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How to prove that if $\mathbb{Z}_p$ is the set of $p$-adic integers then $\displaystyle{\mathbb{Z}_p=\varprojlim\mathbb{Z}/p^n\mathbb{Z}}$ where the limit denotes the inverse limit?

$\mathbb{Z}_p$ is the inverse limit of the inverse system $(\mathbb{Z}/p^n\mathbb{Z}, f_{mn})_{\mathbb{N}}$, but I don't know what the $f_{mn}$ are.

Can someone help me?

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    The $f_{mn}$ are the "obvious" maps $\mathbb{Z}/p^m\mathbb{Z}\to\mathbb{Z}/p^n\mathbb{Z}$, where $m\gt n$. Note that $p^m\mathbb{Z}\subseteq p^n\mathbb{Z}$, so that $\mathbb{Z}/p^n\mathbb{Z}$ is a quotient of $\mathbb{Z}/p^n\mathbb{Z}$. The connective maps are the quotient maps.2012-06-11
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    How are you defining the $p$-adics, if not as the inverse limit? I'm asking because for many, the $p$-adics **are** the inverse limit (by definition), so asking how to prove that they are the same would be asking how to "prove" a definition. So obviously, you must have a different definition of $\mathbb{Z}_p$, and you probably should specify it.2012-06-11
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    Now that I think about it, after you state your definition of $\Bbb Z_p$ (I guess it is with power series expansions), you can also mention if you want to show isomorphisms as additive groups, as rings, and if you want topology in the mix as well. I was hasty adding the (topological-groups) tag; I imagine you just want an isomorphism as bare groups.2012-06-11
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    You can view a $p$-adic integer $a_0 + a_1 p + ...$ as a sequence $(..., a_2 p^2 + a_1p + a_0, a_1 p + a_0, a_0)$, which is an element of the inverse limit. This is how the isomorphism goes I believe.2012-06-11

2 Answers 2

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Let $p$ be a prime number. Let $x$ be a non-zero element in $\mathbb{Z}$. There exist integers $n$ and $a$ such that $x = p^na$, $(p, a)$ = 1. $n$ is uniquely determined by $x$. We denote $|x|_p = p^{-n}$. We define $|0|_p = 0$. For $x, y\in \mathbb{Z}$, we denote $d(x, y) = |x - y|_p$. $d$ is a metric on $\mathbb{Z}$. With this metric $d$, $\mathbb{Z}$ becomes a topological ring. We define $\mathbb{Z_p}$ as the completion of $\mathbb{Z}$ with respect to $d$. $\mathbb{Z_p}$ is a topological ring which contains $\mathbb{Z}$ as a dense subring.

$(p^n\mathbb{Z_p}), n = 1, 2, ...$ is a fundamental system of neighbourhoods of 0 in $\mathbb{Z_p}$. Each $\mathbb{Z_p}/p^n\mathbb{Z_p}$ is isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$. Hence it suffices to prove that $\mathbb{Z_p} \cong \varprojlim\mathbb{Z_p}/p^n\mathbb{Z_p}$. Let $f_n: \mathbb{Z_p} \rightarrow \mathbb{Z_p}/p^n\mathbb{Z_p}$ be the canonical map for each n. When $n ≧ m$, $p^n\mathbb{Z_p} ⊂ p^m\mathbb{Z_p}$. Hence we can define a ring homomorphism $f_{mn}: \mathbb{Z_p}/p^n\mathbb{Z_p} \rightarrow \mathbb{Z_p}/p^m\mathbb{Z_p}$ by $f_{mn}(f_n(x)) = f_m(x)$. Let $A = \varprojlim\mathbb{Z_p}/p^n\mathbb{Z_p}$. Since $f_{mn}f_n = f_m$, we can define a map $f: \mathbb{Z_p} \rightarrow A$ by $f(x) = (f_n(x))$ for each $x \in \mathbb{Z_p}$. It's easy to see that $f$ is a continuous ring homomorphism. Since $∩p^n\mathbb{Z_p} = 0$, $f$ is injective.

