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

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.

• 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 3 at 8:37
• 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 3 at 10:45
• @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 3 at 15:45
• @user10354138 user334732 That's quite right, I deleted that comment. – Torsten Schoeneberg Jun 3 at 15:48
• @user334732 When $p$ is composite (and not a prime power), then zero-divisors exist. – rukhin Jun 4 at 3:20

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).
• 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? – samerivertwice Jun 3 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 $\lvert 2\rvert_8=1/2$. – samerivertwice Jun 3 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 3 at 17:34
• 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 3 at 17:44
• ... 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 – Torsten Schoeneberg Jun 3 at 20:12