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?

  • 2
    $\begingroup$ $v$ 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 $v(r_1+\cdots+r_n)=v(r_n)$. $\endgroup$ – user26857 Feb 24 at 18:45

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