2 Explicit Chabauty–Kim at depth 2

We begin with a brief review of a few essential results from the Chabauty–Kim method [Reference KimKim05, Reference KimKim09]. Associated to a pointed curve $(X,b)$ over $\mathbb{Q}$ , with good reduction outside a finite set $T_{0}$ , and a prime $p$ of good reduction, we have a map

$$\begin{eqnarray}j_{n}:X(\mathbb{Q})\rightarrow H^{1}(G_{T},U_{n}),\end{eqnarray}$$

where $T:=T_{0}\,\cup \,\{p\}$ , $G_{T}$ is the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside $T$ , and $U_{n}$ is the maximal $n$ -unipotent quotient of the $\mathbb{Q}_{p}$ pro-unipotent completion of $\unicode[STIX]{x1D70B}_{1}^{\acute{\text{e}}\text{t}}(X_{\overline{\mathbb{Q}}},b)$ . We also have local maps

$$\begin{eqnarray}j_{n,v}:X(\mathbb{Q}_{v})\rightarrow H^{1}(G_{\mathbb{ Q}_{v}},U_{n})\end{eqnarray}$$

for $v$ in $T_{0}$ and

$$\begin{eqnarray}j_{n,p}:X(\mathbb{Q}_{p})\rightarrow H^{1}(G_{\mathbb{ Q}_{p}},U_{n})\end{eqnarray}$$

(for the definition of $H_{f}^{1}(G_{\mathbb{Q}_{p}},U_{n})$ see [Reference KimKim05]). We define

$$\begin{eqnarray}X(\mathbb{Q}_{p})_{n}:=j_{n,p}^{-1}\biggl(\operatorname{loc}_{p}\biggl(\mathop{\bigcap }_{v\in T_{0}}\operatorname{loc}_{v}^{-1}(X(\mathbb{Q}_{v}))\biggr)\biggr)\subset X(\mathbb{Q}_{p}),\end{eqnarray}$$

where $\operatorname{loc}_{v}$ is the localisation map from $H^{1}(G_{T},U_{n})$ to $H^{1}(G_{v},U_{n})$ . By construction, the set of rational points $X(\mathbb{Q})$ is a subset of $X(\mathbb{Q}_{p})_{n}$ for all $n$ .

The behaviour of the maps $j_{n,v}$ is fundamentally different depending on whether or not $v=p$ . In the $v\neq p$ case, we have the following theorem, due to Kim and Tamagawa [Reference Kim and TamagawaKT08, Corollary 0.2].

In fact, one can bound the image of $j_{n,v}$ in terms of the reduction data of the curve as follows. Let $L$ be a finite extension of $\mathbb{Q}_{v}$ over which $X_{L}$ acquires stable reduction. Let ${\mathcal{X}}/{\mathcal{O}}_{L}$ be a regular semistable model, and let $V({\mathcal{X}}_{k_{L}})$ denote the set of irreducible components of the special fibre. Since the model is regular, specialisation induces a well-defined map

$$\begin{eqnarray}r_{v}:X(\mathbb{Q}_{v})\rightarrow V({\mathcal{X}}_{k_{L}}).\end{eqnarray}$$

For $v$ a prime of bad reduction, we define $n_{v}\in \mathbb{Z}_{{>}0}$ to be the size of the image of $X(\mathbb{Q}_{v})$ under $j_{2,v}$ in $H^{1}(G_{v},U_{2})$ .

Lemma 2.2. With notation as above,

$$\begin{eqnarray}n_{v}\leqslant \operatorname{im}(r_{v}).\end{eqnarray}$$

A detailed proof of this lemma will appear in forthcoming work of the second named author and Alex Betts. However, for the sake of completeness, we briefly indicate the method of proof. First, if $L|\mathbb{Q}_{v}$ is a finite extension, then it is easy to show that $H^{1}(G_{\mathbb{Q}_{v}},U_{n})\rightarrow H^{1}(G_{L},U_{n})$ is injective, hence one reduces to the case where $X$ has stable reduction. In this case, one can use the description of the action of $G_{L}$ on $\unicode[STIX]{x1D70B}_{1}^{\acute{\text{e}}\text{t},(v^{\prime })}(X_{\overline{\mathbb{Q}}_{v}},b)$ (the maximal prime-to- $v$ quotient of $\unicode[STIX]{x1D70B}_{1}^{\acute{\text{e}}\text{t}}(X_{\overline{\mathbb{Q}}_{v}},b)$ ) in terms of the dual graph of a regular semistable model given in [Reference OdaOda95] to deduce that if points $b_{1}$ and $b_{2}$ lie on a common irreducible component of $V({\mathcal{X}}_{k_{L}})$ , then the class of $[\unicode[STIX]{x1D70B}_{1}^{\acute{\text{e}}\text{t},(v^{\prime })}(X_{\overline{\mathbb{Q}}_{v}};b_{1},b_{2})]$ in $H^{1}(G_{L},\unicode[STIX]{x1D70B}_{1}^{\acute{\text{e}}\text{t},(v^{\prime })}(X_{\overline{\mathbb{Q}}_{v}},b_{1}))$ is trivial. This straightforwardly implies the lemma.

The finiteness of the maps $j_{n,v}$ allows us to partition the set $X(\mathbb{Q}_{p})_{n}$ as follows. We refer to a tuple $\unicode[STIX]{x1D6FC}=(\unicode[STIX]{x1D6FC}_{v})_{v}$ in $\prod _{v\in T_{0}}\operatorname{im}(j_{n,v})$ as a collection of local conditions, and define

$$\begin{eqnarray}X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}:=j_{n,p}^{-1}\biggl(\operatorname{loc}_{p}\biggl(\mathop{\bigcap }_{v\in T_{0}}\operatorname{loc}_{v}^{-1}(\unicode[STIX]{x1D6FC}_{v})\biggr)\biggr)\subset X(\mathbb{Q}_{p}).\end{eqnarray}$$

By construction, $X(\mathbb{Q}_{p})_{n}$ is the disjoint union of the $X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}$ for $\unicode[STIX]{x1D6FC}$ a collection of local conditions. The bound in Theorem 1.1 comes from a bound on $\#X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}$ in the case of $n=2$ , together with a bound on the number of local conditions, that is, on the size of $\prod _{v\in T_{0}}j_{2,v}X(\mathbb{Q}_{v})$ .

2.1 Local structure at $p$

The power series $G$ in the introduction is from the following result of [Reference Balakrishnan and DograBD17].

Proposition 2.3 [Reference Balakrishnan and DograBD17, Proposition 6.4].

Let $X/\mathbb{Q}$ be a curve of genus $g>1$ . Suppose $X$ satisfies Condition A or Condition B. Let $\unicode[STIX]{x1D714}_{0},\ldots ,\unicode[STIX]{x1D714}_{2g-1}\in H^{0}(X,\unicode[STIX]{x1D6FA}(D))$ be differentials of the second kind forming a basis of $H_{\operatorname{dR}}^{1}(X)$ , where $D$ is an effective divisor. Then, for all local conditions $\unicode[STIX]{x1D6FC}$ , there are constants $a_{ij}$ and $a_{i}$ , a differential of the third kind $\unicode[STIX]{x1D702}$ , and a function $h\in H^{0}(X,{\mathcal{O}}(2D))$ such that

$$\begin{eqnarray}X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}\subset \{z\in X(\mathbb{Q}_{p}):G(z)=0\},\end{eqnarray}$$

where

$$\begin{eqnarray}G(z):=\mathop{\sum }_{0\leqslant i,j<2g}a_{ij}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}+\mathop{\sum }_{0\leqslant i<2g}a_{i}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}+\int _{b}^{z}\unicode[STIX]{x1D702}+h(z).\end{eqnarray}$$

2.2 Proof of Corollary 1.2

In this subsection we prove that Theorem 1.1 implies Corollary 1.2.

Lemma 2.4. Let $v\neq p$ be a prime of potential good reduction. Then, for all $n$ , the map

$$\begin{eqnarray}j_{n,v}:X(\mathbb{Q}_{v})\rightarrow H^{1}(G_{v},U_{n})\end{eqnarray}$$

is trivial.

