Proposition 9

Suppose that $p \neq 0$. (i) Let $a, b$ be non-zero ideals of $A$ such that $b \subset a \subset m$. In the group $G(a)/G(b)$, every element has order a power of $p$. (ii) Suppose that $v(p) = 1$. If $x \in G$ is such that $\omega(x) > \frac{1}{p-1}$, then

By § 5, formula (4), for all $x \in G$, where $c_\alpha \in A'$ for all $\alpha$. Even for proving (i) it can be assumed that $v(p) = 1$. Then if $\omega(x) \geq 1$, it follows that $\omega(x^p) \geq \omega(x) + 1$ and hence $\omega(x^{p^n})$ tends to $+\infty$ as $n$ tends to $+\infty$; this proves (i). As $\binom{p}{i}$ is divisible by $p$ for $1 \leq i \leq p - 1$, Proposition 2 of § 5, no. 3, proves that $c_\alpha \in pA'$ for and hence On the other hand, if $|\alpha| \geq p$, $\omega(c_\alpha x^\alpha) \geq p \omega(x)$ and $p \omega(x) > \omega(x) + 1$ if $\omega(x) > \frac{1}{p - 1}$. This proves (ii).