Explicit Serre weights for two-dimensional Galois representations

Abstract

We prove the explicit version of the Buzzard–Diamond–Jarvis conjecture formulated by Dembele et al. (Serre weights and wild ramification in two-dimensional Galois representations, Preprint (2016), arXiv:1603.07708 [math.NT]). More precisely, we prove that it is equivalent to the original Buzzard–Diamond–Jarvis conjecture, which was proved for odd primes (under a mild Taylor–Wiles hypothesis) in earlier work of the third author and coauthors.

1 Introduction

The weight part of Serre’s conjecture Hilbert modular forms predicts the weights of the Hilbert modular forms giving rise to a particular modular mod $p$ Galois representation, in terms of the restrictions of this Galois representation to decomposition groups above $p$ . The conjecture was originally formulated in [BDJ10] in the case that $p$ is unramified in the totally real field. Under a mild Taylor–Wiles hypothesis on the image of the global Galois representation, this conjecture has been proved for $p>2$ in a series of papers of the third author and coauthors, culminating in the paper [GLS15], which proves a generalization allowing $p$ to be arbitrarily ramified. We refer the reader to the introduction to [GLS15] for a discussion of these results.

Let $K/\mathbb{Q}_{p}$ be an unramified extension and let $\overline{\unicode[STIX]{x1D70C}}:G_{K}\rightarrow \operatorname{GL}_{2}(\overline{\mathbb{F}}_{p})$ be a (continuous) representation. If $\overline{\unicode[STIX]{x1D70C}}$ is irreducible, then the recipe for predicted weights in [BDJ10] is completely explicit, but in the case where it is a non-split extension of characters, the recipe is in terms of the reduction modulo $p$ of certain crystalline extensions of characters. This description is not useful for practical computations and the recent paper [DDR16] proposed an alternative recipe in terms of local class field theory, along with the Artin–Hasse exponential, which can be made completely explicit in concrete examples (indeed, [DDR16, §§9–10] gives substantial numerical evidence for their conjecture).

In this paper, we prove [DDR16, Conjecture 7.2], which says that the recipes of [BDJ10] and [DDR16] agree. This is a purely local conjecture and our proof is purely local. Our main input is the results of [GLS14] (and their generalization to $p=2$ in [Wan16]). We briefly sketch our approach. Suppose that $\overline{\unicode[STIX]{x1D70C}}\cong (\!\begin{smallmatrix}\unicode[STIX]{x1D712}_{1} & \ast \\ 0 & \unicode[STIX]{x1D712}_{2}\end{smallmatrix}\!)$ , and set $\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{1}\unicode[STIX]{x1D712}_{2}^{-1}$ . For a given Serre weight, the recipes of [BDJ10] and [DDR16] determine subspaces $L_{\text{BDJ}}$ and $L_{\text{DDR}}$ of $H^{1}(G_{K},\unicode[STIX]{x1D712})$ , and we have to prove that $L_{\text{BDJ}}=L_{\text{DDR}}$ .

Let $K_{\infty }/K$ be the (non-Galois) extension obtained by adjoining a compatible system of $p^{n}$ th roots of a fixed uniformizer of $K$ for all $n$ . The restriction map $H^{1}(G_{K},\unicode[STIX]{x1D712})\rightarrow H^{1}(G_{K_{\infty }},\unicode[STIX]{x1D712})$ is injective unless $\unicode[STIX]{x1D712}$ is the mod $p$ cyclotomic character, and [GLS14, Theorem 7.9] allows us to give an explicit description of the image of $L_{\text{BDJ}}$ in $H^{1}(G_{K_{\infty }},\unicode[STIX]{x1D712})$ in terms of Kisin modules. The theory of the field of norms gives a natural isomorphism of $G_{K_{\infty }}$ with $G_{k((u))}$ , where $k$ is the residue field of $K$ , and we obtain a description of the image of $L_{\text{BDJ}}$ in $H^{1}(G_{k((u))},\unicode[STIX]{x1D712})$ in terms of Artin–Schreier theory. On the other hand, we prove a compatibility of the Artin–Hasse exponential with the field of norms construction that allows us to compute the image of $L_{\text{DDR}}$ in $H^{1}(G_{k((u))},\unicode[STIX]{x1D712})$ . We then use an explicit reciprocity law of Schmid [Sch36] to reduce the comparison of $L_{\text{BDJ}}$ and $L_{\text{DDR}}$ to a purely combinatorial problem, which we solve.

It is possible that the conjecture of [DDR16] could be extended to the case that $p$ ramifies in $K$ ; we have not tried to do this, but we expect that if such a generalization exists, it could be proved by the methods of this paper, using the results of [GLS15].

The fourth author’s PhD thesis [Mav16] proved [DDR16, Conjecture 7.2] in generic cases using similar techniques to those of this paper in the setting of $(\unicode[STIX]{x1D711},\unicode[STIX]{x1D6E4})$ -modules (using the results of [CD11] where we appeal to [GLS14]), while the first three authors arrived separately at the strategy presented here for resolving the general case.

2 Notation

We follow the conventions of [GLS15], which are the same as those in the arXiv version of [GLS14] (see [GLS15, Appendix A] for a correction to some of the indices in the published version of [GLS14]). Let $p$ be prime, and let $K/\mathbb{Q}_{p}$ be a finite unramified extension of degree $f$ , with residue field $k$ . Embeddings $\unicode[STIX]{x1D70E}:k{\hookrightarrow}\overline{\mathbb{F}}_{p}$ biject with $\mathbb{Q}_{p}$ -linear embeddings $K{\hookrightarrow}\overline{\mathbb{Q}}_{p}$ , and we choose one such embedding $\unicode[STIX]{x1D70E}_{0}:k{\hookrightarrow}\overline{\mathbb{F}}_{p}$ , and recursively require that $\unicode[STIX]{x1D70E}_{i+1}^{p}=\unicode[STIX]{x1D70E}_{i}$ . Note that $\unicode[STIX]{x1D70E}_{i+f}=\unicode[STIX]{x1D70E}_{i}$ . Note also that this convention is opposite to that of [DDR16], so that their $\unicode[STIX]{x1D70E}_{i}$ is our $\unicode[STIX]{x1D70E}_{-i}$ ; consequently, to compare our formulae to those of [DDR16], one has to negate the indices throughout.

If $\unicode[STIX]{x1D70B}$ is a root of $x^{p^{f}-1}+p=0$ then we have the fundamental character $\unicode[STIX]{x1D714}_{f}:G_{K}\rightarrow k^{\times }$ defined by

$$\begin{eqnarray}\unicode[STIX]{x1D714}_{f}(g)=g(\unicode[STIX]{x1D70B})/\unicode[STIX]{x1D70B}\hspace{0.6em}({\rm mod}\hspace{0.2em}\unicode[STIX]{x1D70B}{\mathcal{O}}_{K(\unicode[STIX]{x1D70B})}).\end{eqnarray}$$

The composite of $\unicode[STIX]{x1D714}_{f}$ with the Artin map $\text{Art}_{K}$ (which we normalize so that a uniformizer corresponds to a geometric Frobenius element) is the homomorphism $K^{\times }\rightarrow k^{\times }$ sending $p$ to $1$ and sending elements of ${\mathcal{O}}_{K}^{\times }$ to their reductions modulo $p$ . For each $\unicode[STIX]{x1D70E}:k{\hookrightarrow}\overline{\mathbb{F}}_{p}$ , we set $\unicode[STIX]{x1D714}_{\unicode[STIX]{x1D70E}}:=\unicode[STIX]{x1D70E}\circ \unicode[STIX]{x1D714}|_{I_{K}}$ and $\unicode[STIX]{x1D714}_{i}:=\unicode[STIX]{x1D714}_{\unicode[STIX]{x1D70E}_{i}}$ so that, in particular, we have $\unicode[STIX]{x1D714}_{i+1}^{p}=\unicode[STIX]{x1D714}_{i}$ .

If $l/k$ is a finite extension, we choose an embedding $\widetilde{\unicode[STIX]{x1D70E}}_{0}:l{\hookrightarrow}\overline{\mathbb{F}}_{p}$ extending $\unicode[STIX]{x1D70E}_{0}$ , and again set $\widetilde{\unicode[STIX]{x1D70E}}_{i}=\widetilde{\unicode[STIX]{x1D70E}}_{i+1}^{p}$ . We have an isomorphism

(2.0.1) $$\begin{eqnarray}l\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}\stackrel{{\sim}}{\longrightarrow }\mathop{\prod }_{\widetilde{\unicode[STIX]{x1D70E}}_{i}}\overline{\mathbb{F}}_{p},\end{eqnarray}$$

with the projection onto the factor labelled by $\widetilde{\unicode[STIX]{x1D70E}}_{i}$ being given by $x\otimes y\mapsto \widetilde{\unicode[STIX]{x1D70E}}_{i}(x)y$ . Under this isomorphism, the automorphism $\unicode[STIX]{x1D711}\otimes \text{id}$ on $l\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}$ becomes identified with the automorphism on $\prod \overline{\mathbb{F}}_{p}$ given by $(y_{i})\mapsto (y_{i-1}).$

If ${\mathcal{M}}$ is an $l\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}$ -module equipped with a $\unicode[STIX]{x1D711}$ -linear endomorphism $\unicode[STIX]{x1D711}$ , then the isomorphism (2.0.1) induces a corresponding decomposition ${\mathcal{M}}\stackrel{{\sim}}{\longrightarrow }\prod _{i}{\mathcal{M}}_{i},$ and the endomorphism $\unicode[STIX]{x1D711}$ of ${\mathcal{M}}$ induces $\overline{\mathbb{F}}_{p}$ -linear morphisms $\unicode[STIX]{x1D711}:{\mathcal{M}}_{i-1}\rightarrow {\mathcal{M}}_{i}$ .

