Are the 2-adic valuations for $2^{-n/m}$ algebraically reliable?

Question

Interpret each of the base $p$ Lyndon Words $W$ of length $n$ as a base $p$ number and let it represent the $p-$adic number which it naturally represents:

$\displaystyle{\sum_{k=0}^\infty}p^{kn}W$

So e.g. $1_2$ represents the number $-1$ and $01_2$ represents $-\frac13$

The function $x\mapsto(x-2^{\nu_2(x)})$ acts on 2-adic numbers by orbiting the cycles of each Lyndon word if we give ourselves a little leeway in ignoring some trailing zeroes.

Letting $m$ count the number of $1$'s in any base $2$ Lyndon word, then $f(x)=\dfrac{x-2^{\nu_2(x)}}{2^{n/m}}$ deals with any trailing zeroes such that $f$ is now periodic with $f^m(x)=x$ provided the 2-adic valuations are reliable.

By $\nu_2(x)$ being "algebraically reliable" I mean for all possible valuations $\nu$ then $f(\nu x)=\nu f(x)$. Note that $f$ is a special case of $ax+b$.

Question

Are these 2-adic valuations for $2^{-n/m}$ algebraically reliable?

I've seen extensions I think for $2^{-1/2^n}$ and for roots of unity but I don't know how far this principle extends and whether it extends to $p^{-n/m}$.

EDIT

I have this much which I need to parse:

an element $α$ whose minimal polynomial over $\Bbb Q_2$ has degree $d$ is given the value $|N_{\mathbb{Q}_2(\alpha)|\mathbb{Q}_2}(\alpha)|^{1/d}$. This is treated extensively in any book on p-adics, local fields etc.

From this, I think if $x=2^{m/n}$ then its minimal polynomial is the unique irreducible monic polynomial of smallest degree $p(x)$ with rational coefficients such that $p(x)=0$ and whose leading coefficient is $1$.

So if I simplify initially to find for $m=1$ then by Eisenstein's irreducibility criterion I think I want $\alpha$ to be the solution to a polynomial $x^n-p$, i.e. $x^n-2$.

But then I can't parse $|N_{\mathbb{Q}_2(\alpha)|\mathbb{Q}_2}(\alpha)|^{1/d}$

Answer 1

It seems that you want something like: For any $a, b \in K$, the function $f: x \mapsto ax + b p^{v_p(x)}$ "should" satisfy

$$f(\color{red}{p^{\frac{m}{n}}} x) \stackrel{?}= \color{red}{p^{\frac{m}{n}}} f(x).$$

To make you aware of a big problem in this:

Good news: The standard $p$-adic valuation $v_p : \mathbb Q_p \rightarrow \mathbb Z \cup \{\infty\}$, and $p$-adic absolute value $\lvert \cdot \rvert_p : \mathbb Q_p \rightarrow p^{\mathbb Z} \cup \{0\}$, extend uniquely all the way up to $\mathbb C_p$. We get maps

$v_p : \mathbb C_p \rightarrow \mathbb Q \cup \{\infty\}$, and

$\lvert \cdot \rvert_p : \mathbb C_p \rightarrow p^{\mathbb Q} \cup \{0\}$.

But beware very bad news: This last codomain $p^{\mathbb Q}$ is a subset of the real numbers $\mathbb R$. In particular, while

$$v_p (\color{red}{p^{\frac{m}{n}}}x) = \frac{m}{n} + v_p(x)$$

holds true if $\color{red}{p^{\frac{m}{n}}}$ denotes any of the (up to $n$ distinct) elements of $K$ whose $n$-th power is $p^m$, to write

$$ \lvert \color{red}{p^{\frac{m}{n}}} x \rvert_p = \color{blue}{p^{-\frac{m}{n}}} \lvert x \rvert_p,$$

as true and tempting as it is, is extremely treacherous. Because: The red $p^{\frac{m}{n}}$, as said, can stand for any of the up to $n$ elements $\color{red}{\text{in } K}$ such that (...), while the blue $p^{\frac{m}{n}}$ denotes the unique (if $n$ is even: unique positive) number $\color{blue}{\text{in }\mathbb R}$ whose $n$-th power is $p^m$. Those red numbers have very little in common with that blue number, as we are in the trap exposed in my little essay here.

To pin it down: To even think that writing $$x \mapsto p^{v_p(x)}$$ defines a map, as soon as the valuation $v_p$ takes non-integer values, runs into a problem:

• If we mean it as a map $K \rightarrow \mathbb R$, (which in standard notation one would write $x \rightarrow \lvert x \rvert_p^{-1}$), then its codomain is not its domain. Which implies that you cannot compose it with itself, i.e. expressions like $f^{\circ k}$ make no sense. Neither would something like $\color{red}{ax+b}\color{blue}{p^{\frac{m}{n}}}$, because $a,x,b \in K$ while $p^{\frac{m}{n}} \in \mathbb R$.

• If we mean it as a map $K \rightarrow K$, i.e. want $K$ to be the codomain, then it is not clear which of the up to $n$ elements in $K$ we would mean by writing "$p^{\frac{m}{n}}$". And even if we somehow agree on choices here, such a "re-interpreted" value map would no longer share any of the useful properties of the actual absolute value $\lvert \cdot \rvert_p$ (I am not even sure if one could make such a choice multiplicative, in any case the triangle inequality can not even be stated now!)