Proof. We denote by $\mathcal {J}$ the set of all arrays $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \varepsilon _1 & \dots & \varepsilon _n \\ s_1 & \dots & s_n \end {pmatrix} $ for some n such that $s_1, \dots , s_n \leq q$ .

The second statement follows at once from the first statement. We prove the first statement by induction on $\operatorname {\mathrm {depth}}(\mathfrak {s})$ . For $\operatorname {\mathrm {depth}}(\mathfrak {s}) = 1$ , we let $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \varepsilon \\ s \end {pmatrix} $ with $s \le q$ . It follows from special cases of power sums in [Reference Thakur37, §3.3] that for all $d \in \mathbb {Z}$ , $ S_{d} \begin {pmatrix} \varepsilon \\ s \end {pmatrix} = \dfrac {\varepsilon ^d}{\ell ^s_d} = \operatorname {\mathrm {Si}}_{d} \begin {pmatrix} \varepsilon \\ s \end {pmatrix}. $ Suppose that the first statement holds for all arrays $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} \in \mathcal {J}$ with $\operatorname {\mathrm {depth}}(\mathfrak {s}) = n - 1$ and for all $ d \in \mathbb {Z}$ . Let $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \varepsilon _1 & \dots & \varepsilon _n \\ s_1 & \dots & s_n \end {pmatrix} $ be an element of $\mathcal {J}$ . Note that if $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} \in \mathcal {J}$ , then $ \begin {pmatrix} \boldsymbol {\varepsilon }_{-} \\ \mathfrak {s}_{-} \end {pmatrix} \in \mathcal {J}$ . It follows from induction hypothesis and the fact $s_1 \leq q$ that for all $d \in \mathbb {Z}$

$$ \begin{align*} S_d \begin{pmatrix} \boldsymbol{\varepsilon} \\ \mathfrak{s} \end{pmatrix} = S_d \begin{pmatrix} \varepsilon_1 \\ s_1 \end{pmatrix} S_{<d} \begin{pmatrix} \boldsymbol{\varepsilon}_{-} \\ \mathfrak{s}_{-} \end{pmatrix} = \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon_1 \\ s_1 \end{pmatrix} \operatorname{\mathrm{Si}}_{<d} \begin{pmatrix} \boldsymbol{\varepsilon}_{-} \\ \mathfrak{s}_{-} \end{pmatrix} = \operatorname{\mathrm{Si}}_d \begin{pmatrix} \boldsymbol{\varepsilon} \\ \mathfrak{s} \end{pmatrix}. \end{align*} $$

This proves the lemma.

Proof. The proof follows the same line as in [Reference Ngo Dac31, Proposition 2.1]. We omit the proof here and refer the reader to [Reference Im, Kim, Le, Ngo Dac and Pham25, Proposition 1.3] for more details.

Proof. From the definition and Equation (2.6), we have $\mathcal {B}^*_{\sigma _n,v_n}(R)$ is of the form

$$ \begin{align*} \sum \limits_i a_i \operatorname{\mathrm{Si}}_d \begin{pmatrix} \sigma_n & \boldsymbol{\varepsilon}_i \\ v_n& \mathfrak{s}_i \end{pmatrix} + \sum \limits_i b_i \operatorname{\mathrm{Si}}_d \begin{pmatrix} \sigma_n & \boldsymbol{\epsilon}_i \\ v_n& \mathfrak{t}_i \end{pmatrix} + \sum \limits_i b_i \operatorname{\mathrm{Si}}_d \begin{pmatrix} \sigma_n \epsilon_{i1} & {\boldsymbol{\epsilon}_i}_- \\ v_n + t_{i1} & {\mathfrak{t}_i}_- \end{pmatrix} = 0. \end{align*} $$

Apply the operator $\mathcal B^*_{\sigma _1,v_1} \circ \dots \circ \mathcal B^*_{\sigma _{n - 1},v_{n - 1}}$ to $\mathcal {B}^*_{\sigma _n,v_n}(R)$ , the result then follows from the definition.

Proof. We see that $\mathcal C_{\Sigma ,V}(R_{\varepsilon })$ is of the form

$$ \begin{align*} \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon \\ q \end{pmatrix} \operatorname{\mathrm{Si}}_{d} \begin{pmatrix} \Sigma \\ V \end{pmatrix} + \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon \\ q \end{pmatrix} \operatorname{\mathrm{Si}}_{<d} \begin{pmatrix} \Sigma \\ V \end{pmatrix} + \varepsilon^{-1} D_1 \operatorname{\mathrm{Si}}_{d+1} \begin{pmatrix} \varepsilon& 1 \\ 1 & q-1 \end{pmatrix} \operatorname{\mathrm{Si}}_{<d+1} \begin{pmatrix} \Sigma \\ V \end{pmatrix} = 0. \end{align*} $$

It follows from Equation (2.6) and Proposition 2.3 that

$$ \begin{align*} \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon \\ q \end{pmatrix} \operatorname{\mathrm{Si}}_{d} \begin{pmatrix} \Sigma \\ V \end{pmatrix} + \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon \\ q \end{pmatrix} \operatorname{\mathrm{Si}}_{<d} \begin{pmatrix} \Sigma \\ V \end{pmatrix} &= \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon\sigma_1 & \Sigma_{-} \\ q + v_1 & V_{-} \end{pmatrix} + \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon& \Sigma \\ q & V \end{pmatrix}, \\ \varepsilon^{-1}D_1 \operatorname{\mathrm{Si}}_{d+1} \begin{pmatrix} \varepsilon& 1 \\ 1 & q-1 \end{pmatrix} \operatorname{\mathrm{Si}}_{<d+1} \begin{pmatrix} \Sigma \\ V \end{pmatrix} &= \sum \limits_i b_i \operatorname{\mathrm{Si}}_{d+1} \begin{pmatrix} \varepsilon& \boldsymbol{\epsilon}_i \\ 1 & \mathfrak{t}_i \end{pmatrix} , \end{align*} $$

where $b_i \in A$ are divisible by $D_1$ and $ \begin {pmatrix} \boldsymbol {\epsilon }_i \\ \mathfrak {t}_i \end {pmatrix} $ are arrays satisfying $ \begin {pmatrix} \boldsymbol {\epsilon }_i \\ \mathfrak {t}_i \end {pmatrix} \leq \begin {pmatrix} 1 \\ q - 1 \end {pmatrix} + \begin {pmatrix} \Sigma \\ V \end {pmatrix} $ for all i. This proves the proposition.

