Modular forms and some cases of the Inverse Galois Problem

Abstract

We prove new cases of the Inverse Galois Problem by considering the residual Galois representations arising from a fixed newform. Specific choices of weight $3$ newforms will show that there are Galois extensions of ${\mathbb Q}$ with Galois group $\operatorname {PSL}_2({\mathbb F}_p)$ for all primes p and $\operatorname {PSL}_2({\mathbb F}_{p^3})$ for all odd primes $p \equiv \pm 2, \pm 3, \pm 4, \pm 6 \ \pmod {13}$ .

1 Introduction

The Inverse Galois Problem asks whether every finite group is isomorphic to the Galois group of some extension of ${\mathbb Q}$ . There has been much work on using modular forms to realize explicit simple groups of the form $\operatorname {PSL}_2({\mathbb F}_{\ell ^r})$ as Galois groups of extensions of ${\mathbb Q}$ (cf. [4–6, 13, 14]). For example, [6, Section 3.2] shows that $\operatorname {PSL}_2({\mathbb F}_{\ell ^2})$ occurs as a Galois group of an extension of ${\mathbb Q}$ for all primes $\ell $ in an explicit set of density $1-1/2^{10}$ (and for primes $\ell \leq 5,000,000$ ). Furthermore, it is shown in [6] that $\operatorname {PSL}_2({\mathbb F}_{\ell ^4})$ occurs as a Galois group of an extension of ${\mathbb Q}$ when $\ell \equiv 2,3 \ \pmod {5}$ or $\ell \equiv \pm 3,\pm 5,\pm 6,\pm 7 \ \pmod {17}$ .

The goal of this paper is to try to realize more groups of the form $\operatorname {PSL}_2({\mathbb F}_{\ell ^r})$ for odd r. We will achieve this by working with newforms of odd weight; the papers mentioned above focus on even weight modular forms (usual weight $2$ ). We will give background and describe the general situation in Section 1.1. In Sections 1.2 and 1.3, we will use specific newforms of weight $3$ to realize many groups of the form $\operatorname {PSL}_2({\mathbb F}_{\ell ^r})$ with r equal to $1$ and $3$ , respectively.

Throughout the paper, we fix an algebraic closure ${\overline {\mathbb Q}}$ of ${\mathbb Q}$ and define the group $G:= \operatorname {\mathrm {Gal}}({\overline {\mathbb Q}}/{\mathbb Q})$ . For a ring R, we let $\operatorname {PSL}_2(R)$ and $\operatorname {PGL}_2(R)$ be the quotients of $\operatorname {SL}_2(R)$ and $\operatorname {GL}_2(R)$ , respectively, by its subgroup of scalar matrices (in particular, this notation may disagree with the R-points of the corresponding group scheme $\operatorname {PSL}_2$ or $\operatorname {PGL}_2$ ).

1.1 General results

Fix a newform $f(\tau )=\sum _{n=1}^\infty a_n q^n$ of weight $k>1$ on $\Gamma _1(N)$ without complex multiplication, where the $a_n$ are complex numbers and $q=e^{2\pi i \tau }$ with $\tau $ a variable of the complex upper half-plane. Let $\varepsilon \colon\ ({\mathbb Z}/N{\mathbb Z})^\times \to {\mathbb C}^\times $ be the nebentypus of f.

Let E be the subfield of ${\mathbb C}$ generated by the coefficients $a_n$ ; it is also generated by the coefficients $a_p$ with primes $p\nmid N$ . The field E is a number field, and all the $a_n$ are known to lie in its ring of integers ${\mathcal O}$ . The image of $\varepsilon $ lies in $E^\times $ . Let K be the subfield of E generated by the algebraic integers $r_p:=a_p^2/\varepsilon (p)$ for primes $p\nmid N$ ; denote its ring of integers by R.