Proof. Let $L|\mathbb{Q}_{v}$ be an extension over which $X$ acquires good reduction. Then the map

$$\begin{eqnarray}j_{n,L}:X(L)\rightarrow H^{1}(G_{L},U_{n})\end{eqnarray}$$

has trivial image. Recall from [Reference SerreSer97, I.5.8] that, given a profinite group $G$ , closed normal subgroup $H$ , and $G$ -group $A$ , we get an exact sequence of pointed sets

$$\begin{eqnarray}H^{1}(G/H,A^{H})\rightarrow H^{1}(G,A)\stackrel{\operatorname{Res}}{\longrightarrow }H^{1}(H,A).\end{eqnarray}$$

We apply this when $G=G_{v}$ , $H=G_{L}$ . We claim $U_{n}^{G_{L}}=1$ . To see this, note that it is enough to show that the graded pieces $U_{n}[i]$ of $U_{n}$ with respect to the central series filtration satisfy $U_{n}[i]^{G_{w}}=1$ , which follows from the fact that $U_{n}[i]$ is an unramified representation of $G_{L}$ of weight $-i$ .

Hence we deduce that the restriction map

$$\begin{eqnarray}H^{1}(G_{v},U_{n})\stackrel{\operatorname{Res}}{\longrightarrow }H^{1}(G_{L},U_{n})\end{eqnarray}$$

is injective. The lemma thus follows from commutativity of the following diagram.

Now let $X$ be as in Corollary 1.2. Then all the $n_{v}$ are 1. Taking $p=3$ and using the Hasse–Weil estimate $\#X(\mathbb{F}_{3})\leqslant 4+2g\sqrt{3}$ and the trivial bound $\#W(\mathbb{F}_{3})\leqslant 4$ , we deduce

$$\begin{eqnarray}\#X(\mathbb{Q})\leqslant (8g^{3}+(64+4\sqrt{3})g^{2}+(64+4\sqrt{3})g+24)\unicode[STIX]{x1D705}_{3},\end{eqnarray}$$

from which the corollary follows.

3 Bounding the number of zeros via a differential operator

In this section, we explain how to bound the zeros of a power series $G$ by finding a bound on ${\mathcal{D}}(G)$ for a suitably ‘nice’ (in a way we will make precise shortly) differential operator ${\mathcal{D}}$ . The construction of a nice differential operator in the case when $G$ is the Coleman function from Proposition 2.3 will be given in the next section.

We begin by fixing notation. We denote by $v$ the $p$ -adic valuation homomorphism $\mathbb{Q}_{p}^{\times }\rightarrow \mathbb{Z}$ . We fix a point $b$ and a rational function $x$ which is a uniformising parameter at $b$ . We let $]b[$ denote the tube (or residue disk) of $b$ , that is, the set of points reducing to $b$ mod $p$ (we also denote this set as $]\overline{b}[$ , where $\overline{b}$ is the mod $p$ reduction of $b$ ). Given an analytic function $F$ on $]b[$ , we let $N_{b}(F)$ denote the number of $\mathbb{C}_{p}$ -valued zeros of $F$ in $]b[$ , counted with multiplicity.

Let $C_{i}$ denote the function $\mathbb{Q}_{p}[\![x]\!]\rightarrow \mathbb{Q}_{p}$ sending a power series to its $x^{i}$ coefficient. By a differential operator we will simply mean an element of the non-commutative ring $\mathbb{Q}_{p}[\![x]\!][d/dx]$ . By an algebraic differential operator we will mean a differential operator in the image of $\mathbb{Q}_{p}(X)[d/dx]$ , where $\mathbb{Q}_{p}(X)$ denotes the function field of $X$ over $\mathbb{Q}_{p}$ . The order of a differential operator will refer to its degree as a polynomial in $d/dx$ , when given in the form $\sum _{i=0}^{N}a_{i}(d/dx)^{i}$ , for $a_{i}\in \mathbb{Q}_{p}[\![x]\!]$ .

The main result of this section is the following proposition, which shows that one may use nice differential operators to bound the zeros of power series, in analogy with Coleman’s use of differentiation.

The proof of this proposition will occupy the remainder of the section. Before giving the proof, we remark on how it is applied in the proof of Theorem 1.1. To deduce part (i) of the theorem, it will be enough to deduce (under the assumptions of the theorem) that, for each $b\in X(\mathbb{F}_{p})$ and each set of local conditions $\unicode[STIX]{x1D6FC}$ , we may choose a non-zero effective divisor $D$ disjoint from $]b[$ , and differentials $\unicode[STIX]{x1D714}_{0},\ldots ,\unicode[STIX]{x1D714}_{2g-1}$ and a nice differential operator ${\mathcal{D}}$ of degree $N$ , such that

$$\begin{eqnarray}N_{b}({\mathcal{D}}(G))+N\leqslant 16g^{3}+15g^{2}-16g+10,\end{eqnarray}$$

where $G$ is the function from Proposition 2.3. In practice, we bound the number of zeros of ${\mathcal{D}}(G)$ in $]b[$ by the degree of ${\mathcal{D}}(G)$ . In the hyperelliptic case, we choose $D$ , $\unicode[STIX]{x1D714}_{i}$ , and $G$ in a more uniform way (more precisely, they are the same for all points in $X-W$ ), and for this reason, we get an improved bound

$$\begin{eqnarray}\mathop{\sum }_{x\in (X-W)(\mathbb{F}_{p})}N_{b}({\mathcal{D}}(G))+N\leqslant \deg ({\mathcal{D}}(G))+N\#(X-W)(\mathbb{F}_{p}).\end{eqnarray}$$

Now let $G$ be a power series in $\mathbb{Q}_{p}[\![x]\!]$ , with ${\mathcal{D}}(G)\in H^{0}(X,{\mathcal{O}}(D))$ , with $D$ an effective divisor (in our intended application, $G$ will be the power series in Proposition 2.3). Let $M$ denote the length of the slope ${\leqslant}-1$ part of the Newton polygon. Write ${\mathcal{D}}$ as $\sum _{i=1}^{N}g_{i}(d/dx)^{i}$ , where $g_{N}\in \mathbb{Z}_{p}[\![x]\!]^{\times }$ . Recall the following well-known lemma.

Lemma 3.4. For any $n_{1}\leqslant n_{2}$ ,

$$\begin{eqnarray}v\biggl(\frac{n_{2}!}{n_{1}!}\biggr)\leqslant \log _{p}(n_{1})+\frac{n_{2}-n_{1}}{p-1}.\end{eqnarray}$$

Proof. Using Legendre’s formula $v(n!)=(n-s(n))/(p-1),$ where $s(n)$ is the sum of digits in base $p$ , it follows that

$$\begin{eqnarray}\displaystyle v\biggl(\frac{n_{2}!}{n_{1}!}\biggr)=\frac{n_{2}-n_{1}}{p-1}+\frac{s(n_{1})-s(n_{2})}{p-1} & {\leqslant} & \displaystyle \frac{n_{2}-n_{1}}{p-1}+\frac{s(n_{1})}{p-1}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{n_{2}-n_{1}}{p-1}+\frac{(p-1)\log _{p}(n_{1})}{p-1}\nonumber\\ \displaystyle & = & \displaystyle \frac{n_{2}-n_{1}}{p-1}+\log _{p}(n_{1}).\Box \nonumber\end{eqnarray}$$

Lemma 3.5. Let $M$ be the length of the slope ${\leqslant}-1$ part of the Newton polygon of $G$ . Suppose that $M>1$ , and $i\leqslant M$ satisfies

$$\begin{eqnarray}v(C_{i}(G))\leqslant v(C_{M}(G))+v(M!/i!).\end{eqnarray}$$

Then $i\geqslant \unicode[STIX]{x1D705}_{p}^{-1}M.$

Proof. Since $i\leqslant M$ and $M$ is the length of the slope ${\leqslant}-1$ part of the Newton polygon, we have

$$\begin{eqnarray}M-i\leqslant v(C_{i}(G))-v(C_{M}(G))\leqslant v(M!/i!).\end{eqnarray}$$

This implies $\log _{p}(i)+(M-i)/(p-1)\geqslant M-i$ , by the previous lemma. Using the inequality $\log _{p}(i)\leqslant i/\log (p)$ , we get

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{p}i\geqslant M.\Box\end{eqnarray}$$

Given a power series $F$ , let $S(F)=\{i\geqslant 0:v(C_{i}(F))=\min \{v(C_{j}(F)):j\geqslant 0\}\}$ if this minimum exists, and take $S$ to be empty otherwise.

