Introduction

The goal of this work is to complete the proof of an old conjecture of Geyer–Jarden in characteristic 0. The conjecture deals with a finitely generated field $K$ of $\mathbb{Q}$ . We fix an algebraic closure $\tilde{K}$ of $K$ . Then, the absolute Galois group $\text{Gal}(K)=\text{Gal}(\tilde{K}/K)$ of $K$ is a profinite group. It is equipped with a unique Haar measure $\unicode[STIX]{x1D707}_{K}$ with $\unicode[STIX]{x1D707}_{K}(\text{Gal}(K))=1$ [Reference Fried and JardenFrJ08, p. 378, Section 18.5]. For each positive integer $e\geqslant 1$ , the group $\text{Gal}(K)^{e}$ is equipped with the product measure, which we also denote by $\unicode[STIX]{x1D707}_{K}$ . We say that a certain statement holds for almost all $\boldsymbol{\unicode[STIX]{x1D70E}}\in \text{Gal}(K)^{e}$ if the set of $\boldsymbol{\unicode[STIX]{x1D70E}}\in \text{Gal}(K)^{e}$ for which that statement holds has $\unicode[STIX]{x1D707}_{K}$ -measure 1. For each $\boldsymbol{\unicode[STIX]{x1D70E}}=(\unicode[STIX]{x1D70E}_{1},\ldots ,\unicode[STIX]{x1D70E}_{e})\in \text{Gal}(K)^{e}$ , we consider the field

$$\begin{eqnarray}\tilde{K}(\boldsymbol{\unicode[STIX]{x1D70E}})=\{x\in \tilde{K}\mid \unicode[STIX]{x1D70E}_{i}x=x,\,i=1,\ldots ,e\}.\end{eqnarray}$$

Given an abelian variety $A$ over $K$ and a positive integer $m$ , we denote the kernel of the multiplication of $A$ by $m$ with $A_{m}$ . For a prime number $l$ , we write $A_{l^{\infty }}=\bigcup _{i=1}^{\infty }A_{l^{i}}$ .

B. Previous results. Conjecture A along with its analog to positive characteristics has been proved in [Reference Geyer and JardenGeJ78, p. 259, Theorem 1.1] when $A$ is an elliptic curve. The analog of the conjecture is true for an arbitrary abelian variety over a finite field [Reference Jacobson and JardenJaJ84, p. 114, Proposition 4.2]. Note that the latter paper contains a proof of Part (a) of Conjecture A and its analog to positive characteristic. Unfortunately, that proof is false as indicated in [Reference Jacobson and JardenJaJ85].

Part (c) of Conjecture A along with its analog to positive characteristic and Part (b) of the conjecture appear in [Reference Jacobson and JardenJaJ01, Main Theorem].

The main result of [Reference Geyer and JardenGeJ05] considers a nonzero abelian variety $A$ over a number field $K$ and says that there exists a finite Galois extension $L$ of $K$ such that for almost all $\unicode[STIX]{x1D70E}\in \text{Gal}(L)$ there exist infinitely many primes $l$ with $A_{l}(\tilde{K}(\unicode[STIX]{x1D70E}))\neq 0$ .

Finally, Zywina [Reference ZywinaZyw16] improves [Reference Geyer and JardenGeJ05] by proving Part (a) of Conjecture A for a number field $K$ not only for almost all $\unicode[STIX]{x1D70E}\in \text{Gal}(L)$ for some $L$ as [Reference Geyer and JardenGeJ05] does but for almost all $\unicode[STIX]{x1D70E}\in \text{Gal}(K)$ .

We generalize Zywina’s result to an arbitrary finitely generated extension $K$ of $\mathbb{Q}$ .

D. On the proof. Let $g=\dim (A)$ . For each prime number $l$ let

$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{A,l}:\text{Gal}(K)\rightarrow \text{GL}_{2g}(\mathbb{F}_{l})\end{eqnarray}$$

be the $l$ -ic representation (also called the mod- $l$ representation) of $\text{Gal}(K)$ induced by the action of $\text{Gal}(K)$ on the vector space $A_{l}$ over $\mathbb{F}_{l}$ of dimension  $2g$ .

D1. Serre’s theorem. The proof of [Reference Geyer and JardenGeJ05] uses the main result of [Reference SerreSer86]. That result deals with a number field $K$ . Among others, it gives a finite Galois extension $L$ of $K$ , a positive integer $n$ , and for each $l$ a connected reductive subgroup $H_{l}$ of $\text{GL}_{2g,\mathbb{F}_{l}}$ such that $(H_{l}(\mathbb{F}_{l}):\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L)))$ divides $n$ . In addition, the fields $L(A_{l})$ with $l$ ranging over all prime numbers are linearly disjoint over $L$ . Another important feature of Serre’s theorem is the existence of a set $\unicode[STIX]{x1D6EC}$ of prime numbers of positive Dirichlet density, such that $H_{l}$ splits over $\mathbb{F}_{l}$ for each $l\in \unicode[STIX]{x1D6EC}$ .

D2. Borel–Cantelli lemma. For each $l$ let

$$\begin{eqnarray}S_{l}=\{\unicode[STIX]{x1D70E}\in \text{Gal}(L)\mid \unicode[STIX]{x1D70C}_{A,l}(\unicode[STIX]{x1D70E})~\text{has eigenvalue}~1\}.\end{eqnarray}$$

Then, [Reference Geyer and JardenGeJ05] proves the existence of a positive constant $c$ and a set $\unicode[STIX]{x1D6EC}$ of positive Dirichlet density such that $\unicode[STIX]{x1D707}_{L}(S_{l})>c/l$ for each $l\in \unicode[STIX]{x1D6EC}$ . Thus, $\sum _{l\in \unicode[STIX]{x1D6EC}}\unicode[STIX]{x1D707}_{L}(S_{l})=\infty$ . In addition, by D1, the sets $S_{l}$ with $l$ ranging over $\unicode[STIX]{x1D6EC}$ are $\unicode[STIX]{x1D707}_{L}$ -independent. It follows from the Borel–Cantelli lemma, that almost all $\unicode[STIX]{x1D70E}\in \text{Gal}(L)$ lie in infinitely many $S_{l}$ ’s with $l\in \unicode[STIX]{x1D6EC}$ . Thus, for almost all $\unicode[STIX]{x1D70E}\in \text{Gal}(L)$ there exist infinitely many $l$ ’s such that $A_{l}(\tilde{K}(\unicode[STIX]{x1D70E}))\neq 0$ , which is the desired result over  $L$ .

D3. Zywina’s combinatorial approach. Zywina makes a more careful use of the Borel–Cantelli lemma. In [Reference ZywinaZyw16] he chooses a set $B$ of representatives of $\text{Gal}(K)$ modulo $\text{Gal}(L)$ . For each $l$ and every $\unicode[STIX]{x1D6FD}\in B$ he considers the set

$$\begin{eqnarray}U_{\unicode[STIX]{x1D6FD},l}=\{\unicode[STIX]{x1D70E}\in \unicode[STIX]{x1D6FD}\text{Gal}(L)\mid \unicode[STIX]{x1D70C}_{A,l}(\unicode[STIX]{x1D70E})~\text{has eigenvalue}~1\}.\end{eqnarray}$$

Then, he constructs a positive constant $c$ and a set $\unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FD}}$ of prime numbers having positive Dirichlet density such that

(1) $$\begin{eqnarray}\unicode[STIX]{x1D707}_{K}(U_{\unicode[STIX]{x1D6FD},l})\geqslant \frac{c}{l}\quad \text{for each}~l\in \unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FD}}.\end{eqnarray}$$

Again, by the Borel–Cantelli lemma, this leads to the conclusion that the $\unicode[STIX]{x1D707}_{K}$ -measure of the set $U_{\unicode[STIX]{x1D6FD}}$ of all $\unicode[STIX]{x1D70E}\in \text{Gal}(K)$ that belong to infinitely many $U_{\unicode[STIX]{x1D6FD},l}$ is $\frac{1}{[L:K]}$ . Since the $U_{\unicode[STIX]{x1D6FD}}$ ’s with $\unicode[STIX]{x1D6FD}\in B$ are disjoint, it follows that for almost all $\unicode[STIX]{x1D70E}\in \text{Gal}(K)$ there are infinitely many $l$ ’s such that $A_{l}(\tilde{K}(\unicode[STIX]{x1D70E}))\neq 0$ .

D4. Function Fields. Now assume that $K$ is a finitely generated extension of $\mathbb{Q}$ of positive transcendence degree and choose a subfield $E$ of $K$ such that $K/E$ is a regular extension of transcendence degree 1. We wish to find a place of $K/E$ with residue field $\overline{K}$ that induces a good reduction of $A$ onto an abelian variety $\overline{A}$ over $\overline{K}$ such that

(2) $$\begin{eqnarray}\text{Gal}(K(A_{l})/K)\cong \text{Gal}(\overline{K}(\overline{A}_{l})/\overline{K})\end{eqnarray}$$

for at least every $l$ in a set of positive Dirichlet density.

D5. Hilbert irreducibility theorem. The first idea that comes into mind is to use the Hilbert irreducibility theorem. However, that theorem can take care of only finitely many prime numbers, so it is of no use for our problem.

D6. Openness theorem. Instead, we choose a smooth curve $S$ over $E$ whose function field is $K$ such that $A$ has a good reduction along $S$ and set $\hat{K}=\prod _{l\in \mathbb{L}}K(A_{l})$ , where $\mathbb{L}$ is the set of all prime numbers. Using a combination of results of Anna Cadoret and Akio Tamagawa that goes under the heading “openness theorem” (Proposition 1.6), we find a closed point $\mathbf{s}$ of $S$ with an open decomposition group in $\text{Gal}(\hat{K}/K)$ . Let $\overline{K}_{\mathbf{s}}$ be the residue field of $K$ at $\mathbf{s}$ and $\hat{K}_{\mathbf{s}}=\prod _{l\in \mathbb{L}}\overline{K}_{\mathbf{s}}(A_{\mathbf{s},l})$ , where $A_{\mathbf{s}}$ is the reduction of $A$ at $\mathbf{s}$ . Then, there exists a finite extension $K^{\prime }$ of $K$ in $\hat{K}$ such that the reduction modulo $\mathbf{s}$ induces an isomorphism $\text{Gal}(\hat{K}/K^{\prime })\cong \text{Gal}(\hat{K}_{\mathbf{s}}/\overline{K}_{\mathbf{s}})$ . This gives the desired isomorphism (2) for $K^{\prime }$ rather than for $K$ and for all prime numbers  $l$ .