Proof. The proof follows the same line as in [Reference Ngo Dac31, Proposition 2.12 and 2.13]. We outline the proof here and refer the reader to [Reference Im, Kim, Le, Ngo Dac and Pham25] for more details. For Part 1, since $\operatorname {\mathrm {Init}}(\mathfrak {s}) = (s_1, \dots , s_{k-1})$ , we get $s_k> q$ . Set $ \begin {pmatrix} \Sigma \\ V \end {pmatrix} = \begin {pmatrix} \varepsilon ^{-1} \varepsilon _{k} & \varepsilon _{k+1} &\dots & \varepsilon _n \\ s_k - q & s_{k+1} &\dots & s_n \end {pmatrix} $ . By Proposition 2.6, $\mathcal {C}_{\Sigma ,V}(R_{\varepsilon })$ is of the form

(2.8) $$ \begin{align} \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon_{k} & \dots & \varepsilon_n \\ s_{k} & \dots & s_n \end{pmatrix} + \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon & \varepsilon^{-1}\varepsilon_{k} & \varepsilon_{k+1} & \dots & \varepsilon_n \\ q & s_k- q & s_{k+1} & \dots & s_n \end{pmatrix} + \sum \limits_i b_i \operatorname{\mathrm{Si}}_{d+1} \begin{pmatrix} \varepsilon & \boldsymbol{\epsilon}_i \\ 1 & \mathfrak{t}_i \end{pmatrix} =0, \end{align} $$

where $b_i \in A$ divisible by $D_1$ and $ \begin {pmatrix} \boldsymbol {\epsilon }_i \\ \mathfrak {t}_i \end {pmatrix} $ are arrays satisfying for all i,

$$ \begin{align*} \begin{pmatrix} \boldsymbol{\epsilon}_i \\ \mathfrak{t}_i \end{pmatrix} \leq \begin{pmatrix} 1 \\ q - 1 \end{pmatrix} + \begin{pmatrix} \Sigma \\ V \end{pmatrix} = \begin{pmatrix} \varepsilon^{-1} \varepsilon_{k} & \varepsilon_{k+1} &\dots & \varepsilon_n \\ s_k - 1 & s_{k+1} &\dots & s_n \end{pmatrix}. \end{align*} $$

For $m \in \mathbb {N}$ , we denote by $q^{\{m\}}$ the sequence of length m with all terms equal to q. We agree by convention that $q^{\{0\}}$ is the empty sequence. Setting $s_0 = 0$ , we may assume that there exists a maximal index j with $0 \leq j \leq k-1$ such that $s_j < q$ , hence $\operatorname {\mathrm {Init}}(\mathfrak {s}) = (s_1, \dots , s_j, q^{\{k-j-1\}})$ . Then the operator $ \mathcal {BC}_{\varepsilon _{j+1},q} \circ \dots \circ \mathcal {BC}_{\varepsilon _{k-1},q}$ applied to the relation (2.8) gives

$$ \begin{align*} \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon_{j+1} & \dots & \varepsilon_{k-1} & \varepsilon_{k} & \dots & \varepsilon_n \\ q & \dots & q & s_{k} & \dots & s_n \end{pmatrix} & + \operatorname{\mathrm{Si}}_d \begin{pmatrix} \varepsilon_{j+1} & \dots & \varepsilon_{k-1} & \varepsilon & \varepsilon^{-1}\varepsilon_{k} & \epsilon_{k+1} & \dots & \epsilon_n \\ q & \dots & q & q & s_k- q & s_{k+1} & \dots & s_n \end{pmatrix} \\ & + \sum \limits_i b_{i_1 \dots i_{k-j}} \operatorname{\mathrm{Si}}_{d+1} \begin{pmatrix} \varepsilon_{j+1} & \boldsymbol{\epsilon}_{i_1 \dots i_{k-j}} \\ 1 & \mathfrak{t}_{i_1 \dots i_{k-j}} \end{pmatrix} =0, \end{align*} $$

where $b_{i_1 \dots i_{k-j}} \in A$ are divisible by $D_1$ and

(2.9) $$ \begin{align} \begin{pmatrix} \boldsymbol{\epsilon}_{i_1 \dots i_{k-j}} \\ \mathfrak{t}_{i_1 \dots i_{k-j}} \end{pmatrix} \leq \begin{pmatrix} \varepsilon_{j+2}& \dots & \varepsilon_{k-1} & \varepsilon & \varepsilon^{-1}\varepsilon_{k} & \varepsilon_{k+1}& \dots & \varepsilon_n \\ q& \dots & q & q & s_{k} - 1 & s_{k+1} &\dots & s_n \end{pmatrix}. \end{align} $$

Set $ \begin {pmatrix} \Sigma ' \\ V' \end {pmatrix} = \begin {pmatrix} \varepsilon _{1} &\dots & \varepsilon _j \\ s_1 &\dots & s_j \end {pmatrix} $ . Applying $\mathcal {B}^*_{\Sigma ',V'}$ to the above relation and using Lemma 2.5, we can deduce that

(2.10) $$ \begin{align} \operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\varepsilon} \\ \mathfrak{s} \end{pmatrix} =& - \operatorname{\mathrm{Li}} \begin{pmatrix} \varepsilon_1 & \dots & \varepsilon_{k-1} & \varepsilon & \varepsilon^{-1}\varepsilon_{k} & \varepsilon_{k+1} & \dots & \varepsilon_n \\ s_1 & \dots & s_{k-1} & q & s_k- q & s_{k+1} & \dots & s_n \end{pmatrix} \\ \notag & - \sum \limits_i b_{i_1 \dots i_{k-j}} \operatorname{\mathrm{Li}} \begin{pmatrix} \varepsilon_1 & \dots & \varepsilon_j & \varepsilon_{j+1} & \boldsymbol{\epsilon}_{i_1 \dots i_{k-j}} \\ s_1 & \dots & s_j & 1 & \mathfrak{t}_{i_1 \dots i_{k-j}} \end{pmatrix} \\\notag & - \sum \limits_i b_{i_1 \dots i_{k-j}} \operatorname{\mathrm{Li}} \begin{pmatrix} \varepsilon_1 & \dots & \varepsilon_{j-1} & \varepsilon_j \varepsilon_{j+1} & \boldsymbol{\epsilon}_{i_1 \dots i_{k-j}} \\ s_1 & \dots & s_{j-1} & s_j+1 & \mathfrak{t}_{i_1 \dots i_{k-j}} \end{pmatrix}. \end{align} $$

