Reductions of points on algebraic groups, II

TL;DR

This paper extends previous work on the reduction of points on algebraic groups by analyzing the density of primes where a point's order is coprime to a given integer, expressing it through $ ext{l}$-adic integrals and proving its rationality.

Contribution

It generalizes earlier results to arbitrary positive integers, providing a formula for the density and showing it is a rational number with bounded denominator.

Findings

Density expressed as sum of $ ext{l}$-adic integrals

Density is a rational number with bounded denominator

Extends previous prime-specific results to general integers

Abstract

Let $A$ be the product of an abelian variety and a torus over a number field $K$, and let $m$ be a positive integer. If $\alpha \in A(K)$ is a point of infinite order, we consider the set of primes $\mathfrak p$ of $K$ such that the reduction $(\alpha \bmod \mathfrak p)$ is well defined and has order coprime to $m$. This set admits a natural density, which we are able to express as a finite sum of products of $\ell$-adic integrals, where $\ell$ varies in the set of prime divisors of $m$. We deduce that the density is a rational number, whose denominator is bounded (up to powers of $m$) in a very strong sense. This extends the results of the paper "Reductions of points on algebraic groups" by Davide Lombardo and the second author, where the case $m$ prime is established.

Equations

$${\mathrm{Dens}}_m(\alpha )=\prod _{\ell \mid m}{\mathrm{Dens}}_{\ell }(\alpha )$$

$$I_{H,f}:={\int }_{H}fd{\mu }_H.$$

$$H^{\prime }:=H\cap G^{\prime }$$

$$H^{\prime }=\prod _{\ell }{H}_{\ell }^{\prime }$$

$${\mu }_{H^{\prime }}=\prod _{\ell }{\mu }_{H_{\ell }^{\prime }}.$$

$${\mu }_{H(x)}=(H:H^{\prime }){\mu }_H,$$

$$I_{H,f}=\sum _{x\in H/H^{\prime }}{\int }_{H(x)}fd{\mu }_H=\frac{1}{(H:H^{\prime })}\sum _{x\in H/H^{\prime }}{\int }_{H(x)}fd{\mu }_{H(x)}.$$

$$H(x)=\prod _{\ell }{H}_{\ell }(x),$$

$$f=\prod _{\ell }{f}_{x,\ell }$$

$$I_{H,f}=\frac{1}{(H:H^{\prime })}\sum _{x\in H/H^{\prime }}\prod _{\ell }{\int }_{H_{\ell }(x)}{f}_{x,\ell }d{\mu }_{H_{\ell }(x)}.$$

$$f=\prod _{\ell }{f}_{\ell }$$

$$I_{H,f}=\prod _{\ell }{\int }_{H_{\ell }}{f}_{\ell }d{\mu }_{H_{\ell }}.$$

$$K_{m^{-n}}:=K(A[m^n])\text{for }n⩾1$$

$$K_{m^{-\infty }}:=\bigcup _{n⩾1}{K}_{m^{-n}}.$$

$$m^{-\infty }\alpha =\underset{n⩾1}{\underleftarrow{\lim }}{m}^{-n}\alpha .$$

$$K_{m^{-n}\alpha }:=K(m^{-n}\alpha )\text{for }n⩾1$$

$$K_{m^{-\infty }\alpha }:=\bigcup _{n⩾1}{K}_{m^{-n}\alpha }.$$

$$\begin{array}[]{cccc}\omega_{\alpha}:&\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)&\longrightarrow&T_{m}A\rtimes\operatorname{Aut}(T_{m}A)\\ &\sigma&\longmapsto&(t,M)\end{array}$$

$$T_{m}A⋊\mathrm{Aut}(T_{m}A)\cong \prod _{\ell }{\mathbb{Z}}_{\ell }^{b_A}⋊\prod _{\ell }{\mathrm{GL}}_{b_A}({\mathbb{Z}}_{\ell })\cong \prod _{\ell }({\mathbb{Z}}_{\ell }^{b_A}⋊{\mathrm{GL}}_{b_A}({\mathbb{Z}}_{\ell })).$$

$$[K_{m^{-N}}:K_{m^{-n}}]=m^{d_A(N-n)}.$$

$$[K_{m^{-N}\alpha }:K_{m^{-n}\alpha }]=(m^{b_A+d_A}{)}^{N-n}.$$

$$[K_{m^{-N}\alpha }:K_{m^{-N}}(m^{-n}\alpha )]=m^{b_A(N-n)}.$$

$${\mathrm{C}}_m:=m^{b_{A}n}/[K_{m^{-n}\alpha }:K_{m^{-n}}]$$

$$C_{\alpha }\in H^1(\mathrm{Gal}(K_{m^{-\infty }\alpha }/K),T_{m}A).$$

$$\begin{array}[]{cccc}c_{\beta}:&\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)&\longrightarrow&T_{m}A\\ &\sigma&\longmapsto&\sigma(\beta)-\beta\,.\end{array}$$

$$\operatorname{Res}_{\sigma}:H^{1}\bigl{(}\operatorname{Gal}(K_{m^{-\infty}\alpha}/K),T_{m}A\bigr{)}\longrightarrow H^{1}({\langle\sigma\rangle},T_{m}A)\,.$$

$$S_{\alpha }:=\{\sigma :C_{\alpha }\in \ker ({\mathrm{Res}}_{\sigma })\}=\{\sigma :\sigma (\beta )=\beta \text{ for some β∈m−∞α}\}.$$

$${\mathcal{W}}_{m^n}(M):=\{t\in A[m^n]\mid (t,M)\in \mathrm{Gal}(K_{m^{-n}\alpha }/K)\}$$

$$\mathrm{w}_{m^{n}}(M):=\frac{\#\bigl{(}\operatorname{Im}(M-I)\cap{\mathcal{W}}_{m^{n}}(M)\bigr{)}}{\#\operatorname{Im}(M-I)}\in\mathbb{Q}\,.$$

$$\iota_{\ell^{n}}\colon\operatorname{Gal}(K_{\ell^{-n}\alpha}K_{m^{-1}}/K)\hookrightarrow\bigl{(}A[\ell^{n}]\rtimes\mathcal{G}(\ell^{n})\bigr{)}\times\mathcal{G}(m).$$

Full text

1. Introduction

This article is the continuation of the paper Reductions of points on algebraic groups by Davide Lombardo and the second author [4]. We refer to this other work for the history of the problem and further references.

Let $A$ be the product of an abelian variety and a torus over a number field $K$, and let $m$ be a positive square-free integer. If $\alpha\in A(K)$ is a point of infinite order, we consider the set of primes $\mathfrak{p}$ of $K$ such that the reduction $(\alpha\bmod\mathfrak{p})$ is well defined and has order coprime to $m$. This set admits a natural density (see Theorem 7), which we denote by $\operatorname{Dens}_{m}(\alpha)$.

The main question is whether we can write

[TABLE]

where $\ell$ varies over the prime divisors of $m$. Let $K(A[m])$ be the $m$-torsion field of $A$. We prove that (1) holds if $K(A[m])=K$ (i.e. if $A(K)$ contains all $m$-torsion points) or, more generally, if the degree $[K(A[\ell]):K]$ is a power of $\ell$ (see Corollary 19). Indeed, (1) holds if the torsion/Kummer extensions of $\alpha$ related to different prime divisors of $m$ are linearly disjoint over $K$. In general, (1) does not hold; see Section 7.2 for an explicit example.

