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let $Z_p$ denote the $p$-adic integers, then it has a topology as a subspace of $\prod_nZ/p^nZ$, where $Z/p^nZ$ is given the discrete topology.

(reference I posted before: Why $Z_p$ is closed.)

Now let $v_p(x)=n$ if $x=p^nu$ with $u$ not divisible by $p$. (That is, $v_p(x)$ is the $p$-adic valuation.) Then the topology on $Z_p$ can also be defined by the distance $d(x,y)=e^{-v_p(x-y)}.$

Why are those two topologies equivalent?

I think the basis of $Z_p$ is $\{Z_p \bigcap \pi^{-1}_n(s)\}_{s\in Z/p^nZ,\, n\in \mathbb{N}}$ as subspace topology, ($\pi_n$ the projection map from $\prod_nZ/p^nZ$ to $Z/p^nZ$). But it's hard to check each $\{Z_p \bigcap \pi^{-1}_n(s)\}$ is open as distance topology.

And the open ball in distance topology is also hard to verify as union of basis mentioned above.

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  • $\begingroup$ In fact the two topologies are not only equivalent, but they are in fact equal, see my proof below. $\endgroup$
    – Olórin
    Jan 22 '15 at 12:40
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I will give an elementary proof of something stronger : the equality of the two topologies.

Note (and it is elementary to prove) that $v_p : \mathbf{Z}_p \backslash\{0\} \to \mathbf{N}$ has the following properties : $v_p (xy) = v_p(x) v_p(y)$ and $v_p(x+y)\geq \inf(v_p(x), v_p(y))$ for each $x,y$ such that $v_p (xy)$ and $v_p(x+y)$ have a meaning. (If you want you can extend $v_p$ at $0$ by setting $v_p (0) = +\infty$.) This obviously implies that (the ring) $\mathbf{Z}_p$ is a domain.

Now, the ideals $p^n \mathbf{Z}_p$ form a fundamental system of neighbourhoods of $0$ in $\mathbf{Z}_p$ for the product topology. (By abuse of language I call product topology on $\mathbf{Z}_p$ the topology induced on $\mathbf{Z}_p$ by the product topology on the product $\prod_{n\in\mathbf{N}} \mathbf{Z}/p^n \mathbf{Z}$ of which $\mathbf{Z}_p$ is a subspace.) Indeed, let $V$ be a neighbourhood of $0$ for the product topology. $V$ contains therefore an open (for the product topology...) $U$ containing $0$. By definition, $U = \mathbf{Z}_p \cap U'$ where $U'$ is open in $\prod_{n\in\mathbf{N}} \mathbf{Z}/p^n \mathbf{Z}$ of which $\mathbf{Z}_p$ for the product topology. By definition, $U'$ is union of sets of the form $\prod_{n\in\mathbf{N}} A_n$ where for each $n$ the set $A_n$ is open in $\mathbf{Z}/p^n \mathbf{Z}$, that is, is just a subset of $\mathbf{Z}/p^n \mathbf{Z}$ as the latter is endowed with the discrete topology, and where $A_n \not= \mathbf{Z}/p^n \mathbf{Z}$ only for finitely many values of $n$. Obviously one can assume that $U'$ is of this form, $U' = \prod_{n\in\mathbf{N}} A_n$, with the hypothesis previously made on the $A_n$'s kept. Let $d$ be the last integer for which $A_d \not= \mathbf{Z}/p^d \mathbf{Z}$. Then $A_n = \mathbf{Z}/p^n \mathbf{Z}$ contains $p^d (\mathbf{Z}/p^n \mathbf{Z})$ when $n > d$, and then $U'$ contains $p^d \prod_{n\in\mathbf{N}} \mathbf{Z}/p^n \mathbf{Z}$, so that $U$ contains $p^d\mathbf{Z}_p$. We have shown that each neighbourhood of $0$ contains a suitable $p^d\mathbf{Z}_p$, showing therefore that the ideals $p^n \mathbf{Z}_p$ form indeed a fundamental system of neighbourhoods of $0$ in $\mathbf{Z}_p$. But $x\in p^d\mathbf{Z}_p$ if and only if $v_p(x)\geq n$, if and only if $d(0,x) \leq e^{-n}$ if and only if $d(0,x) < e^{-(n-1)}$ so that the $p^n \mathbf{Z}_p$ are open balls in the metric topology.

The operations (sum and product) are continuous in the various $\mathbf{Z}/p^n \mathbf{Z}$ as these rings have the discrete topology, so that the "compositions" of operations of $\mathbf{Z}_p$ with the projections $\mathbf{Z}_p \to \mathbf{Z}/p^n \mathbf{Z}$ are all continuous, and by definition of the product topology, this means that the operations of $\mathbf{Z}_p$ are continuous.

The continuity of the operations implies that translations in $\mathbf{Z}_p$ are continuous, and this implies that for each $x\in\mathbf{Z}_p$ the sets $x+p^n \mathbf{Z}_p$ (which is the open ball of center $x$ and radius $e^{-(n+1)}$ as well as the closed ball of center $x$ and radius $e^{-n}$) form a fundamental basis of neighbourhoods of $x$ in both topopolgies (product and metric) on $\mathbf{Z}_p$. This shows that the product topology and that the metric topology are in fact equal. (Not only equivalent.)