We now prove a key lemma which gives a quantitative relation between the Newton polygon of $G$ and the Newton polygon of ${\mathcal{D}}(G)$ , when ${\mathcal{D}}$ is a nice differential operator. The idea of the proof is as follows. Let $s$ denote the least $i$ such that $v(C_{i}({\mathcal{D}}(G)))$ attains its minimum. We would like to say that if the valuation of $C_{M}(G)$ is smaller than the valuation of $C_{i}(G)$ for all $i<M$ , then the valuation of $C_{M-N}({\mathcal{D}}(G))$ is smaller than that of $C_{i}({\mathcal{D}}(G))$ for all $i<M-N$ (and hence $s\geqslant M-N$ ).

This is not quite true, because when we apply $(d/dx)^{N}$ to $C_{M}(G)x^{M}$ , we increase the valuation by $v(M!/(M-N)!)$ , so it may happen that there is some cancellation. However, for such cancellation to occur, there must be an $M_{1}<M$ for which $v(C_{M_{1}}(G))$ is within $v(M!/(M-N)!)$ of $v(C_{M}(G))$ . Similarly, if $v(C_{M_{1}-N}({\mathcal{D}}(G)))$ is not smaller than $v(C_{i}({\mathcal{D}}(G)))$ for all $i<M_{1}-N$ , then there must be some $M_{2}<M_{1}$ such that $v(C_{M_{2}}(G))$ is close to $v(C_{M_{1}}(G))$ , and so on, giving a sequence $M,M_{1},\ldots$ until we get to $M_{n}\leqslant s+N$ . By construction, the $v(C_{M_{i}}(G))$ are ‘close together’, but since $M$ is the endpoint of the slope ${\leqslant}-1$ part of the Newton polygon they are also ‘far apart’, and comparing these two conditions gives the lemma.

Note that, without any additional conditions on ${\mathcal{D}}$ or $G$ , to prove a result of the form ‘ ${\mathcal{D}}(G)$ has small slopes implies $G$ has small slopes’, it is necessary to assume $p>2$ (consider, for example, the case ${\mathcal{D}}=(d/dx)-1$ and $G=\exp (x)+1$ ). Note that, by Lemma 3.3, if $F$ is algebraic without poles on $]b[$ , then $\min S(F)\leqslant \operatorname{ord}_{\overline{b}}(\operatorname{red}_{p}(F))$ . Hence the following lemma implies Proposition 3.2.

Lemma 3.6. Let $p>2$ , and let $M$ be the length of the slope ${\leqslant}-1$ part of the Newton polygon of $G$ . Suppose $S({\mathcal{D}}(G))$ is non-empty. Then

$$\begin{eqnarray}M<\unicode[STIX]{x1D705}_{p}(N+\min S({\mathcal{D}}(G))).\end{eqnarray}$$

Proof. For integers $i\leqslant j$ , let

$$\begin{eqnarray}q(i,j):=\left\{\begin{array}{@{}ll@{}}v(i!/(j-N)!), & \text{if}\;i\geqslant j-N,\\ 0, & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

For $k\geqslant 0$ , let

$$\begin{eqnarray}T(k)=\{0\leqslant i\leqslant k:v(C_{i}(G))+q(i,k)\leqslant v(C_{k}(G))+q(k,k)\}.\end{eqnarray}$$

Clearly, for all $k$ , we have $k\in T(k)$ . Suppose $N\leqslant k\leqslant M$ , and $T(k)=\{k\}$ . It follows that

(1) $$\begin{eqnarray}v(C_{k-N}({\mathcal{D}}(G)))=v(C_{k}(G))+q(k,k).\end{eqnarray}$$

Indeed, writing ${\mathcal{D}}=\sum g_{i}(d^{i}/dx^{i})$ , we have that

(2) $$\begin{eqnarray}C_{k-N}({\mathcal{D}}(G))=\mathop{\sum }_{i,j,m\in \{0,\ldots ,N\}\times \mathbb{Z}_{{\geqslant}0}^{2}:j+m-i=k-N}\frac{m!}{(m-i)!}C_{m}(G)C_{j}(g_{i}).\end{eqnarray}$$

Note that $v((k!/(k-N)!)C_{k}(G)C_{0}(g_{N}))=q(k,k)+v(C_{k}(G))$ by assumption. For $i$ in $\{0,\ldots ,N\}$ and $j,m$ in $\{0,\ldots k\}$ such that $j+m-i=k-N$ , by our assumption on $T(k)$ we have

$$\begin{eqnarray}\displaystyle v\biggl(\frac{m!}{(m-i)!}C_{m}(G)C_{j}(g_{i})\biggr) & {\geqslant} & \displaystyle v(C_{m}(G))+v(m!)-v((m-i)!)\geqslant v(C_{m}(G))+q(m,k)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle v(C_{k}(G))+q(k,k),\nonumber\end{eqnarray}$$

with simultaneous equalities if and only if $(i,j,m)=(N,0,k)$ .

For all $0\leqslant a<k-N$ , by expanding out ${\mathcal{D}}(G)$ as in (2), we have

$$\begin{eqnarray}\displaystyle v(C_{a}({\mathcal{D}}(G))) & {\geqslant} & \displaystyle \min \{v(C_{i}(G))+q(i,N+i-j):0\leqslant i,j,i+j\leqslant a\}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \min \{v(C_{i}(G))+q(i,k):0\leqslant i\leqslant a\},\nonumber\end{eqnarray}$$

since $i-j\leqslant k-N$ implies $q(i,k)\leqslant q(i,N+i-j)$ . Hence by our assumption on $T(k)$ , we have

$$\begin{eqnarray}v(C_{a}({\mathcal{D}}(G)))>v(C_{k}(G))+q(k,k).\end{eqnarray}$$

We deduce that if $k$ is less than or equal to $M$ and satisfies $T(k)=\{k\}$ , then

(3) $$\begin{eqnarray}\min (S({\mathcal{D}}(G)))+N\geqslant k.\end{eqnarray}$$

In particular, if $T(M)=\{M\}$ , then the lemma follows.

Now suppose that $T(M)$ has cardinality larger than 1. We define a decreasing sequence $M_{0},\ldots ,M_{n}$ of positive integers as follows. Let $M_{0}:=M$ , and define $M_{1}=\min T(M_{0})$ . If $T(M_{1})=\{M_{1}\}$ , this is the end of the sequence, otherwise we define $M_{2}$ as the minimum, and so on. Let $M_{n}$ be the last term in the sequence. Since $T(M_{n})=\{M_{n}\}$ , by (3) we have

$$\begin{eqnarray}\min (S({\mathcal{D}}(G)))+N\geqslant M_{n}.\end{eqnarray}$$

Note that for each $i$ , we have

$$\begin{eqnarray}v(C_{M_{i+1}}(G))+q(M_{i+1},M_{i})\leqslant v(C_{M_{i}}(G))+q(M_{i},M_{i}).\end{eqnarray}$$

Hence

$$\begin{eqnarray}v(C_{M_{n}}(G))-v(C_{M_{0}}(G))\leqslant \mathop{\sum }_{i}(q(M_{i},M_{i})-q(M_{i+1},M_{i}))\leqslant v(M_{0}!/M_{n}!).\end{eqnarray}$$

Since they lie in the slope ${\leqslant}-1$ part of the Newton polygon, this implies that $M_{n}$ and $M=M_{0}$ satisfy the inequality $M-M_{n}\leqslant v(M!/M_{n}!)$ , which by Lemma 3.4 is less than or equal to $\log _{p}(M_{n})+(M-M_{n})/(p-1)$ . Thus we deduce

$$\begin{eqnarray}\min (S({\mathcal{D}}(G)))+N\geqslant M_{n}\geqslant M-\frac{p-1}{p-2}\log _{p}(M_{n}).\end{eqnarray}$$