D7. Serre’s theorem over $K$ . Now we use a result of [Reference Gajda and PetersenGaP13] and find a finite Galois extension $L$ of $K$ that contains $K^{\prime }$ and satisfies the same reduction conditions that $K^{\prime }$ does and in addition the fields $L(A_{l})$ , with $l$ ranging over all prime numbers, are linearly disjoint over  $L$ .

Note that $\overline{K}_{\mathbf{s}}$ is again finitely generated over $\mathbb{Q}$ and the transcendence degree of $\overline{K}_{\mathbf{s}}$ over $\mathbb{Q}$ is one less than that of  $K$ . Starting with Serre’s theorem for number fields mentioned above and using induction on the transcendence degree over $\mathbb{Q}$ , we now prove the theorem of Serre mentioned in D1 over our current field  $K$ .

D8. Strongly regular points. Having Serre’s theorem for our function field $K$ at our disposal, we now follow the proof of [Reference ZywinaZyw16] to obtain the estimates (1) for our abelian variety $A/K$ . The proof contains a careful analysis of regular points of the reductive groups $H_{l}$ mentioned in Serre’s theorem for $l\in \unicode[STIX]{x1D6EC}$ . It uses Zywina’s crucial observation that if $T$ is an $\mathbb{F}_{l}$ -split maximal torus of $H_{l}$ and $\mathbf{t}\in T(\mathbb{F}_{l})$ , then $\mathbf{t}^{n!}\in \unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L))$ . Moreover, if $\mathbf{t}$ is a regular element of $H_{l}$ and $T$ is the unique maximal torus of $H_{l}$ that contains $\mathbf{t}$ , then the number of points $\mathbf{t}^{\prime }\in T(\mathbb{F}_{l})$ with $(\mathbf{t}^{\prime })^{n!}=\mathbf{t}^{n!}$ is at most $(n!)^{r}$ , where $r=\text{rank}(H_{l})=\dim (T)$ . Finally, still following [Reference ZywinaZyw16], we make use of the Lang–Weil estimates (or rather the more accurate version of these estimates that [Reference ZywinaZyw16] provides) to prove that “most of the points” of $\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(K))$ are regular points of $H_{l}$ whose characteristic polynomials have “maximal numbers of roots in $\mathbb{F}_{l}$ .” (We may refer to these points as “strongly regular.”)

D9. Serre’s density theorem. At some point of the proof, [Reference ZywinaZyw16] uses the Chebotarev density theorem for number fields to choose a prime of $K$ whose Artin class is equal to a previously chosen conjugacy class in $\text{Gal}(L(A_{l})/K)$ (where $L$ is the number field mentioned in Serre’s theorem for number field). Instead, we use Serre’s generalization of the Chebotarev density theorem (Proposition 3.5) to our function field $K$ in order to find a prime $\mathfrak{p}$ of $K$ with the same properties as above.

1 Adelic openness

Let $K$ be a finitely generated transcendental extension of $\mathbb{Q}$ , and let $A$ be an abelian variety over $K$ . We consider $K$ as a function field of one variable over a field $E$ . Using results of Cadoret and Tamagawa, we prove that there exists a finite extension $K^{\prime }$ of $K$ in $\hat{K}=\prod _{l}K(A_{l})$ , with $l$ ranging over all prime numbers, such that the reduction modulo “almost every valuation $v$ of $K^{\prime }$ over $E$ ” maps the group $\text{Gal}(K^{\prime }(A_{l})/K^{\prime })$ , for each $l$ , isomorphically onto the corresponding group with respect to the reduced objects.

To be more specific, let $E$ be a finitely generated extension of $\mathbb{Q}$ , $S$ an absolutely integral smooth curve over $E$ , $K=E(S)$ the function field of $S$ , and $A$ an abelian variety over $K$ of dimension $g>0$ with good reduction along $S$ [Reference ShimuraShi98, p. 95, Proposition 25]. Let $A(\tilde{K})$ be the abelian group of all $\tilde{K}$ -rational points of $A$ . For each $m\in \mathbb{N}$ , let $A_{m}$ be the kernel of multiplication of $A$ by $m$ . By [Reference Milne, Cornell and SilvermanMil85, p. 116, Remark 8.4], $A_{m}(\tilde{K})$ is a free $\mathbb{Z}/m\mathbb{Z}$ -module of rank $2g$ . Moreover, since $A$ is defined over $K$ , each $\unicode[STIX]{x1D70E}\in \text{Gal}(K)$ gives rise to an automorphism of $A(\tilde{K})$ that leaves $A_{m}(\tilde{K})$ invariant.

We denote the set of all prime numbers by $\mathbb{L}$ . For each $l\in \mathbb{L}$ , let $T_{l}(A)=\varprojlim A_{l^{i}}(\tilde{K})$ be the Tate module of $A$ associated with $l$ . Then, $A_{l}(\tilde{K})\cong \mathbb{F}_{l}^{2g}$ and $T_{l}(A)\cong \mathbb{Z}_{l}^{2g}$ , so $\text{Aut}(A_{l})\cong \text{GL}_{2g}(\mathbb{F}_{l})$ and $\text{Aut}(T_{l}(A))\cong \text{GL}_{2g}(\mathbb{Z}_{l})$ . Thus, the action of $\text{Gal}(K)$ on $A(\tilde{K})$ mentioned in the preceding paragraph gives rise to homomorphisms

(1) $$\begin{eqnarray}\unicode[STIX]{x1D70C}_{A,l}:\text{Gal}(K)\rightarrow \text{GL}_{2g}(\mathbb{F}_{l}),\qquad \unicode[STIX]{x1D70C}_{A,l^{\infty }}:\text{Gal}(K)\rightarrow \text{GL}_{2g}(\mathbb{Z}_{l}).\end{eqnarray}$$

Since $\text{Ker}(\unicode[STIX]{x1D70C}_{A,l})=\text{Gal}(K(A_{l}))$ and

$$\begin{eqnarray}\text{Ker}(\unicode[STIX]{x1D70C}_{A,l^{\infty }})=\text{Gal}(K(A_{l^{\infty }}))=\text{Gal}\biggl(\mathop{\bigcup }_{i=1}^{\infty }K(A_{l^{i}})\biggr),\end{eqnarray}$$

the homomorphism $\unicode[STIX]{x1D70C}_{A,l}$ (resp.  $\unicode[STIX]{x1D70C}_{A,l^{\infty }}$ ) (also called the $l$ -ic and the $l$ -adic representations of $\text{Gal}(K)$ ) induces (under an abuse of notation) a homomorphism $\unicode[STIX]{x1D70C}_{A,l}:\text{Gal}(N/K)\rightarrow \text{GL}_{2g}(\mathbb{F}_{l})$ (resp.  $\unicode[STIX]{x1D70C}_{A,l^{\infty }}:\text{Gal}(N/K)\rightarrow \text{GL}_{2g}(\mathbb{Z}_{l})$ ) for each Galois extension $N$ of $K$ that contains $K(A_{l})$ (resp.  $K(A_{l^{\infty }})$ ).

We denote the set of closed points of $S$ by $S_{\text{closed}}$ . By Hilbert Nullstellensatz, $S_{\text{closed}}$ is an infinite set.

Since $S$ is a smooth curve, each $\mathbf{s}\in S_{\text{closed}}$ induces a discrete valuation $v_{\mathbf{s}}$ of $K$ with residue field $\overline{K}_{\mathbf{s}}$ , which is a finite extension of $E$ in $\tilde{E}$ [Reference LangLan58, p. 151, Theorem 1] and where $\tilde{E}$ is the algebraic closure of $E$ in $\tilde{K}$ .

Let $K_{\text{ur}}=K_{\text{ur},S}$ be the maximal Galois extension of $K$ , which is unramified along $S$ , and observe that $\tilde{E}\subseteq K_{\text{ur}}$ , because $\text{char}(E)=0$ . Thus, $\text{Gal}(K_{\text{ur}}/K)$ is the étale fundamental group of $S$ . Since $\text{char}(\overline{K}_{\mathbf{s}})=0$ for each $\mathbf{s}\in S_{\text{closed}}$ , [Reference Serre and TateSeT68, Theorem 1] implies that

(2) $$\begin{eqnarray}K(A_{m})\subseteq K_{\text{ur}}~\text{for each}~m\in \mathbb{N}.\end{eqnarray}$$

By what we said above, $\unicode[STIX]{x1D70C}_{A,l}$ and $\unicode[STIX]{x1D70C}_{A,l^{\infty }}$ give rise to homomorphisms

$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{l}:\text{Gal}(K_{\text{ur}}/K)\rightarrow \text{Aut}(A_{l}),\qquad \unicode[STIX]{x1D70C}_{l^{\infty }}:\text{Gal}(K_{\text{ur}}/K)\rightarrow \text{Aut}(T_{l}(A)).\end{eqnarray}$$

Writing $\unicode[STIX]{x1D70B}_{l}:\text{Aut}(T_{l}(A))\rightarrow \text{Aut}(A_{l})$ for the epimorphism defined by the reduction $\text{GL}_{2g}(\mathbb{Z}_{l})\rightarrow \text{GL}_{2g}(\mathbb{F}_{l})$ modulo $l$ , we have $\unicode[STIX]{x1D70C}_{l}=\unicode[STIX]{x1D70B}_{l}\circ \unicode[STIX]{x1D70C}_{l^{\infty }}$ . Further, the products of the $\unicode[STIX]{x1D70C}_{l}$ ’s, the $\unicode[STIX]{x1D70C}_{l^{\infty }}$ ’s, and the $\unicode[STIX]{x1D70B}_{l}$ ’s, with $l$ ranging over $\mathbb{L}$ , give rise to homomorphisms that fit into the following commutative diagram:

(3)

Next we consider a point $\mathbf{s}\in S_{\text{closed}}$ and choose an extension $v_{\mathbf{s},\text{ur}}$ of $v_{\mathbf{s}}$ to $K_{\text{ur}}$ . Since $\tilde{E}\subseteq K_{\text{ur}}$ , the residue field of $v_{\mathbf{s},\text{ur}}$ is $\tilde{E}$ . For each Galois extension $L$ of $K$ in $K_{\text{ur}}$ we consider the decomposition group of $v_{\mathbf{s},\text{ur}}|_{L}$ over  $K$ ,

$$\begin{eqnarray}D_{\mathbf{s},L/K}=\{\unicode[STIX]{x1D70E}\in \text{Gal}(L/K)\mid ~\text{for all}~x\in L:v_{\mathbf{s},\text{ur}}(\unicode[STIX]{x1D70E}x)\geqslant 0\Leftrightarrow v_{\mathbf{s},\text{ur}}(x)\geqslant 0\}.\end{eqnarray}$$

Since $v_{s,\text{ur}}/v_{\mathbf{s}}$ is unramified, reduction modulo the prime ideal of the valuation ring of $v_{\mathbf{s},\text{ur}}$ gives rise to an isomorphism $\unicode[STIX]{x1D711}_{\mathbf{s}}:D_{\mathbf{s},K_{\text{ur}}/K}\rightarrow \text{Gal}(\overline{K}_{\mathbf{s}})$ [Reference Engler and PrestelEnP10, second paragraph of p. 123 and the “first exact sequence” on p. 124].