Remark. I have shown that $p^n \mathbf{Z}_p$ are neighbourhoods of $0$ in the product topology. They are in fact of course open in this topology. Indeed, we saw that they are open balls for the metric topology, and that the metric and product topology where equal. But can we prove this directly ? Yes ! $x \in p^n \mathbf{Z}_p$, then $x + p^n \mathbf{Z}_p \subseteq p^n \mathbf{Z}_p$, and $x + p^n \mathbf{Z}_p$ is a neighbourhood of $x$ in the product topology. (We saw this fact earlier, with continuity of translations.) This means that $x$ in in the interior of $p^n \mathbf{Z}_p$ for the product topology. As it is true for all $x$, this implies that $p^n \mathbf{Z}_p$ is open in the product topology. Actually, and it is simple algebra also, one can prove that $p^n \mathbf{Z}_p$ is the kernel of the $n$-coordinate $\mathbf{Z}_p \rightarrow \mathbf{Z} / p^n \mathbf{Z}$ which is obviously continuous, and as $\{0\}$ is open in the target, its inverse image, $p^n \mathbf{Z}_p$, is also open.

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  • $\begingroup$ Why is $p^n Z_p$ itself open in product topology? Otherwise we can only conclude that open sets in product topology are open in metric topology. (BTW, I think $d(0,x)<e^{n+1}$ should be $d(0,x)<e^{n-1}$, though it's not important here) $\endgroup$
    – CYC
    Jan 23 '15 at 0:51
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    $\begingroup$ @CYC For the $n$, $n-1$ stuff you are right, I made a typo, and just edited my answer to correct it, but in your comment you forgot minus signs in front of them in the exponentials. For your comment's question, you have $p^n \mathbf{Z}_p = \mathbf{Z}_p \cap (\underbrace{\{0\} \times\ldots\times \{0\}}_{n\textrm{ times}} \times \prod_{k > n} p^n \mathbf{Z}/p^k\mathbf{Z})$ and $\underbrace{\{0\} \times\ldots\times \{0\}}_{n\textrm{ times}} \times \prod_{k > n} p^n \mathbf{Z}/p^k\mathbf{Z}$ is open in $\prod_{k\in\mathbf{N}^{*}} \mathbf{Z}/p^k\mathbf{Z}$ as each of its factor are open... $\endgroup$
    – Olórin
    Jan 23 '15 at 1:07
  • $\begingroup$ @CYC ... the $\mathbf{Z}/p^k\mathbf{Z}$ being endowed with the discrete topology. $\endgroup$
    – Olórin
    Jan 23 '15 at 1:10
  • $\begingroup$ @CYC All of this is very straightforward, but somehow very mildly tedious to check. I followed all this way when I read JP Serre's "Cours d'arithmérique". ;-) Anyway, if you're happy with my answer, don't hesitate to upvote it ! ;-) $\endgroup$
    – Olórin
    Jan 23 '15 at 1:14
  • $\begingroup$ I think $p^nZ_p = Z_p \cap (\{0\}....\times \prod_{k>n} Z/p^kZ)$ is correct, since in $Z_p$ the (n+1)-th component must be divided by $p^n$ for its n-th component is 0, and it's done. But for $p^nZ/p^kZ$, is infinite product of open sets still open? isn't it need to be all but finite $\neq$ whole space(product factors)? $\endgroup$
    – CYC
    Jan 23 '15 at 1:24
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Here's one possible answer, although it may not be the one you are looking for. I think you can show that with the profinite topology, $\text{Frac}(\mathbb{Z}_p)$ is complete with the $v_p$-metric, which clearly extends to $p$-adic metric on $\mathbb{Q}$. Also, it's clear that $\mathbb{Z}$ is dense in $\mathbb{Z}_p$, and so $\mathbb{Q}$ (with a little more work) is dense in $\text{Frac}(\mathbb{Z}_p)$. This tells you that $\text{Frac}(\mathbb{Z}_p)$ with your defined metric is the completion of $\mathbb{Q}$ with the $p$-adic absolute value, thus by the uniqueness of completion, it's isomorphic to $\mathbb{Q}_p$. Since valuation ring goes to valuation ring, your problem is solved.

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  • $\begingroup$ If you look closer to your answer, you will see that you haven't shown anything except the fact that $\mathbf{Z}_p$ constructed as projective limit, endowed with the metric topology defined by vp is isometric to $\mathbf{Z}_p$ defined as completion of $\mathbf{Z}$ (or, it is the same, as the unit ball of the completion of $\mathbf{Q}$) for the $p$-adic absolute value or $v_p$. This does not imply at all that the product and the metric topologies on $\mathbf{Z}_p$ seen as projective limit are equivalent, which is what the OP is asking for. $\endgroup$
    – Olórin
    Jan 22 '15 at 14:40
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    $\begingroup$ @RobertGreen I see now that this is technically true, but I thought this was the more 'obvious' (should be read 'well known' part) of the question--the inverse limit topology is just the one inherited from the product topology. $\endgroup$ Jan 23 '15 at 3:35

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