The first term, the second term and the third term on the right-hand side of Equation (2.10) are referred to as type 1, type 2 and type 3, respectively. From Equation (2.9) and Remark 2.1, one verifies that the arrays of type 1, type 2 and type 3 satisfy the desired conditions. We have proved Part 1.

The proof of Part $2$ follows the same arguments as that of Part 1. We first begin with the relation $R_{\varepsilon _k}$ . Next, we apply $ \mathcal {BC}_{\varepsilon _{j+1},q} \circ \dots \circ \mathcal {BC}_{\varepsilon _{k-1},q}$ to $R_{\varepsilon _k}$ and then apply $\mathcal {B}^*_{\Sigma ',V'}$ . We can deduce that

(2.11) $$ \begin{align} \operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\varepsilon} \\ \mathfrak{s} \end{pmatrix} =& - \sum \limits_i b_{i_1 \dots i_{k-j}} \operatorname{\mathrm{Li}} \begin{pmatrix} \varepsilon_1 & \dots & \varepsilon_j & \varepsilon_{j+1} & \boldsymbol{\epsilon}_{i_1 \dots i_{k-j}} \\ s_1 & \dots & s_j & 1 & \mathfrak{t}_{i_1 \dots i_{k-j}} \end{pmatrix} \\ \notag & - \sum \limits_i b_{i_1 \dots i_{k-j}} \operatorname{\mathrm{Li}} \begin{pmatrix} \varepsilon_1 & \dots & \varepsilon_{j-1} & \varepsilon_j \varepsilon_{j+1} & \boldsymbol{\epsilon}_{i_1 \dots i_{k-j}} \\ s_1 & \dots & s_{j-1} & s_j+1 & \mathfrak{t}_{i_1 \dots i_{k-j}} \end{pmatrix}, \end{align} $$

where $b_{i_1 \dots i_{k-j}} \in A$ are divisible by $D_1$ and

(2.12) $$ \begin{align} \begin{pmatrix} \boldsymbol{\epsilon}_{i_1 \dots i_{k-j}} \\ \mathfrak{t}_{i_1 \dots i_{k-j}} \end{pmatrix} \leq \begin{pmatrix} \varepsilon_{j+2}& \dots & \varepsilon_{k} & 1 \\ q& \dots & q & q - 1 \end{pmatrix}. \end{align} $$

The first term and the second term on the right-hand side of Equation (2.11) are referred to as type 2 and type 3, respectively. From Equation (2.12) and Remark 2.1, one verifies that the arrays of type 2 and type 3 satisfy the desired conditions. We finish the proof.

Proof. The proof follows the same line as that of [Reference Ngo Dac31, Proposition 3.2]. We outline the proof here and refer the reader to [Reference Im, Kim, Le, Ngo Dac and Pham25] for more details. We consider the following statement: $(H_k) \quad $ For all arrays $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} $ , we can express $\operatorname {\mathrm {Li}} \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} $ as a K-linear combination of $\operatorname {\mathrm {Li}} \begin {pmatrix} \boldsymbol {\epsilon } \\ \mathfrak {t} \end {pmatrix} $ ’s of the same weight such that $\mathfrak {t}$ is k-admissible.

We will show that $(H_k)$ holds for all $k \in \mathbb {N}$ by induction on k. For $k = 1$ , we consider all the cases for the first component $s_1$ of $\mathfrak {s}$ . If $s_1 \leq q$ , then either $\mathfrak {s}$ is $1$ -admissible, or $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \varepsilon \\ q \end {pmatrix}$ . We deduce from the relation $R_{\varepsilon }$ that $(H_1)$ holds for the case $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \varepsilon \\ q \end {pmatrix}$ . If $s_1> q$ , we assume that $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \varepsilon _1 & \cdots & \varepsilon _n \\ s_1 & \cdots & s_n \end {pmatrix}$ . Set $\begin {pmatrix} \Sigma \\ V \end {pmatrix} = \begin {pmatrix} \varepsilon _1 & \varepsilon _2 & \cdots & \varepsilon _n \\ s_1 - q & s_2 & \cdots & s_n \end {pmatrix}$ . Applying $C_{\Sigma ,V}$ to the relation $R_{1}$ and using Proposition 2.6, we can deduce that

$$ \begin{align*} \operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\varepsilon} \\ \mathfrak{s} \end{pmatrix} = - \operatorname{\mathrm{Li}} \begin{pmatrix} 1& \varepsilon_1 & \varepsilon_2 & \cdots & \varepsilon_n \\ q & s_1 - q & s_2 & \cdots & s_n \end{pmatrix} - \sum\limits_i b_i \operatorname{\mathrm{Li}} \begin{pmatrix} 1 & \boldsymbol{\epsilon}_i \\ 1 & \mathfrak{t}_i \end{pmatrix}, \end{align*} $$

where $b_i \in K$ for all i. It proves that $(H_1)$ holds.

We next assume that $(H_{k - 1})$ holds. We need to show that $(H_k)$ holds. By using the induction hypothesis of $(H_{k - 1})$ , we can restrict our attention to the array $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \varepsilon _1 & \cdots & \varepsilon _n \\ s_1 & \cdots & s_n \end {pmatrix}$ , where $\mathfrak {s}$ is not k-admissible and $\operatorname {\mathrm {depth}}(\mathfrak {s}) \geq k$ . We will prove that $(H_k)$ holds for the array $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix}$ by induction on $s_1 + \cdots + s_k$ . The case $s_1 + \cdots + s_k = 1$ is a simple check. Assume that $(H_k)$ holds when $s_1 + \cdots + s_k < s$ . We need to show that $(H_k)$ holds when $s_1 + \cdots + s_k = s$ . To do so, we give the following algorithm:

Algorithm: We begin with an array $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix}$ where $\mathfrak {s}$ is not k-admissible, $\operatorname {\mathrm {depth}}(\mathfrak {s}) \geq k$ and $s_1 + \cdots + s_k = s$ .