Take any nonzero prime ideal $\Lambda $ of ${\mathcal O}$ and denote by $\ell =\ell (\Lambda )$ the rational prime lying under $\Lambda $ . Let $E_\Lambda $ and ${\mathcal O}_\Lambda $ be the completions of E and ${\mathcal O}$ , respectively, at $\Lambda $ . From [3], we know that there is a continuous representation

$$\begin{align*}\rho_\Lambda \colon G \to \operatorname{GL}_2({\mathcal O}_\Lambda) \end{align*}$$

such that for each prime $p\nmid N\ell $ , the representation $\rho _\Lambda $ is unramified at p and satisfies

(1.1) $$ \begin{align} \operatorname{\mathrm{tr}}(\rho_\Lambda(\operatorname{\mathrm{Frob}}_p)) = a_p \quad \text{ and } \quad \det(\rho_\Lambda(\operatorname{\mathrm{Frob}}_p))= \varepsilon(p) p^{k-1}. \end{align} $$

The representation $\rho _\Lambda $ is uniquely determined by the conditions (1.1) up to conjugation by an element of $\operatorname {GL}_2(E_\Lambda )$ . By composing $\rho _\Lambda $ with the natural projection arising from the reduction map ${\mathcal O}_\Lambda \to {\mathbb F}_\Lambda :={\mathcal O}/\Lambda $ , we obtain a representation

Composing

with the natural quotient map $\operatorname {GL}_2({\mathbb F}_\Lambda )\to \operatorname {PGL}_2({\mathbb F}_\Lambda )$ , we obtain a homomorphism

Define the field ${\mathbb F}_\lambda :=R/\lambda $ , where $\lambda := \Lambda \cap R$ . There are natural injective homomorphisms $\operatorname {PSL}_2({\mathbb F}_\lambda ) \hookrightarrow \operatorname {PGL}_2({\mathbb F}_\lambda ) \hookrightarrow \operatorname {PGL}_2({\mathbb F}_\Lambda )$ and $\operatorname {PSL}_2({\mathbb F}_\Lambda ) \hookrightarrow \operatorname {PGL}_2({\mathbb F}_\Lambda )$ that we shall view as inclusions.

The main task of this paper is to describe the group for all $\Lambda $ outside of some explicit set. The following theorem of Ribet gives two possibilities for for all but finitely many $\Lambda $ .

Theorem 1.1 (Ribet)

There is a finite set S of nonzero prime ideals of R such that if $\Lambda $ is a nonzero prime ideal of ${\mathcal O}$ with $\lambda :=R\cap \Lambda \notin S$ , then the group is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to either $\operatorname {PSL}_2({\mathbb F}_\lambda )$ or $\operatorname {PGL}_2({\mathbb F}_\lambda )$ .

Proof As noted in Section 3 of [7], this is an easy consequence of [16].

We will give a proof of Theorem 1.1 in Section 4 that allows one to compute such a set S. There are several related results in the literature; for example, Billerey and Dieulefait [1] give a version of Theorem 1.1 when the nebentypus $\varepsilon $ is trivial.

We now explain how to distinguish the two possibilities from Theorem 1.1. Let $L\subseteq {\mathbb C}$ be the extension of K generated by the square roots of the values $r_p=a_p^2/\varepsilon (p)$ with $p\nmid N$ ; it is a finite extension of K (moreover, it is contained in a finite cyclotomic extension of E).

Theorem 1.2 Let $\Lambda $ be a nonzero prime ideal of ${\mathcal O}$ such that is conjugate to $\operatorname {PSL}_2({\mathbb F}_\lambda )$ or $\operatorname {PGL}_2({\mathbb F}_\lambda )$ , where $\lambda =\Lambda \cap R$ . After conjugating , we may assume that . Let $\ell $ be the rational prime lying under $\Lambda $ .

1. (i) If k is odd, then if and only if $\lambda $ splits completely in L.

2. (ii) If k is even and $[{\mathbb F}_\lambda :{\mathbb F}_\ell ]$ is even, then if and only if $\lambda $ splits completely in L.

3. (iii) If k is even, $[{\mathbb F}_\lambda :{\mathbb F}_\ell ]$ is odd, and $\ell \nmid N$ , then .