3 Results

3.1 Fields of norms

We briefly recall (following [Kis09, §1.1.12]) the theory of the field of norms and of étale $\unicode[STIX]{x1D711}$ -modules, adapted to the case at hand. For each $n$ , let $(-p)^{1/p^{n}}$ be a choice of the $p^{n}$ th root of $-p$ , chosen so that $((-p)^{1/p^{n+1}})^{p}=(-p)^{1/p^{n}}$ , and let $K_{n}=K((-p)^{1/p^{n}})$ . Write $K_{\infty }=\bigcup _{n}K_{n}$ . Then, by the theory of the field of norms,

$$\begin{eqnarray}\underset{N_{K_{n+1}/K_{n}}}{\varprojlim }\,K_{n}\end{eqnarray}$$

(the transition maps being the norm maps) can be identified with $k((u))$ , with $((-p)^{1/p^{n}})_{n}$ corresponding to $u$ . If $F$ is a finite extension of $K$ (inside some given algebraic closure of $K$ containing $K_{\infty }$ ), then $F_{\infty }:=FK_{\infty }$ is a finite extension of $K_{\infty }$ , and applying the field of norms construction to $F_{\infty }$ , we obtain a finite separable extension

$$\begin{eqnarray}{\mathcal{F}}:=\underset{N_{FK_{n}/FK_{n-1}}}{\varprojlim }\,FK_{n},\end{eqnarray}$$

of $k((u))$ . If $F$ is Galois over $K$ , then $F_{\infty }$ is Galois over $K_{\infty }$ , and ${\mathcal{F}}$ is also Galois over $k((u))$ , and there is a natural isomorphism of Galois groups

(3.1.1) $$\begin{eqnarray}\text{Gal}({\mathcal{F}}/k((u)))\stackrel{{\sim}}{\longrightarrow }\text{Gal}(F_{\infty }/K_{\infty }),\end{eqnarray}$$

and, composing with the canonical homomorphism $\text{Gal}(F_{\infty }/K_{\infty })\rightarrow \text{Gal}(F/K),$ a natural homomorphism of Galois groups

(3.1.2) $$\begin{eqnarray}\text{Gal}({\mathcal{F}}/k((u)))\rightarrow \text{Gal}(F/K).\end{eqnarray}$$

Every finite extension of $K_{\infty }$ arises as such an $F_{\infty }$ and, in this manner, we obtain a functorial bijection between finite extensions of $K_{\infty }$ and finite separable extensions ${\mathcal{F}}$ of $k((u))$ . In particular, the various isomorphisms (3.1.1) piece together to induce a natural isomorphism of absolute Galois groups

(3.1.3) $$\begin{eqnarray}G_{K_{\infty }}=G_{k((u))}.\end{eqnarray}$$

The utility of the isomorphism (3.1.3) arises from the fact that there is an equivalence of abelian categories between the category of finite-dimensional $\overline{\mathbb{F}}_{p}$ -representations $V$ of $G_{k((u))}$ and the category of étale $\unicode[STIX]{x1D711}$ -modules. The latter are, by definition, finite $k((u))\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}$ -modules ${\mathcal{M}}$ equipped with a $\unicode[STIX]{x1D711}$ -semilinear map $\unicode[STIX]{x1D711}:{\mathcal{M}}\rightarrow {\mathcal{M}}$ , with the property that the induced $k((u))\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}$ -linear map $\unicode[STIX]{x1D711}^{\ast }{\mathcal{M}}\rightarrow {\mathcal{M}}$ is an isomorphism. This equivalence of categories preserves lengths in the obvious sense, and is given by the functors

$$\begin{eqnarray}T:{\mathcal{M}}\rightarrow (k((u))^{\text{sep}}\otimes _{k((u))}{\mathcal{M}})^{\unicode[STIX]{x1D711}=1}\end{eqnarray}$$

(where $k((u))^{\text{sep}}$ is a separable closure of $k((u))$ ) and

$$\begin{eqnarray}V\mapsto (k((u))^{\text{sep}}\otimes _{\mathbb{F}_{p}}V)^{G_{k((u))}}.\end{eqnarray}$$

The isomorphism (3.1.3) then allows us to describe finite-dimensional representations of $G_{K_{\infty }}$ over $\overline{\mathbb{F}}_{p}$ via étale $\unicode[STIX]{x1D711}$ -modules. In the § 3.3 we make this description completely explicit in the context of (the restriction to $K_{\infty }$ of) the crystalline extensions of characters that arise in the conjecture of [BDJ10].

The above isomorphisms of Galois groups are compatible with local class field theory in a natural way. Namely, if $F/K$ and ${\mathcal{F}}/k((u))$ are as above, then the projection map $k((u))=\mathop{\varprojlim }\nolimits_{N_{K_{n+1}/K_{n}}}K_{n}\rightarrow K$ induces a natural map

(3.1.4) $$\begin{eqnarray}k((u))^{\times }/N_{{\mathcal{F}}/k((u))^{\times }}{\mathcal{F}}^{\times }\rightarrow K^{\times }/N_{F/K}F^{\times },\end{eqnarray}$$

and we have the following result.

Lemma 3.1.5. If $F/K$ is a finite abelian extension, then the following diagram commutes.

Proof. This is easily checked directly, and is a special case of [AJ12, Proposition 5.2], which proves a generalization to higher-dimensional local fields; see also [Lau88], where the analogous result is proved for general APF extensions (strictly speaking, the result of [Lau88] does not apply as written in our situation, as the extension $K_{\infty }/K$ is not Galois; but, in fact, the argument still works). In brief, it is enough to check separately the cases that $F/K$ is either unramified or totally ramified; in the former case the result is immediate, while the latter case follows from Dwork’s description of Artin’s reciprocity map for totally ramified abelian extensions [Ser79, XIII §5 Corollary to Theorem 2].◻

3.2 Compatibility of pairings

It will be convenient to establish a further compatibility between various natural pairings. For a field $M$ , let $M^{(p)}/M$ denote the maximal exponent $p$ abelian extension (inside some fixed algebraic closure). If $M_{\infty }/M$ is an extension, then we have a diagram as follows (where $\operatorname{pr}$ is the natural map given by restriction of automorphisms of $M_{\infty }^{(p)}$ to $M^{(p)}$ ).

Lemma 3.2.1. The diagram commutes, in the sense that $\langle \operatorname{pr}\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}\rangle =\langle \unicode[STIX]{x1D6FC},\unicode[STIX]{x1D704}\unicode[STIX]{x1D6FD}\rangle$ .

Proof. Since $H^{1}(G_{M},\overline{\mathbb{F}}_{p})=\text{Hom}(G_{M},\overline{\mathbb{F}}_{p})$ (and similarly for $M_{\infty }$ ), since the pairings are given by evaluation, and since $\unicode[STIX]{x1D704}$ is the natural restriction map, this is clear.◻

Suppose now that $M$ is a finite extension of $\mathbb{Q}_{p}$ with residue field $l$ , and that $\unicode[STIX]{x1D70B}$ is a uniformizer of $M$ . If $M_{\infty }/M$ is the extension given by a compatible choice of $p$ -power roots of $\unicode[STIX]{x1D70B}$ , then

$$\begin{eqnarray}\text{Gal}(M_{\infty }^{(p)}/M_{\infty })\simeq l((u))^{\times }\otimes \mathbb{F}_{p}\end{eqnarray}$$

via the field of norms construction together with local class field theory (applied to $l((u))$ ).