The lemma then follows from the elementary estimate

$$\begin{eqnarray}\log _{p}(M_{n})<M_{n}/\log (p).\Box\end{eqnarray}$$

3.1 Example: integral points on elliptic curves

Before describing a general method for constructing suitable nice differential operators, we illustrate how Proposition 3.2 can be used to prove effective versions of known finiteness results in the quadratic Chabauty method by considering the simplest possible case: that of integral points on a rank 1 elliptic curve. By work of Kim [Reference KimKim10], we know that integral points on rank 1 elliptic curves are contained in the zeros of

$$\begin{eqnarray}G(z)=\int _{t}^{z}\unicode[STIX]{x1D714}_{0}\unicode[STIX]{x1D714}_{1}+a\int _{t}^{z}\unicode[STIX]{x1D714}_{0}\unicode[STIX]{x1D714}_{0}+b_{i}\end{eqnarray}$$

for some constants $a,b_{i}\in \mathbb{Q}_{p}$ , where the number of $b_{i}$ is determined by the Tamagawa numbers at bad primes. In this case, finding a differential operator is quite simple: if we take ${\mathcal{D}}=(d/\unicode[STIX]{x1D714}_{0})^{2}$ , then

$$\begin{eqnarray}{\mathcal{D}}(G)=x+a.\end{eqnarray}$$

Hence, in the notation of Proposition 3.2, we may take $N=2$ , and $\sum _{b\in (E-O)(\mathbb{F}_{p})}N_{b}({\mathcal{D}}(G))=2$ , giving the bound

$$\begin{eqnarray}\#X(\mathbb{Z}_{p})_{2}<2\unicode[STIX]{x1D705}_{p}\#E(\mathbb{F}_{p})\biggl(\mathop{\prod }_{v}m_{v}\biggr).\end{eqnarray}$$

4 Differential operators for rational points: general case

To use Proposition 3.2 to bound $X(\mathbb{Q}_{p})_{2}$ , it remains to give a construction of a nice differential operator ${\mathcal{D}}$ such that ${\mathcal{D}}(G)$ is an algebraic function whose divisor can be controlled when $G$ is the iterated integral function from Proposition 2.3. The construction of the operator ${\mathcal{D}}$ is elementary. First we make some preliminary notes about calculating the divisor of ${\mathcal{D}}(F)$ when $F$ and ${\mathcal{D}}$ are algebraic.

Lemma 4.1. Let $D=\sum n_{i}P_{i}$ be an effective divisor and let $F(x)$ be a function in $H^{0}(X,{\mathcal{O}}(D))$ . Suppose $dx$ is an algebraic differential with divisor $W-W^{\prime }$ , with $W,W^{\prime }$ effective, and $W=\sum m_{i}Q_{i}$ . Define $D_{0}:=\sum P_{i}$ and $W_{0}:=\sum Q_{i}$ . Then, for all $j>0$ ,

$$\begin{eqnarray}\frac{d^{j}F}{dx^{j}}\in H^{0}(X,{\mathcal{O}}(jW+(j-1)W_{0}+D+jD_{0})).\end{eqnarray}$$

In particular, $d^{j}F/dx^{j}\in H^{0}(X,{\mathcal{O}}((2j-1)W+(j+1)D))$ .

We now restrict to our specific case of interest. Fix a point $\overline{z}$ in $X(\mathbb{F}_{p})$ . Let $D$ be an effective divisor on $X$ whose support is disjoint from $]\overline{z}[$ . Let $\unicode[STIX]{x1D714}_{0},\ldots ,\unicode[STIX]{x1D714}_{2g-1}\in H^{0}(X,\unicode[STIX]{x1D6FA}^{1}(D))$ be a set of differentials of the second kind forming a basis of $H_{\operatorname{dR}}^{1}(X)$ . Let $x\in \mathbb{Q}_{p}[\![t]\!]$ be a formal parameter at some point $z_{0}\in ]\overline{z}[$ , such that $dx$ is algebraic with divisor $D_{1}-D_{0}$ (where $D_{1}$ and $D_{0}$ are effective). Let $f_{i}:=\unicode[STIX]{x1D714}_{i}/dx\in H^{0}(X,{\mathcal{O}}(D+D_{1})).$ Finally, let $\unicode[STIX]{x1D702}$ be a differential in $H^{0}(X,\unicode[STIX]{x1D6FA}^{1}(D))$ , and let

$$\begin{eqnarray}G(z):=\sum a_{ij}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}+\sum a_{i}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}+\int _{b}^{z}\unicode[STIX]{x1D702}+h(z)\end{eqnarray}$$

be the Coleman function from Proposition 2.3.

The first step in constructing a differential operator satisfying the hypotheses of Proposition 3.2 is to reduce to constructing a nice differential operator which kills all the  $f_{i}$ .

Lemma 4.2. Suppose ${\mathcal{D}}_{1}=\sum _{i=0}^{N}g_{i}(d/dx)^{i}$ is a nice differential operator of degree $N$ , with coefficients in $H^{0}(X,{\mathcal{O}}(E))$ , for an effective divisor $E$ such that

$$\begin{eqnarray}{\mathcal{D}}_{1}(f_{i})=0\end{eqnarray}$$

for all i. Then ${\mathcal{D}}:={\mathcal{D}}_{1}(d/dx)$ is a nice differential operator with

$$\begin{eqnarray}{\mathcal{D}}(G)\in \left\{\begin{array}{@{}l@{}}H^{0}(X,{\mathcal{O}}(E+3(N-1)D_{1}+(N+3)D)),\quad N\geqslant 4\quad \\ H^{0}(X,{\mathcal{O}}(E+(2N+1)D_{1}+(N+3)D)),\quad N=2,3.\quad \end{array}\right.\end{eqnarray}$$

Proof. The operator ${\mathcal{D}}$ is nice because its leading coefficient is the same as that of ${\mathcal{D}}_{1}$ . We deal with the $\int \unicode[STIX]{x1D714}_{i},\int \unicode[STIX]{x1D702},h$ and $\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}$ terms of $G$ separately. First, we have

$$\begin{eqnarray}{\mathcal{D}}\biggl(\int \unicode[STIX]{x1D714}_{i}\biggr)={\mathcal{D}}_{1}(f_{i})=0.\end{eqnarray}$$

For $\int \unicode[STIX]{x1D702}$ , note that $(d/dx)(\int \unicode[STIX]{x1D702})=\unicode[STIX]{x1D702}/dx\in H^{0}(X,{\mathcal{O}}(D+D_{1}))$ . Thus by Lemma 4.1,

(4) $$\begin{eqnarray}{\mathcal{D}}\biggl(\int \unicode[STIX]{x1D702}\biggr)\in H^{0}(X,{\mathcal{O}}(E+(2N+1)D_{1}+(N+1)D)).\end{eqnarray}$$

For $h$ , by Lemma 4.1 we have, for all $k>0$ ,

$$\begin{eqnarray}\frac{d^{k}h}{dx^{k}}\in H^{0}(X,{\mathcal{O}}((2k-1)D_{1}+(k+2)D)),\end{eqnarray}$$

hence

(5) $$\begin{eqnarray}{\mathcal{D}}(h)\in H^{0}(X,{\mathcal{O}}((2N+1)D_{1}+(N+3)D+E)).\end{eqnarray}$$

