Proof. This follows from (1) and the super-additivity of $f^r$.

Proof. Let $x\in \mathcal {E}^r$. After multiplying by a power of $p$ and a power of $T$ we may assume that $x \in \mathcal {O}_{\mathcal {E}^r}$ and $w_0(x)= 0$. Then for $c$ sufficiently large, we have $x \in \mathcal {O}_{\mathcal {E}}^{r,c}$. The proposition follows from Lemma 3.1.

Proof. Let $F_\infty /F$ be the Abelian $p$-adic Lie extension corresponding to $\psi$. By local class field theory, $F_\infty /F$ corresponds to an open subgroup $H$ of $\mathcal {O}_F^\times$ such that $\mathcal {O}_F^\times / H \cong G_{F_\infty /F}$. The image of the subgroup $U^s = 1 + T^s\mathcal {O}_F$ in $\mathcal {O}_F^\times / H$ corresponds to the subgroup $G_{F_\infty /F}^s$. For any group $A$ we let $A^{\times p}$ denote $\{ a^p\mid a \in A\}$. As we are in characteristic $p$, we have $(U^s)^{\times p} \subset U^{ps}$ for any $s$. On the Galois side of the correspondence this means $(G_{F_\infty /F}^s)^{\times p} \subset G_{F_\infty /F}^{ps}$. However, because $G_{F_\infty /F}$ is isomorphic to $\mathbb {Z}_p^d$ for some $d$, we know that $(G_{F_\infty /F}^{s_{k}(\rho )})^{\times p} = G_{F_\infty /F}^{s_{k+1}(\rho )}$. The multiplicative case is identical.

Proof. Recall the inverse Hasse–Herbrand function

(3)\begin{equation} \psi_{K/F}(s) = \int_0^s [G_{K/F}^0:G_{K/F}^x] \, {{d} x}. \end{equation}

Let $F_\infty$ (respectively, $K_\infty$) be the fixed field of $\rho ^{-1}(\{1\})$ (respectively, $\rho |_{G_K}^{-1}(\{1\})$). We claim that for $s$ sufficiently large, the map $G_{K_\infty /K} \to G_{F_\infty /F}$ restricts to an isomorphism

(4)\begin{equation} G_{K_{\infty}/K}^{\psi_{K/F}(s)} \cong G_{F_\infty/F}^s. \end{equation}

To see this, let $K_1$ be a finite Galois extension of $F$ containing $K$. Using [Reference SerreSer79, § 4, Proposition 15], we see that

\begin{align*} G_{K_1/F}^s \cap G_{K_1/K} &= (G_{K_1/F})_{\psi_{K_1/F}(s)} \cap G_{K_1/K} \\ &= (G_{K_1/K})_{\psi_{K_1/F}(s)} \\ &= (G_{K_1/K})_{\psi_{K_1/K} \circ \psi_{K/F}(s)} \\ &= (G_{K_1/K})^{\psi_{K/F}(s)}. \end{align*}

By taking a limit along the finite subextensions of $K_\infty$ over $F$ we obtain

(5)\begin{equation} G_{K_\infty/F}^s \cap G_{K_\infty/K} = G_{K_\infty/K}^{\psi_{K/F}(s)}. \end{equation}

As $K/F$ is a finite extension we have $G_{K_\infty /F}^s \subset G_{K_\infty /K}$ for large $s$. In addition, for $s$ large the restriction map $G_{K_\infty /F}^s \to G_{F_\infty /F}^s$ is an isomorphism. Combining this with (5) proves (4). From (4) we see that $s_k(\rho |_{G_K})=\psi _{K/F}(s_k(\rho ))$ for $k$ sufficiently large. The proposition follows from (3).

Proof. This follows from Lemma 5.2 and Proposition 5.5.