Step 1: From Proposition 2.7, we can decompose $\operatorname {\mathrm {Li}} \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} $ as follows:

(2.13) $$ \begin{align} \operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\varepsilon} \\ \mathfrak{s} \end{pmatrix} = \underbrace{ - \operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\varepsilon}' \\ \mathfrak{s}' \end{pmatrix} }_{\text{type 1}} + \underbrace{\sum\limits_i b_i\operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\epsilon}_i' \\ \mathfrak{t}^{\prime}_i \end{pmatrix} }_{\text{type 2}} + \underbrace{\sum\limits_i c_i\operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\mu}_i \\ \mathfrak{u}_i \end{pmatrix} }_{\text{type 3}} , \end{align} $$

where $ b_i, c_i \in A$ . The term of type $1$ disappears when $\operatorname {\mathrm {Init}}(\mathfrak {s}) = \mathfrak {s}$ and $s_n = q$ .

Step 2: For all arrays $ \begin {pmatrix} \boldsymbol {\epsilon } \\ \mathfrak {t} \end {pmatrix} $ appearing on the right-hand side of Equation (2.13), if $\mathfrak {t}$ is either k-admissible or $\mathfrak {t}$ satisfies the condition $t_1 + \cdots + t_k < s$ , then we deduce from the induction hypothesis that $(H_k)$ holds for the array $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix}$ , and hence we stop the algorithm. Otherwise, there exists an array $\begin {pmatrix} \boldsymbol {\varepsilon }_1 \\ \mathfrak {s}_1 \end {pmatrix}$ where $\mathfrak {s}_1$ is not k-admissible, $\operatorname {\mathrm {depth}}(\mathfrak {s}_1) \geq k$ and $s_{11} + \cdots + s_{1k} = s$ . For such an array, we repeat the algorithm for $\begin {pmatrix} \boldsymbol {\varepsilon }_1 \\ \mathfrak {s}_1 \end {pmatrix}$ .

It remains to show that the above algorithm stops after a finite number of steps. Indeed, assume that the above algorithm does not stop. Then there exists a sequence of arrays $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \begin {pmatrix} \boldsymbol {\varepsilon }_0 \\ \mathfrak {s}_0 \end {pmatrix}, \begin {pmatrix} \boldsymbol {\varepsilon }_1 \\ \mathfrak {s}_1 \end {pmatrix}, \begin {pmatrix} \boldsymbol {\varepsilon }_2 \\ \mathfrak {s}_2 \end {pmatrix}, \dotsc $ such that $\mathfrak {s}_i$ is not k-admissible and $\operatorname {\mathrm {depth}}(\mathfrak {s}_i) \geq k$ for all $i \geq 0$ . Using Proposition 2.7, one verifies that $\begin {pmatrix} \boldsymbol {\varepsilon }_{i + 1} \\ \mathfrak {s}_{i + 1} \end {pmatrix}$ is of type $3$ with respect to $\begin {pmatrix} \boldsymbol {\varepsilon }_i \\ \mathfrak {s}_i \end {pmatrix}$ for all $i \geq 0$ , hence we obtain an infinite sequence $\operatorname {\mathrm {Init}}(\mathfrak {s}_0) \prec \operatorname {\mathrm {Init}}(\mathfrak {s}_1) \prec \operatorname {\mathrm {Init}}(\mathfrak {s}_2) \prec \cdots $ . For all $i \geq 0$ , since $\mathfrak {s}_i$ is not k-admissible and $\operatorname {\mathrm {depth}}(\mathfrak {s}_i) \geq k$ , we have $\operatorname {\mathrm {depth}}(\operatorname {\mathrm {Init}}(\mathfrak {s}_i)) \leq k$ , hence $\operatorname {\mathrm {Init}}(\mathfrak {s}_i) \preceq q^{\{k\}}$ . This shows that $\operatorname {\mathrm {Init}}(\mathfrak {s}_i) = \operatorname {\mathrm {Init}}(\mathfrak {s}_{i + 1})$ for all i sufficiently large, which is a contradiction.

Proof. The result follows immediately from Proposition 2.9 in the case of $k = w$ .

Proof. We recall that $\mathcal {AT}_w$ is the set of all ACMPL's $\operatorname {\mathrm {Li}} \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix}$ with $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} \in \mathcal {AJ}_w$ .

Let $\text {Li} \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} = \text {Li}\begin {pmatrix} \varepsilon _1 & \dots & \varepsilon _n \\ s_1 & \dots & s_n \end {pmatrix} \in \mathcal {AT}_w$ . Then $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} \in \mathcal {AJ}_w$ , which implies $s_1, \dotsc , s_{n - 1} \leq q$ and $s_n < q$ . We express $\begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix}$ in the following form

$$ \begin{align*} \bigg( \underbrace{\begin{pmatrix} \varepsilon_1 & \dots & \varepsilon_{h_1 - 1}\\ q & \dots & q \end{pmatrix}}_{h_1 - 1\ \text{times}} \begin{pmatrix} \varepsilon_{h_1} \\ r_1 \end{pmatrix} \dots \underbrace{\begin{pmatrix} \varepsilon_{h_1 + \cdots + h_{\ell - 1} + 1} & \dots & \varepsilon_{h_1 + \cdots + h_{\ell - 1} + (h_\ell - 1)}\\ q & \dots & q \end{pmatrix}}_{h_\ell - 1\ \text{times}} \begin{pmatrix} \varepsilon_{h_1 + \cdots + h_{\ell - 1} + h_\ell} \\ r_\ell \end{pmatrix} \bigg), \end{align*} $$

where $h_1, \dotsc , h_\ell \geq 1$ , $h_1 + \cdots + h_\ell = n$ and $0 < r_1, \dotsc , r_\ell < q$ . Then we set

$$ \begin{align*} \begin{pmatrix} \boldsymbol{\varepsilon}' \\ \mathfrak{s}' \end{pmatrix} = \begin{pmatrix} \varepsilon^{\prime}_1 & \dots & \varepsilon^{\prime}_\ell \\ s^{\prime}_1 & \dots & s^{\prime}_\ell \end{pmatrix}, \end{align*} $$

where $\varepsilon ^{\prime }_i = \varepsilon _{h_1 + \cdots + h_{i - 1} + 1} \cdots \varepsilon _{h_1 + \cdots + h_{i - 1} + h_i}$ and $s^{\prime }_i = (h_i - 1)q + r_i$ for $1 \leq i \leq \ell $ . We note that $\iota (\mathfrak {s}')=\mathfrak {s}$ .