We are able to express $\operatorname{Dens}_{m}(\alpha)$ as an integral over the image of the $m$-adic representation (see Theorem 17), and also as a finite sum of products of $\ell$-adic integrals (see Theorems 20 and 21). The latter decomposition allows us to prove that $\operatorname{Dens}_{m}(\alpha)$ is a rational number whose denominator is uniformly bounded in a very strong sense (see Corollary 22).

Finally, we study Serre curves in detail in Section 6. With our results one can explicitly compute $\operatorname{Dens}_{m}(\alpha)$ if the $m^{n}$-Kummer extensions of $\alpha$ (defined in Section 3) have maximal degree for all $n$ or, more generally, if the degrees of these extensions are known and are the same with respect to the base fields $K$ and $K(A[m])$.

2. Integration on profinite groups

For every compact topological group $G$, we write $\mu_{G}$ for the normalised Haar measure on $G$.

Let $G$ be a profinite group, and let $H$ be a closed subgroup of finite index in $G$. For every integrable function $f\colon H\to\mathbb{C}$, we define

[TABLE]

Let $\ell$ vary in a finite set of prime numbers, and suppose that we have $G=\prod_{\ell}G_{\ell}$, where each $G_{\ell}$ is a profinite group containing a pro-$\ell$-group $G^{\prime}_{\ell}$ as a closed subgroup of finite index. Note that we may assume that $G^{\prime}_{\ell}$ is normal in $G_{\ell}$ (up to replacing $G^{\prime}_{\ell}$ by the intersection of its finitely many conjugates in $G_{\ell}$).

The profinite group $G^{\prime}:=\prod_{\ell}G^{\prime}_{\ell}$ has finite index in $G$, and the profinite group

[TABLE]

is a closed subgroup of finite index in each of $G$, $H$ and $G^{\prime}$. Because the $G^{\prime}_{\ell}$ are pro-$\ell$-groups for pairwise different $\ell$, every closed subgroup of $G^{\prime}$ is similarly a product of pro-$\ell$-groups. We can therefore write

[TABLE]

where each $H^{\prime}_{\ell}$ is a closed subgroup of finite index in $G^{\prime}_{\ell}$. The normalised Haar measures on $H^{\prime}$ and the $H^{\prime}_{\ell}$ are thus related by

[TABLE]

For each $x\in H/H^{\prime}$, we write $H(x)$ for the fibre over $x$ of the quotient map $H\to H/H^{\prime}$. Then each $H(x)$ is a left coset of the subgroup $H^{\prime}\subseteq H$. We restrict the Haar measure on $H$ to a measure on the open subset $H(x)$, which we still denote by $\mu_{H}$. The normalised Haar measure on $H(x)$ is then

[TABLE]

and we can decompose (2) as

[TABLE]

By (3), and since $H(x)$ is a torsor under $H^{\prime}$, we can write

[TABLE]

where each $H_{\ell}(x)$ is a torsor under $H^{\prime}_{\ell}$. We equip each $H_{\ell}(x)$ with the normalised Haar measure $\mu_{H_{\ell}(x)}$ coming from the $H^{\prime}_{\ell}$-torsor structure; then $\mu_{H(x)}$ is the product of the $\mu_{H_{\ell}(x)}$.

Theorem 1**.**

Suppose that over $H(x)$ there is a product decomposition

[TABLE]

where the $f_{x,\ell}:H_{\ell}(x)\to\mathbb{C}$ are locally constant functions that are integrable with respect to $\mu_{H_{\ell}(x)}$. We then have

[TABLE]

Proof.

The assertion follows from (4) because we have a product decomposition for $f$ and because $\mu_{H(x)}$ is the product of the $\mu_{H_{\ell}(x)}$. ∎

Proposition 2**.**

Suppose that we have $H=\prod_{\ell}H_{\ell}$, where $H_{\ell}\subseteq G_{\ell}$ (this happens for example if $H\subseteq G^{\prime}$). If there is a product decomposition

[TABLE]

over $H$, where the $f_{\ell}:H_{\ell}\to\mathbb{C}$ are locally constant functions that are integrable with respect to $\mu_{H_{\ell}}$, then we have

[TABLE]

Proof.

The assertion follows from the product decomposition for $f$ and the fact that $\mu_{H}$ is the product of the $\mu_{H_{\ell}}$. ∎

3. The arboreal representation

Let $K$ be a number field, and let $\overline{K}$ be an algebraic closure of $K$. Let $A$ be a connected commutative algebraic group over $K$, and let $b_{A}$ be the first Betti number of $A$. We fix a square-free positive integer $m$, and we let $\ell$ vary in the set of prime divisors of $m$. We also fix a point $\alpha\in A(K)$.

We define $T_{m}A$ as the projective limit of the torsion groups $A[m^{n}]$ for $n\geqslant 1$; we can write $T_{m}A=\prod_{\ell}T_{\ell}A$, where the Tate module $T_{\ell}A$ is a free $\mathbb{Z}_{\ell}$-module of rank $b_{A}$.

We define the torsion fields

[TABLE]

and

[TABLE]

The Galois action on the $m$-power torsion points of $A$ gives the $m$-adic representation of $A$, which maps $\operatorname{Gal}(\overline{K}/K)$ to the automorphism group of $T_{m}A$. We can also speak of the mod $m^{n}$ representation, which describes the Galois action on $A[m^{n}]$. Choosing a $\mathbb{Z}_{\ell}$-basis for $T_{\ell}A$, we can identify the image of the $m$-adic representation with a subgroup of $\prod_{\ell}\operatorname{GL}_{b_{A}}(\mathbb{Z}_{\ell})$ and the image of the mod $m^{n}$ representation with a subgroup of $\prod_{\ell}\operatorname{GL}_{b_{A}}(\mathbb{Z}/\ell^{n}\mathbb{Z})$.

For $n\geqslant 1$, let $m^{-n}\alpha$ be the set of points in $A(\overline{K})$ whose $m^{n}$-th multiple equals $\alpha$. We also write

[TABLE]

This is the set of sequences $\beta=\{\beta_{n}\}_{n\geqslant 1}$ such that $m\beta_{1}=\alpha$ and $m\beta_{n+1}=\beta_{n}$ for every $n\geqslant 1$; it is a torsor under $T_{m}A$. We note that $m^{-n}0=A[m^{n}]$ and $m^{-\infty}0=T_{m}A$.

We define the fields

[TABLE]

and

[TABLE]

We call the field extension $K_{m^{-n}\alpha}/K_{m^{-n}}$ the $m^{n}$-Kummer extension defined by the point $\alpha$. We view the $m$-adic representation as a representation of $\operatorname{Gal}({K_{m^{-\infty}\alpha}}/K)$.

We fix an element $\beta\in m^{-\infty}\alpha$, and define the arboreal representation

[TABLE]

where $M$ is the image of $\sigma$ under the $m$-adic representation and $t=\sigma(\beta)-\beta$. The arboreal representation is an injective homomorphism of profinite groups identifying $\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)$ with a subgroup of

[TABLE]

Similarly, for $\sigma\in\operatorname{Gal}(K_{m^{-n}\alpha}/K)$ we can consider $t\in A[m^{n}]\cong\prod_{\ell}(\mathbb{Z}/\ell^{n}\mathbb{Z})^{b_{A}}$ and $M\in\operatorname{Aut}A[m^{n}]\cong\prod_{\ell}\operatorname{GL}_{b_{A}}(\mathbb{Z}/\ell^{n}\mathbb{Z})$.

