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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.

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    $\begingroup$ 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 $\mathbb{Z}_2$. $\endgroup$ – user10354138 Jun 3 at 8:37
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    $\begingroup$ 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. $\endgroup$ – user10354138 Jun 3 at 10:45
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    $\begingroup$ @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 $1/v$. $\endgroup$ – user10354138 Jun 3 at 15:45
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    $\begingroup$ @user10354138 user334732 That's quite right, I deleted that comment. $\endgroup$ – Torsten Schoeneberg Jun 3 at 15:48
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    $\begingroup$ @user334732 When $p$ is composite (and not a prime power), then zero-divisors exist. $\endgroup$ – rukhin Jun 4 at 3:20
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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).

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  • $\begingroup$ 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? $\endgroup$ – samerivertwice Jun 3 at 16:02
  • $\begingroup$ 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 $\lvert 2\rvert_8=1/2$. $\endgroup$ – samerivertwice Jun 3 at 16:27
  • $\begingroup$ 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. $\endgroup$ – Torsten Schoeneberg Jun 3 at 17:34
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    $\begingroup$ 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 $(\sqrt{-7}, 1/2) \in \Bbb Z_2 \times \Bbb Z_3$? $\endgroup$ – Torsten Schoeneberg Jun 3 at 17:44
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    $\begingroup$ ... 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: math.stackexchange.com/q/2298779/96384 $\endgroup$ – Torsten Schoeneberg Jun 3 at 20:12

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