Remark 1.3

1. (i) From Theorem 1.2, we see that it is more challenging to produce Galois extensions of ${\mathbb Q}$ with Galois group $\operatorname {PSL}_2({\mathbb F}_{\ell ^r})$ with odd r if we focus solely on newforms with k even. However, it is still possible to obtain such groups in the excluded case $\ell | N$ .

2. (ii) Parts (ii) and (iii) of Theorem 1.2 are included for completeness (see [4, Proposition 1.5] for an equivalent version in the case $k=2$ due to Dieulefait). Surprisingly, there has been very little attention in the literature given to the case where k is odd (commenting on a preprint of this work, Dieulefait has shared several explicit examples worked out with Tsaknias and Vila). In Sections 1.2 and 1.3, we give examples with $k=3$ and $L=K$ (so $\lambda $ splits in L for any $\lambda $ ).

1.2 An example realizing the groups $\text {PSL}_2({\mathbb F}_\ell )$

We now give an example that realizes the simple groups $\operatorname {PSL}_2({\mathbb F}_\ell )$ as Galois groups of an extension of ${\mathbb Q}$ for all primes $\ell \geq 7$ . Let $f=\sum _{n=1}^\infty a_n q^n$ be a newform of weight $3$ , level $N=27$ , and nebentypus $\varepsilon (a)=\big (\frac {-3}{a}\big )$ ; it is non-CM, i.e., it does not have complex multiplication. We can choose f so that ¹

$$ \begin{align*} f =q &+ 3iq^2 - 5q^4 - 3iq^5 + 5q^7 - 3iq^8 + 9q^{10} - 15iq^{11} - 10q^{13} + \cdots; \end{align*} $$

the other possibility for f is its complex conjugate .

The subfield E of ${\mathbb C}$ generated by the coefficients $a_n$ is ${\mathbb Q}(i)$ . Take any prime $p\neq 3$ . We will see that . Therefore, $a_p$ or $i a_p$ belongs to ${\mathbb Z}$ when $\varepsilon (p)$ is $1$ or $-1$ , respectively, and hence $r_p=a_p^2/\varepsilon (p)$ is a square in ${\mathbb Z}$ . Therefore, $L=K={\mathbb Q}$ .

In Section 6.1, we shall verify that Theorem 1.1 holds with $S=\{2,3,5\}$ . Take any prime $\ell \geq 7$ and prime $\Lambda \subseteq {\mathbb Z}[i]$ dividing $\ell $ . Theorem 1.2 with $L=K={\mathbb Q}$ implies that is isomorphic to $\operatorname {PSL}_2({\mathbb F}_\ell )$ . The following theorem is now an immediate consequence (it is easy to prove directly for the group $\operatorname {PSL}_2({\mathbb F}_5)\cong A_5$ ).

Theorem 1.4 For each prime $\ell \geq 5$ , there is a Galois extension $K/{\mathbb Q}$ such that $\operatorname {\mathrm {Gal}}(K/{\mathbb Q})$ is isomorphic to the simple group $\operatorname {PSL}_2({\mathbb F}_\ell )$ .■

Remark 1.5

1. (i) In Section 5.5 of [17], Serre describes the image of and proves that it gives rise to a $\operatorname {PSL}_2({\mathbb F}_7)$ -extension of ${\mathbb Q}$ ; however, he does not consider the image modulo other primes. Note that Serre was actually giving an example of his conjecture, so he started with the $\operatorname {PSL}_2({\mathbb F}_7)$ -extension and then found the newform f.

2. (ii) Theorem 1.4 was first proved by the author in [18] by considering the Galois action on the second $\ell $ -adic étale cohomology of a specific surface. One can show that the Galois extensions of [18] could also be constructed by first starting with an appropriate newform of weight $3$ and level $32$ .

1.3 Another example

We now give an example with $K\neq {\mathbb Q}$ . Additional details will be provided in Section 6.2. Let $f=\sum _n a_n q^n$ be a non-CM newform of weight $3$ , level $N=160$ , and nebentypus $\varepsilon (a)=\big (\frac {-5}{a}\big )$ .