Denote by $\mathcal{G}({m^{\infty}})\subseteq\prod_{\ell}\operatorname{GL}_{b_{A}}(\mathbb{Z}_{\ell})$ the image of the $m$-adic representation and by $\mathcal{G}({m^{n}})$ the image of the mod $m^{n}$ representation. Consider the image of the $\ell$-adic representation in $\operatorname{GL}_{b_{A}}(\mathbb{Z}_{\ell})$: the dimension of its Zariski closure in $\operatorname{GL}_{b_{A},\mathbb{Q}_{\ell}}$ is independent of $\ell$ and $K$, and we denote it by $d_{A}$. For example, if $A$ is an elliptic curve, we have $d_{A}=2$ if $A$ has complex multiplication, and $d_{A}=4$ otherwise.

Definition 3**.**

We say that $(A/K,m)$ satisfies eventual maximal growth of the torsion fields if there exists a positive integer $n_{0}$ such that for all $N\geqslant n\geqslant n_{0}$ we have

[TABLE]

We say that $(A/K,m,\alpha)$ satisfies eventual maximal growth of the Kummer extensions if there exists a positive integer $n_{0}$ such that for all $N\geqslant n\geqslant n_{0}$ we have

[TABLE]

Remark 4**.**

Condition (9) means that there is eventual maximal growth of the torsion fields, that $K_{m^{-n}\alpha}$ and $K_{m^{-N}}$ are linearly disjoint over $K_{m^{-n}}$, and that we have

[TABLE]

If there is eventual maximal growth of the Kummer extensions, the rational number

[TABLE]

is independent of $n$ for $n\geqslant n_{0}$.

Proposition 5**.**

If $A$ is a semiabelian variety, then $(A/K,m)$ satisfies eventual maximal growth of the torsion fields. If $A$ is the product of an abelian variety and a torus and $\mathbb{Z}\alpha$ is Zariski dense in $A$, then $(A/K,m,\alpha)$ satisfies eventual maximal growth of the Kummer extensions.

Proof.

By [4, Lemma 10], if $A$ is a semiabelian variety and $\ell$ is a prime divisor of $m$, then $(A/K,\ell,\alpha)$ satisfies eventual maximal growth of the torsion fields. We also know that the degree $[K_{\ell^{-n}}:K_{\ell^{-1}}]$ is a power of $\ell$. Therefore the extensions $K_{m^{-1}}K_{\ell^{-n}}$ for $\ell\mid n$ are linearly disjoint over $K_{m^{-1}}$ and the first assertion follows. By [4, Remark 7], the second assertion holds for $(A/K,\ell,\alpha)$, where $\ell$ is any prime divisor of $m$. We conclude because these Kummer extensions have order a power of $\ell$. ∎

4. Relating the density and the arboreal representation

4.1. The existence of the density

From now on, we assume that $(A/K,m,\alpha)$ satisfies eventual maximal growth of the Kummer extensions.

Remark 6**.**

This is not a restriction if $A$ is the product of an abelian variety and a torus by Proposition 5. Indeed, consider the number of connected components of the Zariski closure of $\mathbb{Z}\alpha$. If this number is not coprime to $m$, then the density $\operatorname{Dens}_{m}(\alpha)$ is zero by [5, Main Theorem] while if it is coprime to $m$ we may replace $\alpha$ by a multiple to reduce to the case where the Zariski closure of $\mathbb{Z}\alpha$ is connected. Finally, we may replace $A$ and reduce to the case where $\mathbb{Z}\alpha$ is Zariski dense.

The $T_{m}A$-torsor $m^{-\infty}\alpha$ from Section 3 defines a Galois cohomology class

[TABLE]

For any choice of $\beta\in m^{-\infty}\alpha$, this is the class of the cocycle

[TABLE]

We also consider the restriction map with respect to the cyclic subgroup generated by some element $\sigma$:

[TABLE]

Theorem 7**.**

If $(A/K,m,\alpha)$ satisfies eventual maximal growth of the Kummer extensions, then the density $\operatorname{Dens}_{m}(\alpha)$ exists and equals the normalized Haar measure in $\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)$ of the subset

[TABLE]

Proof.

The generalisations of [2, Theorem 3.2] and [4, Theorem 5] to the composite case are straightforward. ∎

Similarly to [4, Remark 19], we may equivalently consider $S_{\alpha}$ as a subset of either $\operatorname{Gal}(\overline{K}/K)$ or $\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)$ with their respective normalised Haar measures.

Proposition 8**.**

If $L/K$ is any Galois extension that is linearly disjoint from $K_{m^{-\infty}\alpha}$ over $K$, then we have $\operatorname{Dens}_{L}(\alpha)=\operatorname{Dens}_{K}(\alpha)$.

Proof.

The generalisation of [4, Proposition 20] to the composite case is straightforward. ∎

4.2. Counting elements in the image of the arboreal representation

Definition 9**.**

For $M\in\mathcal{G}(m^{n})$ we define

[TABLE]

and

[TABLE]

For every prime divisor $\ell$ of $m$ and every $n\geqslant 1$, we consider the Galois group of the compositum $K_{\ell^{-n}\alpha}K_{m^{-1}}$ over $K$ and the inclusion

[TABLE]

Definition 10**.**

For all $x\in\mathcal{G}(m)$ and $V\in\mathcal{G}(\ell^{n})$, we define

[TABLE]

and

[TABLE]

We denote by $\pi_{*}$ the projection onto $\mathcal{G}(*)$.

Proposition 11**.**

If $M\in\mathcal{G}(m^{n})$ is such that $\pi_{m}M=x$, then we have

[TABLE]

Proof.

Decomposing $A[m^{n}]=\bigoplus_{\ell}A[\ell^{n}]$, we can write $t=\sum_{\ell}t_{\ell}$. We have $\operatorname{Im}(M-I)=\bigoplus_{\ell}\operatorname{Im}(\pi_{\ell^{n}}M-I)$, and since the extensions $K_{\ell^{-n}\alpha}K_{m^{-1}}/K_{m^{-1}}$ have pairwise coprime degrees and hence are linearly disjoint, we have $t\in{\mathcal{W}}_{m^{n}}(M)$ if and only if for every $\ell$ we have $t_{\ell}\in{\mathcal{W}}_{x,\ell^{n}}(\pi_{\ell^{n}}M)$. This implies the claim. ∎

Lemma 12**.**

For all $x\in\mathcal{G}(m)$ and $V\in\mathcal{G}(\ell^{\infty})$, the value $\mathrm{w}_{x,\ell^{n}}(V)$ is constant for $n$ sufficiently large.

Proof.

This is proved as in [4, Lemma 23]. ∎

By Lemma 12, we can define

[TABLE]

From (15) we deduce that for all $M\in\mathcal{G}(m^{\infty})$, the value $\mathrm{w}_{m^{n}}(M)$ is also constant for $n$ sufficiently large, so we can analogously define

[TABLE]

Proposition 13**.**

If $M\in\mathcal{G}(m^{\infty})$ is such that $\pi_{m}M=x$, then we have

[TABLE]

Proof.

