Proposition 1

Let $r$ and $s$ be two integers $\geqslant 0$. If $H_{r,s} = \sum_{b \in B} \lambda_b e_b$, where $\lambda_b \in \mathbf{Q}$, is the decomposition of $\mathbf{H}$ with respect to the basis $(e_b)_{b \in B}$, then

The ring $\mathbf{A}_{\mathbf{Z}_{(p)}}(I)$ is identified with the sub-$\mathbf{Z}_{(p)}$-module of $\mathbf{A}_{\mathbf{Q}}(I)$ generated by the words $w \in \mathrm{Mo}(I)$. As $\mathbf{L}_{\mathbf{Z}_{(p)}}(I)$ is a direct factor of $\mathbf{A}_{\mathbf{Z}_{(p)}}(I)$,

Let $f$ be the integer such that $f \leq (r + s - 1)\theta < f + 1$. Relation (4) is equivalent to $v_p(\lambda_b) \geq -f$ for all $b \in B$, that is $H_{r,s} \in p^{-f}L_{\mathbf{Z}(p)}$. But this is equivalent also by (5) to $H_{r,s} \in p^{-f}A_{\mathbf{Z}(p)}(I)$.

By formula (11) of § 6, no. 4, it suffices to show that, for every integer $m \geq 1$ and all integers $r_1, \ldots, r_m, s_1, \ldots, s_m$ such that

we have But by Lemma 2, $v_p(r_i!s_i!) \leq (r_i + s_i - 1)\theta$ and $v_p(m!) \leq v_p(m!) \leq (m - 1)\theta$; the left hand side of (7) is therefore bounded above by as it is an integer, it is $\leq f$, which completes the proof.