functor on $\mathbb Z - \mathrm{\mathbf{Mod}}$, let $A_k = \alpha^{-1}(p^k B)$. Then the following sequence is exact: $$ 0 \to A_k \hookrightarrow A \xrightarrow{\pi_k} A/A_k \to 0$$

But for the inverse limit we get $$ 0 \to 0 \hookrightarrow A = \bigoplus_n \mathbb Z / p \mathbb Z \xrightarrow{\pi} \prod_n \mathbb Z / p \mathbb Z $$

where $\pi$ cannot be surjective since it is a group homomorphism mapping zero in the $n$-th component to zero.

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  • $\begingroup$ I don't understand what map $\pi$ is. I think the argument is ok as it is but if someone could tell me what $\pi$ is I'd be very grateful. $\endgroup$ Commented Aug 2, 2012 at 6:25
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    $\begingroup$ Dear Matt, The map $\pi$ is just the inclusion of the direct sum in the direct product. Regards, $\endgroup$
    – Matt E
    Commented Aug 2, 2012 at 11:50
  • $\begingroup$ Dear @MattE, thank you! I'd briefly thought of this possibility but then discarded it because I didn't know how $\pi$ could be the limit of $\pi_k$ where $\pi_k$ are not inclusions. $\endgroup$ Commented Aug 2, 2012 at 11:52
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    $\begingroup$ Reading this quickly, everything looks pretty good to me. It might be good to just write down once and for all what the maps are — how $\varprojlim_n X_n$ is isomorphic to $\prod_n \mathbb Z/p\mathbb Z$ and so one. But I believe that you can do this. One quibble is that $\varprojlim_n$ is a functor on the category of these sequences of modules $(X_n)_n$ together with the transition maps $X_{n+1} \to X_n$. One defines, and I believe that A–M does this, a notion of exactness for this category. $\endgroup$ Commented Aug 25, 2012 at 18:04
  • $\begingroup$ @DylanMoreland Thanks a lot for your comment! What do you mean to say by your last sentence? I'm not sure which bit the quibble is. $\endgroup$ Commented Aug 25, 2012 at 18:40
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The $p$-adic completion should mean “complete with respect to the ideal generated by $p$”. I guess the discussion is restricted to groups, so you could instead think about completing with respect to the subgroup filtration $(p^nG)$. I would forget about the metric for the time being.

For $A$ we have $pA = 0$, so to complete we're taking the inverse limit \[ A \leftarrow A \leftarrow \cdots \] in which all of the transition maps are the identity. This is clearly isomorphic to $A$.

Now we want to see what happens if we complete $A$ with respect to the filtration $(C_n) = (p^nB \cap A)$. So what you should do is write out $p^nB$ and $\alpha(A)$, and then compute these $C_n$. It should come out that $A/C_n = \bigoplus_{i = 1}^n \mathbb Z/p\mathbb Z$, and that the transition map $A/C_{n + 1} \to A/C_n$ forgets the $(n + 1)$-th component. Does the result seem plausible now?

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    $\begingroup$ Note that I'm not being very careful about the distinction between $A$ and $\alpha(A)$. In case you get stuck, another push:$$ p^nB = \{0\} \oplus \cdots \oplus \{0\} \oplus p^n\mathbb Z/p^{n + 1}\mathbb Z \oplus p^{n + 1}\mathbb Z/p^{n + 2}\mathbb Z \oplus \cdots$$ $\endgroup$ Commented Jul 26, 2012 at 8:12
  • $\begingroup$ Thanks a lot! I'm not stuck yet, what I'm doing is reading a bit more of the notes since I haven't read about filtrations. (The question is from an old exercise sheet of mine. I'm doing some of the exercises I didn't do then.) $\endgroup$ Commented Jul 26, 2012 at 8:37
  • $\begingroup$ Thanks Dylan. I'm very stuck: I thought I understood inverse limits but I struggle to understand your answer. I think of inverse limits as sequences such that the $n$-th entry of the sequence is in the $n$-th (sub)group of the group we're taking the limit of. But that doesn't seem to help because then I have a product of direct sums. $\endgroup$ Commented Jul 30, 2012 at 19:42
  • $\begingroup$ Hm... I think we cannot restrict the discussion to groups, we need multiplication to have $p^nG$ defined, so we need either modules or rings. $\endgroup$ Commented Jul 31, 2012 at 9:23
  • $\begingroup$ @Matt An abelian group is always a $\mathbb Z$-module. I will try to catch up on your other comments. $\endgroup$ Commented Aug 25, 2012 at 17:51

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