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I'm reading Number Theory $1$ written by Kato, Kurokawa and Saito. I have a question about one of the proposition in Chapter 2: Conics and $p$-adic Numbers.

Definition: $\mathbb{Z}_{(p)}=\{\frac{a}{b} \mid a,b \in \mathbb{Z}, p \nmid b \}$

Definition: $\mathbb{Z}_p=\{a \in \mathbb Q_p\mid \mathrm{ord}_{p}(a) \geq 0\}$.

There's a Lemma on page 67:

(5): $\mathbb{Z}_p$ is the closure of $\mathbb{Z}_{(p)}$ in $\mathbb Q_p$.

Suppose $a$ is a limit point for a $p$-adically convergent series: $(x_n)_{n \geq 1 }$ in $\mathbb{Z}_{(p)}$. Let $\varepsilon = \frac{1}{p^m}$ for a natural $m$. There exists $N$ so that $\forall i>N: |a-x_i|_{p}< \varepsilon\Rightarrow\frac{1}{p^{\mathrm{ord}_{p}(a-x_i)}} \leq \frac{1}{p^m}$. Thus, there exists a natural $r \geq m $ and some $u \in \mathbb{Z}_{(p)}^ \times $ so that $a-x_i= p^{r}u$ hence, $a = x_{i}+p^{r}u$. Since $x_{i}$ belongs to $\mathbb{Z}_{(p)}$, from the previous equation we conclude that $a$ belongs to $\mathbb{Z}_{(p)}$. $\mathbb{Z}_{(p)}$ contains the limit points of its $p$-adically convergent series with respect to $p$-adic metric. $\ast $

On the other hand, $\mathbb Q_p$ contains classes of $p$-adically convergent series each of which contain a constant series, which is a member of $\mathbb{Z}_{(p)}$ according to $\ast$.

Can we say $\mathbb{Z}_{(p)}$ is closed with $p$-adic metric?

user26857
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  • The closure of $\Bbb Z_{(p)}$ in $\Bbb Q_p$ is $\Bbb Z_p$, like it says in the book. – Angina Seng Apr 27 '20 at 06:17
  • Consider the sequence $x_n=p^n-1$ – J. W. Tanner Apr 27 '20 at 06:20
  • @AnginaSeng Isn't $\mathbb{Z}_{(p)}$ closed with $p$-adic metric? –  Apr 27 '20 at 06:22
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    You get $\Bbb{Z}p$ as the closure of $\Bbb{Z}$ already. Given that $\Bbb{Z}{(p)}$ contains $\Bbb{Z}$ and is contained in $\Bbb{Z}_p$, the claim follows. – Jyrki Lahtonen Apr 27 '20 at 06:27
  • @J.W.Tanner isn't its limit 1? it is contained in $\mathbb{Z}_{(p)}$ –  Apr 27 '20 at 06:29
  • Ok. This Proposition probably comes earlier. That is, before the result I described has been proven. – Jyrki Lahtonen Apr 27 '20 at 06:31
  • What about $x_n=p^n$? – J. W. Tanner Apr 27 '20 at 06:34
  • The sequence $(x_n){n\ge1}$ of partial sums like $$x_n=1+\sum{k=0}^n p^{2^k}$$ converges $p$-adically, we have $x_n\in\Bbb{Z}{(p)}$, but the limit $$\lim{n\to\infty}x_n=1+\sum_{k=0}^\infty p^{2^k}$$is not a rational number. – Jyrki Lahtonen Apr 27 '20 at 06:35
  • @JyrkiLahtonen Aren't $\mathbb{Z}p$ and $\mathbb{Z}{(p)}$ isomorphic? Each classes of $\mathbb{Z}p$ can be represented by their limit point which is a member of $\mathbb{Z}{(p)}$ –  Apr 27 '20 at 06:35
  • No, Nafise! Elements of $\Bbb{Z}_{(p)}$ are rational numbers. The `limit points' usually are not. – Jyrki Lahtonen Apr 27 '20 at 06:37
  • May be this thread has something that helps you? – Jyrki Lahtonen Apr 27 '20 at 06:41
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    A counting argument may be the most compelling. The set $\Bbb{Z}_{(p)}$ is countable as a subset of $\Bbb{Q}$. But it is easy to exhibit uncountably many pairwise non-equivalent $p$-adic Cauchy-sequences. – Jyrki Lahtonen Apr 27 '20 at 06:48
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    Or consider this. The sequence $2, 2^p, 2^{p^2}, 2^{p^3},\ldots,$ that is, $x_n=2^{p^n}$, is a $p$-adic Cauchy sequence. Its limit point $$x=\lim_{n\to\infty}2^{p^n}$$ is thus in the closure of $\Bbb{Z}$. Furthermore, $x$ is $p$-adically close to $2$ (I am assuming $p>3$ here). More importantly, $x_n^p=x_{n+1}$ for all $n$, implying that $$x^p=\lim_{n\to\infty}x_{n+1}=x.$$ As $x\neq0$ this means that $x^{p-1}=1$. There are no elements of $\Bbb{Z}_{(p)}\setminus{1}$ satisfying the equation $x^{p-1}=1$. – Jyrki Lahtonen Apr 27 '20 at 06:54
  • @JyrkiLahtonen Thank you very much! It was extremely helpful. I understood my mistake, not only members of $\mathbb{Z}_{(p)}$ but also rational numbers and complex numbers can be limit points of $p$-adically convergent series. –  Apr 27 '20 at 06:55
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    Warning: It is anything but clear how to view $p$-adics as complex numbers. There are no natural ways of doing that. The $5$-adic square root of $-1$ I constructed (sort of) in that old thread is not a complex number. It is a 5-adic number. We are really constructing something else entirely! At least it is best to think of it that way. The 5-adic square root of $-1$ is a totally different animal from the $13$-adic square root of $-1$ and from the complex square root of $-1$. – Jyrki Lahtonen Apr 27 '20 at 07:01
  • A priori, they is no universal set $\Omega$ of numbers that would have all of $\Bbb{R}$, $\Bbb{C}$, $\Bbb{Q}_p$ for all $p$ as subsets. We can construct such an $\Omega$ by resorting to Zorn's lemma and such, but that process is not natural. It is more often than not better to think of $\Bbb{Q}_p$ as something totally unrelated to $\Bbb{C}$. The only thing they "organically" have in common is $\Bbb{Q}$. – Jyrki Lahtonen Apr 27 '20 at 07:03
  • One sentence summary: The limit points in the $p$-adic context are completely new objects - not something you have seen earlier in your study of mathematics. – Jyrki Lahtonen Apr 27 '20 at 07:09
  • @JyrkiLahtonen Thank you very much it has now been so much clarified –  Apr 27 '20 at 07:17
  • @JyrkiLahtonen I understood my last mistake about one of the examples that I asked, I removed my question. Thank you so much again! Especially the examples really helped me. –  Apr 27 '20 at 07:24

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