Taking the limit as $n\to\infty$ in (15) yields (18). ∎

4.3. Partitioning the image of the $m$-adic representation

We view elements of $\mathcal{G}({m^{\infty}})$ as automorphisms of $A[m^{\infty}]=\bigcup_{n\geqslant 1}A[m^{n}]$. We then classify elements $M\in\mathcal{G}({m^{\infty}})$ according to the group structure of $\ker(M-I)$ and according to the projection $\pi_{m}(M)\in\mathcal{G}({m})$. Note that if $\ker(M-I)$ is finite, then it is a product over the primes $\ell\mid m$ of finite abelian $\ell$-groups that have at most $b_{A}$ cyclic components.

For every group $\mathrm{F}$ of the form $\prod_{\ell}\mathrm{F}_{\ell}$, where $\mathrm{F}_{\ell}$ is a finite abelian $\ell$-group with at most $b_{A}$ cyclic components, we define the sets

[TABLE]

and

[TABLE]

We denote by $\mathcal{M}_{\mathrm{F}}(*)$ and $\mathcal{M}_{x,\mathrm{F}}(*)$, respectively, the images of these sets under the reduction map $\mathcal{G}({m^{\infty}})\to\mathcal{G}(*)$. We also write

[TABLE]

the union being taken over all $x\in\mathcal{G}(m)$ and over all groups $\mathrm{F}=\prod_{\ell}\mathrm{F}_{\ell}$ as above, up to isomorphism.

Proposition 14**.**

The following holds:

1. (1)

The sets $\mathcal{M}_{x,\mathrm{F}}$ are measurable in $\mathcal{G}({m^{\infty}})$. 2. (2)

If $n>v_{\ell}(\exp\mathrm{F})$ for all $\ell\mid m$, then we have

[TABLE] 3. (3)

We have $\mu_{\mathcal{G}(m^{\infty})}(\mathcal{M}_{x,\mathrm{F}})=0$ if and only if $\mathcal{M}_{x,\mathrm{F}}=\emptyset$. 4. (4)

The set $\mathcal{M}$ of (21) is measurable in $\mathcal{G}({m^{\infty}})$. 5. (5)

If $(A/K,m)$ satisfies eventual maximal growth of the torsion fields, then we have

[TABLE]

Proof.

This is proved as in [4, Lemma 21]. ∎

5. The density as an integral

By Theorem 7, computing $\operatorname{Dens}_{m}(\alpha)$ comes down to computing the Haar measure of $S_{\alpha}$ in $\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)$. The generalisation of [4, Remark 17] to the composite case gives

[TABLE]

In view of (21), we consider the sets

[TABLE]

By Proposition 14, the set $S_{\alpha}$ is the disjoint union of the sets $S_{x,\mathrm{F}}$ up to a set of measure [math]. To see that the Haar measure of $S_{x,\mathrm{F}}$ is well defined and to compute it, we define for every $n\geqslant 1$ the set

[TABLE]

Proposition 15**.**

Suppose $n>n_{0}$ and $n>\max_{\ell}\{v_{\ell}(\exp\mathrm{F})\}$ for every $\ell$, where $n_{0}$ is as in Definition 3. Then the set $S_{x,\mathrm{F},m^{n}}$ is the image of $S_{x,\mathrm{F}}$ under the projection to $\operatorname{Gal}(K_{m^{-n}\alpha}/K)$.

Proof.

The set $S_{x,\mathrm{F},m^{n}}$ clearly contains the reduction modulo $m^{n}$ of $S_{x,\mathrm{F}}$. To prove the other inclusion, consider $(t_{m^{n}},M_{m^{n}})\in S_{x,\mathrm{F},m^{n}}$ and a lift $(t,M)\in\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)$. Since $n$ is sufficiently large with respect to $F$, we have $\ker(M-I)\cong F$. Clearly $M_{m^{n}}$ and $M$ have the same projection $x\in\mathcal{G}(m)$. To conclude, it suffices to ensure $t\in\operatorname{Im}(M-I)$. Take $\tau_{m^{n}}\in A[m^{n}]$ satisfying $(M_{m^{n}}-I)(\tau_{m^{n}})=t_{m^{n}}$, and some lift $\tau$ of $\tau_{m^{n}}$ to $T_{m}(A)$: we may replace $t$ by $(M-I)\tau$ because the difference is in $m^{n}T_{m}(A)$ and since $n>n_{0}$ we know that $\operatorname{Gal}(K_{m^{-\infty}\alpha}/K)$ contains $m^{n}T_{m}(A)\times\{I\}$. ∎

Theorem 16**.**

We have

[TABLE]

where $\mathrm{C}_{m}$ is the constant of (10) and $\mathrm{w}_{m^{\infty}}$ is as in (17).

Proof.

Choose $n$ large enough so that $n>n_{0}$ and $n>\max_{\ell}\{v_{\ell}(\exp\mathrm{F})\}$ for every $\ell$, where $n_{0}$ is as in Definition 3. By (11), we can write

[TABLE]

By (12), we can express the summand as

[TABLE]

so from (10) we deduce

[TABLE]

By (9) the left-hand side is a non-increasing function of $n$, and therefore it admits a limit for $n\rightarrow\infty$, which is $\mu(S_{x,\mathrm{F}})$. The right-hand side is an integral over $\mathcal{M}_{x,\mathrm{F}}(m^{n})$ with respect to the normalized counting measure of ${\mathcal{G}}({m^{n}})$, and the matrices in $\mathcal{M}_{x,\mathrm{F}}$ are exactly the matrices in $\mathcal{G}({m^{\infty}})$ whose reduction modulo $m^{n}$ lies in $\mathcal{M}_{x,\mathrm{F}}(m^{n})$. Taking the limit in $n$ we thus find the formula in the statement. ∎

Theorem 17**.**

We have

[TABLE]

where the function $\mathrm{w}_{m^{\infty}}$ is as in (17), and the constant $\mathrm{C}_{m}$ is as in (10).

Recalling the sets $\mathcal{M}_{\mathrm{F}}$ from (19), and that by Proposition 14 their union has measure $1$ in $\mathcal{G}(m^{\infty})$, we can rewrite (25) as

[TABLE]

where $F$ varies over the products over the primes $\ell\mid m$ of finite abelian $\ell$-groups with at most $b_{A}$ cyclic components.

Proof.

We have just shown that we may prove (26) instead. Notice that we have $\mathcal{M}_{\mathrm{F}}=\cup_{x}\mathcal{M}_{x,\mathrm{F}}$. By Theorem 7 we may write $\operatorname{Dens}_{m}(\alpha)=\mu(S_{\alpha})=\sum_{x,F}\mu(S_{x,\mathrm{F}})$ and then it suffices to apply (24). ∎

Corollary 18** ([4, Theorem 1 and Remark 25]).**

In the special case $m=\ell$ we have

[TABLE]

where $F$ varies among the finite abelian $\ell$-groups with at most $b_{A}$ cyclic components. We can also write

[TABLE]

Proof.

It suffices to specialize the notation of (25) and (26) to $m=\ell$. ∎

Notice that we have $\#\ker(M-I)=\ell^{v_{\ell}(\det(M-I))}$ for every $M\in\mathcal{G}(\ell^{\infty})$.