On the other hand, taking Galois cohomology of the short exact sequence

$$\begin{eqnarray}0\rightarrow \overline{\mathbb{F}}_{p}\rightarrow l((u))^{\text{sep}}\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}\,\mathop{\longrightarrow }^{\unicode[STIX]{x1D713}\otimes \text{id}}\,l((u))^{\text{sep}}\otimes _{\mathbb{ F}_{p}}\overline{\mathbb{F}}_{p}\rightarrow 0,\end{eqnarray}$$

where $\unicode[STIX]{x1D713}:l((u))^{\text{sep}}\rightarrow l((u))^{\text{sep}}$ is the Artin–Schreier map defined by $\unicode[STIX]{x1D713}(x)=x^{p}-x$ , yields an isomorphism

$$\begin{eqnarray}\displaystyle H^{1}(G_{M_{\infty }},\overline{\mathbb{F}}_{p})=H^{1}(G_{l((u))},\overline{\mathbb{F}}_{p})=\text{Hom}(G_{l((u))},\overline{\mathbb{F}}_{p})\simeq (l((u))/\unicode[STIX]{x1D713}l((u)))\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}; & & \displaystyle \nonumber\end{eqnarray}$$

concretely, the element $a\in l((u))$ corresponds to the homomorphism $f_{a}:G_{l((u))}\rightarrow \mathbb{F}_{p}$ given by $f_{a}(g)=g(x)-x$ , where $x\in l((u))^{\text{sep}}$ is chosen so that $\unicode[STIX]{x1D713}(x)=a$ . (See e.g. [Ser79, X §3(a)] for more details.)

Theorem 3.2.2. Let $\unicode[STIX]{x1D70E}_{b}\in \text{Gal}(M_{\infty }^{(p)}/M_{\infty })$ be the Galois element corresponding via the local Artin map to an element $b\in l((u))^{\times }\otimes \mathbb{F}_{p}$ , and let $f_{a}$ be the element of $H^{1}(G_{M_{\infty }},\overline{\mathbb{F}}_{p})$ corresponding to an element $a\in (l((u))/\unicode[STIX]{x1D713}l((u)))\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}$ . Then

$$\begin{eqnarray}\langle f_{a},\unicode[STIX]{x1D70E}_{b}\rangle =\operatorname{Tr}_{l\otimes _{\mathbb{ F}_{p}}\overline{\mathbb{F}}_{p}/\overline{\mathbb{F}}_{p}}\biggl(\operatorname{Res}a\cdot \frac{db}{b}\biggr).\end{eqnarray}$$

Proof. This was first proved in [Sch36]; for a more modern proof, see [Ser79, XIV Corollary to Proposition 15]. ◻

3.3 Crystalline extension classes and $L_{\text{BDJ}}$

We begin by briefly recalling some of the main results of [GLS14]. For each $0\leqslant i\leqslant f-1$ we fix an integer $r_{i}\in [1,p]$ ; we then define $r_{i}$ for all integers $i$ by demanding that $r_{i+f}=r_{i}$ . We let $J$ be a subset of $\{0,\ldots ,f-1\}$ , and we assume that $J$ is maximal in the sense of [DDR16, §7.2]; in other words, we assume that:

1. (i) if for some $i>j$ we have $(r_{j},\ldots ,r_{i})=(1,p-1,\ldots ,p-1,p)$ , and $j+1,\ldots ,i\notin J$ , then $j\notin J$ ; and

2. (ii) if all the $r_{i}$ are equal to $p-1$ , or if $p=2$ and all of the $r_{i}$ are equal to $2$ , then $J$ is non-empty.

We let $\unicode[STIX]{x1D712}:G_{K}\rightarrow \overline{\mathbb{F}}_{p}^{\times }$ be a character with the property that

$$\begin{eqnarray}\unicode[STIX]{x1D712}|_{I_{K}}=\mathop{\prod }_{j\in J}\unicode[STIX]{x1D714}_{j}^{r_{j}}\mathop{\prod }_{j\notin J}\unicode[STIX]{x1D714}_{j}^{-r_{j}}.\end{eqnarray}$$

We let $L_{\text{BDJ}}$ denote the subset of $H^{1}(G_{K},\unicode[STIX]{x1D712})$ consisting of those classes corresponding to extensions of the trivial character by $\unicode[STIX]{x1D712}$ that arise as the reductions of crystalline representation whose $\unicode[STIX]{x1D70E}_{i}$ -labelled Hodge–Tate weights are $\{0,(-1)^{i\notin J}r_{i}\}$ , where $(-1)^{i\notin J}$ is $1$ if $i\in J$ and $-1$ otherwise. The subsequent points follow from the proof of [GLS14, Theorem 9.1], together with [GLS14, Lemmas 9.3 and 9.4] and (in the case that $p=2$ ) the results of [Wan16].

1. (i) The subset $L_{\text{BDJ}}$ is an $\overline{\mathbb{F}}_{p}$ -subspace of $H^{1}(G_{K},\unicode[STIX]{x1D712})$ .

2. (ii) An extension class is in $L_{\text{BDJ}}$ if and only if it admits a reducible crystalline lift whose $\unicode[STIX]{x1D70E}_{i}$ -labelled Hodge–Tate weights are $\{0,(-1)^{i\notin J}r_{i}\}$ .

3. (iii) If $J=\{0,\ldots ,f-1\}$ and all $r_{i}=p$ , then $L_{\text{BDJ}}=H^{1}(G_{K},\unicode[STIX]{x1D712})$ .

4. (iv) Assume that we are not in the case of the previous point. Then $\dim _{\overline{\mathbb{F}}_{p}}L_{\text{BDJ}}=|J|$ , unless $\unicode[STIX]{x1D712}=1$ , in which case $\dim _{\overline{\mathbb{F}}_{p}}L_{\text{BDJ}}=|J|+1$ .

We recall below from [DDR16] the definition of another subspace of $H^{1}(G_{K},\unicode[STIX]{x1D712})$ , denoted by $L_{\text{DDR}}$ ; our main result, then, is that $L_{\text{BDJ}}=L_{\text{DDR}}$ . We begin with an easy special case.

Lemma 3.3.1. If $J=\{0,\ldots ,f-1\}$ and every $r_{i}=p$ , then $L_{\text{BDJ}}=L_{\text{DDR}}$ .

Proof. In this case we have $L_{\text{DDR}}=H^{1}(G_{K},\unicode[STIX]{x1D712})$ by definition (see Definition 3.4.1 below), and we already noted above that $L_{\text{BDJ}}=H^{1}(G_{K},\unicode[STIX]{x1D712})$ .◻

We can and do exclude the case covered by Lemma 3.3.1 from now on; that is, in addition to the assumptions made above, we assume that:

• ∙ if every $r_{i}$ is equal to $p$ , then $J\neq \{0,\ldots ,f-1\}$ .

If $\unicode[STIX]{x1D712}=\overline{\unicode[STIX]{x1D716}}$ , then the peu ramifié subspace of $H^{1}(G_{K},\overline{\unicode[STIX]{x1D716}})$ is, by definition, the codimension one subspace spanned by the classes corresponding via Kummer theory to elements of ${\mathcal{O}}_{K}^{\times }$ . Since we have excluded the cases covered by Lemma 3.3.1, $L_{\text{BDJ}}$ is contained in the peu ramifié subspace of $H^{1}(G_{K},\overline{\unicode[STIX]{x1D716}})$ by [DS15, Theorem 4.9].

By [GLS15, Lemma 5.4.2], for any $\unicode[STIX]{x1D712}\neq \overline{\unicode[STIX]{x1D716}}$ the natural restriction map $H^{1}(G_{K},\unicode[STIX]{x1D712})\rightarrow H^{1}(G_{K_{\infty }},\unicode[STIX]{x1D712})$ is injective, while if $\unicode[STIX]{x1D712}=\overline{\unicode[STIX]{x1D716}}$ , then the kernel is spanned by the tres ramifié class corresponding to $-p$ ; in particular, the restriction of this map to $L_{\text{BDJ}}$ is injective. The following theorem describes the image of $L_{\text{BDJ}}$ ; before stating it, we introduce some notation that we will use throughout the paper.

Write $\unicode[STIX]{x1D712}$ as a power of $\unicode[STIX]{x1D714}_{0}$ times an unramified character $\unicode[STIX]{x1D707}:\text{Gal}(L/K)\rightarrow \overline{\mathbb{F}}_{p}^{\times }$ , and write $\unicode[STIX]{x1D707}(\text{Frob}_{K})=a$ , so that $a^{[l:k]}=1$ ; here $\text{Frob}_{K}\in \text{Gal}(L/K)$ denotes the arithmetic Frobenius. For each $\unicode[STIX]{x1D70E}:k{\hookrightarrow}\overline{\mathbb{F}}_{p}$ , we let $\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D707}}$ be the element $(1,a^{-1},\ldots ,a^{1-[l:k]})\in l\otimes _{k,\unicode[STIX]{x1D70E}}\overline{\mathbb{F}}_{p}$ , so that $\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D707}}$ is a basis of the one-dimensional $\overline{\mathbb{F}}_{p}$ -vector space $(l\otimes _{k,\unicode[STIX]{x1D70E}}\overline{\mathbb{F}}_{p})^{\text{Gal}(L/K)=\unicode[STIX]{x1D707}}$ . Similarly, we let $\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D707}^{-1}}$ be the element $(1,a,\ldots ,a^{[l:k]-1})\in l\otimes _{k,\unicode[STIX]{x1D70E}}\overline{\mathbb{F}}_{p}$ .