Finally,

$$\begin{eqnarray}\displaystyle {\mathcal{D}}\biggl(\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}\biggr) & = & \displaystyle {\mathcal{D}}_{1}\biggl(f_{i}\int \unicode[STIX]{x1D714}_{j}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k\leqslant N}g_{k}\biggl(\frac{d}{dx}\biggr)^{k}\biggl(f_{i}\int \unicode[STIX]{x1D714}_{j}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k\leqslant N}g_{k}\mathop{\sum }_{0\leqslant m\leqslant k}\binom{k}{m}\biggl(\frac{d}{dx}\biggr)^{m}(f_{i})\biggl(\frac{d}{dx}\biggr)^{k-m}\biggl(\int \unicode[STIX]{x1D714}_{j}\biggr)\nonumber\\ \displaystyle & = & \displaystyle {\mathcal{D}}_{1}(f_{i})\int \unicode[STIX]{x1D714}_{j}+\mathop{\sum }_{k\leqslant N}g_{k}\mathop{\sum }_{0\leqslant m<k}\binom{k}{m}\biggl(\frac{d}{dx}\biggr)^{m}(f_{i})\biggl(\frac{d}{dx}\biggr)^{k-m-1}(f_{j})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k\leqslant N}g_{k}\mathop{\sum }_{0\leqslant m<k}\binom{k}{m}\biggl(\frac{d}{dx}\biggr)^{m}(f_{i})\biggl(\frac{d}{dx}\biggr)^{k-m-1}(f_{j})\nonumber\\ \displaystyle & & \displaystyle \nonumber\end{eqnarray}$$

since ${\mathcal{D}}(f_{1})=0$ . By Lemma 4.1, for all $k\leqslant N$ , and all $m<k$ ,

$$\begin{eqnarray}\biggl(\frac{d}{dx}\biggr)^{m}(f_{i})\biggl(\frac{d}{dx}\biggr)^{k-m-1}(f_{j})\in H^{0}(X,{\mathcal{O}}((2k-4)D_{1}+(k+1)D_{2})),\end{eqnarray}$$

where $D_{2}:=D_{1}+D$ . Hence

(6) $$\begin{eqnarray}{\mathcal{D}}\biggl(\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}\biggr)\in H^{0}(X,{\mathcal{O}}(3(N-1)D_{1}+(N+1)D+E)).\end{eqnarray}$$

Putting (4), (5) and (6) together, we find

$$\begin{eqnarray}{\mathcal{D}}(G)\in H^{0}(X,{\mathcal{O}}(E+(N+3)D+\max \{3(N-1),2N+1\}D_{1})).\Box\end{eqnarray}$$

4.1 Finding ${\mathcal{D}}_{1}$ : the general case

By the previous lemma, to get a bound on the number of zeros of $G$ , we need to construct a nice differential operator (with algebraic coefficients we can control) which annihilates all the $f_{i}:=\unicode[STIX]{x1D714}_{i}/dx$ . In general, given $m$ functions $F_{1},\ldots ,F_{m}$ , it is an elementary exercise to construct a non-trivial differential operator of order at most $m$ which annihilates all the $F_{i}$ . Hence the non-trivial question is how to find a nice differential operator.

First we introduce some notation. Let $F_{1},\ldots ,F_{2g}$ be elements of a formal power series algebra $\mathbb{Q}_{p}[\![x]\!]$ . Let $S$ be subset of $\mathbb{Z}_{{\geqslant}0}$ of size $2g+1$ . Write $S=\{n_{1},\ldots ,n_{2g+1}\}$ with $n_{i}<n_{i+1}$ . Let $A(S,F_{1},\ldots ,F_{2g})$ denote the $2g\times (2g+1)$ matrix with entries in $\mathbb{Q}_{p}[\![x]\!]$ whose $(i,j)$ th entry is $(1/n_{j}!)(d^{n_{j}}/dx^{n_{j}})(F_{i})$ . Let $A^{(j)}(S,F_{1},\ldots ,F_{2g})$ denote the $2g\times 2g$ matrix obtained by deleting the $j$ th column. Let ${\mathcal{D}}={\mathcal{D}}_{S,F_{1},\ldots ,F_{2g}}\in \mathbb{Q}_{p}[\![x]\!][d/dx]$ denote the differential operator

$$\begin{eqnarray}{\mathcal{D}}_{S,F_{1},\ldots ,F_{2g}}:=\mathop{\sum }_{i=1}^{2g+1}(-1)^{i+1}\frac{n_{2g+1}!}{n_{i}!}\det (A^{(i)})\frac{d^{n_{i}}}{dx^{n_{i}}}.\end{eqnarray}$$

We first note that ${\mathcal{D}}$ is always a differential operator which annihilates the $F_{i}$ , and then show that the set $S$ can be chosen so that ${\mathcal{D}}$ is nice.

Lemma 4.3. For any choice of $S$ , and all i,

$$\begin{eqnarray}{\mathcal{D}}(F_{i})=0.\end{eqnarray}$$

Proof. For any power series $f$ ,

$$\begin{eqnarray}{\mathcal{D}}_{S,F_{1},\ldots ,F_{2g}}(f)=n_{2g+1}!\det \left(\begin{array}{@{}ccc@{}}\displaystyle \frac{1}{n_{1}!}\frac{d^{n_{1}}}{dx^{n_{1}}}(F_{1}) & \ldots & \displaystyle \frac{1}{n_{2g+1}!}\frac{d^{n_{2g+1}}}{dx^{n_{2g+1}}}(F_{1})\\ \vdots & \ddots & \vdots \\ \displaystyle \frac{1}{n_{1}!}\frac{d^{n_{1}}}{dx^{n_{1}}}(F_{2g}) & \ldots & \displaystyle \frac{1}{n_{2g+1}!}\frac{d^{n_{2g+1}}}{dx^{n_{2g+1}}}(F_{2g})\\ \displaystyle \frac{1}{n_{1}!}\frac{d^{n_{1}}}{dx^{n_{1}}}(f) & \ldots & \displaystyle \frac{1}{n_{2g+1}!}\frac{d^{n_{2g+1}}}{dx^{n_{2g+1}}}(f)\end{array}\right).\end{eqnarray}$$

When $f=F_{i}$ , the matrix does not have full rank.◻

We now apply this construction in our case of interest. Let $D,D_{0},D_{1},\unicode[STIX]{x1D714}_{i},f_{i}$ be as defined earlier in this section.

Lemma 4.4. There exists an $S\in \mathbb{Z}_{{\geqslant}0}^{2g+1}$ with $\max S\leqslant \deg (D)+2g-1$ such that ${\mathcal{D}}_{S,f_{0},\ldots ,f_{2g-1}}$ is nice.

Proof. By construction, for any choice of $S$ , the differential operator ${\mathcal{D}}_{S,f_{0},\ldots ,f_{2g-1}}$ lies in $\mathbb{Z}_{p}[\![x]\!][d/dx]$ , hence the only non-trivial condition is that the leading coefficient is in $\mathbb{Z}_{p}^{\times }$ . Note that

$$\begin{eqnarray}\frac{1}{n_{i}!}\frac{d^{n_{i}}}{dx^{n_{i}}}f|_{x=0}=C_{n_{i}}(f),\end{eqnarray}$$

hence requiring that the leading coefficient is in $\mathbb{Z}_{p}^{\times }$ is equivalent to requiring that

$$\begin{eqnarray}\det (C_{i}(f_{j}))_{1\leqslant i\leqslant 2g,0\leqslant j\leqslant 2g-1}\in \mathbb{Z}_{p}^{\times }.\end{eqnarray}$$

Therefore, by definition, the least $N$ such that there exists a subset $S$ with $\max S\leqslant N+1$ for which ${\mathcal{D}}_{S,f_{0},\ldots ,f_{2g-1}}$ is nice is exactly the least $N$ such that the $f_{i}$ remain $\mathbb{F}_{p}$ -linearly independent after reduction mod $(p,x^{N})$ . Suppose that for all subsets $S$ of $\{0,\ldots ,N\}$ of size $2g+1$ , ${\mathcal{D}}_{S,f_{0},\ldots ,f_{2g-1}}$ is not nice. Then there is a non-trivial $\mathbb{F}_{p}$ -linear combination of $\operatorname{red}_{p}\unicode[STIX]{x1D714}_{0},\ldots ,\operatorname{red}_{p}\unicode[STIX]{x1D714}_{2g-1}$ which has a zero of order $N$ . This gives an element of $H^{0}(X_{\mathbb{F}_{p}},\unicode[STIX]{x1D6FA}^{1}(D))$ with a zero of order $N$ , which completes the proof.◻

Proof of Theorem 1.1 part (i).