1. (4) Let $\unicode[STIX]{x1D713}_{\mathbf{s}}:\text{Gal}(\overline{K}_{\mathbf{s}})\rightarrow D_{\mathbf{s},K_{\text{ur}}/K}$ be the inverse of $\unicode[STIX]{x1D711}_{\mathbf{s}}$ . For each $l\in \mathbb{L}$ we consider the homomorphism $\unicode[STIX]{x1D70C}_{l^{\infty },\mathbf{s}}=\unicode[STIX]{x1D70C}_{l^{\infty }}\circ \unicode[STIX]{x1D713}_{\mathbf{s}}:\text{Gal}(\overline{K}_{\mathbf{s}})\rightarrow \text{Aut}(T_{l}(A))$ . It satisfies $\unicode[STIX]{x1D70C}_{l^{\infty },\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}}))=\unicode[STIX]{x1D70C}_{l^{\infty }}(D_{\mathbf{s},K_{\text{ur}}/K})$ . We also consider the homomorphisms

   $$\begin{eqnarray}\unicode[STIX]{x1D70C}_{s}=\unicode[STIX]{x1D70C}\circ \unicode[STIX]{x1D713}_{\mathbf{s}}:\text{Gal}(\overline{K}_{\mathbf{s}})\rightarrow \mathop{\prod }_{l\in \mathbb{L}}\text{Aut}(A_{l})\end{eqnarray}$$
   and
   $$\begin{eqnarray}\unicode[STIX]{x1D70C}_{\infty ,\mathbf{s}}=\unicode[STIX]{x1D70C}_{\infty }\circ \unicode[STIX]{x1D713}_{\mathbf{s}}:\text{Gal}(\overline{K}_{\mathbf{s}})\rightarrow \mathop{\prod }_{l\in \mathbb{L}}\text{Aut}(T_{l}(A)).\end{eqnarray}$$
   They satisfy
   $$\begin{eqnarray}\unicode[STIX]{x1D70C}_{s}(\text{Gal}(\overline{K}_{\mathbf{s}}))=\unicode[STIX]{x1D70C}(D_{\mathbf{s},K_{\text{ur}}/K})~\quad \text{and}\quad ~\unicode[STIX]{x1D70C}_{\infty ,\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}}))=\unicode[STIX]{x1D70C}_{\infty }(D_{\mathbf{s},K_{\text{ur}}/K}).\end{eqnarray}$$

The following result of Anna Cadoret is the main theorem of [Reference CadoretCad15], rewritten in our notation:

Our goal is to prove the assumption of Proposition 1.1, hence to make the consequence of that theorem valid. To this end, we combine two theorems of Cadoret and Tamagawa.

Proposition 1.2. (Cadoret–Tamagawa)

Given $l\in \mathbb{L}$ and $d\in \mathbb{N}$ , we set $S^{(d)}=\{\mathbf{s}\in S_{\text{closed}}\mid [\overline{K}_{\mathbf{s}}:E]\leqslant d\}$ and consider the set

$$\begin{eqnarray}S_{l}=\{\mathbf{s}\in S_{\text{closed}}\mid \unicode[STIX]{x1D70C}_{l^{\infty },\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}}))~\text{is not open in}~\unicode[STIX]{x1D70C}_{l^{\infty }}(\text{Gal}(K_{\text{ur}}/K))\}.\end{eqnarray}$$

Then, $S_{l}\cap S^{(d)}$ is finite.

Proof. Let $\mathbf{s}$ be a point in $S_{\text{closed}}$ that satisfies the conclusion of Corollary 1.3. Then, by (4), the commutative diagram (3) extends to a commutative diagram

(5)

In particular,

$$\begin{eqnarray}\unicode[STIX]{x1D70B}(\unicode[STIX]{x1D70C}_{\infty ,\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}})))=\unicode[STIX]{x1D70C}_{\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}}))\end{eqnarray}$$

and

$$\begin{eqnarray}\unicode[STIX]{x1D70B}(\unicode[STIX]{x1D70C}_{\infty }(\text{Gal}(K_{\text{ur}}/K)))=\unicode[STIX]{x1D70C}(\text{Gal}(K_{\text{ur}}/K)).\end{eqnarray}$$

By Corollary 1.3, $\unicode[STIX]{x1D70C}_{\infty ,\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}}))$ is open in $\unicode[STIX]{x1D70C}_{\infty }(\text{Gal}(K_{\text{ur}}/K))$ . By [Reference Fried and JardenFrJ08, p. 5], $\unicode[STIX]{x1D70B}$ is an open map. Therefore, $\unicode[STIX]{x1D70C}_{\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}}))$ is open in $\unicode[STIX]{x1D70C}(\text{Gal}(K_{\text{ur}}/K))$ .◻

Proof. Observe that $\unicode[STIX]{x1D70C}:\text{Gal}(K_{\text{ur}}/K)\rightarrow \prod _{l\in \mathbb{L}}\text{Aut}(A_{l})$ naturally decomposes as $\unicode[STIX]{x1D70C}=\hat{\unicode[STIX]{x1D70C}}\circ \text{res}_{K_{\text{ur}}/\hat{K}}$ , where $\hat{\unicode[STIX]{x1D70C}}:\text{Gal}(\hat{K}/K)\rightarrow \prod _{l\in \mathbb{L}}\text{Aut}(A_{l})$ is defined by the action of $\text{Gal}(\hat{K}/K)$ on the $A_{l}$ ’s. Since $\hat{K}=\prod _{l\in \mathbb{L}}K(A_{l})$ , the homomorphism $\hat{\unicode[STIX]{x1D70C}}$ is injective.

Similarly, we write $\unicode[STIX]{x1D70C}_{\mathbf{s}}^{\prime }:\text{Gal}(\hat{K}_{\mathbf{s}}/\overline{K}_{\mathbf{s}})\rightarrow \prod _{l\in \mathbb{L}}\text{Aut}(A_{\mathbf{s},l})$ for the corresponding monomorphism associated with $\overline{K}_{\mathbf{s}}$ and $A_{\mathbf{s}}$ . It fits into the following commutative diagram:

(6)

Note that in the notation of Corollary 1.4, $\unicode[STIX]{x1D70C}_{\mathbf{s}}^{\prime }\circ \text{res}_{\widetilde{\overline{K}_{\boldsymbol{ s}}}/\hat{K}_{\boldsymbol{ s}}}=\unicode[STIX]{x1D70C}_{\mathbf{s}}$ . We use Corollary 1.4 in order to choose $\mathbf{s}\in S(\tilde{E})$ such that the group $\unicode[STIX]{x1D70C}_{\mathbf{s}}(\text{Gal}(\overline{K}_{\mathbf{s}}))$ is open in $\unicode[STIX]{x1D70C}(\text{Gal}(K_{\text{ur}}/K))$ . Since both restriction maps in (6) are surjective, $\unicode[STIX]{x1D6FC}^{-1}(\unicode[STIX]{x1D70C}_{\mathbf{s}}^{\prime }(\text{Gal}(\hat{K}_{\mathbf{s}}/\overline{K}_{\mathbf{s}})))$ is open in $\hat{\unicode[STIX]{x1D70C}}(\text{Gal}(\hat{K}/K))$ . Since $\hat{\unicode[STIX]{x1D713}}_{\mathbf{s}}$ is injective, since $\unicode[STIX]{x1D6FC}$ is bijective, and since both $\unicode[STIX]{x1D70C}_{\mathbf{s}}^{\prime }$ and $\hat{\unicode[STIX]{x1D70C}}$ are injective, the group $D_{\mathbf{s},\hat{K}/K}=\hat{\unicode[STIX]{x1D713}}_{\mathbf{s}}(\text{Gal}(\hat{K}_{\mathbf{s}}/\overline{K}_{\mathbf{s}}))$ is open in $\text{Gal}(\hat{K}/K)$ . It follows that $K^{\prime }$ , which is the fixed field of $D_{\mathbf{s},\hat{K}/K}$ in $\hat{K}$ , is a finite extension of $K$ in $\hat{K}$ , as claimed.◻

2 Independent homomorphisms

Let $\unicode[STIX]{x1D6E4}$ be a profinite group, and let $I$ be a set. For each $i\in I$ let $\unicode[STIX]{x1D70C}_{i}$ be a homomorphism of $\unicode[STIX]{x1D6E4}$ into a profinite group $\unicode[STIX]{x1D6E4}_{i}$ . Here we follow the usual convention and always assume that a homomorphism between profinite groups is continuous. In addition, every finite group is equipped with the discrete topology. Let $\unicode[STIX]{x1D70C}=\prod _{i\in I}\unicode[STIX]{x1D70C}_{i}$ be the direct product of the $\unicode[STIX]{x1D70C}_{i}$ ’s. That is, $\unicode[STIX]{x1D70C}$ is the homomorphism from $\unicode[STIX]{x1D6E4}$ to $\prod _{i\in I}\unicode[STIX]{x1D6E4}_{i}$ defined by $\unicode[STIX]{x1D70C}(x)=(\unicode[STIX]{x1D70C}_{i}(x))_{i\in I}$ . Following [Reference SerreSer13] and [Reference Gajda and PetersenGaP13], we say that the family $(\unicode[STIX]{x1D70C}_{i})_{i\in I}$ is independent if $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4})=\prod _{i\in I}\unicode[STIX]{x1D70C}_{i}(\unicode[STIX]{x1D6E4})$ .

Note that if a family $(\unicode[STIX]{x1D70C}_{i})_{i\in I}$ of homomorphisms as in the preceding paragraph is independent and $\unicode[STIX]{x1D6FC}:\unicode[STIX]{x1D6E4}^{\prime }\rightarrow \unicode[STIX]{x1D6E4}$ is an epimorphism of profinite groups, then the family $(\unicode[STIX]{x1D70C}_{i}\circ \unicode[STIX]{x1D6FC})_{i\in I}$ is also independent.