Theorem 3.3.2. The subspace $L_{\text{BDJ}}$ of $H^{1}(G_{K},\unicode[STIX]{x1D712})$ consists of precisely those classes whose restrictions to $H^{1}(G_{K_{\infty }},\unicode[STIX]{x1D712})$ can be represented by étale $\unicode[STIX]{x1D711}$ -modules ${\mathcal{M}}$ of the following form.

Set $h_{i}=r_{i}$ if $i\in J$ and $h_{i}=0$ if $i\not \in J$ . Then we can choose bases $e_{i},f_{i}$ of the ${\mathcal{M}}_{i}$ so that $\unicode[STIX]{x1D711}$ has the form

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}(e_{i-1}) & = & \displaystyle u^{r_{i}-h_{i}}e_{i},\nonumber\\ \displaystyle \unicode[STIX]{x1D711}(f_{i-1}) & = & \displaystyle (a)_{i}u^{h_{i}}f_{i}+x_{i}e_{i}.\nonumber\end{eqnarray}$$

Here $(a)_{i}=1$ for $i\neq 0$ , and equals $a=\unicode[STIX]{x1D707}(\text{Frob}_{K})$ for $i=0$ ; and we have $x_{i}=0$ if $i\not \in J$ and $x_{i}\in \overline{\mathbb{F}}_{p}$ if $i\in J$ , except in the case that $\unicode[STIX]{x1D712}=1$ .

If $\unicode[STIX]{x1D712}=1$ then $a=1$ , and if we fix some $i_{0}\in J$ , then $x_{i_{0}}$ is allowed to be of the form $x_{i_{0}}^{\prime }+x_{i_{0}}^{\prime \prime }u^{p}$ with $x_{i_{0}}^{\prime },x_{i_{0}}^{\prime \prime }\in \overline{\mathbb{F}}_{p}$ (while the other $x_{i}$ are in $\overline{\mathbb{F}}_{p}$ ).

In every case, the $x_{i}$ are uniquely determined by ${\mathcal{M}}$ .

Proof. In the case $p>2$ , this is an immediate consequence of [GLS14, Theorem 7.9] (which describes the corresponding Kisin modules, which are just lattices in ${\mathcal{M}}$ ; the set $J^{\prime }$ appearing there can be taken to be our $J$ by [GLS14, Proposition 8.8] and our assumption that $J$ is maximal) and the proof of [GLS14, Theorem 9.1] (which shows that the different $x_{i}$ give rise to different Galois representations), while if $p=2$ , then the result follows from the results of [Wan16].◻

As in § 2, we let $\unicode[STIX]{x1D70B}$ be a choice of $(p^{f}-1)$ th root of $-p$ . Write $M:=L(\unicode[STIX]{x1D70B})$ , where $L/K$ is an unramified extension of degree prime to $p$ , chosen so that $\unicode[STIX]{x1D712}|_{G_{M}}$ is trivial (in [DDR16] a slightly more general choice of $M$ is permitted, but it is shown there that their constructions are independent of this choice, and this choice is convenient for us). Then $M/K$ is an abelian extension of degree prime to $p$ . Since $(p^{f}-1)$ is prime to $p$ , for each $n\geqslant 1$ there is a unique $p^{n}$ th root $\unicode[STIX]{x1D70B}^{1/p^{n}}$ of $\unicode[STIX]{x1D70B}$ such that $(\unicode[STIX]{x1D70B}^{1/p^{n}})^{(p^{f}-1)}=(-p)^{1/p^{n}}$ , and we set $M_{n}=M(\unicode[STIX]{x1D70B}^{1/p^{n}})$ , $M_{\infty }=\bigcup _{n}M_{n}$ .

If ${\mathcal{M}}$ is an étale $\unicode[STIX]{x1D711}$ -module with corresponding $G_{K_{\infty }}$ -representation $T({\mathcal{M}})$ , then it is easy to check that the étale $\unicode[STIX]{x1D711}$ -module corresponding to $T({\mathcal{M}})|_{G_{M_{\infty }}}$ is

$$\begin{eqnarray}{\mathcal{M}}_{M}:=l((u))\otimes _{k((u)),u\mapsto u^{p^{f}-1}}{\mathcal{M}}.\end{eqnarray}$$

Applying this to one of the étale $\unicode[STIX]{x1D711}$ -modules arising in the statement of Theorem 3.3.2, it follows that (with the obvious choice of basis $e_{i},f_{i}$ for ${\mathcal{M}}_{M}$ ) the matrix of $\unicode[STIX]{x1D711}:{\mathcal{M}}_{M,i-1}\rightarrow {\mathcal{M}}_{M,i}$ is

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}u^{(r_{i}-h_{i})(p^{f}-1)} & x_{i}\\ 0 & (a)_{i}u^{h_{i}(p^{f}-1)}\end{array}\right)\end{eqnarray}$$

whereas above $h_{i}=r_{i}$ if $i\in J$ and $h_{i}=0$ if $i\notin J$ , and $x_{i}$ is zero if $i\notin J$ . Furthermore, $x_{i}\in \overline{\mathbb{F}}_{p}$ , except that if $\unicode[STIX]{x1D712}=1$ , we have fixed a choice of $i_{0}\in J$ , and $x_{i_{0}}$ is allowed to be of the form $x_{i_{0}}^{\prime }+x_{i_{0}}^{\prime \prime }u^{p(p^{f}-1)}$ with $x_{i_{0}}^{\prime },x_{i_{0}}^{\prime \prime }\in \overline{\mathbb{F}}_{p}$ . (Here the ${\mathcal{M}}_{M,i}$ are periodic with period $f[l:k]$ , but of course the $r_{i}$ , $h_{i}$ and $x_{i}$ depend only on $i$ modulo $f$ .)

We now make a change of basis, setting $e_{i}^{\prime }=u^{\unicode[STIX]{x1D6FC}_{i}}e_{i}$ and $f_{i}^{\prime }=a^{\lfloor i/f\rfloor }u^{\unicode[STIX]{x1D6FD}_{i}}f_{i}$ (where $0\leqslant i\leqslant f[l:k]-1$ ), so that the matrix of $\unicode[STIX]{x1D711}:{\mathcal{M}}_{M,i-1}\rightarrow {\mathcal{M}}_{M,i}$ becomes

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}u^{(r_{i}-h_{i})(p^{f}-1)+p\unicode[STIX]{x1D6FC}_{i-1}-\unicode[STIX]{x1D6FC}_{i}} & a^{\lfloor i-1/f\rfloor }x_{i}u^{p\unicode[STIX]{x1D6FD}_{i-1}-\unicode[STIX]{x1D6FC}_{i}}\\ 0 & u^{h_{i}(p^{f}-1)+p\unicode[STIX]{x1D6FD}_{i-1}-\unicode[STIX]{x1D6FD}_{i}}\end{array}\right)\!.\end{eqnarray}$$

We choose the $\unicode[STIX]{x1D6FC}_{i},\unicode[STIX]{x1D6FD}_{i}$ so that the entries on the diagonal become trivial; concretely, this means that we set

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{i}=-\mathop{\sum }_{j=0}^{f-1}(r_{i+j+1}-h_{i+j+1})p^{f-1-j},\quad \unicode[STIX]{x1D6FD}_{i}=-\mathop{\sum }_{j=0}^{f-1}h_{i+j+1}p^{f-1-j}.\end{eqnarray}$$

Write $\unicode[STIX]{x1D709}_{i}:=\unicode[STIX]{x1D6FC}_{i}-p\unicode[STIX]{x1D6FD}_{i-1}$ , so that we have

$$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}=\mathop{\sum }_{j=0}^{f-1}(-1)^{i+j+1\notin J}r_{i+j+1}p^{f-1-j}+\unicode[STIX]{x1D6FF}_{i\in J}r_{i}(p^{f}-1),\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}_{i\in J}=1$ if $i\in J$ and $0$ otherwise.

With the obvious basis for ${\mathcal{M}}_{M}$ as an $l((u))\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}$ -module, $\unicode[STIX]{x1D719}_{{\mathcal{M}}_{M}}$ is given by the matrix

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}1 & (x_{i}a^{-1}\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}u^{-\unicode[STIX]{x1D709}_{i}})_{i=0,\ldots ,f-1}\\ 0 & 1\end{array}\right)\end{eqnarray}$$

where $\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}$ is the element of $l\otimes _{k,\unicode[STIX]{x1D70E}_{i}}\overline{\mathbb{F}}_{p}$ that we defined above. Then $T({\mathcal{M}}_{M})$ is an extension of the trivial representation by itself, and thus corresponds to an element of $\text{Hom}(G_{l((u))},\overline{\mathbb{F}}_{p})$ . By the definition of $T$ , the kernel of this homomorphism corresponds to the Artin–Schreier extension of $l((u))$ determined by $(x_{i}\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}u^{-\unicode[STIX]{x1D709}_{i}})_{i=0,\ldots ,f-1}$ . We have therefore proved the following result.

