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For the sake of a simple example I can work with, let $\Bbb Q_2[\zeta]$ be some extension of a set of 2-adic numbers to include a square root of two.

I interpret this comment to mean these numbers can be written with radix $\{0,1,2,3\}$ and in general $m^{th}$ roots of $p^n$ with $m<n$ can be represented with the radix $\{0,\ldots p^n-1\}$. But what is the base here? Is it still two, or is it now some square root of two (or in general some $m^{th}$ root of $p^n$?

The term "representative map" appears to be in general usage but I can't find a definition. I'm assuming this is similar to picking Teichmuller representatives, right?

How do I lift the square roots of two to representatives? From my limited superficial experience I'm guessing the process is something like: take a polynomial $P(x)=x^2-2$ whose zeroes are roots of two and keep taking residues mod $p^n$ for increasing $n$. But I'm not clear how this works in base $\sqrt2$ and with radix $\{0,1,2,3\}$

Although my specific query here is in base $2$ I think the general question is What's a representative map?

Robert Frost
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In my comment you link to, all I am saying is that if $x \in \mathbb Z_p$, then an expression like

$$ x- (x \text{ mod } p^n)$$

(which you write in your question there) does not make sense, because $x \text{ mod } p^n$ is an element of the quotient $\mathbb Z_p / (p^n)$, and you cannot add or subtract elements of $\mathbb Z_p / (p^n)$ to or from elements of $\mathbb Z_p$.

From the context I guessed what you actually meant there was the truncation map

$$\sum_{k \ge 0} a_k p^k \mapsto \sum_{k \ge n+1} a_k p^k$$

where I further guessed that you use the popular although non-canonical choice of the $a_k$ being chosen from the natural numbers $\{0, \dots, p-1\}$. Which are representatives of the elements of the quotient $\mathbb Z_p/(p)$ in $\mathbb Z_p$. More generally, the natural numbers $\{0, \dots, p^n-1\}$ form a set of representatives of $\mathbb Z_p/(p^n)$ in $\mathbb Z_p$, and if

$$[ \cdot ] : \mathbb Z_p/(p^n) \rightarrow \mathbb Z_p$$

denotes the "representative map" which to each element of the quotient $$0 \text{ mod } p^n, 1 \text{ mod } p^n, \dots, p^n-1 \text{ mod } p^n$$ assigns the representative $$0, 1, \dots p^n -1,$$ then you can write the above truncation map conveniently as:

$$x \mapsto x - [x\text{ mod } p^n]$$

This is also convenient in that it shows that your truncation map, maybe unbeknownst to you, depends on that choice of representatives, and any different choice of a representative map would lead to a different "truncation" map which shares some but not all properties with the one you care about.

As a final note, for what it's worth, nowhere in here do I leave the $p$-adic integers. I know you want to adjoin some $\zeta$, but so far this issue here has absolutely nothing to do with that.

  • ok thank-you for the clarification. – Robert Frost Dec 13 '22 at 18:56
  • I do value this level of formalism despite it not coming naturally to me, because it's like keeping your units in physics. You were indeed right that the representatives looked canonical to me at first sight so I will invest some time in learning how they are not and what the alternative choices might be. – Robert Frost Dec 15 '22 at 15:42
  • Your notation $\mathbb Z_p / (p^n)$ would be the same thing as $\mathbb Z_p / p^n\Bbb Z_p$, so is analogous to $\Bbb Z/p^n\Bbb Z$, correct? (except of course with possibly infinite tail not necessarily ending in $\overline{r}:r\in{0\ldots p-1}$). – Robert Frost Dec 16 '22 at 12:30
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    Same thing, yes. For any (commutative) ring $R$, and any $a\in R$, the notations $aR$ and $(a)$ are interchangeable for denoting the principal ideal generated by $a$. – Torsten Schoeneberg Dec 16 '22 at 16:34