One of the ingredients of the proof of the following lemma appears in [Reference Gajda and PetersenGaP13, Remark 3.2(b)(ii)].

Remark 2.3. Let $K$ be a field. For each $i\in I$ let $\unicode[STIX]{x1D70C}_{i}:\text{Gal}(K)\rightarrow G_{i}$ be a homomorphism of profinite groups, and let $K_{i}$ be the fixed field of $\text{Ker}(\unicode[STIX]{x1D70C}_{i})$ in $\tilde{K}$ . Consider a Galois extension $\hat{K}$ of $K$ in $\tilde{K}$ that contains each $K_{i}$ and let $\hat{\unicode[STIX]{x1D70C}}_{i}:\text{Gal}(\hat{K}/K)\rightarrow G_{i}$ be the homomorphism induced by $\unicode[STIX]{x1D70C}_{i}$ . As noticed in [Reference Gajda and PetersenGaP13, Remark 3.1], the family $(K_{i})_{i\in I}$ is linearly disjoint over $K$ if and only if the restriction maps

$$\begin{eqnarray}\hat{\unicode[STIX]{x1D70C}}_{i}:\text{Gal}(\hat{K}/K)\rightarrow G_{i},\quad i\in I\end{eqnarray}$$

are independent (see also [Reference Fried and JardenFrJ08, Lemma 2.5.6]).

3 Serre’s density theorem

We give in this section an account of a generalization of the Chebotarev density theorem to finitely generated extensions of $\mathbb{Q}$ due to Jean-Pierre Serre. We call this generalization “Serre’s density theorem.”

3.2 Regular rings

Let $K$ be a finitely generated extension of $\mathbb{Q}$ and let $L$ be a finite Galois extension of $K$ , and choose a transcendence base $(t_{1},\ldots ,t_{r})$ for $K/\mathbb{Q}$ . By Section 3.1, $R_{0}=\mathbb{Z}[t_{1},\ldots ,t_{r}]$ is a Nagata ring and the Krull dimension, $\dim (R_{0})$ , of $R_{0}$ is $r+1=\text{trans.deg}(K/\mathbb{Q})+1$ . Therefore, the integral closure $R$ of $R_{0}$ in $K$ is a finitely generated $R_{0}$ -module with $\text{Quot}(R)=K$ . Thus, $R=\mathbb{Z}[x_{1},\ldots ,x_{k}]$ for some $x_{1},\ldots ,x_{k}\in K$ and $R$ is a Nagata ring with $\dim (R)=\dim (R_{0})=\text{trans.deg}(K/\mathbb{Q})+1$ . The set

$$\begin{eqnarray}U=\{\mathfrak{p}\in \text{Spec}(R)\mid R_{\mathfrak{p}}~\text{is a regular ring}\}\end{eqnarray}$$

is open in $\text{Spec}(R)$ [Reference GrothendieckGro65, p. 166, Corollary 6.12.6]. Moreover, $U$ is nonempty, because it contains the generic point of $\text{Spec}(R)$ . Therefore, there exists a nonzero element $f\in R$ such that $\text{Spec}(R[f^{-1}])\subseteq U$ . In particular, the ring $R[f^{-1}]$ is regular. Adding $f^{-1}$ to the set $\{x_{1},\ldots ,x_{k}\}$ , if necessary, we may assume that $\text{Spec}(R)$ is smooth, so $R$ is a regular ring.

Since $R$ is a Nagata ring, its integral closure $R_{L}$ in $L$ is a finitely generated $R$ -module, hence a finitely generated ring extension of $\mathbb{Z}$ . Moreover, the fixed ring of $R_{L}$ under $\text{Gal}(L/K)$ is $R$ . Hence, $\text{Spec}(R)$ is isomorphic to the quotient scheme of $\text{Spec}(R_{L})$ modulo $\text{Gal}(L/K)$ , where $\text{Gal}(L/K)$ acts on $\text{Spec}(R_{L})$ in the natural way [Reference Görtz and WedhornGoW10, p. 331, Proposition 12.27]. Finally, we replace $R$ and $R_{L}$ by $R[u]$ and $R_{L}[u]$ , if necessary, where $u$ is an appropriate element of $K^{\times }$ , to assume that $R_{L}$ is a ring cover of $R$ in the terminology of [Reference Fried and JardenFrJ08, p. 109, Remark 6.1.5]. This means that $R_{L}=R[z]$ , where $\text{discr}(\text{irr}(z,K))$ is a unit of $R$ and $\text{irr}(z,K)$ is the monic irreducible polynomial of $z$ over $K$ . In particular, $R_{L}$ is standard étale over $R$ [Reference RaynaudRay70, p. 19, (2)].

3.3 Dirichlet density

We denote the set of maximal ideals of $R$ by $\text{Max}(R)$ . For each $\mathfrak{p}\in \text{Max}(R)$ , the residue ring $R/\mathfrak{p}$ is a finite field (see [Reference SerreSer65, p. 83, Section 1.3] or [Reference EisenbudEis95, p. 132, Theorem 4.19]) and we set $N\mathfrak{p}=|R/\mathfrak{p}|$ . Note that $\text{Max}(R)$ (resp.  $\text{Max}(R_{L})$ ) is the set of closed points of $\text{Spec}(R)$ (resp.  $\text{Spec}(R_{L})$ ).

This allows us to use the notation and the results of [Reference SerreSer65, Section 2.7] for $X=\text{Spec}(R_{L})$ , $G=\text{Gal}(L/K)$ , and $Y=\text{Spec}(R)$ . Let $d=\dim (Y)$ . Then, $d=\text{trans.deg}(K)+1$ . Accordingly, the Dirichlet density of a subset $B$ of $\text{Max}(R)$ is defined as the limit

(1) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}(B)=\lim _{s\rightarrow d^{+}}\frac{\mathop{\sum }_{\mathfrak{p}\in B}(1/N\mathfrak{p}^{s})}{\mathop{\sum }_{\mathfrak{p}\in \text{Max}(R)}(1/N\mathfrak{p}^{s})},\end{eqnarray}$$

if it exists. By [Reference SerreSer65, p. 84, Corollary 2], the denominator of the fraction in (1) diverges as $s\rightarrow d^{+}$ . Hence, $\unicode[STIX]{x1D6FF}(B)=0$ if $B$ is finite. In other words, if $\unicode[STIX]{x1D6FF}(B)>0$ , then $B$ is infinite.

4 Images of $l$ -ic representations

Let $A$ be an abelian variety over a number field $K$ . Using previous results of Faltings and Nori, Serre proved the existence of a finite Galois extension $L$ of $K$ with a great amount of information about the groups $\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L))$ . We use Proposition 1.6 to generalize Serre’s result to finitely generated extensions of $\mathbb{Q}$ , but limit our generalization only to properties we need in what follows.

Proof. First suppose that $K$ is a number field. By Serre, there exist positive integers $n_{0},r,l_{0}$ and for each $l\geqslant l_{0}$ there exists a connected reductive subgroup $H_{l}$ of $\text{GL}_{2g,\mathbb{F}_{l}}$ of rank $r$ such that (a) holds with $K_{0}=K$ , $A_{0}=A$ , $L_{0}=L$ , and $n=n_{0}$ (see [Reference ZywinaZyw16, Theorem 3.1] for the statement). A full account of statement (a) and the proof can be found in [Reference SerreSer86] (see also letters from Serre to M.-F. Vignéra [Reference SerreSer00, #137] and K. Ribet [Reference SerreSer00, #138]). Finally, the statement about the independence of the family $(\unicode[STIX]{x1D70C}_{A,l}|_{l\geqslant l_{0}})$ is proved in [Reference SerreSer13, Theorem 1].

Thus, (a) and (b) hold when $K$ is a number field.

Now assume that the transcendence degree of $K$ over $\mathbb{Q}$ is positive. In Section 1 and in particular in Setup 1.5 we have introduced the following objects: $E$ is a finitely generated extension of $\mathbb{Q}$ , $S$ is a smooth curve over $E$ whose function field is $K$ , $\mathbf{s}$ is a closed point of $S$ , $K_{\text{ur}}$ is the maximal unramified extension of $K$ along $S$ (it contains $K(A_{l})$ for each $l\in \mathbb{L}$ ), $\overline{K}_{\mathbf{s}}$ is the residue field of $K$ at $\mathbf{s}$ (it is a finite extension of $E$ with $\text{trans.deg}(\overline{K}_{\mathbf{s}}/\mathbb{Q})=\text{trans.deg}(K/\mathbb{Q})-1$ ), $A_{\mathbf{s}}$ is an abelian variety over $\overline{K}_{\mathbf{s}}$ of dimension $g$ , $K^{\prime }$ is a finite extension of $K$ (Proposition 1.6), $\hat{\unicode[STIX]{x1D713}}_{\mathbf{s}}:\text{Gal}(\hat{K}_{\mathbf{s}}/\overline{K}_{\mathbf{s}})\rightarrow \text{Gal}(\hat{K}/K^{\prime })$ is an isomorphism, and $\hat{\unicode[STIX]{x1D711}}_{\mathbf{s}}:\text{Gal}(\hat{K}/K^{\prime })\rightarrow \text{Gal}(\hat{K}_{\mathbf{s}}/\overline{K}_{\mathbf{s}})$ is the inverse of $\hat{\unicode[STIX]{x1D713}}_{\mathbf{s}}$ .

An induction hypothesis on the transcendence degree over $\mathbb{Q}$ applied to $\overline{K}_{\mathbf{s}}$ and $A_{\mathbf{s}}$ gives a number field $K_{0}$ , an abelian variety $A_{0}$ over $K_{0}$ , a finite Galois extension $L_{0}$ of $K_{0}$ and positive integers $n_{0},r,l_{0}$ such that

1. (1a) For each prime number $l\geqslant l_{0}$ there is a connected reductive subgroup $H_{l}$ of $\text{GL}_{2g,\mathbb{F}_{l}}$ such that $\unicode[STIX]{x1D70C}_{A_{0},l}(\text{Gal}(L_{0}))$ is a subgroup of $H_{l}(\mathbb{F}_{l})$ of index ${\leqslant}n_{0}$ . Moreover $H_{l}$ is of rank $r$ and contains the center $\mathbb{G}_{m}$ of $\text{GL}_{2g,\mathbb{F}_{l}}$ .

1. (1b) The family $(\unicode[STIX]{x1D70C}_{A_{0},l}|_{\text{Gal}(L_{0})})_{l\geqslant l_{0}}$ of homomorphisms is independent.