Corollary 3.3.3. The image of $L_{\text{BDJ}}$ in $H^{1}(G_{M_{\infty }},\overline{\mathbb{F}}_{p})=\text{Hom}(G_{l((u))},\overline{\mathbb{F}}_{p})$ is spanned by the classes $f_{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}u^{-\unicode[STIX]{x1D709}_{i}}}$ corresponding via Artin–Schreier theory to the elements

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}u^{-\unicode[STIX]{x1D709}_{i}}\in l\otimes _{k,\unicode[STIX]{x1D70E}_{i}}\overline{\mathbb{F}}_{p}\subseteq l\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p},\end{eqnarray}$$

for $i\in J$ , together with the class $f_{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i_{0}},\unicode[STIX]{x1D707}^{-1}}u^{p(p^{f}-1)-\unicode[STIX]{x1D709}_{i_{0}}}}$ if $\unicode[STIX]{x1D712}=1$ .

As in [DDR16, §3.2], we may write $\unicode[STIX]{x1D712}|_{I_{K}}=\unicode[STIX]{x1D714}_{0}^{n_{0}}$ for some unique $n_{0}$ of the form $n_{0}=\sum _{j=1}^{f}a_{j}p^{f-j}$ with each $a_{j}\in [1,p]$ and at least one $a_{j}\neq p$ . We set

$$\begin{eqnarray}n_{i}=\mathop{\sum }_{j=1}^{f}a_{i+j}p^{f-j},\end{eqnarray}$$

so we have $\unicode[STIX]{x1D712}|_{I_{K}}=\unicode[STIX]{x1D714}_{i}^{n_{i}}$ , and for all $i,j$ we have

$$\begin{eqnarray}p^{-i}n_{i}\equiv p^{-j}n_{j}\hspace{0.6em}({\rm mod}\hspace{0.2em}p^{f}-1).\end{eqnarray}$$

Note that we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D712}|_{I_{K}} & = & \displaystyle \mathop{\prod }_{j\in J}\unicode[STIX]{x1D714}_{j}^{r_{j}}\mathop{\prod }_{j\notin J}\unicode[STIX]{x1D714}_{j}^{-r_{j}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\prod }_{j=0}^{f-1}\unicode[STIX]{x1D714}_{i}^{-(-1)^{i+j+1\in J}r_{i+j+1}p^{f-1-j}}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D714}_{i}^{\unicode[STIX]{x1D6FC}_{i}-p\unicode[STIX]{x1D6FD}_{i-1}}=\unicode[STIX]{x1D714}_{i}^{\unicode[STIX]{x1D709}_{i}},\nonumber\end{eqnarray}$$

so that, in particular, we have

(3.3.4) $$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}\equiv n_{i}\hspace{0.6em}({\rm mod}\hspace{0.2em}p^{f}-1).\end{eqnarray}$$

3.4 The Artin–Hasse exponential and $L_{\text{DDR}}$

We now recall some of the definitions made in [DDR16, §5.1]. In particular, for each $i$ we define an embedding $\unicode[STIX]{x1D70E}_{i}^{\prime }$ and an integer $n_{i}^{\prime }$ as follows. If $a_{i-1}\neq p$ , then we set $\unicode[STIX]{x1D70E}_{i}^{\prime }=\unicode[STIX]{x1D70E}_{i-1}$ and $n_{i}^{\prime }=n_{i-1}$ . If $a_{i-1}=p$ , then we let $j$ be the greatest integer less than $i$ such that $a_{j-1}\neq p-1$ , and we set $\unicode[STIX]{x1D70E}_{i}^{\prime }=\unicode[STIX]{x1D70E}_{j-1}$ and $n_{i}^{\prime }=n_{j-1}-(p^{f}-1)$ . Note that we always have $n_{i}^{\prime }>0$ .

We let $E(x)=\exp (\sum _{m\geqslant 0}x^{p^{m}}/p^{m})\in \mathbb{Z}_{p}[[x]]$ denote the Artin–Hasse exponential. For any $\unicode[STIX]{x1D6FC}\in \mathfrak{m}_{M}$ , we define the homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D6FC}}:l\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}\rightarrow {\mathcal{O}}_{M}^{\times }\otimes _{\mathbb{ F}_{p}}\overline{\mathbb{F}}_{p}\end{eqnarray}$$

by $\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D6FC}}(a\otimes b):=E([a]\unicode[STIX]{x1D6FC})\otimes b$ , where $[\cdot ]:l\rightarrow W(l)$ is the Teichmüller lift. Then we set

$$\begin{eqnarray}u_{i}:=\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D70B}^{n_{i}^{\prime }}}(\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i}^{\prime },\unicode[STIX]{x1D707}})\in {\mathcal{O}}_{M}^{\times }\otimes \overline{\mathbb{F}}_{p}.\end{eqnarray}$$

In the case that $\unicode[STIX]{x1D712}=1$ , we also set $u_{\operatorname{triv}}:=\unicode[STIX]{x1D70B}\otimes 1\in M^{\times }\otimes \overline{\mathbb{F}}_{p}$ , and in the case that $\unicode[STIX]{x1D712}=\overline{\unicode[STIX]{x1D716}}$ , the mod $p$ cyclotomic character, we set $u_{\operatorname{cyc}}:=\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D70B}^{p(p^{f}-1)/(p-1)}}(b\otimes 1)$ , where $b\in l$ is any element with $\operatorname{Tr}_{l/\mathbb{F}_{p}}(b)\neq 0$ . It is shown in [DDR16, §5] that the $u_{i}$ , together with $u_{\operatorname{triv}}$ if $\unicode[STIX]{x1D712}=1$ , and $u_{\operatorname{cyc}}$ if $\unicode[STIX]{x1D712}=\overline{\unicode[STIX]{x1D716}}$ , are a basis of the $\overline{\mathbb{F}}_{p}$ -vector space

$$\begin{eqnarray}U_{\unicode[STIX]{x1D712}}:=(M^{\times }\otimes \overline{\mathbb{F}}_{p}(\unicode[STIX]{x1D712}^{-1}))^{\text{Gal}(M/K)}.\end{eqnarray}$$

Via the Artin map $\text{Art}_{M}$ , we may write

$$\begin{eqnarray}H^{1}(G_{K},\unicode[STIX]{x1D712})\cong \text{Hom}_{\text{Gal}(M/K)}(M^{\times },\overline{\mathbb{F}}_{p}(\unicode[STIX]{x1D712}))\end{eqnarray}$$

and, thus, identify $H^{1}(G_{K},\unicode[STIX]{x1D712})$ with the $\overline{\mathbb{F}}_{p}$ -dual of $U_{\unicode[STIX]{x1D712}}$ . We then define a basis of $H^{1}(G_{K},\unicode[STIX]{x1D712})$ by letting $c_{i}$ , $c_{\operatorname{triv}}$ (if $\unicode[STIX]{x1D712}=1$ ) and $c_{\operatorname{cyc}}$ (if $\unicode[STIX]{x1D712}=\overline{\unicode[STIX]{x1D716}}$ ) denote the dual basis to that given by the $u_{i}$ , $u_{\operatorname{triv}}$ and $u_{\operatorname{cyc}}$ .

Recall from [DDR16, §7.1] the definition of the set $\unicode[STIX]{x1D707}(J)$ . It is defined as follows: $\unicode[STIX]{x1D707}(J)=J$ , unless there is some $i\notin J$ for which we have $a_{i-1}=p$ , $a_{i-2}=p-1,\ldots ,a_{i-s}=p-1,a_{i-s-1}\neq p-1$ , and at least one of $i-1,i-2,\ldots ,i-s$ is in $J$ . If this is the case, we let $x$ be minimal such that $i-x\in J$ , and we consider the set obtained from $J$ by replacing $i-x$ with $i$ . Then $\unicode[STIX]{x1D707}(J)$ is the set obtained by simultaneously making all such replacements (that is, making these replacements for all possible $i$ ).

Definition 3.4.1. We define $L_{\text{DDR}}$ to be the subspace of $H^{1}(G_{K},\unicode[STIX]{x1D712})$ spanned by the classes $c_{i}$ for $i\in \unicode[STIX]{x1D707}(J)$ , together with the class $c_{\operatorname{triv}}$ if $\unicode[STIX]{x1D712}=1$ , and the class $c_{\operatorname{cyc}}$ if $\unicode[STIX]{x1D712}=\overline{\unicode[STIX]{x1D716}}$ , $J=\{0,\ldots ,f-1\}$ and every $r_{i}=p$ .

3.5 The comparison of $L_{\text{BDJ}}$ and $L_{\text{DDR}}$

In this section, we prove that the classes in $L_{\text{BDJ}}$ are orthogonal to certain $u_{i}$ . We begin with a computation that will allow us to compare the constructions underlying the definition of $L_{\text{DDR}}$ , which involve the Artin–Hasse exponential, with the field of norms constructions underlying the description of $L_{\text{BDJ}}$ .

Lemma 3.5.1. For any $n\geqslant 1$ , $a\in l$ and $r\geqslant 1$ with $(r,p)=1$ we have $N_{K_{n}/K}E([a^{1/p^{n}}](\unicode[STIX]{x1D70B}^{1/p^{n}})^{r})=E([a]\unicode[STIX]{x1D70B}^{r})$ .