From Proposition 2.7 and Proposition 2.10, we can decompose $\text {Li}\begin {pmatrix} \boldsymbol {\varepsilon }' \\ \mathfrak {s}' \end {pmatrix}$ as follows:

$$ \begin{align*} \operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\varepsilon}' \\ \mathfrak{s}' \end{pmatrix} = \sum a_{\boldsymbol{\epsilon},\mathfrak t}^{\boldsymbol{\varepsilon}',\mathfrak{s}'} \operatorname{\mathrm{Li}} \begin{pmatrix} \boldsymbol{\epsilon} \\ \mathfrak{t} \end{pmatrix} , \end{align*} $$

where $ \begin {pmatrix} \boldsymbol {\epsilon } \\ \mathfrak {t} \end {pmatrix} $ ranges over all elements of $\mathcal {AJ}_w$ and $a_{\boldsymbol {\epsilon },\mathfrak t}^{\boldsymbol {\varepsilon }',\mathfrak {s}'} \in A$ satisfying

$$ \begin{align*} a_{\boldsymbol{\epsilon},\mathfrak t}^{\boldsymbol{\varepsilon}',\mathfrak{s}'} \equiv \begin{cases} \pm 1 \ (\text{mod } D_1) & \text{if } \begin{pmatrix} \boldsymbol{\epsilon} \\ \mathfrak{t} \end{pmatrix} = \begin{pmatrix} \boldsymbol{\varepsilon} \\ \mathfrak{s} \end{pmatrix},\\ 0 \ \ (\text{mod } D_1) & \text{otherwise}. \end{cases} \end{align*} $$

Note that $\operatorname {\mathrm {Li}} \begin {pmatrix} \boldsymbol {\varepsilon }' \\ \mathfrak {s}' \end {pmatrix} \in \mathcal {AS}_w$ . Thus, the transition matrix from the set consisting of such $\operatorname {\mathrm {Li}} \begin {pmatrix} \boldsymbol {\varepsilon }' \\ \mathfrak {s}' \end {pmatrix} $ as above (we allow repeated elements) to the set consisting of $\operatorname {\mathrm {Li}} \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} $ with $ \begin {pmatrix} \boldsymbol {\varepsilon } \\ \mathfrak {s} \end {pmatrix} \in \mathcal {AJ}_w$ is invertible. It then follows again from Proposition 2.10 that $\mathcal {AS}_w$ is a set of generators for $\mathcal {AL}_w$ , as desired.

Proof. The proof will be divided into two steps.

Step 1. We first construct a dual t-motive to which we will apply the Anderson–Brownawell–Papanikolas criterion. We recall $a_i(t):=a_i \rvert _{\theta =t} \in \mathbb F_q[t]$ .

For each pair $(\mathfrak {s}_i;\mathfrak {Q}_i)$ , we have attached to it a matrix $\Phi _{\mathfrak {s}_i,\mathfrak {Q}_i}$ . For $\mathfrak {s}_i=(s_{i1},\dots ,s_{i \ell _i}) \in \mathbb N^{\ell _i}$ and $\mathfrak {Q}_i=(Q_{i1},\dots ,Q_{i\ell _i}) \in (\mathbb F_q^\times )^{\ell _i}$ we recall

$$\begin{align*}I(\mathfrak{s}_i;\mathfrak{Q}_i)=\{\emptyset,(s_{i1};Q_{i1}),\dots,(s_{i1},\dots,s_{i (\ell_i-1)};Q_{i1},\dots,Q_{(\ell_i-1)})\}, \end{align*}$$

and $I:=\cup _i I(\mathfrak {s}_i;\mathfrak {Q}_i)$ .

We now construct a new matrix $\Phi '$ indexed by elements of I, say

$$\begin{align*}\Phi'=\left(\Phi^{\prime}_{(\mathfrak t;\mathfrak{Q}),(\mathfrak t';\mathfrak{Q}')}\right)_{(\mathfrak t;\mathfrak{Q}),(\mathfrak t';\mathfrak{Q}') \in I} \in \operatorname{\mathrm{Mat}}_{|I|}(\overline K[t]). \end{align*}$$

For the row which corresponds to the empty pair $\emptyset $ , we put

$$ \begin{align*} \Phi^{\prime}_{\emptyset,(\mathfrak t';\mathfrak{Q}')}= \begin{cases} (t-\theta)^w & \text{if } (\mathfrak t';\mathfrak{Q}')=\emptyset, \\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

For the row indexed by $(\mathfrak t;\mathfrak {Q})=(s_{i1},\dots ,s_{i j};Q_{i1},\dots ,Q_{ij})$ for some i and $1 \leq j \leq \ell _i-1$ , we put

$$ \begin{align*} \Phi^{\prime}_{(\mathfrak t;\mathfrak{Q}),(\mathfrak t';\mathfrak{Q}')}= \begin{cases} (t-\theta)^{w-w(\mathfrak t')} & \text{if } (\mathfrak t';\mathfrak{Q}')=(\mathfrak t;\mathfrak{Q}), \\ Q_{ij}^{(-1)} (t-\theta)^{w-w(\mathfrak t')} & \text{if } (\mathfrak t';\mathfrak{Q}')=(s_{i1},\dots,s_{i (j-1)};Q_{i1},\dots,Q_{i (j-1)}), \\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

Note that $\Phi _{\mathfrak {s}_i,\mathfrak {Q}_i}'=\left (\Phi ^{\prime }_{(\mathfrak t;\mathfrak {Q}),(\mathfrak t';\mathfrak {Q}')}\right )_{(\mathfrak t;\mathfrak {Q}),(\mathfrak t';\mathfrak {Q}') \in I(\mathfrak {s}_i;\mathfrak {Q}_i)}$ for all i.