Moreover, there exists a finite Galois extension $L_{\mathbf{s}}$ of $\overline{K}_{\mathbf{s}}$ and a positive integral multiple $n_{\mathbf{s}}$ of $n_{0}$ with the following properties:

1. (2a) For all $l\geqslant l_{0}$ , $\unicode[STIX]{x1D70C}_{A_{0,l}}(\text{Gal}(L_{\mathbf{s}}))\leqslant H_{l}(\mathbb{F}_{l})$ , and $(H_{l}(\mathbb{F}_{l}):\unicode[STIX]{x1D70C}_{A_{\mathbf{s}},l}(\text{Gal}(L_{\mathbf{s}})))\leqslant n_{\mathbf{s}}$ .

1. (2b) The family $(\unicode[STIX]{x1D70C}_{A_{\mathbf{s}},l}|_{\text{Gal}(L_{\boldsymbol{ s}})})_{l\geqslant l_{0}}$ of homomorphisms is independent.

Let $L^{\prime }$ be the fixed field in $K_{\text{ur}}$ of $\unicode[STIX]{x1D713}_{\mathbf{s}}(\text{Gal}(L_{\mathbf{s}}))$ . Then, $L^{\prime }$ is a finite Galois extension of $K^{\prime }$ in $K_{\text{ur}}$ and, by the last statement of Setup 1.5, we have the following commutative diagram:

Since $\unicode[STIX]{x1D711}_{\mathbf{s}}$ maps $\text{Gal}(K_{\text{ur}}/L^{\prime })$ surjectively onto $\text{Gal}(L_{\mathbf{s}})$ , it follows from (2b) that the family $(\unicode[STIX]{x1D70C}_{A,l}|_{\text{Gal}(L^{\prime })})_{l\geqslant l_{0}}$ of homomorphisms is independent (Section 2, second paragraph).

Since $K^{\prime }$ is a finite extension of $K$ , so is $L^{\prime }$ . However, $L^{\prime }$ need not be Galois over $K$ . Nevertheless, by Lemma 2.2, $K$ has a finite Galois extension $L$ in $K_{\text{ur}}$ that contains $L^{\prime }$ such that the family $(\unicode[STIX]{x1D70C}_{A,l}|_{\text{Gal}(L)})_{l\geqslant l_{0}}$ is independent. Moreover, for each $l\geqslant l_{0}$ we have $\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L))\leqslant \unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L^{\prime }))\leqslant H_{l}(\mathbb{F}_{l})$ and, by (2a),

$$\begin{eqnarray}\displaystyle & & \displaystyle (H_{l}(\mathbb{F}_{l}):\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L)))\nonumber\\ \displaystyle & & \displaystyle \quad =(H_{l}(\mathbb{F}_{l}):\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L^{\prime })))\cdot (\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L^{\prime })):\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L)))\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (H_{l}(\mathbb{F}_{l}):\unicode[STIX]{x1D70C}_{A_{\mathbf{s}},l}(\text{Gal}(L_{\mathbf{s}})))\cdot [L:L^{\prime }]\leqslant n_{\mathbf{s}}[L:L^{\prime }].\nonumber\end{eqnarray}$$

Thus, $L$ satisfies Conditions (a) and (b) of the proposition with $n=n_{\mathbf{s}}[L:L^{\prime }]$ .◻

4.2 Tori in reductive groups

Let $H$ be a connected reductive group over a field $F$ and let $T$ be a maximal torus of $H$ over $F$ . Then, $T(\tilde{F})$ is isomorphic to $(\tilde{F}^{\times })^{r}$ for some positive integer $r$ , called the rank of $H$ [Reference SpringerSpr98, p. 117, Section 7.2.1]. In particular, $T$ is absolutely integral. By [Reference SpringerSpr98, p. 108, Proposition 6.4.2], all maximal tori of $H$ are conjugate, so the rank of $H$ is independent of  $T$ .

We say that $T$ $F$ -splits if $T$ is isomorphic over $F$ to the group $\mathbb{D}_{r}$ of diagonal matrices. Thus, in this case $T(F)\cong (F^{\times })^{r}$ . We say that $H$ $F$ -splits if $H$ has a maximal $F$ -split torus [Reference SpringerSpr98, p. 271, Section 16.2.1]. By [Reference SpringerSpr98, p. 256, Theorem 15.2.6], all $F$ -split maximal tori are conjugate by an element of $H(F)$ .

A result of Zywina gives additional information on the groups $H_{l}$ mentioned in Proposition 4.1.

Lemma 4.3. Let $A$ , $g$ , $K$ , $L$ , $n,r,l_{0}$ , and $H_{l}$ with $l\geqslant l_{0}$ , be as in Proposition 4.1. Then, there is a finite Galois extension $M$ of $\mathbb{Q}$ such that if $l\in \mathbb{L}$ splits completely in $M$ and is sufficiently large, then the following holds:

1. (a) The reductive group $H_{l}$ $\mathbb{F}_{l}$ -splits.

2. (b) Let $(X_{ij},Y)_{1\leqslant i,j\leqslant 2g}$ be independent variables. We identify $\text{GL}_{2g}(\mathbb{F}_{l})$ with the closed subvariety of $\text{Spec}(\mathbb{F}_{l}[X_{ij},Y]_{1\leqslant i,j\leqslant 2g})=\mathbb{A}_{\mathbb{F}_{l}}^{4g^{2}+1}$ defined by the equation $\det ((X_{ij})_{1\leqslant i,j\leqslant 2g})Y=1$ .

   Let $T$ be an $\mathbb{F}_{l}$ -split maximal torus of $H_{l}$ . Then, the torus $T$ , viewed as a closed subvariety of $\mathbb{A}_{\mathbb{F}_{l}}^{4g^{2}+1}$ , is defined by at most $c_{1}$ polynomials of degree at most $c_{2}$ , where $c_{1}$ and $c_{2}$ are constants that do not depend on  $l$ .

Proof. By Proposition 4.1, the subgroups $H_{l}$ satisfy Conditions (a) and (b) of that proposition with respect to an abelian variety $A_{0}$ of dimension $g$ defined over a number field $K_{0}$ and with respect to a finite Galois extension $L_{0}$ of $K_{0}$ . Therefore, our lemma follows from [Reference ZywinaZyw16, Lemma 3.2] (in which $A=A_{0}$ , $K=K_{0}$ , and $L=L_{0}$ ).◻

Another auxiliary tool that we quote from [Reference ZywinaZyw16] is the following variant of a theorem of Lang–Weil.

Proposition 4.4. [Reference ZywinaZyw16, Theorem 2.1]

Let $q$ be a power of a prime number and consider a Zariski-closed subset $V$ of $\mathbb{A}_{\mathbb{F}_{q}}^{k}$ with $k>1$ defined by the simultaneous vanishing of $s$ polynomials $f_{1},\ldots ,f_{s}$ in $\mathbb{F}_{q}[X_{1},\ldots ,X_{k}]$ , each of which is of degree at most $e$ . Let $V_{1},\ldots ,V_{m}$ be the irreducible components of $V_{\tilde{\mathbb{F}}_{q}}$ , which have the same dimension as  $V$ . Then,

( a ) $$\begin{eqnarray}|V(\mathbb{F}_{q})|\leqslant mq^{\dim (V)}+6(3+se)^{k+1}2^{s}q^{\dim (V)-\frac{1}{2}}.\end{eqnarray}$$

If all of the components $V_{1},\ldots ,V_{m}$ are defined over $\mathbb{F}_{q}$ , then

( b ) $$\begin{eqnarray}||V(\mathbb{F}_{q})|-mq^{\dim (V)}|\leqslant 6(3+se)^{k+1}2^{s}q^{\dim (V)-\frac{1}{2}}.\end{eqnarray}$$

In the rest of this section we bound the constant $m$ that appears in Proposition 4.4 in terms of the degrees of $f_{1},\ldots ,f_{s}$ .

Lemma 4.5. Let $F$ be an algebraically closed field, $Y$ an irreducible algebraic variety in $\mathbb{A}_{F}^{n}$ , $H$ a hypersurface in $\mathbb{A}_{F}^{n}$ , and $Z_{1},\ldots ,Z_{s}$ the irreducible components of $Y\cap H$ . Then, $\sum _{j=1}^{s}\deg (Z_{j})\leqslant \deg (Y)\deg (H)$ .

Proof. The degrees of $Y$ and $H$ do not change by taking the Zariski closures of these varieties in $\mathbb{P}_{F}^{n}$ . The number of the components of $Y\cap H$ may only increase. Hence, we may assume that $Y$ and $H$ are projective.

If $Y\subseteq H$ , then $Y=Y\cap H$ is the unique irreducible component of $Y\cap H$ and $\deg (Y)\leqslant \deg (Y)\deg (H)$ . If on the other hand $Y\not \subseteq H$ , then by [Reference HartshorneHar77, p. 53, Theorem 7.7],

(3) $$\begin{eqnarray}\mathop{\sum }_{j=1}^{s}i(Y,H;Z_{j})\deg (Z_{j})=\deg (Y)\deg (H).\end{eqnarray}$$

Since the intersection multiplicities $i(Y,H;Z_{j})$ are positive integers, the conclusion of the lemma follows from (3).◻

Lemma 4.6. Let $F$ be an algebraically closed field and let $f_{1},\ldots ,f_{k}\in F[X_{1},\ldots ,X_{n}]$ be nonzero polynomials. Let

$$\begin{eqnarray}V=V(f_{1},\ldots ,f_{k})=\text{Spec}(F[X_{1},\ldots ,X_{n}]/\mathop{\sum }_{i=1}^{n}F[X_{1},\ldots ,X_{n}]f_{i})\end{eqnarray}$$

be the algebraic variety in $\mathbb{A}_{F}^{n}$ defined by $f_{1},\ldots ,f_{k}$ and let $Z_{1},\ldots ,Z_{m}$ be the irreducible components of $V$ . Then, $m\leqslant \sum _{i=1}^{m}\deg (Z_{i})\leqslant \prod _{i=1}^{k}\deg (f_{i})$ .

Proof. Since $\deg (Z_{i})\geqslant 1$ for all $i$ , the left inequality is clear. We prove the right inequality.