Proof. Let $\unicode[STIX]{x1D701}$ be a primitive $p^{n}$ th root of unity. Then

$$\begin{eqnarray}\displaystyle N_{K_{n}/K}E([a^{1/p^{n}}](\unicode[STIX]{x1D70B}^{1/p^{n}})^{r})= & \displaystyle ~\mathop{\prod }_{k=0}^{p^{n}-1}E([a^{1/p^{n}}](\unicode[STIX]{x1D70B}^{1/p^{n}})^{r}\unicode[STIX]{x1D701}^{k}) & \displaystyle \\ \displaystyle = & \displaystyle ~\mathop{\prod }_{k=0}^{p^{n}-1}\exp \biggl(\mathop{\sum }_{m\geqslant 0}\frac{[a^{1/p^{n}}]^{p^{m}}(\unicode[STIX]{x1D70B}^{1/p^{n}})^{rp^{m}}\unicode[STIX]{x1D701}^{kp^{m}}}{p^{m}}\biggr)\\ \displaystyle = & \displaystyle ~\exp \biggl(\mathop{\sum }_{k=0}^{p^{n}-1}\mathop{\sum }_{m\geqslant 0}\frac{[a^{1/p^{n}}]^{p^{m}}(\unicode[STIX]{x1D70B}^{1/p^{n}})^{rp^{m}}\unicode[STIX]{x1D701}^{kp^{m}}}{p^{m}}\biggr)\\ \displaystyle = & \displaystyle ~\exp \biggl(\mathop{\sum }_{m\geqslant 0}\frac{[a^{1/p^{n}}]^{p^{m}}(\unicode[STIX]{x1D70B}^{1/p^{n}})^{rp^{m}}}{p^{m}}\mathop{\sum }_{k=0}^{p^{n}-1}\unicode[STIX]{x1D701}^{kp^{m}}\biggr).\end{eqnarray}$$

Now the sum over roots of unity is $0$ if $\unicode[STIX]{x1D701}^{p^{m}}\neq 1$ (equivalently, $m<n$ ) and $p^{n}$ if $\unicode[STIX]{x1D701}^{p^{m}}=1$ (equivalently, $m\geqslant n$ ). Hence,

$$\begin{eqnarray}\displaystyle N_{K_{n}/K}E([a^{1/p^{n}}](\unicode[STIX]{x1D70B}^{1/p^{n}})^{r}) & = & \displaystyle \exp \biggl(\mathop{\sum }_{m\geqslant n}\frac{[a^{1/p^{n}}]^{p^{m}}(\unicode[STIX]{x1D70B}^{1/p^{n}})^{rp^{m}}p^{n}}{p^{m}}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \exp \biggl(\mathop{\sum }_{m\geqslant 0}\frac{[a^{1/p^{n}}]^{p^{n+m}}(\unicode[STIX]{x1D70B}^{1/p^{n}})^{rp^{n+m}}p^{n}}{p^{m+n}}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \exp \biggl(\mathop{\sum }_{m\geqslant 0}\frac{[a]^{p^{m}}(\unicode[STIX]{x1D70B}^{r})^{p^{m}}}{p^{m}}\biggr)=E([a]\unicode[STIX]{x1D70B}^{r}).\nonumber\end{eqnarray}$$

For each $r\geqslant 1$ have a homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D716}_{u^{r}}:l\otimes \overline{\mathbb{F}}_{p}\rightarrow l((u))^{\times }\otimes \mathbb{F}_{p}\end{eqnarray}$$

defined by $\unicode[STIX]{x1D716}_{u^{r}}(a\otimes b)=E(au^{r})\otimes b$ . Then, for each $i$ , we set

$$\begin{eqnarray}\tilde{u} _{i}:=\unicode[STIX]{x1D716}_{u^{n_{i}^{\prime }}}(\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i}^{\prime },\unicode[STIX]{x1D707}})\in l((u))^{\times }\otimes \mathbb{F}_{p}.\end{eqnarray}$$

Lemma 3.5.2. Let $r\geqslant 1$ be coprime to $p$ . Then under the homomorphism (3.1.4) (with $M$ in place of $K$ ), the image of $E([a]u^{r})$ is equal to $E([a]\unicode[STIX]{x1D70B}^{r})$ ; consequently, for each $i$ , the image of $\tilde{u} _{i}$ is $u_{i}$ .

Proof. This is an immediate consequence of Lemma 3.5.1, taking into account Lemma 3.6.1 below, which shows that $n_{i}^{\prime }$ is coprime to $p$ .◻

We now state and prove our main result, which establishes [DDR16, Conjecture 7.2], by reducing the equality $L_{\text{DDR}}=L_{\text{BDJ}}$ to a purely combinatorial problem that is solved in § 3.6.

Theorem 3.5.3. We have $L_{\text{BDJ}}=L_{\text{DDR}}$ .

Proof. Since we have $\dim _{\overline{\mathbb{F}}_{p}}L_{\text{BDJ}}=\dim _{\overline{\mathbb{F}}_{p}}L_{\text{DDR}}=|J|+\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D712}=1}$ , it is enough to prove that $L_{\text{BDJ}}\subseteq L_{\text{DDR}}$ . By the definition of $L_{\text{DDR}}$ , it is equivalent to prove that the image of every class in $L_{\text{BDJ}}$ in $H^{1}(G_{M},\overline{\mathbb{F}}_{p})$ is orthogonal under the pairing of § 3.2 to the elements $u_{j}\in U_{\unicode[STIX]{x1D712}}$ , $j\notin \unicode[STIX]{x1D707}(J)$ .

In the case that $\unicode[STIX]{x1D712}=\overline{\unicode[STIX]{x1D716}}$ , we also need to show that the classes are orthogonal to $u_{\operatorname{cyc}}$ ; to see this, note that, as explained in [DDR16, §6.4] the classes $c_{i}$ (together with $c_{\operatorname{triv}}$ if $p=2$ ) span the space of classes which are (equivalently) flatly or typically ramified in the sense of [DDR16, §3.3], which are exactly the peu ramifié classes; in other words, the classes orthogonal to $u_{\operatorname{cyc}}$ are exactly the peu ramifié classes. As we recalled in § 3.3, it follows from [DS15, Theorem 4.9] that every class in $L_{\text{BDJ}}$ is peu ramifié.

Combining Lemmas 3.1.5 and 3.2.1, Theorem 3.2.2, Lemma 3.5.2 and Corollary 3.3.3, we see that we must show that for all $i\in J$ , $j\notin \unicode[STIX]{x1D707}(J)$ , the residue

(3.5.4) $$\begin{eqnarray}\operatorname{Tr}_{l\otimes _{\mathbb{ F}_{p}}\overline{\mathbb{F}}_{p}/\overline{\mathbb{F}}_{p}}\operatorname{Res}(\text{dlog}(\tilde{u} _{j})\cdot \unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}u^{-\unicode[STIX]{x1D709}_{i}})\end{eqnarray}$$

vanishes. (If $\unicode[STIX]{x1D712}=1$ , then we must also show that the pairing with $\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i_{0}},\unicode[STIX]{x1D707}^{-1}}u^{p(p^{f}-1)-\unicode[STIX]{x1D709}_{i_{0}}}$ vanishes.)

Since

$$\begin{eqnarray}\text{dlog}E(X)=(X+X^{p}+X^{p^{2}}+\cdots \,)\,\text{dlog}X\end{eqnarray}$$

and $\text{dlog}(\unicode[STIX]{x1D706}u^{n})=n\cdot u^{-1}$ , the pairing (3.5.4) evaluates to

$$\begin{eqnarray}\operatorname{Tr}_{l\otimes _{\mathbb{ F}_{p}}\overline{\mathbb{F}}_{p}/\overline{\mathbb{F}}_{p}}\operatorname{Res}\biggl(\mathop{\sum }_{m\geqslant 0}n_{j}^{\prime }(\unicode[STIX]{x1D711}\otimes 1)^{m}(\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{j}^{\prime },\unicode[STIX]{x1D707}})u^{n_{j}^{\prime }p^{m}-1}\cdot \unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}u^{-\unicode[STIX]{x1D709}_{i}}\biggr).\end{eqnarray}$$

(Here $\unicode[STIX]{x1D711}\otimes 1:l\otimes \overline{\mathbb{F}}_{p}\rightarrow l\otimes \overline{\mathbb{F}}_{p}$ is the $p$ th power map on $l$ .)