We define $\Phi \in \operatorname {\mathrm {Mat}}_{|I|+1}(\overline K[t])$ by

$$ \begin{align*} \Phi=\begin{pmatrix} \Phi' & 0 \\ \mathbf{v} & 1 \end{pmatrix} \in \operatorname{\mathrm{Mat}}_{|I|+1}(\overline K[t]), \quad \mathbf{v}=(v_{\mathfrak t,\mathfrak{Q}})_{(\mathfrak t;\mathfrak{Q}) \in I} \in \operatorname{\mathrm{Mat}}_{1 \times|I|}(\overline K[t]), \end{align*} $$

where

$$\begin{align*}v_{\mathfrak t;\mathfrak{Q}}=\sum_i a_i(t) Q_{i \ell_i}^{(-1)} (t-\theta)^{w-w(\mathfrak t)}. \end{align*}$$

Here, the sum runs through the set of indices i such that $(\mathfrak t;\mathfrak {Q})=(s_{i1},\dots ,s_{i (\ell _i-1)};Q_{i1},\dots ,Q_{i (\ell _i-1)})$ , and the empty sum is defined to be zero.

We now introduce a rigid analytic trivialization matrix $\Psi $ for $\Phi $ . We define $\Psi '=\left (\Psi ^{\prime }_{(\mathfrak t;\mathfrak {Q}),(\mathfrak t';\mathfrak {Q}')}\right )_{(\mathfrak t;\mathfrak {Q}),(\mathfrak t';\mathfrak {Q}') \in I} \in \text {GL}_{|I|}(\mathbb T)$ as follows. For the row which corresponds to the empty pair $\emptyset $ , we define

$$ \begin{align*} \Psi^{\prime}_{\emptyset,(\mathfrak t';\mathfrak{Q}')}= \begin{cases} \Omega^w & \text{if } (\mathfrak t';\mathfrak{Q}')=\emptyset, \\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

For the row indexed by $(\mathfrak t;\mathfrak {Q})=(s_{i1},\dots ,s_{i j};Q_{i1},\dots ,Q_{i j})$ for some i and $1 \leq j \leq \ell _i-1$ , we put

$$ \begin{align*} &\Psi^{\prime}_{(\mathfrak t;\mathfrak{Q}),(\mathfrak t';\mathfrak{Q}')}= \\ & \begin{cases} \mathfrak L(\mathfrak t;\mathfrak{Q}) \Omega^{w-w(\mathfrak t)} & \text{if } (\mathfrak t';\mathfrak{Q}')=\emptyset, \\ \mathfrak L(s_{i(k+1)},\dots,s_{ij}; & \\ \quad Q_{i(k+1)},\dots,Q_{ij}) \Omega^{w-w(\mathfrak t)} & \text{if } (\mathfrak t';\mathfrak{Q}')=(s_{i1},\dots,s_{i k};Q_{i1},\dots,Q_{i k}) \text{ for some } 1 \leq k \leq j, \\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

Note that $\Psi _{\mathfrak {s}_i,\mathfrak {Q}_i}'=\left (\Psi ^{\prime }_{(\mathfrak t;\mathfrak {Q}),(\mathfrak t';\mathfrak {Q}')}\right )_{(\mathfrak t;\mathfrak {Q}),(\mathfrak t';\mathfrak {Q}') \in I(\mathfrak {s}_i;\mathfrak {Q}_i)}$ for all i.

We define $\Psi \in \text {GL}_{|I|+1}(\mathbb T)$ by

$$ \begin{align*} \Psi=\begin{pmatrix} \Psi' & 0 \\ \mathbf{f} & 1 \end{pmatrix} \in \text{GL}_{|I|+1}(\mathbb T), \quad \mathbf{f}=(f_{\mathfrak t;\mathfrak{Q}})_{\mathfrak t \in I} \in \operatorname{\mathrm{Mat}}_{1 \times|I|}(\mathbb T). \end{align*} $$

Here, we recall (see Equation (3.8))

$$ \begin{align*} f_{\mathfrak t;\mathfrak{Q}}= \sum_i a_i(t) \mathfrak L(s_{i(k+1)},\dots,s_{i \ell_i};Q_{i(k+1)},\dots,Q_{i \ell_i}), \end{align*} $$

where the sum runs through the set of indices i such that $(\mathfrak t;\mathfrak {Q})=(s_{i1},\dots ,s_{i k};Q_{i1},\dots ,Q_{i k})$ for some $0 \leq k \leq \ell _i-1$ . In particular, $f_{\emptyset }= \sum _i a_i(t) \mathfrak L(\mathfrak {s}_i;\mathfrak {Q}_i)$ .

By construction and by §3.2, we get $\Psi ^{(-1)}=\Phi \Psi $ , that means $\Psi $ is a rigid analytic trivialization for $\Phi $ .

Step 2. Next, we apply the Anderson–Brownawell–Papanikolas criterion (see Theorem 3.2) to prove Theorem 3.4.

In fact, we define

$$ \begin{align*} \widetilde \Phi=\begin{pmatrix} 1 & 0 \\ 0 & \Phi \end{pmatrix} \in \operatorname{\mathrm{Mat}}_{|I|+2}(\overline K[t]) \end{align*} $$

and consider the vector constructed from the first column vector of $\Psi $

$$ \begin{align*} \widetilde \psi=\begin{pmatrix} 1 \\ \Psi_{(\mathfrak t;\mathfrak{Q}),\emptyset}' \\ f_\emptyset \end{pmatrix}_{(\mathfrak t;\mathfrak{Q}) \in I}. \end{align*} $$

Then we have $\widetilde \psi ^{(-1)}=\widetilde \Phi \widetilde \psi $ .