First we consider the case where $k=1$ . Let $f_{1}=cg_{1}^{d_{1}}\cdots g_{m}^{d_{m}}$ be the decomposition of $f_{1}$ into a product of powers of irreducible polynomials in $F[X_{1},\ldots ,X_{n}]$ , no one of which is a product of the other with an element of $K^{\times }$ , and $c\in K^{\times }$ . Then, $V(g_{1}),\ldots ,V(g_{m})$ are the irreducible components of $V(f_{1})$ . By [Reference HartshorneHar77, p. 52, Proposition 7.6(d)], we have $\sum _{i=1}^{m}\deg (V(g_{i}))=\sum _{i=1}^{m}\deg (g_{i})\leqslant \sum _{i=1}^{m}d_{i}\deg (g_{i})=\deg (f_{1})$ .

Now we assume that $k\geqslant 2$ , set $V_{k-1}=V(f_{1},\ldots ,f_{k-1})$ , and let $W_{1},\ldots ,W_{m^{\prime }}$ be the irreducible components of $V_{k-1}$ . An induction assumption implies that

(4) $$\begin{eqnarray}\mathop{\sum }_{i=1}^{m^{\prime }}\deg (W_{i})\leqslant \deg (f_{1})\cdots \deg (f_{k-1}).\end{eqnarray}$$

For each $1\leqslant i\leqslant m^{\prime }$ let $Z_{i,1},\ldots ,Z_{i,m_{i}^{\prime }}$ be the irreducible components of $W_{i}\cap V(f_{k})$ . Then, the $Z_{ij}$ with $i=1,\ldots ,m^{\prime }$ and $j=1,\ldots ,m_{i}^{\prime }$ are the irreducible components of $V$ (eventually with repetitions). By Lemma 4.5, $\sum _{j=1}^{m_{i}^{\prime }}\deg (Z_{ij})\leqslant \deg (W_{i})\deg (f_{k})$ . It follows from (4) that

$$\begin{eqnarray}\mathop{\sum }_{i=1}^{m^{\prime }}\mathop{\sum }_{j=1}^{m_{i}^{\prime }}\deg (Z_{ij})\leqslant \mathop{\sum }_{i=1}^{m^{\prime }}\deg (W_{i})\deg (f_{k})\leqslant \deg (f_{1})\cdots \deg (f_{k-1})\deg (f_{k}),\end{eqnarray}$$

and this implies the desired inequality. ◻

5 Good reduction of abelian varieties

We generalize results of Serre and Tate in [Reference Serre and TateSeT68] about good reduction of abelian schemes over discrete valuation rings to results about good reduction of abelian schemes over more general integral domains.

5.4 Reduction modulo $\mathfrak{p}$

We consider a prime ideal $\mathfrak{p}\in \text{Spec}(R)$ and compose each $\unicode[STIX]{x1D6FD}\in \text{Hom}_{R}(B,R)$ with the quotient map $R\rightarrow R/\mathfrak{p}$ followed by the inclusion $R/\mathfrak{p}\rightarrow \overline{K}_{\mathfrak{p}}$ to get a homomorphism $\unicode[STIX]{x1D6FD}_{\mathfrak{p}}$ as in the following commutative diagram

The map $\unicode[STIX]{x1D6FD}\mapsto \unicode[STIX]{x1D6FD}_{\mathfrak{p}}$ gives rise to a reduction map modulo $\mathfrak{p}$ :

(1) $$\begin{eqnarray}\text{Hom}_{R}(B,R)\rightarrow \text{Hom}_{R}(B,\overline{K}_{\mathfrak{p}}).\end{eqnarray}$$

There is a natural bijection $\text{Hom}_{R}(B,R)\rightarrow \text{Mor}_{R}(\text{Spec}(R),\text{Spec}(B))$ that maps each $\unicode[STIX]{x1D6FD}\in \text{Hom}_{R}(B,R)$ onto the $R$ -morphism $\text{Spec}(R)\rightarrow \text{Spec}(B)$ that maps each prime ideal of $R$ onto its inverse image in $B$ under $\unicode[STIX]{x1D6FD}$ [Reference LiuLiu06, p. 48, Proposition 2.3.25]. By definition, ${\mathcal{A}}_{m}(R)=\text{Mor}_{R}(\text{Spec}(R),\text{Spec}(B))$ . An analogous rule applies to $\overline{K}_{\mathfrak{p}}$ rather than to  $R$ . This gives a commutative diagram

(2)

where the vertical arrows are bijections. Note that if

$$\begin{eqnarray}s\in \text{Mor}_{R}(\text{Spec}(R),\text{Spec}(B)),\end{eqnarray}$$

then $f_{\mathfrak{p}}(s)=s\circ i_{\mathfrak{p}}$ , where $i_{\mathfrak{p}}$ is the natural map $\text{Spec}(\overline{K}_{\mathfrak{p}})\rightarrow \text{Spec}(R/\mathfrak{p})\rightarrow \text{Spec}(R)$ that maps the zero ideal of $\overline{K}_{\mathfrak{p}}$ onto $\mathfrak{p}$ . Thus,

(3) $$\begin{eqnarray}\text{if}~s,s^{\prime }\in \text{Mor}_{R}(\text{Spec}(R),\text{Spec}(B))~\text{and}~f_{\mathfrak{p}}(s)=f_{\mathfrak{p}}(s^{\prime }),\text{then}~s(\mathfrak{p})=s^{\prime }(\mathfrak{p}).\end{eqnarray}$$

Lemma 5.5. Let $R$ and ${\mathcal{A}}$ be as in Section 5.1 and let $m$ be a positive integer. Consider a prime ideal $\mathfrak{p}$ of $R$ and let $\mathfrak{o}$ be the zero ideal of $R$ . Then, the following statements about the objects introduced in this section are true:

1. (a) the map $f_{\mathfrak{o}}:{\mathcal{A}}_{m}(R)\rightarrow A_{m}(K)$ is bijective;

2. (b) let $\mathfrak{p}$ be a prime ideal of $R$ such that $\text{char}(\overline{K}_{\mathfrak{p}})\nmid m$ . Then, the map $f_{\mathfrak{p}}:{\mathcal{A}}_{m}(R)\rightarrow A_{\mathfrak{p},m}(\overline{K}_{\mathfrak{p}})$ is injective, hence

3. (c) the specialization map  is injective.

Proof of (a).

We consider Diagram (2) in the case where $\mathfrak{p}$ is the zero ideal $\mathfrak{o}$ of $R$ . In this case, $\overline{K}_{\mathfrak{p}}=K$ . Let $\unicode[STIX]{x1D704}:R\rightarrow K$ be the inclusion map. By the commutativity of that diagram, it suffices to prove that the map $\text{Hom}_{R}(B,R)\rightarrow \text{Hom}_{R}(B,K)$ defined by $\unicode[STIX]{x1D6FC}\mapsto \unicode[STIX]{x1D704}\circ \unicode[STIX]{x1D6FC}$ is bijective.

Indeed, the map $\unicode[STIX]{x1D6FC}\mapsto \unicode[STIX]{x1D704}\circ \unicode[STIX]{x1D6FC}$ is injective, because $\unicode[STIX]{x1D704}$ is injective. In order to prove that the map is surjective it suffices to prove that $\unicode[STIX]{x1D6FD}(B)\subseteq R$ for each $\unicode[STIX]{x1D6FD}\in \text{Hom}_{R}(B,K)$ .

Indeed, if $x\in B$ , then $x$ is integral over $R$ (by Section 5.2). Hence, so is $\unicode[STIX]{x1D6FD}(x)$ . Since $R$ is integrally closed, $\unicode[STIX]{x1D6FD}(x)\in R$ , as has to be proved.

Proof of (b).

Since $\text{char}(\overline{K}_{\mathfrak{p}})\nmid m$ , we have $\text{char}(K)\nmid m$ . Hence, we may consider the integrally closed integral domain $R^{\prime }=R[m^{-1}]$ . Then we make a base change from $R$ to $R^{\prime }$ and consider the prime ideal $\mathfrak{p}^{\prime }=\mathfrak{p}R^{\prime }$ of $R^{\prime }$ . We also set ${\mathcal{A}}^{\prime }={\mathcal{A}}_{R^{\prime }}$ and $B^{\prime }=B[m^{-1}]$ . Then, $\text{Quot}(R^{\prime })=\text{Quot}(R)=K$ , $\text{Quot}(R^{\prime }/\mathfrak{p}^{\prime })=\text{Quot}(R/\mathfrak{p})=\overline{K}_{\mathfrak{p}}$ , ${\mathcal{A}}_{m}^{\prime }=\text{Spec}(B^{\prime })$ . Finally, we may identify $\text{Hom}_{R}(B,\overline{K}_{\mathfrak{p}})$ with $\text{Hom}_{R^{\prime }}(B^{\prime },\overline{K}_{\mathfrak{p}})$ . Hence, by Diagram (2), we may identify $A_{\mathfrak{p},m}(\overline{K}_{\mathfrak{p}})$ with $A_{\mathfrak{p}^{\prime },m}^{\prime }(\overline{K}_{\mathfrak{p}})$ .

By (a) (applied to $R$ and to $R^{\prime }$ ), we may identify ${\mathcal{A}}_{m}(R)$ and ${\mathcal{A}}_{m}^{\prime }(R^{\prime })$ with $A_{m}(K)$ ; hence we may identify ${\mathcal{A}}_{m}(R)$ with ${\mathcal{A}}_{m}(R^{\prime })$ . Let $f_{\mathfrak{p}^{\prime }}^{\prime }:{\mathcal{A}}_{m}^{\prime }(R^{\prime })\rightarrow A_{\mathfrak{p},m}(\overline{K}_{\mathfrak{p}})=A_{\mathfrak{p}^{\prime },m}^{\prime }(\overline{K}_{\mathfrak{p}})$ be the analogous map to $f_{\mathfrak{p}}:{\mathcal{A}}_{m}(R)\rightarrow A_{\mathfrak{p},m}(\overline{K}_{\mathfrak{p}})$ . Then, the following diagram commutes:

Thus, it suffices to prove that the morphism $f_{\mathfrak{p}^{\prime }}^{\prime }$ is injective.

Let $s,t$ be elements of ${\mathcal{A}}_{m}^{\prime }(R^{\prime })$ such that $f_{\mathfrak{p}^{\prime }}^{\prime }(s)=f_{\mathfrak{p}^{\prime }}^{\prime }(t)$ . By (3) for $R^{\prime }$ rather than for $R$ , $s(\mathfrak{p}^{\prime })=t(\mathfrak{p}^{\prime })$ . Diagram (2) identifies both $s$ and $t$ as elements of $\text{Mor}_{R^{\prime }}(\text{Spec}(R^{\prime }),\text{Spec}(B^{\prime }))$ , that is, as sections of the morphism $h:\text{Spec}(B^{\prime })\rightarrow \text{Spec}(R^{\prime })$ induced from the inclusion $R^{\prime }\subseteq B^{\prime }$ . Since $m$ is a unit of $R^{\prime }$ , Remark 5.3 implies that $B^{\prime }$ is étale over $R^{\prime }$ . Since $h$ is affine, it is separated [Reference LiuLiu06, p. 100, Proposition 3.3.4]. Hence, by [Reference MilneMil80, p. 25, Corollary 3.12] or [Reference GrothendieckGro71, Exposé 1, p. 6, Corollary 5.3], $s=t$ , as claimed.