This residue is given by the coefficient of $u^{-1}$ , so we see that this pairing can be non-zero only when $\unicode[STIX]{x1D709}_{i}=p^{m}n_{j}^{\prime }$ for some $m\geqslant 0$ (if $\unicode[STIX]{x1D712}=1$ , then we must also consider the possibility that $\unicode[STIX]{x1D709}_{i}-p(p^{f}-1)=p^{m}n_{j}^{\prime }$ , but this is excluded by Lemma 3.6.6 below). If this holds, then the pairing evaluates to

$$\begin{eqnarray}n_{j}^{\prime }\operatorname{Tr}_{l\otimes _{\mathbb{F}_{p}}\overline{\mathbb{F}}_{p}/\overline{\mathbb{F}}_{p}}(\unicode[STIX]{x1D711}\otimes 1)^{m}(\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{j}^{\prime },\unicode[STIX]{x1D707}})\cdot \unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}.\end{eqnarray}$$

Now, we have

$$\begin{eqnarray}(\unicode[STIX]{x1D711}\otimes 1)^{m}(\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{j}^{\prime },\unicode[STIX]{x1D707}})\cdot \unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i},\unicode[STIX]{x1D707}^{-1}}=(\unicode[STIX]{x1D711}\otimes 1)^{m}(\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{j}^{\prime },\unicode[STIX]{x1D707}}\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D70E}_{i-m},\unicode[STIX]{x1D707}^{-1}})\end{eqnarray}$$

which is non-zero if and only if $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{i-m}$ , in which case its trace to $\overline{\mathbb{F}}_{p}$ is equal to $[l:k]$ .

In conclusion, we have seen that in order for the pairing to be non-zero, we require:

1. (i) $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{i-m}$ ; and

2. (ii) $\unicode[STIX]{x1D709}_{i}=p^{m}n_{j}^{\prime }$ .

(In fact, although we do not need this stronger statement, we observe that the pairing is non-zero if and only if these conditions hold, because $n_{j}^{\prime }$ is always a unit by Lemma 3.6.1, while $[l:k]$ is prime to $p$ .) By Proposition 3.6.7 below, these conditions imply that $j\in \unicode[STIX]{x1D707}(J)$ , as required.◻

Remark 3.5.5. It is clear that the method of the proof of Theorem 3.5.3 could be used to compare the bases of $L_{\text{BDJ}}$ and $L_{\text{DDR}}$ that we have been working with. We have checked that in suitably generic cases the bases are the same (up to scalars), but that in exceptional cases they may differ.

3.6 Combinatorics

Our main aim in this section is to prove Proposition 3.6.7, which was used in the proof of Theorem 3.5.3. We begin with some simple observations; the following three lemmas give us some control on the quantities $\unicode[STIX]{x1D709}_{i}$ and $n_{i}^{\prime }$ which will be important in the proof of Proposition 3.6.7.

Lemma 3.6.1. The quantity $n_{i}^{\prime }$ is not divisible by $p$ .

Proof. This is automatic if $a_{i-1}\neq p$ because then $n_{i}^{\prime }=n_{i-1}\equiv a_{i-1}\hspace{0.6em}({\rm mod}\hspace{0.2em}p)$ . Assume that $a_{i-1}=p$ , and write that $(a_{i-1},a_{i-2},\ldots ,a_{j})=(p,p-1,\ldots ,p-1)$ , with $a_{j-1}\neq p-1$ . Now

$$\begin{eqnarray}n_{i}^{\prime }:=n_{j-1}-(p^{f}-1)\equiv n_{j-1}+1\equiv a_{j-1}+1\hspace{0.6em}({\rm mod}\hspace{0.2em}p).\end{eqnarray}$$

However, since $a_{j-1}\neq p-1$ and lies in $[1,p]$ , we have $a_{j-1}\not \equiv -1\hspace{0.6em}{\rm mod}\hspace{0.2em}p$ , and so $n_{i}^{\prime }\not \equiv 0\hspace{0.6em}({\rm mod}\hspace{0.2em}p)$ .◻

Lemma 3.6.2. If $i\in J$ , then $0<\unicode[STIX]{x1D709}_{i}<p^{2}(p^{f}-1)/(p-1)$ .

Proof. Since $i\in J,$ we have

(3.6.3) $$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}=p^{f}r_{i}+(-1)^{i+1\notin J}p^{f-1}r_{i+1}+(-1)^{i+2\notin J}p^{f-2}r_{i+2}+\cdots +(-1)^{i-1\notin J}pr_{i-1}.\end{eqnarray}$$

The upper bound is immediate, as we have $r_{j}\leqslant p$ for all $j$ (and in the case that all $r_{j}$ are equal to $p$ , we are not allowing $J^{c}$ to be empty). For the lower bound, if $r_{i}\geqslant 2$ , then $\unicode[STIX]{x1D709}_{i}\geqslant 2p^{f}-(p^{f}+p^{f-1}+\cdots +p^{2})>0$ , so we may assume that $r_{i}=1$ . Suppose that $J\neq \{i\}$ , and let $x\geqslant 0$ be minimal so that $i+x+1\in J$ . Since $r_{i}=1$ and $i\in J$ , it follows from the maximality condition on $J$ that no initial segment of $(r_{i+1},\ldots ,r_{i+x})$ can be $(p-1,p-1,\ldots ,p)$ (which also excludes the degenerate case consisting of a single initial $p$ ). Hence, either all the $r_{j}$ for $j\in [i+1,i+x]$ are at most $p-1$ , in which case

$$\begin{eqnarray}p^{f-1}r_{i+1}+\cdots +p^{f-x}r_{i+x}\leqslant (p^{f-1}+\cdots +p^{f-x})(p-1)=p^{f}-p^{f-x},\end{eqnarray}$$

so that

$$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}\geqslant p^{f-x}+p^{f-x-1}-(p^{f-x-2}+\cdots +p)p=p^{f-x}-p^{f-x-2}-\cdots -p^{2}>0,\end{eqnarray}$$

or for some $y<x$ we have $r_{i+1},\ldots ,r_{i+y}=p-1$ and $r_{i+y+1}<p-1$ , in which case

$$\begin{eqnarray}\displaystyle p^{f-1}r_{i+1}+\cdots +p^{f-x}r_{i+x}\leqslant & \displaystyle \,\,(p^{f-1}+\cdots +p^{f-y})(p-1) & \displaystyle \\ \displaystyle & \displaystyle +(p-2)p^{f-y-1}+p(p^{f-y-2}+\cdots p^{f-x})\\ \displaystyle = & \displaystyle \,\,(p^{f-1}+\cdots +p^{f-x})(p-1)\\ \displaystyle & \displaystyle -p^{f-y-1}+p^{f-y-2}+\cdots +p^{f-x}\\ \displaystyle \leqslant & \displaystyle \,\,(p^{f-1}+\cdots +p^{f-x})(p-1)\\ \displaystyle = & \displaystyle \,\,p^{f}-p^{f-x},\end{eqnarray}$$

and one proceeds as above. Finally, if $J=\{i\}$ , then arguing as above (and, again, using the maximality condition on $J$ ) we see (considering the two cases as above) that $\unicode[STIX]{x1D709}_{i}\geqslant p^{f}-(p^{f-1}+\cdots +p)(p-1)=p>0$ .◻

Lemma 3.6.4. For any value of $i$ , we have $(p^{f}-1)/(p-1)\leqslant n_{i}<(p^{f}-1)+(p^{f}-1)/(p-1)$ .

Proof. This is immediate from the definition of $n_{i}$ .◻

Let $v_{p}(\unicode[STIX]{x1D709}_{i})$ denote the $p$ -adic valuation of $\unicode[STIX]{x1D709}_{i}$ . The following lemma shows that $\unicode[STIX]{x1D709}_{i}$ is in some sense a function of this valuation, and is crucial for our main argument.

Lemma 3.6.5. If $i\in J$ , and if $m:=v_{p}(\unicode[STIX]{x1D709}_{i})$ , then $m\geqslant 1$ . If furthermore $m>1$ , then we have $\unicode[STIX]{x1D709}_{i}=p^{m}(n_{i-m}-(p^{f}-1))$ , while if $m=1$ , then either $\unicode[STIX]{x1D709}_{i}=pn_{i-1}$ or $\unicode[STIX]{x1D709}_{i}=p(n_{i-1}-(p^{f}-1))$ , depending on whether or not $\unicode[STIX]{x1D709}_{i}/p\geqslant (p^{f}-1)/(p-1)$ .

Proof. Equation (3.6.3) shows that $m$ is at least $1$ if $i\in J$ . From (3.3.4), we deduce that $\unicode[STIX]{x1D709}_{i}/p^{m}\equiv n_{i-m}\hspace{0.6em}({\rm mod}\hspace{0.2em}p^{f}-1)$ . By Lemma 3.6.2 we have

$$\begin{eqnarray}0<\unicode[STIX]{x1D709}_{i}/p^{m}<p^{2-m}(p^{f}-1)/(p-1),\end{eqnarray}$$

so that if $m\geqslant 2$ it follows by Lemma 3.6.4 that

$$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}/p^{m}<(p^{f}-1)/(p-1)\leqslant n_{i-m}<(p^{f}-1)+(p^{f}-1)/(p-1).\end{eqnarray}$$

Since $\unicode[STIX]{x1D709}_{i}>0$ by Lemma 3.6.2, the congruence modulo $p^{f}-1$ forces the equality $n_{i-m}-\unicode[STIX]{x1D709}_{i}/p^{m}=(p^{f}-1)$ . If $m=1$ , then we have

$$\begin{eqnarray}0<\unicode[STIX]{x1D709}_{i}/p<(p^{f}-1)+(p^{f}-1)/(p-1)\end{eqnarray}$$

and the claim follows in the same way. ◻

The following simple lemma was used in the proof of Theorem 3.5.3 in the case $\unicode[STIX]{x1D712}=1$ .