To complete the proof of Theorem 1.1, it remains to estimate the degree of the coefficients of a nice ${\mathcal{D}}:={\mathcal{D}}_{S,f_{0},\ldots ,f_{2g-1}}$ . We follow the construction of basis differentials in [Reference Darmon, Daub, Lichtenstein and RotgerDDLR15, § 4.2]. Let $P$ be a non-Weierstrass point of $X(\mathbb{Q}_{p})$ whose mod $p$ reduction is different from $b$ , and let $h\in \mathbb{Q}_{p}(X)$ be a non-constant function in $H^{0}(X,{\mathcal{O}}((g\,+\,1)P))$ . Let $\unicode[STIX]{x1D714}_{0},\ldots ,\unicode[STIX]{x1D714}_{g-1}$ be a basis of $H^{0}(X,\unicode[STIX]{x1D6FA}^{1})$ . Define $\unicode[STIX]{x1D714}_{i+g}=h\unicode[STIX]{x1D714}_{i}$ for $0\leqslant i\leqslant g-1$ . Then $(\unicode[STIX]{x1D714}_{i})_{0\leqslant i\leqslant 2g-1}$ gives a basis of $H_{\operatorname{dR}}^{1}(X)$ .

Let $\unicode[STIX]{x1D714}\in H^{0}(X,\unicode[STIX]{x1D6FA}^{1})$ be a differential which does not vanish mod $p$ . Let $x$ denote the formal parameter on $]z[$ obtained by integrating $\unicode[STIX]{x1D714}$ . Let $D_{1}=(\unicode[STIX]{x1D714})$ . As above, define $f_{i}=\unicode[STIX]{x1D714}_{i}/dx\in H^{0}(X,{\mathcal{O}}(D_{1}+(g+1)P))$ .

Since $D=(g+1)P$ above, we have an $S=\{n_{1},\ldots ,n_{2g+1}\}$ such that ${\mathcal{D}}_{1}={\mathcal{D}}_{S,f_{0},\ldots ,f_{2g-1}}$ is nice and $\max S\leqslant 3g$ . Hence the differential operator ${\mathcal{D}}:={\mathcal{D}}_{1}(d/\unicode[STIX]{x1D714})$ has order at most $3g+1$ . To apply Lemma 4.2, it remains to estimate the degrees of the coefficients of ${\mathcal{D}}_{S,f_{0},\ldots ,f_{2g-1}}$ . By Lemma 4.1, we have, for all $k\geqslant 0$ ,

$$\begin{eqnarray}\biggl(\frac{d^{k}}{dx^{k}}\biggr)f_{i}\in H^{0}(X,{\mathcal{O}}(2kD_{1}+(g+k+1)P)).\end{eqnarray}$$

The $k$ th coefficient of ${\mathcal{D}}$ is hence a sum of functions in

$$\begin{eqnarray}H^{0}\biggl(X,{\mathcal{O}}\biggl(\mathop{\sum }_{1\leqslant i\leqslant 2g+1,i\neq k}(2n_{i}D_{1}+(g+n_{i}+1)P)\biggr)\biggr).\end{eqnarray}$$

Note that

$$\begin{eqnarray}\mathop{\sum }_{i\neq k}n_{i}\leqslant \mathop{\sum }_{i=g+1}^{3g}i=4g^{2}+g.\end{eqnarray}$$

Hence the coefficients of ${\mathcal{D}}_{1}$ are in

$$\begin{eqnarray}H^{0}(X,{\mathcal{O}}((8g^{2}+2g)D_{1}+3g(2g+1)P)).\end{eqnarray}$$

Applying Lemma 4.2 with $g\geqslant 2$ , $N=3g$ , $E=(8g^{2}+2g)D_{1}+3g(2g+1)P$ , and $D=(g+1)P$ , we find that

$$\begin{eqnarray}\displaystyle {\mathcal{D}}(G) & \in & \displaystyle H^{0}(X,{\mathcal{O}}((8g^{2}+2g)D_{1}+3g(2g+1)P+3(3g-1)D_{1}+3(g+1)^{2}P))\nonumber\\ \displaystyle & = & \displaystyle H^{0}(X,{\mathcal{O}}((8g^{2}+11g-3)D_{1}+3(3g^{2}+3g+1)P)).\nonumber\end{eqnarray}$$

Hence ${\mathcal{D}}(G)$ has degree at most

$$\begin{eqnarray}(8g^{2}+11g-3)(2g-2)+3(3g^{2}+3g+1)=16g^{3}+15g^{2}-19g+9.\end{eqnarray}$$

Applying Proposition 3.2, we deduce that on each residue disk, $X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}$ has at most

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{p}(16g^{3}+15g^{2}-16g+10)\end{eqnarray}$$

points. ◻

5 Differential operators for rational points: hyperelliptic case

5.1 The hyperelliptic case: non-Weierstrass disks

In this subsection we prove the second part of Theorem 1.1. Let $X$ be a hyperelliptic curve of genus $g$ with good reduction at $p\neq 2g+1$ and Mordell–Weil rank $g$ . The assumptions on $p$ imply that $X$ has a smooth model over $\mathbb{Z}_{p}$ of the form

$$\begin{eqnarray}y^{2}=f(x)=x^{2g+2}+a_{2g+1}x^{2g+1}+\cdots +a_{0}.\end{eqnarray}$$

Let $W$ denote the subscheme of Weierstrass points.

As mentioned below Proposition 3.2, our strategy for applying Proposition 3.2 is slightly different from the previous section, in that we choose one set of basis differentials and one differential operator ${\mathcal{D}}$ for the whole of $X-W$ . This means we get a bound of the form

$$\begin{eqnarray}\#(X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}\cap \,](X-W)_{\mathbb{F}_{p}}[)<\unicode[STIX]{x1D705}_{p}(\deg ({\mathcal{D}}(G))+N\#(X-W)_{\mathbb{F}_{p}}),\end{eqnarray}$$

and similarly for the Weierstrass disks.

Let $\unicode[STIX]{x1D714}_{i}$ be the differential $x^{i}dx/y$ . We take as a basis of $H_{\operatorname{dR}}^{1}(X)$ a subset of the $\mathbb{Q}$ -span of the differentials $\unicode[STIX]{x1D714}_{i}:=x^{i}dx/y$ , $0\leqslant i\leqslant 2g$ . Hence, changing notation somewhat, we may write $G$ in the form

$$\begin{eqnarray}G(z)=\mathop{\sum }_{0\leqslant i,j\leqslant 2g}a_{ij}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}+\mathop{\sum }_{0\leqslant i\leqslant 2g}a_{i}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}+h(z).\end{eqnarray}$$

Let $\infty :=\infty ^{+}+\infty ^{-}$ denote the degree $2$ divisor of the two points $\infty ^{+},\infty ^{-}$ above infinity. Since all the $\unicode[STIX]{x1D714}_{i}$ are in $H^{0}(X,\unicode[STIX]{x1D6FA}^{1}((g+1)\infty ))$ , $h$ lies in $H^{0}(X,{\mathcal{O}}(2(g+1)\infty ))$ .

First let ${\mathcal{D}}_{0}$ be the differential operator $d/\unicode[STIX]{x1D714}_{0}$ . Then define $G_{1}:={\mathcal{D}}_{0}G$ . Hence

$$\begin{eqnarray}G_{1}=\sum a_{ij}x^{i}\int \unicode[STIX]{x1D714}_{j}+\sum a_{i}x^{i}+h_{1},\end{eqnarray}$$

where $h_{1}:=y(d/dx)h$ . Define ${\mathcal{D}}_{1}=(d/dx)^{2g+1}$ and ${\mathcal{D}}={\mathcal{D}}_{1}{\mathcal{D}}_{0}$ .

Lemma 5.1. Let $W$ denote the degree $2g+2$ divisor of Weierstrass points. Then the power series ${\mathcal{D}}(G)$ lies in $H^{0}(X,{\mathcal{O}}((g+1)\infty +(4g+1)W))$ .