Proof of (c).

This follows from (b) and from (a). ◻

Let $\unicode[STIX]{x1D70B}:{\mathcal{A}}\rightarrow \text{Spec}(R)$ be as in Section 5.1 and let $m$ be a positive integer. We use Lemma 5.5(a) to identify ${\mathcal{A}}_{m}(R)$ and $A_{m}(K)$ .

If $\text{char}(\overline{K}_{\mathfrak{p}})\nmid m$ , then by Lemma 5.5(c), the injective map $f_{\mathfrak{p}}:{\mathcal{A}}_{m}(R)\rightarrow A_{\mathfrak{p},m}(\overline{K}_{\mathfrak{p}})$ can be considered as an injective homomorphism $f_{\mathfrak{p}}:A_{m}(K)\rightarrow A_{\mathfrak{p},m}(\overline{K}_{\mathfrak{p}})$ that we call the reduction map modulo $\mathfrak{p}$ . If $m^{\prime }$ is a multiple of $m$ and $m^{\prime }\notin \mathfrak{p}$ , then the reduction map modulo $\mathfrak{p}$ with respect to $m^{\prime }$ extends the reduction map modulo $\mathfrak{p}$ with respect to  $m$ .

Lemma 5.6. Let $R$ and ${\mathcal{A}}$ be as in Section 5.1. Let $m$ be a positive integer and $N$ an algebraic extension of $K$ that contains $K(A_{m})$ . We denote the integral closure of $R$ in $N$ by $R_{N}$ . Consider a prime ideal $\mathfrak{p}$ of $R$ that does not contain $m$ . Then, for each $\mathfrak{P}\in \text{Spec}(R_{N})$ over $\mathfrak{p}$ , reduction modulo $\mathfrak{P}$ maps $A_{m}(N)$ isomorphically onto $A_{\mathfrak{p},m}(\overline{N}_{\mathfrak{P}})$ .

Proof. By Section 5.1, $\dim (A_{\mathfrak{p}})$ is equal to the relative dimension $g$ of ${\mathcal{A}}$ over $R$ . Hence, by [Reference Milne, Cornell and SilvermanMil85, p. 116, Remark 8.4], $|A_{m}(N)|=m^{2g}$ and $|A_{\mathfrak{P},m}(\overline{N}_{\mathfrak{P}})|=|A_{\mathfrak{p},m}(\overline{N}_{\mathfrak{P}})|\leqslant m^{2g}$ . By Lemma 5.5(c) for $R_{N}$ and $N$ rather than $R$ and $K$ , the reduction map $A_{m}(N)\rightarrow A_{\mathfrak{P},m}(\overline{N}_{\mathfrak{P}})$ is injective. Hence, that map is bijective, so it is an isomorphism.◻

5.7 Good reduction of representations

Again, let $R$ , ${\mathcal{A}}$ , $B$ , and $m$ be as in Section 5.2. In particular, $B=R[x_{1},\ldots ,x_{k}]$ is a finitely generated ring extension of $R$ . Let $I$ be the kernel of the $R$ -homomorphism $R[X_{1},\ldots ,X_{k}]\rightarrow B$ that maps $X_{i}$ onto $x_{i}$ for $i=1,\ldots ,k$ .

We consider again a Galois extension $N$ of $K$ that contains $K(A_{m})$ , a prime ideal $\mathfrak{p}$ of $R$ that does not contain $m$ , and a prime ideal $\mathfrak{P}$ of the integral closure $R_{N}$ of $R$ in $N$ that lies over $\mathfrak{p}$ . Then, $A_{\mathfrak{p},m}(\overline{N}_{\mathfrak{P}})=\text{Hom}_{R}(B,\overline{N}_{\mathfrak{P}})$ . As usual, we identify each element of $\text{Hom}_{R}(B,\overline{N}_{\mathfrak{P}})$ with a  $k$ -tuple

$$\begin{eqnarray}(x_{1,\mathfrak{P}},\ldots ,x_{k,\mathfrak{P}})\end{eqnarray}$$

with coordinates in $\overline{N}_{\mathfrak{P}}$ at which every $h\in I$ vanishes. Then, $x_{1},\ldots ,x_{k}$ lie in $R_{N}$ and reduction modulo $\mathfrak{P}$ maps $x_{1},\ldots ,x_{k}$ onto $x_{1,\mathfrak{P}},\ldots ,x_{k,\mathfrak{P}}$ , respectively. This gives another presentation to the reduction modulo $\mathfrak{P}$ of $A_{m}(N)$ mentioned in Lemma 5.6 and its proof.

Next, we recall that $B$ is étale over $R$ (Section 5.3) and assume that $R_{N}$ is also étale over $R$ , at least locally over $\mathfrak{p}$ (e.g.,  $N=K(A_{m^{\prime }})$ , where $m^{\prime }$ is a multiple of $m$ that does not belong to $\mathfrak{p}$ ). Let $D_{\mathfrak{P}/\mathfrak{p}}=\{\unicode[STIX]{x1D70E}\in \text{Gal}(N/K)\mid \unicode[STIX]{x1D70E}\mathfrak{P}=\mathfrak{P}\}$ be the decomposition group of $\mathfrak{P}$ over $\mathfrak{p}$ . Then, the reduction $x\mapsto \overline{x}$ modulo $\mathfrak{P}$ (with $x\in R_{N}$ ) induces an isomorphism $\unicode[STIX]{x1D70E}\rightarrow \overline{\unicode[STIX]{x1D70E}}$ of $D_{\mathfrak{P}/\mathfrak{p}}$ onto $\text{Gal}(\overline{N}_{\mathfrak{P}}/\overline{K}_{\mathfrak{p}})$ defined by $\overline{\unicode[STIX]{x1D70E}}\,\overline{x}=\overline{\unicode[STIX]{x1D70E}x}$ . Indeed, if $N/K$ is finite, then $R_{N}/R$ is locally standard étale in a Zariski-open neighborhood of $\mathfrak{p}$ [Reference MilneMil80, p. 26, Theorem 3.14]. Thus, there exists $z\in N$ such that $R_{N,\mathfrak{P}}=R_{\mathfrak{p}}[z]$ , where the discriminant of $\text{irr}(z,K)$ is a unit of $R_{\mathfrak{p}}$ (Section 3.2). Now apply [Reference Fried and JardenFrJ08, p. 109, Lemma 6.1.4].

This isomorphism is then compatible with the isomorphism $A_{m}(N)\rightarrow A_{\mathfrak{p},m}(\overline{N}_{\mathfrak{P}})$ given by Lemma 5.6, which leads to the following commutative triangle:

where $\unicode[STIX]{x1D70C}_{A,m}$ is the $m$ -ic representation induced by the action of $\text{Gal}(K)$ on $A_{m}(\tilde{K})$ . If $l$ is a prime number that does not belong to $\mathfrak{p}$ and if $K(A_{l^{\infty }})\subseteq N$ , the preceding diagram applied to $l,l^{2},l^{3},\ldots$ gives rise to a commutative diagram for the $l$ -adic representations:

6 Bounds on degrees

Given a field $F$ and a nonzero polynomial $f\in F[X]$ , we denote the number of distinct roots of $f(X)$ in $\tilde{F}$ by $\unicode[STIX]{x1D708}(f(X))$ . We prove that the condition “ $\unicode[STIX]{x1D708}(f(X))\leqslant d$ ” is equivalent to a “Zariski-closed condition on the coefficients of $f$ .”

Proof. Let $f(\mathbf{t},X)=\prod _{i=1}^{m}(X-x_{i})$ be the decomposition of $f(\mathbf{t},X)$ in $\widetilde{F(\mathbf{t})}$ . We set $\mathbf{x}=(x_{1},\ldots ,x_{m})$ . Then, $F(\mathbf{t},\mathbf{x})/F(\mathbf{t})$ is a finite normal extension of fields. Moreover, $F[\mathbf{t},\mathbf{x}]/F[\mathbf{t}]$ is an integral extension of integral domains. Hence, by [Reference MumfordMum88, p. 171, Proposition II.7.4], the corresponding morphism $\unicode[STIX]{x1D711}:U=\text{Spec}(F[\mathbf{t},\mathbf{x}])\rightarrow \text{Spec}(F[\mathbf{t}])$ is closed.

Let $I$ be the set of all $d$ -tuples $\mathbf{i}=(i_{1},\ldots ,i_{d})$ of integers between $1$ and $m$ . For each $\mathbf{i}\in I$ and $1\leqslant j\leqslant m$ we consider the Zariski-closed subset $W_{\mathbf{i},j}$ of $U$ defined by the equation $(X_{j}-X_{i_{1}})\cdots (X_{j}-X_{i_{d}})=0$ .