Lemma 3.6.6. Suppose that $\unicode[STIX]{x1D712}=1$ and that $i\in J$ . Then there are no solutions to the equation $\unicode[STIX]{x1D709}_{i}-p(p^{f}-1)=p^{m}(p^{f}-1)$ , for any $m\geqslant 0$ .

Proof. Since $\unicode[STIX]{x1D712}=1$ , we have $n_{j}=p^{f}-1$ for all $j$ . From Lemma 3.6.5, we find that either $v_{p}(\unicode[STIX]{x1D709}_{i})\geqslant 2$ , in which case $\unicode[STIX]{x1D709}_{i}=0$ (contradicting Lemma 3.6.2), or $v_{p}(\unicode[STIX]{x1D709}_{i})=1$ , in which case either $\unicode[STIX]{x1D709}_{i}=0$ or $\unicode[STIX]{x1D709}_{i}=p(p^{f}-1)$ . The first case again contradicts Lemma 3.6.2. The second case leads to the equation $0=p^{m}(p^{f}-1)$ , which has no solutions, as required.◻

We now prove our main combinatorial result.

Proposition 3.6.7. Suppose that $i\in J$ , and that for some integers $j,m$ we have:

1. (i) $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{i-m}$ ; and

2. (ii) $\unicode[STIX]{x1D709}_{i}=p^{m}n_{j}^{\prime }$ ;

then $j\in \unicode[STIX]{x1D707}(J)$ .

Proof. By Lemma 3.6.1, we must have $m=v_{p}(\unicode[STIX]{x1D709}_{i})$ . Suppose first that $m=1$ and $\unicode[STIX]{x1D709}_{i}=pn_{i-1}$ . We need to solve the equations $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{i-1}$ and $n_{j}^{\prime }=n_{i-1}$ .

If $a_{j-1}=p$ , then we have $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{s-1}$ and $n_{j}^{\prime }=n_{s-1}-(p^{f}-1)$ , where $s$ is the greatest integer less than $j$ for which $a_{s-1}\neq p-1$ . Since $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{i-1}$ by assumption, we find that $s=i$ . However, then $n_{i-1}=n_{j}^{\prime }=n_{i-1}-(p^{f}-1)$ , which is not possible.

Thus, $a_{j-1}\neq p$ and, hence, we have $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{j-1}$ , so that $j=i$ . We must show that $j=i\in \unicode[STIX]{x1D707}(J)$ . By the definition of $\unicode[STIX]{x1D707}(J)$ , this will be the case unless for some $s>i$ we have $i+1,\ldots ,s\notin J$ , and $(a_{i},\ldots ,a_{s-1})=(p-1,\ldots ,p-1,p)$ . Suppose then that this holds; we must show that we cannot have $\unicode[STIX]{x1D709}_{i}=pn_{i-1}$ after all. Now, by definition and the assumption that $i+1,\ldots ,s\notin J$ , we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}_{i}/p & = & \displaystyle p^{f-1}r_{i}-p^{f-2}r_{i+1}-\cdots +(-1)^{s+1\notin J}p^{f+i-2-s}r_{s+1}+\cdots +(-1)^{i-1\notin J}r_{i-1}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle p^{f}-(p^{f-2}+\cdots +p^{f+i-s-1})+(p^{f+i-2-s}+\cdots +1)p\nonumber\\ \displaystyle & = & \displaystyle p^{f}-(p^{f-2}+\cdots +p^{f+i-s})+(p^{f+i-2-s}+\cdots +p)\nonumber\end{eqnarray}$$

while

$$\begin{eqnarray}\displaystyle n_{i-1} & = & \displaystyle p^{f-1}a_{i}+p^{f-2}a_{i+1}+\cdots +a_{i-1}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle p^{f-1}(p-1)+\cdots +p^{f+i+1-s}(p-1)+p^{f+i-s}p+p^{f+i-1-s}+\cdots +1\nonumber\\ \displaystyle & = & \displaystyle p^{f}+p^{f+i-1-s}+\cdots +1,\nonumber\end{eqnarray}$$

which gives the required contradiction.

Having disposed of the case that $m=1$ and $\unicode[STIX]{x1D709}_{i}=pn_{i-1}$ , it follows from Lemma 3.6.5 that we may assume that $\unicode[STIX]{x1D709}_{i}=p^{m}(n_{i-m}-(p^{f}-1))$ . We show first that we cannot have $a_{j-1}\neq p$ . Indeed, if this occurs, then by definition we have $n_{j}^{\prime }=n_{j-1}$ and $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{i-1}$ , so that the equations we need to solve are $i-m=j-1$ , and $n_{i-m}-(p^{f}-1)=n_{j-1}$ , which are mutually inconsistent, since together they imply that $n_{j-1}-(p^{f}-1)=n_{j-1}$ .

We are thus reduced to the case when $a_{j-1}=p$ and, by the definition of $n_{j}^{\prime }$ , we see (since $\unicode[STIX]{x1D70E}_{j}^{\prime }=\unicode[STIX]{x1D70E}_{i-m}$ ) that $i-m$ must be congruent to the greatest integer $i^{\prime }$ less than $j-1$ with $a_{i^{\prime }}\neq p-1$ . Replacing $i$ by something congruent to its modulo $f$ , we may assume that $i-m=i^{\prime }$ , so that $a_{i-m}\neq p-1$ , $a_{i-m+1}=\cdots =a_{j-2}=p-1$ and $a_{j-1}=p$ . Again, we must show that this implies that $j\in \unicode[STIX]{x1D707}(J)$ . By the definition of $\unicode[STIX]{x1D707}(J)$ , this will be the case unless $i-m+1,\ldots ,j-2,j-1,j\notin J$ . Since we are assuming that $i\in J$ , this implies, in particular, that $j$ is contained in the interval $[i-m,i)$ . We now show that this leads to a contradiction. Consider the equation $\unicode[STIX]{x1D709}_{i}/p^{m}=n_{i-m}-(p^{f}-1)$ . From the definitions and the assumptions we are making, we have

$$\begin{eqnarray}\displaystyle n_{i-m} & = & \displaystyle p^{f-1}a_{i-m+1}+\cdots +p^{f-x}a_{i-m+x}+\cdots +a_{i-m}\nonumber\\ \displaystyle & = & \displaystyle p^{f}+p^{f-m+i-j}a_{j}+\cdots +a_{i-m},\nonumber\end{eqnarray}$$

so that

$$\begin{eqnarray}\displaystyle n_{i-m}-(p^{f}-1) & = & \displaystyle 1+p^{f-m+i-j}a_{j}+\cdots +a_{i-m}\nonumber\\ \displaystyle & {>} & \displaystyle p^{f-m+i-j}+p^{f-m+i-j-1}+\cdots +1.\nonumber\end{eqnarray}$$

Thus,

(3.6.8) $$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}=p^{m}(n_{i-m}-(p^{f}-1))>p^{f+i-j}+p^{f+i-j-1}+\cdots +p^{m}.\end{eqnarray}$$

Since $\unicode[STIX]{x1D709}_{i}\leqslant p^{2}(p^{f}-1)/(p-1)$ by Lemma 3.6.2, we conclude that, in particular,

$$\begin{eqnarray}(p^{f}-1)/(p-1)>\unicode[STIX]{x1D709}_{i}/p^{2}>p^{f+i-j-2}=p^{(f-1)+(i-j-1)},\end{eqnarray}$$

which is only possible if $i=j+1$ . Assume now that this is the case. Then we may rewrite (3.6.8) in the form

(3.6.9) $$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}=p^{m}(n_{i-m}-(p^{f}-1))>p^{f+1}+p^{f}+\cdots +p^{m}.\end{eqnarray}$$

We also find that $i-m+1,\ldots ,i-1\notin J$ , so that, from the definition of $\unicode[STIX]{x1D709}_{i}$ (and taking into account the fact that $i\in J$ ), we compute

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}_{i} & = & \displaystyle p^{f}r_{i}+\cdots +(-1)^{i-m\notin J}p^{m}r_{i-m}-(p^{m-1}r_{i-m+1}+\cdots +pr_{i-1})\nonumber\\ \displaystyle & {\leqslant} & \displaystyle p^{f}r_{i}+\cdots +(-1)^{i-m\notin J}p^{m}r_{i-m}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (p^{f}+\cdots +p^{m})p=p^{f+1}+p^{f}+\cdots +p^{m+1}.\nonumber\end{eqnarray}$$

This contradicts (3.6.9), and completes the argument. ◻