# Valuation of $x^\lambda$ in a complete, $p$-valued group

Suppose for $$p$$ a prime that $$(G,\omega)$$ is a complete $$p$$-valued group, $$x \in G$$ and $$\lambda \in \mathbb Z_p$$ (the $$p$$-adic integers).

Let $$x^\lambda$$ denote the unique element of $$G$$ such that $$\forall t>0, \; x^\lambda G_t = x^{\lambda_t}G_t$$ for all $$\lambda_t \in \mathbb Z$$ such that $$v_p(\lambda - \lambda_t) >t$$, where:

$$G_t = \{x \in G \mid \omega(x) \geq t\}$$ and $$v_p$$ is the $$p$$-adic valuation on $$\mathbb Z$$.

How can I show that: $$\omega(x^\lambda) = \omega(x) + v_p(\lambda)$$?

Using the fact that $$G$$ is complete, we may then consider the identification of $$G$$ with $$\hat G$$ to see that:

$$\omega(x^\lambda) = \text{min}_{t>0}(\omega(x^{\lambda_t}) \mid x^{\lambda_t} \notin G_t)$$

And then for fixed $$t$$ we may write $$\lambda_t = p^{r_t}m_t$$ where $$p \nmid m_t$$ to get:

$$\omega(x^{\lambda_t}) = \omega(x^{m_t}) + r_t$$, using the fact that $$(G,\omega)$$ is a $$p$$-valued group.

It seems now that I want to show $$\omega(x^{m_t}) = \omega(x)$$ but how might I go about showing this?

## 1 Answer

Unfortunately, you do not tell us what definition and properties of of $$p$$-valuations you want to use here. I follow the original definitions of Lazard, Groupes anlytiques p-adiques (here, II(1.1.1), p. 45 and III(2.1.2), p. 81). The books Analytic pro-p Groups by Dixon, duSautoy, Mann, Segal and p-Adic Lie Groups by Schneider might use slightly different axiomatisations, I do not have them before me right now.

Now basically what you have to notice is that

if $$x,y \in G$$ with $$\omega(x) \neq \omega(y)$$, then $$\omega(xy) = \min(\omega(x), \omega(y))$$

and

for all $$x \in G$$, we have $$\omega(x^{p^n}) = \omega(x) +n$$.

The first follows from the axiom $$\omega(xy^{-1}) \ge \min(\omega(x), \omega(y))$$ (which first of all implies $$\omega(x) = \omega(x^{-1})$$) with the same trick one uses to show the analogue for non-archimedean valuations / ultrametrics; the second follows by induction from Lazard's axiom III(2.1.2.3). By the way, note the stronger statement Lazard proves a few lines later in (2.1.4).

Now all you need is that for any $$\lambda \in \Bbb N$$ with $$gcd(p, \lambda) =1$$, there exist $$a,b \in \Bbb Z$$ such that $$b\lambda-pa=1$$ and hence

$$\omega(x^\lambda) \le \omega((x^\lambda)^b) = \omega(x \cdot x^{pa}) =\omega(x)$$

(for the first inequality, use the axiom mentioned, and for the last equality, use the two properties highlighted above). The reverse inequality is again obvious from the axiom.

This answers this question by setting $$\lambda =m_t$$, and your other question https://math.stackexchange.com/q/3114057/96384 which is just the special case $$1\le \lambda \le p-1$$.

The whole procedure, by the way, is modelled after raising e.g. principal units of $$p$$-adic valuation rings, or $$(1+T)$$ in $$\Bbb Z_p[[T]]$$, to $$\Bbb Z_p$$-powers, which I suggest you look up as a basic case; it should be covered in any course that goes beyond the absolute basics of $$p$$-adic theory.