Corollary 19**.**

If the fields $K_{\ell^{-\infty}\alpha}$ are linearly disjoint over $K$ (this happens for example if $K_{m^{-1}}=K$, or more generally if the degree $[K_{\ell^{-1}}:K]$ is a power of $\ell$ for each $\ell$), then we have

[TABLE]

Proof.

Let $\ell$ vary among the prime divisors of $m$. Note that we have $\mathrm{C}_{m}=\prod_{\ell}\mathrm{C}_{\ell}$. By assumption, we also know $\mathcal{G}({m^{\infty}})=\prod_{\ell}\mathcal{G}({\ell^{\infty}})$ and $\mathrm{w}_{m^{\infty}}(M)=\prod_{\ell}\mathrm{w}_{\ell^{\infty}}(\pi_{\ell^{\infty}}M)$. We may then apply Proposition 2 and find that (25) is the product of the expressions (28). ∎

Theorem 20**.**

Write $\mathcal{G}({m^{\infty}})\subseteq\prod_{\ell}\mathcal{G}({\ell^{\infty}})$ as finite disjoint union of sets $H_{i}=\prod_{\ell}H_{i,\ell}$ where $H_{i,\ell}\in\mathcal{G}({\ell^{\infty}})$ are translates of closed subgroups of finite index. Suppose that for every $\ell$ and for every $n\geqslant 1$ we know $[K_{m^{-1}}K_{\ell^{-n}\alpha}:K_{m^{-1}}K_{\ell^{-n}}]=[K_{\ell^{-n}\alpha}:K_{\ell^{-n}}]$. Then we have

[TABLE]

Proof.

We can apply Proposition 2 because there is a product decomposition formula for the function $\mathrm{w}_{m^{\infty}}$. Namely the assumption on the Kummer extensions implies that for every $n\geqslant 1$ we have $\mathrm{w}_{m^{n}}(M)=\prod_{\ell}\mathrm{w}_{\ell^{n}}(\pi_{\ell^{n}}M)$ and hence we find $\mathrm{w}_{m^{\infty}}(M)=\prod_{\ell}\mathrm{w}_{\ell^{\infty}}(\pi_{\ell^{\infty}}(M))$. ∎

Theorem 21**.**

Denote by $H(x)=\prod_{\ell}H_{\ell}(x)$ the set of matrices in $\mathcal{G}({m^{\infty}})\subseteq\prod_{\ell}\mathcal{G}({\ell^{\infty}})$ mapping to $x$ in $\mathcal{G}({m})$. We then have

[TABLE]

where $\mathrm{w}_{x,\ell^{\infty}}$ is as in (16).

Proof.

Write $S_{x}=\bigcup_{\mathrm{F}}S_{x,\mathrm{F}}$ and recall from Proposition 14 that the set of matrices $M$ for which $\ker(M-I)$ is infinite has measure zero in $\mathcal{G}({m^{\infty}})$. By (24) we have

[TABLE]

By (18) the assertion follows from Theorem 1. ∎

Corollary 22**.**

The density $\operatorname{Dens}_{m}(\alpha)$ is a rational number. Moreover, fix $g\geqslant 1$. There exists a polynomial $p_{g}(t)\in\mathbb{Z}[t]$ with the following property: whenever $K$ is a number field and $A$ is the product of an abelian variety and a torus with $\dim(A)=g$, then for all $\alpha\in A(K)$ we have

[TABLE]

where $\ell$ varies over the prime divisors of $m$.

Proof.

Recall that $\mathrm{C}_{m}$ is an integer. In view of Lemma 12, we can consider each $\ell$-adic integral in (31) and proceed as in the proof of [4, Theorem 33]. ∎

6. Serre curves

6.1. Definition of Serre curves

Let $E$ be an elliptic curve over a number field $K$. We choose a Weierstrass equation for $E$ of the form

[TABLE]

where $x_{1},x_{2},x_{3}\in K(E[2])$ are the $x$-coordinates of the points of order $2$. The discriminant of the right-hand side of (33) is $\Delta=\sqrt{\Delta}^{2}$, where

[TABLE]

We thus have $K(\sqrt{\Delta})\subseteq K(E[2])$, and we define a character

[TABLE]

For any choice of basis of the $2$-torsion of $E$, we have the 2-torsion representation

[TABLE]

Let $\psi$ be the unique non-trivial character $\operatorname{GL}_{2}(\mathbb{Z}/2\mathbb{Z})\to\{\pm 1\}$; this corresponds to the sign character under any isomorphism of $\operatorname{GL}_{2}(\mathbb{Z}/2\mathbb{Z})$ with $S_{3}$. The character $\psi_{E}$ factors as

[TABLE]

From now on, we take $K=\mathbb{Q}$. All number fields that we will consider will be subfields of a fixed algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$.

Let $d$ be an element of $\mathbb{Q}^{\times}$. Let $m_{d}$ be the conductor of $\mathbb{Q}(\sqrt{d})$; this is the smallest positive integer such that $\sqrt{d}$ lies in the cyclotomic field $\mathbb{Q}(\zeta_{m_{d}})$. Let $d_{\mathrm{sf}}$ be the square-free part of $d$. We have

[TABLE]

We define a character

[TABLE]

If $\sigma$ is the automorphism of $\mathbb{Q}(\zeta_{m_{d}})$ defined by $\sigma(\zeta_{m_{d}})=\zeta_{m_{d}}^{a}$ with $a\in(\mathbb{Z}/m_{d}\mathbb{Z})^{\times}$, then $\varepsilon_{d}(\sigma)$ equals the Jacobi symbol \bigl{(}\frac{d_{\mathrm{sf}}}{a}\bigr{)}. We view $\varepsilon_{d}$ as a character of $\operatorname{GL}_{2}(\mathbb{Z}/m_{d}\mathbb{Z})$ by composing with the determinant.

For all $n\geqslant 1$, we have a canonical projection

[TABLE]

Fixing a $\widehat{\mathbb{Z}}$-basis for the projective limit of the torsion groups $E[n](\overline{\mathbb{Q}})$, we have a torsion representation

[TABLE]

The image of $\rho_{E}$ is contained in the subgroup

[TABLE]

of index $2$ in $\operatorname{GL}_{2}(\widehat{\mathbb{Z}})$. This expresses the fact that $\sqrt{\Delta}$ is contained in both $\mathbb{Q}(E[2])$ and $\mathbb{Q}(E[m_{\Delta}])$. An elliptic curve is said to be a Serre curve if the image of $\rho_{E}$ is equal to $H_{\Delta}$. As proven by N. Jones [1], almost all elliptic curves over $\mathbb{Q}$ are Serre curves.

6.2. Counting matrices

Let $\ell$ be a prime number. For all integers $a,b\geqslant 0$, we write $\mathcal{M}_{\ell}(a,b)$ for the set of matrices $M\in\operatorname{GL}_{2}(\mathbb{Z}_{\ell})$ such that the kernel of $M-I$ as an endomorphism of $(\mathbb{Q}_{\ell}/\mathbb{Z}_{\ell})^{2}$ is isomorphic to $\mathbb{Z}/\ell^{a}\mathbb{Z}\times\mathbb{Z}/\ell^{a+b}\mathbb{Z}$.