Take E, K, L, R, and ${\mathcal O}$ as in Section 1.1. We will see in Section 6.2 that $E=K(i)$ and that K is the unique cubic field in ${\mathbb Q}(\zeta _{13})$ . We will also observe that $L=K$ .

Take any odd prime $\ell $ congruent to $\pm 2$ , $\pm 3$ , $\pm 4$ , or $\pm 6$ modulo $13$ . Let $\Lambda $ be any prime ideal of ${\mathcal O}$ dividing $\ell $ , and set $\lambda = \Lambda \cap R$ . The assumption on $\ell $ modulo $13$ implies that $\lambda = \ell R$ and that ${\mathbb F}_\lambda \cong {\mathbb F}_{\ell ^3}$ . In Section 6.2, we shall compute a set S as in Theorem 1.1 which does not contain $\lambda $ . Theorem 1.2 with $L=K$ implies that is isomorphic to $\operatorname {PSL}_2({\mathbb F}_\lambda )\cong \operatorname {PSL}_2({\mathbb F}_{\ell ^3})$ . The following is an immediate consequence.

Theorem 1.6 If $\ell $ is an odd prime congruent to $\pm 2$ , $\pm 3$ , $\pm 4$ , or $\pm 6$ modulo $13$ , then the simple group $\operatorname {PSL}_2({\mathbb F}_{\ell ^3})$ occurs as the Galois group of an extension of ${\mathbb Q}$ .

2 The fields K and L

Take a newform f with notation and assumptions as in Section 1.1.

2.1 The field K

Let $\Gamma $ be the set of automorphisms $\gamma $ of the field E for which there is a primitive Dirichlet character $\chi _\gamma $ that satisfies

(2.1) $$ \begin{align} \gamma(a_p) = \chi_\gamma(p) a_p \end{align} $$

for all primes $p\nmid N$ . The set of primes p with $a_p\neq 0$ has density $1$ since f is non-CM, so the image of $\chi _\gamma $ lies in $E^\times $ and the character $\chi _\gamma $ is uniquely determined from $\gamma $ .

Define M to be N or $4N$ if N is odd or even, respectively. The conductor of $\chi _\gamma $ divides M (cf. [12, Remark 1.6]). Moreover, there is a quadratic Dirichlet character $\alpha $ with conductor dividing M and an integer i such that $\chi _\gamma $ is the primitive character coming from $\alpha \varepsilon ^i$ (cf. [12, Lemma 1.5(i)]).

For each prime $p\nmid N$ , we have (cf. [15, p. 21]), so complex conjugation induces an automorphism $\gamma $ of E and $\chi _\gamma $ is the primitive character coming from $\varepsilon ^{-1}$ . In particular, $\Gamma \neq 1$ if $\varepsilon $ is nontrivial.

Remark 2.1 More generally, we could have instead considered an embedding $\gamma \colon E \to {\mathbb C}$ and a Dirichlet character $\chi _\gamma $ such that (2.1) holds for all sufficiently large primes p. This gives the same twists, since $\gamma (E)=E$ and the character $\chi _\gamma $ is unramified at primes $p\nmid N$ (cf. [12, Remark 1.3]).

The set $\Gamma $ is in fact an abelian subgroup of $\operatorname {\mathrm {Aut}}(E)$ (cf. [12, Lemma 1.5(ii)]). Denote by $E^{\Gamma }$ the fixed field of E by $\Gamma $ .

Lemma 2.2

1. (i) We have $K=E^{\Gamma} $ , and hence $\operatorname {\mathrm {Gal}}(E/K)=\Gamma $ .

2. (ii) There is a prime $p\nmid N$ such that $K={\mathbb Q}(r_p)$ .

