# Clarification on a proof about $p$-valued groups.

First we have a few definitions:

• $$(R,v)$$ is a filtered ring if $$v : R \rightarrow \mathbb R^{\geq0} \cup \{\infty\}$$ satisfies:

1. $$v(r-s) \geq \min\{v(r), v(s)\}$$

2. $$v(rs) \geq v(r) + v(s)$$

3. $$v(1) = 0$$

4. $$v(0) = \infty$$

• And additionally we say the filtered ring is separated if $$v(r) = \infty \Rightarrow r = 0$$

• $$(G, \omega)$$ is a filtered group if $$\omega : G \rightarrow \mathbb R^{>0} \cup \{\infty\}$$ satisfies:

1. $$\omega(xy^{-1}) \geq \min\{\omega(x), \omega(y)\}$$

2. $$\omega(x^{-1}y^{-1}xy) \geq \omega(x) + \omega(y)$$

• And additionally we say the filtered group separated if $$\omega(g) = \infty \Rightarrow g = e$$

• A separated, filtered group $$(G, \omega)$$ is a $$p$$-valuation for a prime $$p$$ if:

1. $$\omega(g) > \frac{1}{p-1}$$

2. $$\omega(g^p) = \omega(g) + 1$$

With the necessary definitions out of the way, the Proposition I am confused by is:

Suppose $$(R,v)$$ is a separated, filtered ring such that $$v(pr) = v(r) + 1, \; \forall r \in R$$ for some prime $$p$$.

Let $$G = \{r \in R \mid r \text{ is a unit and } v(r-1) > 0\}$$ and define $$\omega: G \rightarrow \mathbb R^{>0} \cup \{\infty\}$$ by $$\omega(r) = v(r-1)$$. It can be seen that this defines a separated, filtered group $$(G, \omega)$$.

Then $$\omega$$ restricts to a $$p$$-valuation on the subgroup $$G_{\frac{1}{p-1}^+} = \{g \in G \mid \omega(g) > \frac{1}{p-1}\}$$

Below is the proof we are given, wherein $$v_p$$ denotes the $$p$$-adic filtration on the integers.

That $$\omega$$ restricts to a separated filtration on $$G_{\frac{1}{p-1}^+}$$ is easy to check.

Given $$x \in G_{\frac{1}{p-1}^+}$$, we must show that $$\omega(x^p) = \omega(x) + 1$$

But $$x^p - 1 = (1 + (x-1))^p - 1 = \sum_{i = 1}^{p} {p\choose i}(x - 1)^i$$

$$v_p({p \choose i}) \geq 1, \; \forall 1 \leq i \leq p-1$$, so:

$$v(p (p^{-1}{p \choose i})(x-1)^i) \geq 1 + iv(x-1) > 1 + v(x-1)$$ for $$i \geq 2$$

Finally $$v((x-1)^p) \geq pv(x-1) > v(x-1)+ 1$$ since $$v(x-1) > \frac{1}{p-1}$$

So $$v(x^p - 1) = v(x-1) + 1$$

My problem with the proof is I don't see how what we did allows us to make the final conclusion. We showed that almost everything in the sum that gives $$x^p - 1$$ as an element of $$R$$ has value greater than $$v(x-1) + 1$$, with the exception of the term $$p(x-1)$$ which has value exactly $$v(x-1) + 1$$.

But surely because $$(R,v)$$ is a separated, filtered ring, by the properties above we then have that $$v(x^p - 1) \geq \min\{v({p \choose i}(x-1)^i)\} = v(x-1) + 1$$, and thus all we have is a lower bound?

Is the upper bound trivial for some reason that I've missed? I don't entirely see why it's then obvious that we actually have equality here?

• $v$ behaves like a valuation, that is, if $v(r)>v(s)$ then $v(r\pm s)=v(s)$. You can generalize: if $v(r_1)>\cdots>v(r_n)$, then $v(r_1+\cdots+r_n)=v(r_n)$. – user26857 Feb 24 at 18:45