Let $x = (x_n) \in A$. For each n, choose $a_n \in \mathbb{Z_p}$ such that $f_n(a_n) = x_n$. Since $a_n ≡ a_{n+1}$ (mod $p^n\mathbb{Z_p}$), $(a_n)$ is a Cauchy sequence in $\mathbb{Z_p}$. Hence there exists $a$ = lim $a_n$ in $\mathbb{Z_p}$. It's easy to see that $f(a) = x$. Hence $f$ is surjectve.

It remains to prove that $f$ is an open map. Let $π_n:A \rightarrow \mathbb{Z_p}/p^n\mathbb{Z_p}$ be the projection map. Since $f_n = π_nf$, $p^n\mathbb{Z_p} = (f_n)^{-1}(0) = f^{-1}((π_n)^{-1}(0))$. Hence $f(p^n\mathbb{Z_p}) = (π_n)^{-1}(0)$. Hence $f(p^n\mathbb{Z_p})$ is open. QED

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Let me guess the definition of $p$-adic integers you got in mind is some $p$-power series. Now, forget it and let us consider some topology:

We DEFINE the $p$-adic integers to be the completion of certain metric on $\mathbb{Z}$, then the $p$-power series is a way to make completion, and the inverse limit is another way. However, the completion of a metric space is unique up to unique isometry. Now you can check the isometry is a ring isomorphism by hands.

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    Yes, but proving that the inverse limit is the completion of $\mathbb{Z}$ with respect to p-adic metric is not so trivial and you are suppose to do it.2012-06-11
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    @Makoto Kato, by writting explicitly, a small element in the inverse limit is something of the form $(0,0,...,0,a_1,a_2,...)$ and elements in $\mathbb{Z}$ are of the form $(a,a,a...)$ (maybe mod something in the first few terms). Thus $\mathbb{Z}$ is dense in the inverse limit and every Cauchy sequence in $\mathbb{Z}$ converges in the inverse limit, and therefore the inverse limit is a completion. Anyway, you have already gave a detailed proof in your answer, right?2012-06-11
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    *and every Cauchy sequence in $\mathbb Z$ converges in the inverse limit,* How do you prove this?2012-06-14
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    @Makoto Kato, let $A_1,...A_N,...$ be a Cauchy sequence in $\mathbb{Z}$, then for every positive integer $n$, the $n$-th terms in the sequence becoming stable for big enough $N$ in this sequence, denote it by $b_n$. Then $(b_1,b_2,...)$ is an element in the inverse limit and the sequence $A_i$ converges to it.2012-06-15
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    What is $N$? What is the relation between $n$ and $N$?2012-06-15
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    @Makoto Kato, $N$ denotes the index of the Cauchy sequence $A_1,...,A_N,...$ in $\mathbb{Z}$, for each $A_i$, we may write it as an element in the inverse limit $(a_1,a_2,...,a_n,...)$. The point in my last comment is: since $A_1,A_2,..$ converges, the $n$-th terms become stable for sufficiently big enough $N$.2012-06-15
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    Note that since we write $A_i$ as an element in the inverse limit, each $A_i$ has an $n$-th term. If you are confused about how can one write an elemnt in $\mathbb{Z}$ in this form, see the comment of fretty in the original post2012-06-15
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    Could you explain why the n-th terms become stable for sufficiently big enough N?2012-06-15
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    @Makoto Kato, note that $A_i$ and $A_j$ have distance less than $p^{-n}$ (in $p$-adic metric) if and only if their first $n$ terms coincide, now can you see?2012-06-16
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    I was just asking because your "proof" was insufficient, not because I didn't know the proof. If you don't want to write the proof in the first place, it's better that you put your "answer" in the comment section.2012-06-16
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    @Makoto Kato, there are different ways on giving an answer. I think sometimes a conceptual one is more helpful than a dense one with all details. All things we communicated in above comments are elementary metric space computations and can be found in textbooks, so OP can do it himself.2012-06-17
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    I understand your sentiment, but I still think your "answer" is suitable in the comment section. OP doesn't seem to have a clue without detailed explanation. I don't think just saying read a texbook to him can be of any help.2012-06-17
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    @Makoto Kato, "I don't think just saying read a texbook to him can be of any help"--agreed, so I gave a conceptual answer. But I don't accept your judgement about where should my answer been placed or how should it looked like (say, how detail it should be), I DO think those judgements are rather rude. If OP (and you) has confusion on details, he (and you) may ask in comments and I may comment more, otherwise I have done my part.2012-06-17