If $\mathcal{N}$ is a subset of $\mathcal{M}_{\ell}(a,b)$ that is the preimage in $\mathcal{M}_{\ell}(a,b)$ of its reduction modulo $\ell^{n}$, then we have

[TABLE]

by [3, Theorem 27] (where the number of lifts is independent of the matrix).

Proposition 23**.**

If $\mathcal{N}$ is a subset of $\mathcal{M}_{\ell}(a,b)$ that is the preimage in $\mathcal{M}_{\ell}(a,b)$ of its reduction modulo $\ell$, then we have

[TABLE]

Proof.

We are working with $\operatorname{GL}_{2}(\mathbb{Z}_{\ell})$, so we can apply [3, Proposition 33] (see also [3, Definition 19]). This gives the assertion for the set $\mathcal{M}_{\ell}(a,b)$; we can conclude because of (34). ∎

We now collect some results in the case $\ell=2$. From [3, Theorem 2] we know

[TABLE]

We consider the action of $\operatorname{GL}_{2}(\mathbb{Z}/2^{3}\mathbb{Z})$ on $\mathbb{Q}(\zeta_{2^{3}})$ defined by $M\zeta_{2^{3}}=\zeta_{2^{3}}^{\det M}$. The matrices $M\in\operatorname{GL}_{2}(\mathbb{Z}/2^{3}\mathbb{Z})$ that fix $\sqrt{-1}$ are those with $\det(M)=1,5$. The ones that fix $\sqrt{2}$ are those with $\det(M)=1,7$. The ones that fix $\sqrt{-2}$ are those with $\det(M)=1,3$.

For $a,b\in\{0,1,2,3\}$ and $z\in\{-1,2,-2\}$, we write $\mathcal{N}_{2}(a,b,z)$ for the set of matrices in $\mathcal{M}_{2}(a,b)$ that fix $\sqrt{z}$.

Lemma 24**.**

We have

[TABLE]

and

[TABLE]

Proof.

For $a,b\in\{0,1,2,3\}$ and $d\in(\mathbb{Z}/2^{3}\mathbb{Z})^{\times}$, let $h(a,b,d)$ be the number of matrices $M\in\operatorname{GL}_{2}(\mathbb{Z}/2^{3}\mathbb{Z})$ such that $\det(M)=d$ and $\ker(M-I)\cong\mathbb{Z}/2^{a}\mathbb{Z}\times\mathbb{Z}/2^{a+b}\mathbb{Z}$. Using [9] one can easily count these matrices:

• •

$h(0,0,d)=128$, $h(0,1,d)=96$ and $h(0,2,d)=h(0,3,d)=48$ for all $d$;

• •

$h(1,0,d)=32$ for $d=1,5$ and $h(1,0,d)=16$ for $d=3,7$;

• •

for $b=1,2$ we have $h(1,b,d)=12$ for $d=1,5$ and $h(1,b,d)=24$ for $d=3,7$;

• •

$h(2,0,1)=4$, $h(2,0,5)=2$ and $h(2,0,d)=0$ for $d=3,7$;

• •

$h(2,1,1)=3$, $h(2,1,5)=6$ and $h(2,1,d)=0$ for $d=3,7$;

• •

$h(3,0,1)=1$ (the identity matrix) and $h(3,0,d)=0$ for $d=3,5,7$.

This classification and (34) lead to the measures in the statement. ∎

Lemma 25**.**

For all $a,b\geq 0$ and all $M\in\mathcal{M}_{2}(a,b)$, we have

[TABLE]

Proof.

Consider matrices $M\in\operatorname{GL}_{2}(\mathbb{Z}/2\mathbb{Z})$. The matrices

[TABLE]

satisfy $\psi(M)=1$ and $\dim_{\mathbb{F}_{2}}\ker(M-I)\in\{0,2\}$. The matrices

[TABLE]

satisfy $\psi(M)=-1$ and $\dim_{\mathbb{F}_{2}}\ker(M-I)=1$. This implies the claim. ∎

Now let $\ell$ be an odd prime number, and suppose that $M\in\operatorname{GL}_{2}(\mathbb{Z}/\ell\mathbb{Z})$ is not the identity. We compare the two conditions $\varepsilon_{\ell}(M)=1$ and $\ell\mid\det(M-I)$. Note that the characteristic polynomials of $M$ and $M-I$ have roots differing only by a shift by $1$. Write $\chi(M)=x^{2}+\alpha x+\beta$ for the characteristic polynomial of $M$. The condition $\varepsilon_{\ell}(M)=1$ is a condition on $\det(M)=\beta$. More precisely, we want $\beta$ to be a square modulo $\ell$, and in this case we write $\beta=s^{2}$. On the other hand we would like to know whether $\ell\mid\det(M-I)=\chi(1)$, which means $1+\alpha+\beta=0$ (we are working modulo $\ell$). Thus $\alpha=-(1+s^{2})$. We then have

[TABLE]

Lemma 26**.**

Let $M$ vary in $\operatorname{GL}_{2}(\mathbb{Z}/\ell\mathbb{Z})\setminus\{I\}$, where $\ell$ is an odd prime number.

1. (1)

There are $\frac{1}{2}(\ell+1)^{2}(\ell-2)$ matrices $M$ satisfying $\varepsilon_{\ell}(M)=1$ and $\ell\mid\det(M-I)$. 2. (2)

There are $\frac{1}{2}\ell(\ell^{3}-2\ell^{2}-\ell+4)$ matrices $M$ satisfying $\varepsilon_{\ell}(M)=1$ and $\ell\nmid\det(M-I)$. 3. (3)

There are $\frac{1}{2}\ell(\ell^{2}-1)$ matrices $M$ satisfying $\varepsilon_{\ell}(M)=-1$ and $\ell\mid\det(M-I)$. 4. (4)

There are $\frac{1}{2}\ell(\ell^{2}-1)(\ell-2)$ matrices $M$ satisfying $\varepsilon_{\ell}(M)=-1$ and $\ell\nmid\det(M-I)$.

Proof.

1. (1)

In (38), if $\beta=s^{2}\neq 1$ (thus there are $\frac{\ell-1}{2}-1$ possibilities for $\beta$), then the corresponding matrices are diagonal and hence we only have to choose the two distinct eigenspaces; this gives $(\ell+1)\ell$ matrices for every characteristic polynomial. If $\beta=1$, then we have the identity (which we are excluding) and the $\ell^{2}-1$ matrices conjugate to $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}$. Note that (1) can also be obtained from [7, Table 1]. 2. (2)

There are $\frac{1}{2}\#\operatorname{GL}_{2}(\mathbb{Z}/\ell\mathbb{Z})$ matrices satisfying $\varepsilon_{\ell}=1$, and we only need to subtract the identity and the matrices from (1). 3. (3)

There are $\ell^{3}-2\ell$ matrices in $\operatorname{GL}_{2}(\mathbb{Z}/\ell\mathbb{Z})$ having $1$ as an eigenvalue (see for example [3, Proof of Theorem 2]), and we only need to subtract the identity and the matrices from (1). 4. (4)

There are $\frac{1}{2}\#\operatorname{GL}_{2}(\mathbb{Z}/\ell\mathbb{Z})$ matrices satisfying $\varepsilon_{\ell}=-1$, and we only need to subtract the matrices from (3). Alternatively, there are $\#\operatorname{GL}_{2}(\mathbb{Z}/\ell\mathbb{Z})-(\ell^{3}-2\ell)$ matrices that do not have $1$ as eigenvalue, and we only need to subtract the matrices from (2).