Proof Take any $p\nmid N$ . For each $\gamma \in \Gamma $ , we have

$$\begin{align*}\gamma(r_p) = \gamma(a_p^2)/\gamma(\varepsilon(p)) = \chi_\gamma(p)^2 a_p^2/\gamma(\varepsilon(p)) = a_p^2/\varepsilon(p) = r_p, \end{align*}$$

where we have used that $\chi _\gamma (p)^2 = \gamma (\varepsilon (p))/\varepsilon (p)$ (cf. [12, Proof of Lemma 1.5(ii)]). This shows that $r_p$ belong in $E^{\Gamma} $ and hence $K\subseteq E^{\Gamma} $ since $p\nmid N$ was arbitrary. To complete the proof of the lemma, it thus suffices to show that $E^{\Gamma} ={\mathbb Q}(r_p)$ for some prime $p\nmid N$ .

For $\gamma \in \Gamma $ , let $\widetilde \chi _\gamma \colon G \to {\mathbb C}^\times $ be the continuous character such that $\widetilde \chi _\gamma (\operatorname {\mathrm {Frob}}_p)=\chi _\gamma (p)$ for all $p\nmid N$ . Define the group $H = \bigcap _{\gamma \in \Gamma } \ker \widetilde \chi _\gamma $ ; it is an open normal subgroup of G with $G/H$ is abelian. Let ${\mathcal K}$ be the subfield of ${\overline {\mathbb Q}}$ fixed by H; it is a finite abelian extension of ${\mathbb Q}$ .

Fix a prime $\ell $ and a prime ideal $\Lambda | \ell $ of ${\mathcal O}$ . In the proof of Theorem 3.1 of [16], Ribet proved that $E^{\Gamma} ={\mathbb Q}(a_v^2)$ for a positive density set of finite place $v\nmid N\ell $ of ${\mathcal K}$ , where $a_v:=\operatorname {\mathrm {tr}}(\rho _\Lambda (\operatorname {\mathrm {Frob}}_v))$ . There is thus a finite place $v\nmid N\ell $ of ${\mathcal K}$ of degree $1$ such that $E^{\Gamma} ={\mathbb Q}(a_v^2)$ . We have $a_v=a_p$ , where p is the rational prime that v divides, so $E^{\Gamma} ={\mathbb Q}(a_p^2)$ . Since v has degree $1$ and ${\mathcal K}/{\mathbb Q}$ is abelian, the prime p must split completely in ${\mathcal K}$ and hence $\chi _\gamma (p)=1$ for all $\gamma \in \Gamma $ ; in particular, $\varepsilon (p)=1$ . Therefore, $E^{\Gamma} ={\mathbb Q}(r_p)$ .

2.2 The field L

Recall that we defined L to be the extension of K in ${\mathbb C}$ obtained by adjoining the square root of $r_p = a_p^2/\varepsilon (p)$ for all $p\nmid N$ . The following allows one to find a finite set of generators for the extension $L/K$ and gives a way to check the criterion of Theorem 1.2.

Lemma 2.3

1. (i) Choose primes $p_1,\ldots , p_m \nmid N$ that generate the group $({\mathbb Z}/M{\mathbb Z})^\times $ and satisfy $r_{p_i}\neq 0$ for all $1\leq i\leq m$ . Then $L=K(\sqrt {r_{p_1}},\ldots ,\sqrt {r_{p_m}})$ .

2. (ii) Take any nonzero prime ideal $\lambda $ of R that does not divide $2$ . Let $p_1,\ldots , p_m$ be primes as in (i). Then the following are equivalent:

   1. (a) $ \lambda $ splits completely in L.

   2. (b) For all $p\nmid N$ , $r_p$ is a square in $K_\lambda $ .

   3. (c) For all $1\leq i \leq m$ , $r_{p_i}$ is a square in $K_\lambda $ .