We also observe that for all $(\mathfrak t;\mathfrak {Q}) \in I$ we have $\Psi _{(\mathfrak t;\mathfrak {Q}),\emptyset }'=\mathfrak L(\mathfrak t;\mathfrak {Q}) \Omega ^{w-w(\mathfrak t)}$ . Further,

$$ \begin{align*} a+f_\emptyset(\theta)=a+\sum_i a_i \mathfrak L(\mathfrak{s}_i;\mathfrak{Q}_i)(\theta)=0. \end{align*} $$

By Theorem 3.2 with $\rho =(a,0,\dots ,0,1)$ , we deduce that there exists $\mathbf {h}=(g_0,g_{\mathfrak t,\mathfrak {Q}},g) \in \operatorname {\mathrm {Mat}}_{1 \times (|I|+2)}(\overline K[t])$ such that $\mathbf {h} \psi =0$ and that $g_{\mathfrak t,\mathfrak {Q}}(\theta )=0$ for $(\mathfrak t,\mathfrak {Q}) \in I$ , $g_0(\theta )=a$ and $g(\theta )=1 \neq 0$ . If we put $ \mathbf {g}:=(1/g)\mathbf {h} \in \operatorname {\mathrm {Mat}}_{1 \times (|I|+2)}(\overline K(t))$ , then all the entries of $ \mathbf {g}$ are regular at $t=\theta $ .

Now, we have

(3.9) $$ \begin{align} ( \mathbf{g}- \mathbf{g}^{(-1)} \widetilde \Phi) \widetilde \psi= \mathbf{g} \widetilde \psi-( \mathbf{g} \widetilde \psi)^{(-1)}=0. \end{align} $$

We write $ \mathbf {g}- \mathbf {g}^{(-1)} \widetilde \Phi =(B_0,B_{\mathfrak t,\mathfrak {Q}},0)_{\mathfrak t \in I}$ . We claim that $B_0=0$ and $B_{\mathfrak t,\mathfrak {Q}}=0$ for all $(\mathfrak t;\mathfrak {Q}) \in I$ . In fact, expanding Equation (3.9) we obtain

(3.10) $$ \begin{align} B_0+\sum_{\mathfrak t \in I} B_{\mathfrak t,\mathfrak{Q}} \mathfrak L(\mathfrak t;\mathfrak{Q}) \Omega^{w-w(\mathfrak t)}=0. \end{align} $$

By Equation (3.3), we see that for $(\mathfrak t;\mathfrak {Q}) \in I$ and $j \in \mathbb {N}$ ,

$$ \begin{align*} \mathfrak{L}(\mathfrak t;\mathfrak{Q})(\theta^{q^j})=(\mathfrak{L}(\mathfrak t;\mathfrak{Q})(\theta))^{q^j} \end{align*} $$

which is nonzero by Condition $(LW)$ .

First, as the function $\Omega $ has a simple zero at $t=\theta ^{q^k}$ for $k \in \mathbb {N}$ , specializing Equation (3.10) at $t=\theta ^{q^j}$ yields $B_0(\theta ^{q^j})=0$ for $j \geq 1$ . Since $B_0$ belongs to $\overline K(t)$ , it follows that $B_0=0$ .

Next, we put $w_0:=\max _{(\mathfrak t;\mathfrak {Q}) \in I} w(\mathfrak t)$ and denote by $I(w_0)$ the set of $(\mathfrak t;\mathfrak {Q}) \in I$ such that $w(\mathfrak t)=w_0$ . Then dividing Equation (3.10) by $\Omega ^{w-w_0}$ yields

(3.11) $$ \begin{align} \sum_{(\mathfrak t;\mathfrak{Q}) \in I} B_{\mathfrak t,\mathfrak{Q}} \mathfrak L(\mathfrak t;\mathfrak{Q}) \Omega^{w_0-w(\mathfrak t)}=\sum_{(\mathfrak t;\mathfrak{Q}) \in I(w_0)} B_{\mathfrak t,\mathfrak{Q}} \mathfrak L(\mathfrak t;\mathfrak{Q})+\sum_{(\mathfrak t;\mathfrak{Q}) \in I \setminus I(w_0)} B_{\mathfrak t,\mathfrak{Q}} \mathfrak L(\mathfrak t;\mathfrak{Q}) \Omega^{w_0-w(\mathfrak t)}=0. \end{align} $$

Since each $B_{\mathfrak t,\mathfrak {Q}}$ belongs to $\overline K(t)$ , they are defined at $t=\theta ^{q^j}$ for $j \gg 1$ . Note that the function $\Omega $ has a simple zero at $t=\theta ^{q^k}$ for $k \in \mathbb {N}$ . Specializing Equation (3.11) at $t=\theta ^{q^j}$ and using Equation (3.3) yields

$$\begin{align*}\sum_{(\mathfrak t;\mathfrak{Q}) \in I(w_0)} B_{\mathfrak t,\mathfrak{Q}}(\theta^{q^j}) (\mathfrak L(\mathfrak t;\mathfrak{Q})(\theta))^{q^j}=0 \end{align*}$$

for $j \gg 1$ .

We claim that $B_{\mathfrak t,\mathfrak {Q}}(\theta ^{q^j})=0$ for $j \gg 1$ and for all $(\mathfrak t;\mathfrak {Q}) \in I(w_0)$ . Otherwise, we get a nontrivial $\overline K$ -linear relation between $\mathfrak L(\mathfrak t;\mathfrak {Q})(\theta )$ with $(\frak t;\mathfrak {Q}) \in I$ of weight $w_0$ . By Proposition 3.3, we deduce a nontrivial K-linear relation between $\mathfrak L(\mathfrak t;\mathfrak {Q})(\theta )$ with $(\frak t;\mathfrak {Q}) \in I(w_0)$ , which contradicts with Condition $(LW)$ . Now, we know that $B_{\mathfrak t,\mathfrak {Q}}(\theta ^{q^j})=0$ for $j \gg 1$ and for all $(\mathfrak t;\mathfrak {Q}) \in I(w_0)$ . Since each $B_{\mathfrak t,\mathfrak {Q}}$ belongs to $\overline K(t)$ , it follows that $B_{\mathfrak t,\mathfrak {Q}}=0$ for all $(\mathfrak t;\mathfrak {Q}) \in I(w_0)$ .