Proof. Let $F_\infty$ be the fixed field of $\ker (\psi )$. Then $F_\infty /F$ is a $\mathbb {Z}_p$-extension, and thus by Artin–Schreier–Witt theory corresponds to an equivalence class of $W(F)/(1-\nu )W(F)$ (here $\nu$ denotes the standard $p$-power Frobenius endomorphism on the ring of $p$-typical Witt vectors). By [Reference Kosters and WanKW18, Proposition 3.1], for any $b \in \mathfrak {K}$ with $Tr_{\mathfrak {K}/\mathbb {F}_p}(b)\neq 0$, there exists a unique representative $a_0$ in $z$ of the form

\[ a_0 = r[b]+ \sum_{i\geqslant 1,(i,p)=1} r_i T^{-i}, \]

where $r \in \mathbb {Z}_p$ and $r_i \in W(\mathfrak {K})$ (here we regard $\mathcal {E}$ as a subring of $W(F)$ via the map $\iota$ defined at the beginning of § 6.1). Note that by Lemma 6.6, the element $a_0$ is maximal. As in § 6.1.3, from $a_0$ we obtain a surjective character $\psi _0: Gal(F_\infty /F) \to \mathbb {Z}_p$ (take $y_0$ satisfying $y_0^\nu - y_0 = a_0$ and then set $\psi _0(g)=y_0^g - y_0$). Thus, for some $b \in \mathbb {Z}_p$ we have $\psi =b\psi _0$, so that $a=ba_0$ is a maximal Frobenius structure of $\psi$. A proposition due to Kosters–Wan [Reference Kosters and WanKW18, Proposition 3.3] states that $s_n(\psi _0)$ is equal to $-c_n(a_0)$ (note that [Reference Kosters and WanKW18] state their result in terms of conductors, but this easily translates into a statement about higher ramification groups). From here we deduce that $s_n(\psi )=-c_n(a)$.

Proof. Let $\psi _0:G_F \to \mathbb {Z}_p^+$ be the character defined by $b^{-1}\psi$ and let $a_0$ be a maximal Frobenius of $\psi _0$. By Lemma 6.1 we see that $b\sum _{i=0}^{f-1} a_0^{\nu ^i}$ is a Frobenius structure of $\psi$. The result follows from Corollary 6.5 and Theorem 6.9.

Proof. We may replace $L$ with a field that is Galois over $\mathbb {Q}_p$. Let $r=\min _{b \in B}(v_p(x_bb))$ and write $r={i}/{e}+n$ with $0\leqslant i < e$. Let $B'\subset B$ consist of all $b\in B$ such that $v_p(x_bb)=r$. Note that $B'$ is a subset of $B_i$ from Definition 6.11. For each $b\in B'$ we write $b=u_b\pi ^i p^{r_b}$ and $x_b=v_b p^{s_b}$, where $u_b\in \mathcal {O}_L^\times$ and $v_b \in W(\mathfrak {K})^\times$. Let $\overline {A}$ be a matrix whose columns are $[\overline {u}_b,\overline {u}_b^p, \dots, \overline {u}_b^{p^{f-1}}]^T$ for each $b \in B'$. By the definition of nice, we know that the rank of $\overline {A}$ is $|B'|$. Thus, there exists $j$ such that

\[ \sum_{b \in B'} \overline{u_b}^{p^j} \overline{v_b} \neq 0. \]

Let $g$ be an element of $G_{L/\mathbb {Q}_p}$ that reduces to the $p^j$th power map on $\mathcal {O}_L/\pi \mathcal {O}_L$ and let $c ={g(\pi ^i)}/{\pi ^i}$. Then

\[ \sum_{b \in B} x_b b^g \equiv c\pi^ip^n \sum_{b \in B'} \overline{u_b}^{p^j} \overline{v_b} \not \equiv 0\mod \pi^{er+1}. \]

Proof. Let $n=\min _{b \in B} w_k( y_b b)$ and let $x_b\in W(\mathfrak {K})$ denote the coefficient of $T^n$ in $y_b$. Note that $\min _{b \in B} v_p(x_b b) \leqslant k$. By Lemma 6.13, there exists $g \in \Theta$ with $v_p(\sum _{b \in B} x_b b^{g})\leqslant k$. It follows that the $p$-adic valuation of the coefficient of $T^n$ in $\sum _{b \in B} y_b b^{g}$ is less than $k$.