Proof Take any prime $p\nmid N$ . To prove part (i), it suffices to show that $\sqrt {r_p}$ belongs to the field $L':=K(\sqrt {r_{p_1}},\ldots ,\sqrt {r_{p_m}})$ . This is obvious if $r_p=0$ , so assume that $r_p\neq 0$ . Since the $p_i$ generate $({\mathbb Z}/M{\mathbb Z})^\times $ by assumption, there are integers $e_i\geq 0$ such that $p\equiv p_1^{e_1} \cdots p_m^{e_m} \ \pmod {M}$ . Take any $\gamma \in \Gamma $ . Using that the conductor of $\chi _\gamma $ divides M and (2.1), we have

$$\begin{align*}\gamma\Big( \frac{a_p}{{\prod}_i a_{p_i}^{e_i}} \Big) =\frac{\chi_\gamma(p)}{\chi_\gamma({\prod}_i p_i^{e_i})} \cdot \frac{a_p}{{\prod}_i a_{p_i}^{e_i}} = \frac{\chi_\gamma(p)}{\chi_\gamma(p)}\cdot \frac{a_p}{{\prod}_i a_{p_i}^{e_i}} = \frac{a_p}{{\prod}_i a_{p_i}^{e_i}}. \end{align*}$$

Since $E^{\Gamma} =K$ by Lemma 2.2(i), the value $a_p/{\prod }_i a_{p_i}^{e_i}$ belongs to K; it is nonzero since $r_p\neq 0$ and $r_{p_i}\neq 0$ . We have $\varepsilon (p) = \prod _i \varepsilon (p_i)^{e_i}$ since the conductor of $\varepsilon $ divides M. Therefore,

$$\begin{align*}\frac{r_p}{{\prod}_i r_{p_i}^{e_i}} = \frac{a_p^2}{{\prod}_i (a_{p_i}^2)^{e_i}} = \bigg(\frac{a_p}{{\prod}_i a_{p_i}^{e_i}}\bigg)^2 \in (K^\times)^2. \end{align*}$$

This shows that $\sqrt {r_p}$ is contained in $L'$ as desired. This proves (i); part (ii) is an easy consequence of (i).

Remark 2.4 Finding primes $p_i$ as in Lemma 2.3(i) is straightforward since $r_p\neq 0$ for all p outside a set of density $0$ (and the primes representing each class $a\in ({\mathbb Z}/M{\mathbb Z})^\times $ have positive density). Lemma 2.3(ii) gives a straightforward way to check if $\lambda $ splits completely in L. Let $e_i$ be the $\lambda $ -adic valuation of $r_{p_i}$ , and let $\pi $ be a uniformizer of $K_\lambda $ ; then $r_{p_i}$ is a square in $K_\lambda $ if and only if e is even and the image of $r_{p_i}/\pi ^{e_i}$ in ${\mathbb F}_\lambda $ is a square.

3 Proof of Theorem 1.2

We may assume that is $\operatorname {PSL}_2({\mathbb F}_\lambda )$ or $\operatorname {PGL}_2({\mathbb F}_\lambda )$ . For any $n\geq 1$ , the group $\operatorname {GL}_2({\mathbb F}_{2^n})$ is generated by $\operatorname {SL}_2({\mathbb F}_{2^n})$ and its scalar matrices, so $\operatorname {PSL}_2({\mathbb F}_{2^n})=\operatorname {PGL}_2({\mathbb F}_{2^n})$ . The theorem is thus trivial when $\ell =2$ , so we may assume that $\ell $ is odd.

Take any $\alpha \in \operatorname {PGL}_2({\mathbb F}_\lambda )\subseteq \operatorname {PGL}_2({\mathbb F}_\Lambda )$ and choose any matrix $A\in \operatorname {GL}_2({\mathbb F}_\Lambda )$ whose image in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ is $\alpha $ . The value $\operatorname {\mathrm {tr}}(A)^2/\det (A)$ does not depend on the choice of A and lies in ${\mathbb F}_\lambda $ (since we can choose A in $\operatorname {GL}_2({\mathbb F}_\lambda )$ ); by abuse of notation, we denote this common value by $\operatorname {\mathrm {tr}}(\alpha )^2/\det (\alpha )$ .

Lemma 3.1 Suppose that $p\nmid N\ell $ is a prime for which $r_p \not \equiv 0\ \pmod {\lambda }$ . Then is contained in $\operatorname {PSL}_2({\mathbb F}_\lambda )$ if and only if the image of $a_p^2/(\varepsilon (p) p^{k-1})=r_p/p^{k-1}$ in ${\mathbb F}_\lambda ^\times $ is a square.