∎

6.3. Partitioning the image of the $m$-adic representation

Let $E$ be a Serre curve over $\mathbb{Q}$. Let $\Delta$ be the minimal discriminant of $E$, and let $\Delta_{\mathrm{sf}}$ be its square-free part. We write $\Delta_{\mathrm{sf}}=zu$, where $z\in\{1,-1,2,-2\}$ and where $u$ is an odd fundamental discriminant. Then $|u|$ is the odd part of $m_{\Delta}$, and we have $\varepsilon_{\Delta}=\varepsilon_{z}\cdot\varepsilon_{u}$ as characters of $(\mathbb{Z}/m_{\Delta}\mathbb{Z})^{\times}$.

Now let $m$ be a square-free positive integer. If $m=2$, or if $m$ is odd, or if $u$ does not divide $m$, then we have

[TABLE]

If $m\neq 2$ is even and $u$ divides $m$, then $\mathcal{G}(m^{\infty})$ has index $2$ in $\prod_{\ell}\mathcal{G}({\ell}^{\infty})$. The defining condition for the image of the $m$-adic representation is then $\psi=\varepsilon_{\Delta}$, or equivalently

[TABLE]

We may then partition $\mathcal{G}({m}^{\infty})\subseteq\prod_{\ell\mid m}\mathcal{G}({\ell}^{\infty})$ into two sets that are products, namely

[TABLE]

and

[TABLE]

The set $\mathcal{G}(|u|^{\infty})\cap\{\varepsilon_{u}=1\}$ is the disjoint union of sets of the form $\prod_{\ell\mid u}(\mathcal{G}({\ell}^{\infty})\cap\{\varepsilon_{\ell}=\pm 1\})$, choosing an even number of minus signs; for the set $\mathcal{G}(|u|^{\infty})\cap\{\varepsilon_{u}=-1\}$ we have to choose an odd number of minus signs. Since each $\ell\mid u$ is odd, the two sets $\mathcal{G}({\ell}^{\infty})\cap\{\varepsilon_{\ell}=\pm 1\}$ can be investigated with the help of Lemma 26. Finally, the two sets $\mathcal{G}({2}^{\infty})\cap\{\psi\cdot\varepsilon_{z}=\pm 1\}$ can be investigated using Lemmas 24 and 25.

7. Examples

7.1. Example (non-surjective mod 3 representation)

Consider the non-CM elliptic curve

[TABLE]

of discriminant $-3^{19}\cdot 17$ and conductor $153=3^{2}\cdot 17$ over $\mathbb{Q}$ [8, label 153.b2]. The group $E(\mathbb{Q})$ is infinite cyclic and is generated by the point

[TABLE]

The image of the $3$-adic representation is the inverse image of its reduction modulo $3$, the image of the mod $3$ representation is isomorphic to the symmetric group of order $6$, and the $3$-adic Kummer map is surjective [4, Example 6.4]. The image of the mod $3$ representation has a unique subgroup of index $2$, so the field $\mathbb{Q}(E[3])$ contains as its only quadratic subextension the cyclotomic field $\mathbb{Q}(\sqrt{-3})$.

The image of the $2$-adic representation is $\operatorname{GL}_{2}(\mathbb{Z}_{2})$; see [8]. By [2, Theorem 5.2] the $2$-adic Kummer map is surjective: the assumptions of that result are satisfied because the prime $p=941$ splits completely in $E[4]$ but the point $(\alpha\bmod p)$ is not $2$-divisible over $\mathbb{F}_{p}$. Since the image of the mod $2$ representation has a unique subgroup of index $2$, the field $\mathbb{Q}(E[2])$ contains as its only quadratic subextension the field $\mathbb{Q}(\sqrt{-51})$ (the square-free part of the discriminant of $E$ is $-51$).

We have $\mathbb{Q}(E[2])\cap\mathbb{Q}(E[9])=\mathbb{Q}$ because the residual degree modulo $22699$ of the extension $\mathbb{Q}(E[2],E[9])/\mathbb{Q}(E[9])$ is divisible by $3$ and the degree of this extension is even because $\mathbb{Q}(\sqrt{-51})$ is not contained in $\mathbb{Q}(E[3])$. We deduce $\mathbb{Q}(E[2])\cap\mathbb{Q}(E[3^{\infty}])=\mathbb{Q}$ by applying [4, Theorem 12 (i)] (where $K=\mathbb{Q}(E[2])$).

Moreover, we have $\mathbb{Q}(E[3])\cap\mathbb{Q}(E[4])=\mathbb{Q}$ because $\mathbb{Q}(\sqrt{-3})$ is not contained in $\mathbb{Q}(E[4])$: the prime $941$ is not congruent to $1$ modulo $3$ and splits completely in $\mathbb{Q}(E[4])$. By [4, Theorem 12 (i)] we conclude that $\mathbb{Q}(E[3])\cap\mathbb{Q}(E[2^{\infty}])=\mathbb{Q}$.

The $2$-adic Kummer extensions of $\alpha$ have maximal degree also over $\mathbb{Q}(E[3])$, in view of the maximality of the $2$-Kummer extension, because the prime $4349$ splits completely in $\mathbb{Q}(2^{-2}\alpha)$ but not in $\mathbb{Q}(\sqrt{-3})$; see [4, Theorem 12 (ii)] (where $K=\mathbb{Q}(\sqrt{-3})$).

The $3$-adic Kummer extensions of $\alpha$ have maximal degree also over $\mathbb{Q}(E[2])$ because the prime $217981$ splits completely in $\mathbb{Q}(3^{-2}\alpha)$ but $3$ divides the residual degree of $\mathbb{Q}(E[2])$; see [4, Theorem 12 (ii)] (where $K=\mathbb{Q}(E[2])$).

We thus have $\mathcal{G}({6}^{\infty})=\mathcal{G}({2}^{\infty})\times\mathcal{G}({3}^{\infty})$, the $2^{\infty}$ Kummer extensions are independent from $\mathbb{Q}(E[3])$, and the $3^{\infty}$ Kummer extensions are independent from $\mathbb{Q}(E[2])$. We are thus in the situation that the fields $\mathbb{Q}(2^{-\infty}\alpha)$ and $\mathbb{Q}(3^{-\infty}\alpha)$ are linearly disjoint over $\mathbb{Q}$. We deduce that the equality

[TABLE]

holds for $\alpha$ and for its multiples. The $2$-densities can be evaluated by [4, Theorem 32], for the $3$-densities see [4, Example 6.4]. By testing the primes up to $10^{6}$, we have computed an approximation to $\operatorname{Dens}_{6}(\alpha)$ using [9].

[TABLE]

7.2. The Serre curve $y^{2}+y=x^{3}+x^{2}$

The elliptic curve

[TABLE]

of discriminant $-43$ and conductor 43 over $\mathbb{Q}$ [8, label 43.a1] is a Serre curve [6, Example 5.5.7]. The group $E(\mathbb{Q})$ is infinite cyclic and is generated by the point

[TABLE]

A computation shows that $\alpha$ is not divisible by $2$ over the 4-torsion field of $E$. Therefore, by [2, Theorem 5.2], for every prime number $\ell$ and for every $n\geqslant 1$ the degree of the $\ell^{n}$-Kummer extension is maximal, i.e.