Proof. By Lemma 6.1 we know that $\{\log (m_g)\}_{g \in \Theta }$ is a $\Theta$-maximal Frobenius structure for $\log \circ \rho$ and that $c_k(m_g) = c_k(\log (m_g))$ for all $k\geqslant 0$. The proposition follows from Lemma 5.1 and Proposition 6.8.

Proof. After replacing $F$ with a finite extension we may assume $\rho (G_F) \subset 1+p\mathcal {O}_L$. We may also assume that $\rho$ is either harshly ramified or unramified by Corollary 5.6. Let $\{m_g\}_{g \in \Theta }$ be a $\Theta$-maximal Frobenius structure. There exists $c>0$ such that $\{m_g\}_{g \in \Theta } \subset \mathcal {O}_{\mathcal {E}}^{ r,c}$. By Proposition 6.15, we know $\rho$ has $r$-bounded monodromy. The statement about $r<1$ follows from Corollary 5.3.

Proof. The first equation is immediate. Define $S_m=R_a(A; \underbrace {B,\dots,B}_{m \text { times}}, C)$ and note that

\[ S_m=\begin{cases} AS_m^{\sigma^a} + BS_{m-1}^{\sigma^a} & m\geqslant 1, \\ AS_0^{\sigma^a} + C & m=0. \end{cases} \]

Thus, $\sum S_m$ is the solution to the Frobenius equation $x=(A+B)x^{\sigma ^a} + C$, which proves (10).

Proof. Parts (i)–(ii) follow from the definition of $\mathcal {O}_{\mathcal {E}_\infty }^{r,c}$. To prove part (iii), note that $xz^\sigma \in \mathcal {O}_{\mathcal {E}_\infty }^{r,qc}$.

Proof. We proceed by induction on $r=\mathbf {len}(\lambda )$. When $r=1$ the result follows from (13). Let $r\geqslant 1$ and assume the result holds for all $k< r$. Write $\lambda =(\mu _0,\mu _1,\lambda _0)$, where $\mathbf {len}(\lambda _0)=r-2$. By our inductive hypothesis we may assume that $(\mu _1,\lambda _0)$ is $q^a$-prime. If $q^a|\mu _0$, from (13) and (14) we obtain

\[ S_a(\mu_0,\mu_1,\lambda_0) = S_a(\mu_0+q^a\mu_1,q^a\lambda_0) - T^{\mu_0}S_a(\mu_1,q^a\lambda_0) + S_a\biggl(\frac{\mu_0}{q^a},\mu_1,\lambda_0\biggr). \]

Both $S_a(\mu _0+q^a\mu _1,q^a\lambda _0)$ and $T^{\mu _0}S_a(\mu _1,q^a\lambda _0)$ are contained in $\mathcal {N}_a^{(p)}$ by our inductive hypothesis, so it suffices to prove $S_a({\mu _0}/{q^a},\mu _1,\lambda _0)\in \mathcal {N}_a^{(p)}$. Repeating this argument proves the lemma.

Proof. Write $A$ as $q1_d + q^2 A_1$. There exists $c>c_1$ such that $qA_1$ and $C$ have entries in $\mathcal {O}_{\mathcal {E}_\infty }^{1,c}$. After increasing $N_0$, we may assume that the results of § 7.2.1 hold. By Lemma 7.5 and Lemma 7.6, we may also assume the entries of $qA_1$ and $C$ lie in $T^{-1}\mathcal {O}_{E_a}[T^{-1}]\,{\cap}\, \mathcal {O}_{\mathcal {E}_\infty }^{1,c}$. Then, by Corollary 7.3, we know that

\[ R(q1_d; \underbrace{qA_1,\dots,qA_1}_{m \text{ times}},C) \]

is contained in $M_{d\times 1}(\mathcal {O}_{\mathcal {E}_\infty }^{1,c})$ for all $m$. Thus, for $m_0$ sufficiently large, we know that

\[ R(q1_d; \underbrace{q^2A_1,\dots,q^2A_1}_{m \text{ times}},C) = q^mR(q1_d; \underbrace{qA_1,\dots,qA_1}_{m \text{ times}},C) \]