Proof. We have ${\mathcal{D}}(\int \unicode[STIX]{x1D714}_{i})=0$ for all $i$ , so it will be enough to prove this for $h$ and for $\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}$ . Note that we have $\operatorname{div}(dx)=W-2\infty$ , $\operatorname{div}(y)=W-(g+1)\infty$ , and so $\operatorname{div}(\unicode[STIX]{x1D714}_{0})=(g-1)\infty$ . For $h$ , we use Lemma 4.1 to deduce

$$\begin{eqnarray}{\mathcal{D}}_{0}(h)\in H^{0}(X,{\mathcal{O}}((3g+2)\infty )).\end{eqnarray}$$

Since $dx$ has divisor $W-2\infty$ , if $F\in H^{0}(X,{\mathcal{O}}(n\infty +mW))$ , a direct computation gives

$$\begin{eqnarray}\frac{dF}{dx}\in \left\{\begin{array}{@{}ll@{}}H^{0}(X,{\mathcal{O}}((n-1)\infty +(m+2)W)),\quad & m>0,\\ H^{0}(X,{\mathcal{O}}((n-1)\infty +W)),\quad & m=0.\end{array}\right.\end{eqnarray}$$

Hence ${\mathcal{D}}(h)\in H^{0}(X,{\mathcal{O}}((g+1)\infty +(4g+1)W)).$

For $\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}$ , we have

(7) $$\begin{eqnarray}\displaystyle {\mathcal{D}}\biggl(\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}\biggr) & = & \displaystyle {\mathcal{D}}_{1}\biggl(x^{i}\int \unicode[STIX]{x1D714}_{j}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k=0}^{i}\binom{2g+1}{k}\frac{i!}{(i-k)!}x^{i-k}\biggl(\frac{d}{dx}\biggr)^{2g-k+1}\int (\unicode[STIX]{x1D714}_{j})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k=0}^{i}\binom{2g+1}{k}\frac{i!}{(i-k)!}x^{i-k}\biggl(\frac{d}{dx}\biggr)^{2g-k}\biggl(\frac{x^{j}}{y}\biggr).\end{eqnarray}$$

Now we have

$$\begin{eqnarray}\frac{x^{j}}{y}\in H^{0}(X,{\mathcal{O}}((j-g-1)\infty +W)),\end{eqnarray}$$

so

$$\begin{eqnarray}\biggl(\frac{d}{dx}\biggr)^{(2g-k)}\biggl(\frac{x^{j}}{y}\biggr)\in H^{0}(X,{\mathcal{O}}((j-3g+k-1)\infty +(4g-2k+1)W)),\end{eqnarray}$$

which gives that

$$\begin{eqnarray}x^{i-k}\biggl(\frac{d}{dx}\biggr)^{(2g-k)}\biggl(\frac{x^{j}}{y}\biggr)\in H^{0}(X,{\mathcal{O}}((i+j-1-3g)\infty +(4g-2k+1)W)),\end{eqnarray}$$

and thus each summand of (7) is an element of $H^{0}(X,{\mathcal{O}}((g-1)\infty +(4g+1)W))$ .◻

Since the degree of $(g+1)\infty +(4g+1)W$ is $2g+2+(4g+1)(2g+2)=8g^{2}+12g+4$ , applying Proposition 3.2, and summing over all non-Weierstrass residue disks, we find that the number of points of $X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}$ which reduce to non-Weierstrass points away from infinity is at most

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{p}((8g^{2}+12g+4)+(2g+2)\#(X-W-\infty )(\mathbb{F}_{p})).\end{eqnarray}$$

For the two points at infinity, we may apply the same analysis with the equation at infinity

$$\begin{eqnarray}y^{2}=a_{0}x^{2g+2}+a_{1}x^{2g+1}+\cdots +1.\end{eqnarray}$$

We deduce that the number of points of $X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}$ which reduce to non-Weierstrass points is at most

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{p}((16g^{2}+24g+8)+(2g+2)\#(X-W)(\mathbb{F}_{p})).\end{eqnarray}$$

5.2 The hyperelliptic case: Weierstrass points

The computation at Weierstrass disks is carried out in a manner similar to the method developed in § 4. The essential difference is that, instead of trying to find a new nice differential operator ${\mathcal{D}}_{1}$ annihilating the $2g$ functions $\{f_{0},\ldots ,f_{2g-1}\}$ for each residue disk $]b[$ , we find a differential operator ${\mathcal{D}}_{1}$ which annihilates the $2g+1$ functions $\unicode[STIX]{x1D714}_{i}/\unicode[STIX]{x1D714}_{0}$ ( $0\leqslant i\leqslant 2g$ ) at all Weierstrass disks. The price paid for this is that the degree is slightly larger. Let $B\in M_{2g+1}(\mathbb{Q}_{p}(X))$ denote the matrix

$$\begin{eqnarray}B=\biggl(\frac{1}{(2i)!}\biggl(\frac{d}{\unicode[STIX]{x1D714}_{0}}\biggr)^{2i}x^{j}\biggr)_{0\leqslant i,j\leqslant 2g}.\end{eqnarray}$$

Lemma 5.2. For all Weierstrass points $z=(\unicode[STIX]{x1D6FC},0)\in X(\mathbb{F}_{p})$ , $\det (B)\in {\mathcal{O}}_{{\mathcal{X}},z}^{\times }$ .

Proof. Clearly $\det (B)$ is defined at $z$ , so it is sufficient to prove it is non-zero at $z$ . First note that, since $x-\unicode[STIX]{x1D6FC}$ has a zero of order $2$ at $z$ , a linear combination of $1,x,\ldots ,x^{2g}$ can have a zero of order at most $4g$ . Hence the $(2g+1)\times (4g+1)$ matrix $((d^{i}/\unicode[STIX]{x1D714}_{0}^{i})x^{j}|_{z})_{0\leqslant i\leqslant 4g,0\leqslant j\leqslant 2g}$ has rank at least $2g+1$ . On the other hand, for all odd $j$ , $(d^{j}/\unicode[STIX]{x1D714}_{0}^{j})x^{i}$ is an odd function with respect to the hyperelliptic involution, and hence vanishes at $z$ . Hence $B|_{z}$ is invertible in $M_{2g+1}(\mathbb{F}_{p})$ .◻

Proof of Theorem 1.1 part (ii).

We deduce that we can apply the construction of § 4.1 with $A$ taken to be the $(2g+1)\times (2g+2)$ matrix

$$\begin{eqnarray}\biggl(\biggl(\frac{d}{\unicode[STIX]{x1D714}_{0}}\biggr)^{i}x^{j}\biggr)_{0\leqslant j\leqslant 2g,i=0,2,4,\ldots ,4g,4g+1}.\end{eqnarray}$$

The function $(d/\unicode[STIX]{x1D714}_{0})^{i}x^{j}$ lies in $H^{0}(X,{\mathcal{O}}((gi+j)\infty ))$ , hence the differential operator

$$\begin{eqnarray}{\mathcal{D}}_{1}={\mathcal{D}}_{S,\unicode[STIX]{x1D714}_{0}/\unicode[STIX]{x1D714}_{0},\unicode[STIX]{x1D714}_{1}/\unicode[STIX]{x1D714}_{0},\ldots ,\unicode[STIX]{x1D714}_{2g}/\unicode[STIX]{x1D714}_{0}}\end{eqnarray}$$

has coefficients in

$$\begin{eqnarray}H^{0}\biggl(X,{\mathcal{O}}\biggl(\biggl(g\mathop{\sum }_{i=1}^{2g}2i\biggr)+g(4g+1)+\mathop{\sum }_{j=0}^{2g}j\biggr)\biggr)=H^{0}(X,{\mathcal{O}}((4g^{3}+8g^{2}+2g)\infty )).\end{eqnarray}$$

Define ${\mathcal{D}}:={\mathcal{D}}_{1}{\mathcal{D}}_{0}$ , where ${\mathcal{D}}_{0}:=(d/\unicode[STIX]{x1D714}_{0})$ . Applying Lemma 4.2 with $E=(4g^{3}+8g^{2}+2g)\infty$ , $N=4g+1$ , $D_{1}=(g-1)\infty$ , $D=(g+1)\infty$ , we deduce

$$\begin{eqnarray}{\mathcal{D}}(G)\in H^{0}(X,{\mathcal{O}}((4g^{3}+24g^{2}-2g+4)\infty )).\end{eqnarray}$$