Next, we put $w_1:=\max _{(\mathfrak t;\mathfrak {Q}) \in I \setminus I(w_0)} w(\mathfrak t)$ and denote by $I(w_1)$ the set of $(\mathfrak t;\mathfrak {Q}) \in I$ such that $w(\mathfrak t)=w_1$ . Dividing Equation (3.10) by $\Omega ^{w-w_1}$ and specializing at $t=\theta ^{q^j}$ yields

$$\begin{align*}\sum_{(\mathfrak t;\mathfrak{Q}) \in I(w_1)} B_{\mathfrak t,\mathfrak{Q}}(\theta^{q^j}) (\mathfrak L(\mathfrak t;\mathfrak{Q})(\theta))^{q^j}=0 \end{align*}$$

for $j \gg 1$ . Since $w_1<w$ , by Proposition 3.3 and Condition $(LW)$ again we deduce that $B_{\mathfrak t,\mathfrak {Q}}(\theta ^{q^j})=0$ for $j \gg 1$ and for all $(\mathfrak t;\mathfrak {Q}) \in I(w_1)$ . Since each $B_{\mathfrak t,\mathfrak {Q}}$ belongs to $\overline K(t)$ , it follows that $B_{\mathfrak t,\mathfrak {Q}}=0$ for all $(\mathfrak t;\mathfrak {Q}) \in I(w_1)$ . Repeating the previous arguments, we deduce that $B_{\mathfrak t,\mathfrak {Q}}=0$ for all $(\mathfrak t;\mathfrak {Q}) \in I$ as required.

We have proved that $ \mathbf {g}- \mathbf {g}^{(-1)} \widetilde \Phi =0$ . Thus,

$$ \begin{align*} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \text{Id} & 0 \\ g_0/g & (g_{\mathfrak t,\mathfrak{Q}}/g)_{(\mathfrak t;\mathfrak{Q}) \in I} & 1 \end{pmatrix}^{(-1)} \begin{pmatrix} 1 & 0 \\ 0 & \Phi \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \Phi' & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \text{Id} & 0 \\ g_0/g & (g_{\mathfrak t,\mathfrak{Q}}/g)_{(\mathfrak t;\mathfrak{Q}) \in I} & 1 \end{pmatrix}. \end{align*} $$

By [Reference Chang, Papanikolas and Yu11, Prop. 2.2.1], we see that the common denominator b of $g_0/g$ and $g_{\mathfrak t,\mathfrak {Q}}/g$ for $(\mathfrak t,\mathfrak {Q}) \in I$ belongs to $\mathbb F_q[t] \setminus \{0\}$ . If we put $\delta _0=bg_0/g$ and $\delta _{\mathfrak t,\mathfrak {Q}}=b g_{\mathfrak t,\mathfrak {Q}}/g$ for $(\mathfrak t,\mathfrak {Q}) \in I$ which belong to $\overline K[t]$ and $\delta :=(\delta _{\mathfrak t,\mathfrak {Q}})_{\mathfrak t \in I} \in \operatorname {\mathrm {Mat}}_{1 \times |I|}(\overline K[t])$ , then $\delta _0^{(-1)}=\delta _0$ and

(3.12) $$ \begin{align} \begin{pmatrix} \text{Id} & 0 \\ \delta & 1 \end{pmatrix}^{(-1)} \begin{pmatrix} \Phi' & 0 \\ b \mathbf{v} & 1 \end{pmatrix} = \begin{pmatrix} \Phi' & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \text{Id} & 0 \\ \delta & 1 \end{pmatrix}. \end{align} $$

If we put $X:=\begin {pmatrix} \text {Id} & 0 \\ \delta & 1 \end {pmatrix} \begin {pmatrix} \Psi ' & 0 \\ b \mathbf {f} & 1 \end {pmatrix}$ , then $X^{(-1)}=\begin {pmatrix} \Phi ' & 0 \\ 0 & 1 \end {pmatrix} X$ . By [Reference Papanikolas32, §4.1.6], there exist $\nu _{\mathfrak t,\mathfrak {Q}} \in \mathbb F_q(t)$ for $(\mathfrak t,\mathfrak {Q}) \in I$ such that if we set $\nu =(\nu _{\mathfrak t,\mathfrak {Q}})_{(\mathfrak t,\mathfrak {Q}) \in I} \in \operatorname {\mathrm {Mat}}_{1 \times |I|}(\mathbb F_q(t))$ ,

$$ \begin{align*} X=\begin{pmatrix} \Psi' & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \text{Id} & 0 \\ \nu & 1 \end{pmatrix}. \end{align*} $$

Thus, the equation $\begin {pmatrix} \text {Id} & 0 \\ \delta & 1 \end {pmatrix} \begin {pmatrix} \Psi ' & 0 \\ b \mathbf {f} & 1 \end {pmatrix}=\begin {pmatrix} \Psi ' & 0 \\ 0 & 1 \end {pmatrix} \begin {pmatrix} \text {Id} & 0 \\ \nu & 1 \end {pmatrix}$ implies

(3.13) $$ \begin{align} \delta \Psi'+b \mathbf{f}=\nu. \end{align} $$

The left-hand side belongs to $\mathbb T$ , so does the right-hand side. Thus, $\nu =(\nu _{\mathfrak t,\mathfrak {Q}})_{(\mathfrak t,\mathfrak {Q}) \in I} \in \operatorname {\mathrm {Mat}}_{1 \times |I|}(\mathbb F_q[t])$ . For any $j \in \mathbb {N}$ , by specializing Equation (3.13) at $t=\theta ^{q^j}$ and using Equation (3.3) and the fact that $\Omega $ has a simple zero at $t=\theta ^{q^j}$ we deduce that

$$\begin{align*}\mathbf{f}(\theta)=\nu(\theta)/b(\theta). \end{align*}$$

Thus, for all $(\mathfrak t,\mathfrak {Q}) \in I$ , $f_{\mathfrak t;\mathfrak {Q}}(\theta )$ given as in Equation (3.8) belongs to K.