has entries in $\mathcal {O}_{\mathcal {E}_\infty }^{1,c_1}\cap q^{N_0} \mathcal {O}_{\mathcal {E}_\infty }$ for all $m>m_0$. From Lemma 7.1 we obtain

\[ R(A,C) \equiv \sum_{m=0}^{m_0} R(q1_d; \underbrace{q^2A_1,\dots,q^2A_1}_{m \text{ times}},C_0) \mod \mathcal{O}_{\mathcal{E}_\infty}^{1,c_1}\cap q^{N_0} \mathcal{O}_{\mathcal{E}_\infty}. \]

The right-hand side of this equivalence has entries in $\mathcal {N}_a^{(p)}$ by Lemma 7.7 and Corollary 7.10.

Proof. Observe that a set with $a$-stable growth has $ad$-stable growth for $d\in \mathbb {Z}_{\geqslant 1}$.

Proof. We proceed by induction on $n$. The case where $n=0$. Set $n\geqslant 1$. Let $z=R_a(q^a,x)$, so that $z=q^az^{\sigma ^a} + x$. Note that $w_k(z)\leqslant w_k(x)$, because $q^a \nmid w_{k}(x)$ and $q^a|w_k(q^az^{\sigma ^a})$. We have

\begin{align*} w_{k+sn}(z) &\geqslant \min(w_{k+sn}(q^az^{\sigma^a}), w_{k+sn}(x)) \\ &= \min (q^aw_{k+s(n-1)}(z),w_{k+sn}(x)) \\ &= \min(q^{an}w_k(z),w_{k+sn}(x)). \end{align*}

By (15) we know that $q^{an} w_k(z)< w_{k+sn}(x)$, which proves the result.

Proof. Define $\mathbf {L}(\Lambda )$ to be $\max \{{\mathbf {len}(\lambda _i)}\}$. We proceed by induction on $\mathbf {L}(\Lambda )$. When $\mathbf {L}(\Lambda )$ is $1$, then $x=R_a(q^a,y)$ with $y \in \mathcal {O}_{E_a}[T^{-1}] \cap \mathcal {E}_\infty ^{(q^a)}$ and the lemma follows from Lemma 7.15. Now assume the proposition holds for all collections of $q^a$-prime tuples $\Lambda '$ such that $\mathbf {L}(\Lambda ')\leqslant m$. Let $\Lambda =\{\lambda _1,\dots, \lambda _r\}$ be a collection of $q^a$-prime tuples with $\mathbf {L}(\Lambda )=m+1$. There exist tuples $\Lambda '=\{\lambda _1',\dots, \lambda _r'\}$ and negative integers $\{\kappa _1, \dots, \kappa _r\}$ such that $\mathbf {L}(\Lambda ')=m$ and $\lambda _i=(\kappa _i, \lambda _i')$. For each $d<0$ we define

\[ y_d = \begin{cases} 0 & \text{ if }d \neq \kappa_i \text{ for every }i, \\ \sum _{\kappa_i=d} \beta_iS_a(\lambda_i') & \text{ otherwise.} \end{cases} \]

Let $d_{1},\dots, d_{t}$ be the values of $d$ for which $y_{d}\neq 0$. Then we have

\begin{align*} x&= \sum_{i=1}^r R_a(q^a;\beta_iT^{\kappa_i}S_a(\lambda_i')) \\ &= \sum_{j=1}^t R_a(q^a;T^{{d_j}}y_{d_j}). \end{align*}

By our inductive assumption, there exists $m_{j,k}$ such that $w_{k+sn}(y_{d_j})=q^{an}m_{j,k}$ for $n\gg 0$. Without loss of generality, we may assume $d_1 + q^{an} m_{1,k} < \min _{j\geqslant 2} \{d_j + q^{an} m_{j,k} \}$ for $n \gg 0$. This holds because the $d_j$ are distinct. Thus, $w_{k+sn}(x) = d_1 + q^{an} m_{1,k}$, for large $n$. The proposition follows from Lemma 7.15.