The number of points of $X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}$ on all Weierstrass disks is hence, by Proposition 3.2, at most

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{p}((4g+2)\#W(\mathbb{F}_{p})+2(4g^{3}+24g^{2}-2g+4)).\end{eqnarray}$$

Combining with the bounds from the non-Weierstrass residue disks in the previous subsection, we find that

$$\begin{eqnarray}\#X(\mathbb{Q}_{p})_{\unicode[STIX]{x1D6FC}}\leqslant \unicode[STIX]{x1D705}_{p}((2g+2)\#X(\mathbb{F}_{p})+2g\#W(\mathbb{F}_{p})+8g^{3}+64g^{2}+20g+16).\Box\end{eqnarray}$$

6 Integral points for hyperelliptic curves

The proof of the $g>1$ case of Theorem 1.3 follows a similar strategy to the previous section. Let $X$ be a hyperelliptic curve of genus $g>1$ with equation

$$\begin{eqnarray}y^{2}=f(x)=x^{2g+1}+a_{2g}x^{2g}+\cdots +a_{0}\end{eqnarray}$$

and suppose the rank of the Jacobian of $X$ is equal to $g$ . Then the set $X(\mathbb{Z}_{p})_{2}$ is partitioned into a disjoint union of sets $X(\mathbb{Z}_{p})_{\unicode[STIX]{x1D6FC}}$ . Each set $X(\mathbb{Z}_{p})_{\unicode[STIX]{x1D6FC}}$ is contained in the set of zeros of a Coleman function of the form

$$\begin{eqnarray}G(z)=\mathop{\sum }_{0\leqslant i,j<2g}a_{ij}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}+\mathop{\sum }_{0\leqslant i<2g}a_{i}\int _{b}^{z}\unicode[STIX]{x1D714}_{i}+h(z),\end{eqnarray}$$

where $h\in H^{0}(X,{\mathcal{O}}(4g\infty ))$ , with $\infty$ now denoting the divisor of degree 1 consisting of the unique point at infinity. Let $W$ denote the degree $2g+1$ divisor of Weierstrass points away from infinity and define ${\mathcal{D}}_{0}:=d/\unicode[STIX]{x1D714}_{0}$ . For non-Weierstrass points, we take the differential operator to be

$$\begin{eqnarray}{\mathcal{D}}=\biggl(\frac{d}{dx}\biggr)^{2g}{\mathcal{D}}_{0}.\end{eqnarray}$$

Since $\unicode[STIX]{x1D714}_{0}$ has a zero of order $(2g-2)$ at $\infty$ , we have

$$\begin{eqnarray}\frac{dh}{\unicode[STIX]{x1D714}_{0}}\in H^{0}(X,{\mathcal{O}}((6g-1)\infty )).\end{eqnarray}$$

Similar to the case of an even degree model, since $dx$ has divisor $W-3\infty$ , if $F$ is in $H^{0}(X,{\mathcal{O}}(nW+m\infty ))$ then

$$\begin{eqnarray}\frac{dF}{dx}\in \left\{\begin{array}{@{}ll@{}}H^{0}(X,{\mathcal{O}}((n+2)W+(m-2)\infty )),\quad & n>0,\\ H^{0}(X,{\mathcal{O}}(W+(m-2)\infty )),\quad & n=0.\end{array}\right.\end{eqnarray}$$

Hence ${\mathcal{D}}(h)$ is in $H^{0}(X,{\mathcal{O}}((2g-1)\infty +(4g-1)W))$ .

For the remaining term, note that

(8) $$\begin{eqnarray}\biggl(\frac{d}{dx}\biggr)^{k}\biggl(\frac{x^{j}}{y}\biggr)\in H^{0}(X,{\mathcal{O}}((2k+1)W+(2j-2k-2g-1)\infty )).\end{eqnarray}$$

Since

$$\begin{eqnarray}\displaystyle {\mathcal{D}}\biggl(\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j}\biggr) & = & \displaystyle \biggl(\frac{d}{dx}\biggr)^{2g}\biggl(x^{i}\int \unicode[STIX]{x1D714}_{j}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{0\leqslant k<2g}\binom{2g}{k}\frac{i!}{(i-k)!}x^{i-k}\biggl(\frac{d}{dx}\biggr)^{2g-k-1}\biggl(\frac{x^{j}}{y}\biggr),\nonumber\end{eqnarray}$$

equation (8) implies

$$\begin{eqnarray}x^{i-k}\biggl(\frac{d}{dx}\biggr)^{2g-k-1}\biggl(\frac{x^{j}}{y}\biggr)\in H^{0}(X,{\mathcal{O}}((4g-2k-1)W+(2j+2i-6g+1)\infty )).\end{eqnarray}$$

Hence ${\mathcal{D}}(\int \unicode[STIX]{x1D714}_{i}\unicode[STIX]{x1D714}_{j})$ lies in $H^{0}(X,{\mathcal{O}}((4g-1)W+(2g-3)\infty ))$ . Arguing as in § 5.1, we deduce that the number of integral points of $X$ which reduce to non-Weierstrass points mod $p$ is bounded by

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{p}\biggl(\mathop{\prod }_{v\in T_{0}}m_{v}\biggr)(8g^{2}+4g-4+(2g+1)\#(Y-W)(\mathbb{F}_{p})).\end{eqnarray}$$

6.1 Differential operators at Weierstrass points

Let $B\in M_{2g}(\mathbb{Q}_{p}(X))$ be the matrix

$$\begin{eqnarray}\biggl(\frac{1}{(2i)!}\biggl(\frac{d}{\unicode[STIX]{x1D714}_{0}}\biggr)^{2i}x^{j}\biggr)_{0\leqslant i,j<2g}.\end{eqnarray}$$

As in § 5.2, we may show $\det (B)$ is a unit at all points in $W$ , and hence construct a differential operator ${\mathcal{D}}_{1}$ from the matrix

$$\begin{eqnarray}A=\biggl(\frac{1}{i!}\biggl(\frac{d}{\unicode[STIX]{x1D714}_{0}}\biggr)^{i}x^{j}\biggr)_{0\leqslant j<2g,i=0,2,\ldots ,4g-2,4g-1}.\end{eqnarray}$$

Since $\unicode[STIX]{x1D714}_{0}$ has a zero of order $(2g-2)$ at $\infty$ , we find that

$$\begin{eqnarray}\biggl(\frac{d}{\unicode[STIX]{x1D714}_{0}}\biggr)^{i}x^{j}\in H^{0}(X,{\mathcal{O}}((i(2g-1)+2j)\infty )).\end{eqnarray}$$

We deduce that the coefficients of ${\mathcal{D}}_{1}$ lie in $H^{0}(X,{\mathcal{O}}(8g^{3}+4g^{2}-6g+1)\infty ).$ Define ${\mathcal{D}}:={\mathcal{D}}_{1}{\mathcal{D}}_{0}$ , where ${\mathcal{D}}_{0}:=(d/\unicode[STIX]{x1D714}_{0})$ . Applying Lemma 4.2 with $E=(8g^{3}+4g^{2}-6g+1)\infty$ , $D_{1}=2(g-1)\infty$ , $D=2g\infty$ and $N=4g-1$ , we deduce that ${\mathcal{D}}(G)$ lies in $H^{0}(X,{\mathcal{O}}((8g^{3}+36g^{2}-38g+13)\infty ))$ , hence by Proposition 3.2, we find that the number of integral points reducing to Weierstrass points is bounded by

$$\begin{eqnarray}\biggl(\mathop{\prod }_{v}m_{v}\biggr)\unicode[STIX]{x1D705}_{p}(8g^{3}+36g^{2}-38g+13+4g\#W(\mathbb{F}_{p})).\end{eqnarray}$$

We deduce a bound for the total number of integral points of

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{p}\biggl(\mathop{\prod }_{v\in T_{0}}m_{v}\biggr)(8g^{3}+44g^{2}-34g+9+(2g+1)\#Y(\mathbb{F}_{p})+(2g-1)\#W(\mathbb{F}_{p})).\end{eqnarray}$$