Since $F[\mathbf{t},\mathbf{x}]/F[\mathbf{t}]$ is an integral extension, for each $\mathbf{t}^{\prime }\in T(\tilde{F})$ the map $\mathbf{t}\mapsto \mathbf{t}^{\prime }$ extends to an $F$ -homomorphism $F[\mathbf{t},\mathbf{x}]\rightarrow F[\mathbf{t}^{\prime },\mathbf{x}^{\prime }]$ that maps $\mathbf{x}$ onto $\mathbf{x}^{\prime }=(x_{1}^{\prime },\ldots ,x_{m}^{\prime })$ with $x_{1}^{\prime },\ldots ,x_{m}^{\prime }\in \tilde{F}$ such that $f(\mathbf{t}^{\prime },X)=\prod _{i=1}^{m}(X-x_{i}^{\prime })$ . If $\mathbf{x}^{\prime \prime }=(x_{1}^{\prime \prime },\ldots ,x_{m}^{\prime \prime })$ is another $m$ -tuple in $\tilde{F}^{m}$ such that the map $\mathbf{t}\mapsto \mathbf{t}^{\prime }$ extends to an $F$ -homomorphism $F[\mathbf{t},\mathbf{x}]\rightarrow F[\mathbf{t}^{\prime },\mathbf{x}^{\prime \prime }]$ that maps $\mathbf{x}$ onto $\mathbf{x}^{\prime \prime }$ , then there exists $\unicode[STIX]{x1D70E}\in \text{Aut}(\tilde{F}/F(\mathbf{t}^{\prime }))$ such that $(\mathbf{x}^{\prime })^{\unicode[STIX]{x1D70E}}=\mathbf{x}^{\prime \prime }$ . Hence, $\tilde{V}_{\mathbf{i},j}=\{\mathbf{t}^{\prime }\in T(\tilde{F})\mid (x_{j}^{\prime }-x_{i_{1}}^{\prime })\cdots (x_{j}^{\prime }-x_{i_{d}}^{\prime })=0\}$ is a well-defined subset of $T(\tilde{F})$ . It follows that the set $\tilde{V}=\{\mathbf{t}^{\prime }\in T(\tilde{F})\mid \unicode[STIX]{x1D708}(f(\mathbf{t}^{\prime },X))\leqslant d\}$ satisfies the following condition:

(1) $$\begin{eqnarray}\displaystyle \tilde{V} & = & \displaystyle \mathop{\bigcup }_{\mathbf{i}\in I}\{\mathbf{t}^{\prime }\in T(\tilde{F})\mid \{x_{1}^{\prime },\ldots ,x_{m}^{\prime }\}\subseteq \{x_{i_{1}}^{\prime },\ldots ,x_{i_{d}}^{\prime }\}\}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\bigcup }_{\mathbf{i}\in I}\mathop{\bigcap }_{j=1}^{m}\{\mathbf{t}^{\prime }\in T(\tilde{F})\mid x_{j}^{\prime }=x_{i_{1}}^{\prime }\vee \cdots \vee x_{j}^{\prime }=x_{i_{d}}^{\prime }\}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\bigcup }_{\mathbf{i}\in I}\mathop{\bigcap }_{j=1}^{m}\{\mathbf{t}^{\prime }\in T(\tilde{F})\mid (x_{j}^{\prime }-x_{i_{1}}^{\prime })\cdots (x_{j}^{\prime }-x_{i_{d}}^{\prime })=0\}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\bigcup }_{\mathbf{i}\in I}\mathop{\bigcap }_{j=1}^{m}\tilde{V}_{\mathbf{i},j}.\end{eqnarray}$$

Since $\unicode[STIX]{x1D711}$ is a closed map, $V_{\mathbf{i},j}=\unicode[STIX]{x1D711}(W_{\mathbf{i},j})$ is a Zariski-closed subset of $T$ . Moreover, $W_{\mathbf{i},j}(\tilde{F})=\{(\mathbf{t}^{\prime },\mathbf{x}^{\prime })\in U(\tilde{F})\mid (x_{j}^{\prime }-x_{i_{1}}^{\prime })\cdots (x_{j}^{\prime }-x_{i_{d}}^{\prime })=0\}$ , so $V_{\mathbf{i},j}(\tilde{F})=\tilde{V}_{\mathbf{i},j}$ . It follows that $V=\bigcup _{\mathbf{i}\in I}\bigcap _{j=1}^{m}V_{\mathbf{i},j}$ is a Zariski-closed subset of $T$ defined by polynomials with coefficients in $F$ . Moreover, by (1), $V(\tilde{F})=\bigcup _{\mathbf{i}\in I}\bigcap _{j=1}^{m}V_{\mathbf{i},j}(\tilde{F})=\bigcup _{\mathbf{i}\in I}\bigcap _{j=1}^{m}\tilde{V}_{\mathbf{i},j}=\tilde{V}$ , as desired.◻

Given a polynomial $f$ with coefficients in $\mathbb{Z}$ we consider $f$ for each $l$ also as a polynomial with coefficients in $\mathbb{F}_{l}$ with the original coefficients replaced by their residues modulo $l$ . We say that a scheme $S$ over $\mathbb{Z}$ is absolutely integral if $S_{\tilde{\mathbb{Q}}}=S\times _{\mathbb{Z}}\text{Spec}(\tilde{\mathbb{Q}})$ is integral.

Proof. Let $S_{\mathbb{Q}}=S\times _{\mathbb{Z}}\text{Spec}(\mathbb{Q})$ and $T_{\mathbb{Q}}=T\times _{\mathbb{Z}}\text{Spec}(\mathbb{Q})$ be the generic fibers of $S$ and $T$ . We may write $S=\text{Spec}(\mathbb{Z}[\hat{\mathbf{s}}])$ and $T=\text{Spec}(\mathbb{Z}[\hat{\mathbf{t}}])$ , where $\hat{\mathbf{s}}=({\hat{s}}_{1},\ldots ,{\hat{s}}_{e})$ and $\hat{\mathbf{t}}=(\hat{t}_{1},\ldots ,\hat{t}_{k})$ are tuples of some field extension of $\mathbb{Q}$ such that $\mathbb{Q}(\hat{\mathbf{s}})$ and $\mathbb{Q}(\hat{\mathbf{t}})$ are algebraically independent regular extensions of $\mathbb{Q}$ [Reference Fried and JardenFrJ08, p. 175, Corollary 10.2.2(a)]. By [Reference Fried and JardenFrJ08, p. 41, Lemma 2.6.7], $\mathbb{Q}(\hat{\mathbf{s}})$ and $\mathbb{Q}(\hat{\mathbf{t}})$ are linearly disjoint over $\mathbb{Q}$ , so $\mathbb{Z}[\hat{\mathbf{s}}]\otimes \mathbb{Z}[\hat{\mathbf{t}}]\cong \mathbb{Z}[\hat{\mathbf{s}},\hat{\mathbf{t}}]$ . Hence, $S\times _{\mathbb{Z}}T\cong \text{Spec}(\mathbb{Z}[\hat{\mathbf{s}},\hat{\mathbf{t}}])$ and $S_{\mathbb{Q}}\times _{\mathbb{Q}}T_{\mathbb{Q}}\cong \text{Spec}(\mathbb{Q}[\hat{\mathbf{s}},\hat{\mathbf{t}}])$ .

By Lemma 6.1, there exists a Zariski-closed subvariety $V$ of $S_{\mathbb{Q}}\times _{\mathbb{Q}}T_{\mathbb{Q}}$ defined by polynomials $h_{1},\ldots ,h_{r}$ in $\mathbb{Z}[\mathbf{S},\mathbf{T}]$ such that

(2) $$\begin{eqnarray}V(\tilde{\mathbb{Q}})=\{(\mathbf{s},\mathbf{t})\in S(\tilde{\mathbb{Q}})\times T(\tilde{\mathbb{Q}})\mid \unicode[STIX]{x1D708}(f(\mathbf{s},\mathbf{t},X))\leqslant d\}.\end{eqnarray}$$

We also have

(3) $$\begin{eqnarray}V(\tilde{\mathbb{Q}})=\{(\mathbf{s},\mathbf{t})\in S(\tilde{\mathbb{Q}})\times T(\tilde{\mathbb{Q}})\mid h_{1}(\mathbf{s},\mathbf{t})=0,\ldots ,h_{r}(\mathbf{s},\mathbf{t})=0\}.\end{eqnarray}$$

Thus, the following statement about $\tilde{\mathbb{Q}}$ is true.

(4)

Note that Statement (4) is elementary. In other words, the statement is equivalent to a sentence in the language of rings ${\mathcal{L}}(\text{ring},\mathbb{Z})$ with parameters in $\mathbb{Z}$ [Reference Fried and JardenFrJ08, p. 135, Example 7.3.1]. Hence, by a consequence of the quantifier elimination procedure for the theory of algebraically closed fields [Reference Fried and JardenFrJ08, p. 167, Corollary 9.2.2], that statement holds over $\tilde{\mathbb{F}}_{l}$ for every large prime number $l$ . In addition, by [Reference Fried and JardenFrJ08, p. 179, Proposition 10.4.2], $S_{\mathbb{F}_{l}}$ and $T_{\mathbb{F}_{l}}$ are absolutely integral varieties over $\mathbb{F}_{l}$ for each large  $l$ .

For each $l$ as in the preceding paragraph and for every $\mathbf{s}\in S(\tilde{\mathbb{F}}_{l})$ , let $U_{l,\mathbf{s}}$ be the Zariski-closed subset of $T_{\mathbb{F}_{l}(\mathbf{s})}$ defined by the polynomials

$$\begin{eqnarray}h_{1}(\mathbf{s},\mathbf{T}),\ldots ,h_{r}(\mathbf{s},\mathbf{T}).\end{eqnarray}$$

Since (4) holds over $\tilde{\mathbb{F}}_{l}$ , we have

(5) $$\begin{eqnarray}U_{l,\mathbf{s}}(\tilde{\mathbb{F}}_{l})=\{\mathbf{t}\in T(\tilde{\mathbb{F}}_{l})\mid \unicode[STIX]{x1D708}(f(\mathbf{s},\mathbf{t},X))\leqslant d\}.\end{eqnarray}$$

Moreover, the degrees of the polynomials $h_{1}(\mathbf{s},\mathbf{T}),\ldots ,h_{r}(\mathbf{s},\mathbf{T})$ that define $U_{l,\mathbf{s}}$ are at most $\deg _{\mathbf{T}}h_{1}(\mathbf{S},\mathbf{T}),\ldots ,\deg _{\mathbf{T}}h_{r}(\mathbf{S},\mathbf{T})$ , respectively. Since the latter numbers are independent of $\mathbf{s}$ , the second statement of the lemma is also true.◻

7 Counting points

Following [Reference ZywinaZyw16], we find in this section a set $\unicode[STIX]{x1D6EC}$ of prime numbers with positive Dirichlet density such that for all large $l\in \unicode[STIX]{x1D6EC}$ there are “many” points in $H_{l}(\mathbb{F}_{l})$ having $1$ as an eigenvalue, where $H_{l}$ is the reductive subgroup of $\text{GL}_{2g,\mathbb{F}_{l}}$ introduced in Proposition 4.1. Let $n$ be a positive integer and let $L$ be a finite Galois extension of $K$ such that $\unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(L))$ is contained in $H_{l}(\mathbb{F}_{l})$ with index ${\leqslant}n$ (Proposition 4.1(b)). After fixing an $\mathbb{F}_{l}$ -split maximal torus $T$ of $H_{l}$ , each of those points is of the form $\mathbf{b}\mathbf{t}^{n!}$ , where $\mathbf{b}\in \unicode[STIX]{x1D70C}_{A,l}(\text{Gal}(K))$ depends only on $l$ and $\mathbf{t}$ is an $\mathbb{F}_{l}$ -rational point of $T$ such that $\mathbf{t}^{n!}$ is a regular element of  $H_{l}$ .