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How do p-adic fields degenerate for non-prime $p$?

Let $d(x,y)$ be the inverse of the highest power of $4$ that divides $\lvert x-y\rvert$

Then let $\Bbb Z_4$ be the completion of $\Bbb Z$ under this distance.

I expect this to be degenerate to some degree, since non-prime valuations fail to yield unique representations for numbers.

How much understanding do we have of how this degenerates? For example, are there distinct elements in $\Bbb Z$ for which $d(x,y)=0$ and can we define the infinite sequences that connect them?

What numbers, if any, will have multiple representations?

I'm asking this to better understand in general - but also with half an eye on the graph of the Collatz function, so if there is an example of how sequences are arbitrarily close in $\Bbb Z_4$, sequences of the form $S_n=4^n x+\dfrac{4^n-1}3$ which converge to $-\frac13$ and then by proxy all sequences of the form $2^n\cdot S_n:n\in\Bbb Z$, which converge to the set $-\dfrac{2^n}3$ would be the most useful examples to me.

it's a hire car baby
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    The Cauchy sequences in the $\mathbb{Z}_4$-metric are the same as those in $\mathbb{Z}_2$ (agree to high power of $4$ iff agree to high power of $2$), so you just get $\mathbb{Z}_2$. – user10354138 Jun 03 '19 at 08:37
  • What do you mean by “p-adic fields degenerate for non-prime $p$”? When $p$ is composite, you don’t have a field of $p$-adic numbers to begin with. – rukhin Jun 03 '19 at 09:32
  • @user10354138 I wasn't sure about balls of radius $\frac12$ e.g. $2$ is outside the ball of radius $1$ around $0$ in $\Bbb Z_4$. Are you sure this doesn't cause anything to degenerate? – it's a hire car baby Jun 03 '19 at 10:40
  • @rukhin would that include when the composite number is $p^2: p\in\text{prime}$ - would that not be a field, contrary to what user10354138 says? – it's a hire car baby Jun 03 '19 at 10:41
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    it doesn't cause any problem. You just need to fit two balls in there instead of one, and on the opposite direction you can fit two "smaller" balls ("smaller" in set-theoretic, not the metric sense). You still get, e.g., totally disconnected, same open subgroups, etc. – user10354138 Jun 03 '19 at 10:45
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    @TorstenSchoeneberg didn't he do it with $c=1/4$? That was my interpretation of "inverse of the highest power of 4 that divides $\lvert x-y\rvert$", i.e. the number $4^{-v}$ rather than $1/v$. – user10354138 Jun 03 '19 at 15:45
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    @user10354138 user334732 That's quite right, I deleted that comment. – Torsten Schoeneberg Jun 03 '19 at 15:48
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    @user334732 When $p$ is composite (and not a prime power), then zero-divisors exist. – rukhin Jun 04 '19 at 03:20
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    Compare also https://math.stackexchange.com/q/1919274/96384 – Torsten Schoeneberg Feb 23 '24 at 21:46

1 Answers1

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Quite generally, completing $\Bbb Z$ with respect to an $n$-adic metric gives the direct product of the $p$-adic numbers for those prime $p$ which divide $n$: $$\Bbb Z_n := \varprojlim_{k} \Bbb Z /n^k \simeq \quad... \quad\simeq \prod_{p \vert n, \;p \text{ prime }} \Bbb Z_p$$ (fill in the missing steps with the Chinese Remainder Theorem and cofinal index sets in inverse limits).

In particular, if $n$ is the power of one prime $p$, you just get back the $p$-adic integers. One way to easily see that is by noticing that metrics $d_p$ and $d_{p^i}$ induce the same topology (even uniform structure), as each open ball of one contains an open ball of the other. (And that is a reformulation of the directed sets $(p^r)_r$ and $(p^{ir})_r$ in the inverse limits being cofinal.)

Compare also 4-adic numbers and zero divisors and Are the $p^n$-adic numbers isomorphic to the $p$-adic numbers? (where the argument might be a bit too short by making an additional assumption though, see comments).

Torsten Schoeneberg
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  • Then is this a direct product...? e.g. if $n=6$, we get the direct product of a base 2 and base 3 string, which may have multiple representations as a single base $6$ string? – it's a hire car baby Jun 03 '19 at 16:02
  • Professor Lubin has thrown me with his answer which you linked because he appears to state that distance in $\Bbb Z_n:n\not\in\text{prime}$ is no longer the highest power of $n$ that divides $\lvert x-y\rvert$ As only $8^0=1$ divides $2$ so by my definition $\lvert 2\rvert_8=1$ but he gives $\lvert 2\rvert_8=1/2$. – it's a hire car baby Jun 03 '19 at 16:27
  • I think Prof. Lubin might tacitly make the assumption that we are talking of multiplicative values. Whereas if we define an $n$-adic valuation the way we do it here, it is just submultiplicative (and cannot be multiplicative unless $n$ is prime); the good thing is, for $n=p^i$ we still get the $p$-adics by the argument here and in the other linked question. – Torsten Schoeneberg Jun 03 '19 at 17:34
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    Re first comment: Yes, direct product. Assuming by "string" for $x \in \Bbb Z_n$ you mean a sequence of $a_i$'s, each $a_i \in \lbrace 0,...,n-1 \rbrace$, s.t. $x \equiv \sum_{i=0}^{r-1} a_i n^i$ mod $n^r$, then I think these representations are unique, i.e. a pair of a $2$-adic string and a $3$-adic string corresponds to a unique $6$-adic string, and vice versa. It might be a good exercise to do that: What $6$-adic string corresponds to $(\sqrt{-7}, 1/2) \in \Bbb Z_2 \times \Bbb Z_3$? – Torsten Schoeneberg Jun 03 '19 at 17:44
  • I mastered rational number representations but don't know where to start with irrational numbers. – it's a hire car baby Jun 03 '19 at 17:53
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    ... which is fair, as one needs computation for each bit, and should restrict oneself to, say, the first five digits or so. Many approaches tailor-made for $2$-adic square roots are to be found here: https://math.stackexchange.com/q/2298779/96384 – Torsten Schoeneberg Jun 03 '19 at 20:12
  • I was unclear what you meant by "index sets" here. Is an index set the radix? – it's a hire car baby Dec 21 '21 at 12:54
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    E.g. for any $r \in \mathbb N$, the index sets $p^k_{k \in \mathbb N}$ and $(p^r)^k_{k \in \mathbb N}$ are cofinal in each other, hence the inverse limits $$\varprojlim_{p^k} \Bbb Z /p^k (=: \mathbb Z_p)$$ and $$\varprojlim_{(p^r)^k} \Bbb Z /(p^r)^k (= \mathbb Z_{p^r})$$ are isomorphic. (Lazily one might write them as $\varprojlim_{k} \Bbb Z /p^k$ and $\varprojlim_{k} \Bbb Z /(p^r)^k$, but that kind of hides the cofinality in the objects over which the limit is taken.) – Torsten Schoeneberg Dec 21 '21 at 17:09