[TABLE]

The $43$-adic Kummer extensions have maximal degree also over $\mathbb{Q}(E[2])$, i.e.

[TABLE]

because the degree $[\mathbb{Q}(E[2]):\mathbb{Q}]=6$ is coprime to $43$.

The extensions $\mathbb{Q}({2^{-1}\alpha})$ and $\mathbb{Q}(E[2\cdot 43])$ are linearly disjoint over $\mathbb{Q}(E[2])$, as can be seen by investigating the residual degree for the reduction modulo the prime $29327$, which splits completely in $\mathbb{Q}(E[2])$. Indeed, the residual degree of the extension $\mathbb{Q}({2^{-1}\alpha})$ equals $4$ while the residual degree of the extension $\mathbb{Q}(E[2\cdot 43])$ is odd because the prime is congruent to $1$ modulo $43$, and there are points of order $43$ in the reductions (the subgroup of the upper unitriangular matrices in $\operatorname{GL}_{2}(\mathbb{Z}/43\mathbb{Z})$ has order $43$).

The $2$-adic Kummer extensions have maximal degree also over $\mathbb{Q}(E[43])$, i.e.

[TABLE]

To see this, we consider the intersection $L$ of $\mathbb{Q}_{2^{-n}\alpha}$ and $\mathbb{Q}(E[43])$. This is a Galois extension of $\mathbb{Q}$, and the group $G=\operatorname{Gal}(L/\mathbb{Q})$ is a quotient of both $(\mathbb{Z}/2^{n}\mathbb{Z})^{2}\rtimes\operatorname{GL}_{2}(\mathbb{Z}/2^{n}\mathbb{Z})$ and $\operatorname{GL}_{2}(\mathbb{Z}/43\mathbb{Z})$. Because $\operatorname{SL}_{2}(\mathbb{Z}/43\mathbb{Z})$ has no non-trivial quotient that can be embedded into a quotient of $(\mathbb{Z}/2^{n}\mathbb{Z})^{2}\rtimes\operatorname{GL}_{2}(\mathbb{Z}/2^{n}\mathbb{Z})$, the quotient map $\operatorname{GL}_{2}(\mathbb{Z}/43\mathbb{Z})\to G$ factors as

[TABLE]

This implies that $L$ is a subfield of $\mathbb{Q}(\zeta_{43})$. Furthermore, $L$ contains $\mathbb{Q}(\sqrt{-43})$. Using basic Galois theory, we see that the maximal subfield of $\mathbb{Q}(\zeta_{43})$ that can be embedded into $\mathbb{Q}_{2^{-n}\alpha}$ is $\mathbb{Q}(\sqrt{-43})$, and we conclude that $L$ equals $\mathbb{Q}(\sqrt{-43})$.

It follows that for $m=2\cdot 43$ we have the maximal degree $[\mathbb{Q}_{m^{-n}\alpha}:\mathbb{Q}_{m^{-n}}]=m^{2n}$ and, more generally, that for every multiple $P$ of $\alpha$ we have $[\mathbb{Q}_{m^{-n}P}:\mathbb{Q}_{m^{-n}}]=[\mathbb{Q}_{2^{-n}P}:\mathbb{Q}_{2^{-n}}]\cdot[\mathbb{Q}_{43^{-n}P}:\mathbb{Q}_{43^{-n}}]$. We may then apply [4, Example 26] and various results in this paper to compute the following exact densities, and we use [9] to numerically verify them for the primes up to $10^{6}$:

[TABLE]

We conclude by sketching the computations for the point $\alpha$. The $43$-adic representation is surjective and the $43$-Kummer extensions have maximal degree. By parts (3) and (4) of Lemma 26, we find that $\frac{1}{2\cdot 42}$ (respectively, $\frac{41}{2\cdot 42}$) is the counting measure in $\operatorname{GL}_{2}(\mathbb{Z}/43\mathbb{Z})$ of the matrices such that $\varepsilon_{43}=-1$ and that are in $(\mathcal{M}_{43}(0,b)\bmod\ell)$ for some $b>0$ (respectively, for $b=0$). By multiplying this quantity by $43^{-b}\cdot 42$ we obtain by Proposition 23 that $\mu_{\operatorname{GL}_{2}(\mathbb{Z}_{43})}(\mathcal{M}_{43}(0,b))=\frac{1}{2}43^{-b}$ for $b>0$. By [4, Example 26] the contribution to $\operatorname{Dens}_{43}$ coming from the matrices in $\mathcal{G}(43^{\infty})$ such that $\varepsilon_{43}=-1$ is then

[TABLE]

From [4, Theorem 32] we know that $\operatorname{Dens}_{43}(\alpha)=143510179/146927088$, and hence the contribution to $\operatorname{Dens}_{43}(\alpha)$ coming from the matrices in $\mathcal{G}(43^{\infty})$ such that $\varepsilon_{43}=+1$ equals

[TABLE]

Now we work with the $2$-adic representation, which is surjective, and restrict to counting the contribution to $\operatorname{Dens}_{2}(\alpha)$ coming from the matrices satisfying $\psi=-1$. In view of Lemma 25 and Proposition 23, we find $\mu_{\operatorname{GL}_{2}(\mathbb{Z}_{2})}(\mathcal{M}_{2}(0,b))=1/2\cdot 2^{-b}$ for $b>0$. By [4, Example 26] the contribution to $\operatorname{Dens}_{2}(\alpha)$ coming from the matrices in $\mathcal{G}(2^{\infty})$ such that $\psi=-1$ is therefore

[TABLE]

From [4, Theorem 32] we know that $\operatorname{Dens}_{2}(\alpha)=11/21$, and hence the contribution to $\operatorname{Dens}_{2}$ coming from the matrices in $\mathcal{G}(2^{\infty})$ such that $\psi=1$ is

[TABLE]

Finally by the partition in Section 6.3 we can compute the requested density as the combination of the above quantities:

[TABLE]

The factor $2$ accounts for the fact that the integration domain (equipped with its normalised measure) is a subgroup of index $2$ of $\operatorname{GL}_{2}(\mathbb{Z}_{2})\times\operatorname{GL}_{2}(\mathbb{Z}_{43})$, while we compute $\operatorname{Dens}_{2}(\alpha)$ and $\operatorname{Dens}_{43}(\alpha)$ in $\operatorname{GL}_{2}(\mathbb{Z}_{2})$ and $\operatorname{GL}_{2}(\mathbb{Z}_{43})$, respectively.

For the point $2\alpha$, by [4, Example 26] we only need to scale (40) by a factor $2$, giving $1/3$ and $3/7$ as the two contributions to $\operatorname{Dens}_{2}(2\alpha)$ by [4, Theorem 32]. For the point $4\alpha$, we adapt (40) as $2\cdot{1}/{2}\cdot 2^{-2}+\sum_{b>1}4\cdot{1}/{2}\cdot 2^{-2b}$ and obtain $5/12$ and $13/28$ as the two contributions to $\operatorname{Dens}_{2}(4\alpha)$.