Proof Define and ; the image of A in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ is $\alpha $ . The value $\xi _p:=\operatorname {\mathrm {tr}}(\alpha )^2/\det (\alpha ) = \operatorname {\mathrm {tr}}(A)^2/\det (A)$ agrees with the image of $a_p^2/(\varepsilon (p)p^{k-1})= r_p/p^{k-1}$ in ${\mathbb F}_\Lambda $ . Since $r_p \in R$ is nonzero modulo $\lambda $ by assumption, the value $\xi _p$ lies in ${\mathbb F}_\lambda ^\times $ . Fix a matrix $A_0 \in \operatorname {GL}_2({\mathbb F}_\lambda )$ whose image in $\operatorname {PGL}_2({\mathbb F}_\lambda )$ is $\alpha $ ; we have $\xi _p = \operatorname {\mathrm {tr}}(A_0)^2/\det (A_0)$ . Since $\xi _p \neq 0$ , we find that $\xi _p$ and $\det (A_0)$ lie in the same coset in ${\mathbb F}_\lambda ^\times /({\mathbb F}_\lambda ^\times )^2$ .

The determinant gives rise to a homomorphism $d\colon \operatorname {PGL}_2({\mathbb F}_\lambda ) \to {\mathbb F}_\lambda ^\times /( {\mathbb F}_\lambda ^\times )^2$ whose kernel is $\operatorname {PSL}_2({\mathbb F}_\lambda )$ . Define the character

We have $\xi (\operatorname {\mathrm {Frob}}_p) = \det (A_0) \cdot ( {\mathbb F}_\lambda ^\times )^2 = \xi _p \cdot ( {\mathbb F}_\lambda ^\times )^2$ . So $\xi (\operatorname {\mathrm {Frob}}_p)=1$ , equivalently

, if and only if $\xi _p \in {\mathbb F}_\lambda ^\times $ is a square.

Let M be the integer from Section 2.1.

Lemma 3.2 For each $a\in ({\mathbb Z}/M\ell {\mathbb Z})^\times $ , there is a prime $p\equiv a \ \pmod {M\ell }$ such that $r_p \not \equiv 0 \ \pmod {\lambda }$ .

Proof Set ; it is $\operatorname {PSL}_2({\mathbb F}_\lambda )$ or $\operatorname {PGL}_2({\mathbb F}_\lambda )$ by assumption. Let $H'$ be the commutator subgroup of H. We claim that for each coset $\kappa $ of $H'$ in H, there exists an $\alpha \in \kappa $ with $\operatorname {\mathrm {tr}}(\alpha )^2/\det (\alpha )\neq 0$ . If $H'=\operatorname {PSL}_2({\mathbb F}_\lambda )$ , then the claim is easy; note that for any $t\in {\mathbb F}_\lambda $ and $d\in {\mathbb F}_\lambda ^\times $ , there is a matrix in $\operatorname {GL}_2({\mathbb F}_\lambda )$ with trace t and determinant d. When $\#{\mathbb F}_\lambda \neq 3$ , the group $\operatorname {PSL}_2({\mathbb F}_\lambda )$ is non-abelian and simple, so $H'=\operatorname {PSL}_2({\mathbb F}_\lambda )$ . When $\#{\mathbb F}_\lambda =3$ and $H=\operatorname {PGL}_2({\mathbb F}_\lambda )$ , we have $H'=\operatorname {PSL}_2({\mathbb F}_\lambda )$ . It thus suffices to prove the claim in the case where ${\mathbb F}_\lambda ={\mathbb F}_3$ and $H=\operatorname {PSL}_2({\mathbb F}_3)$ . In this case, $H'$ is the unique subgroup of H of index $3$ and the cosets of $H/H'$ are represented by $\left (\begin {smallmatrix} 1 & b \\ 0 & 1 \end {smallmatrix}\right )$ with $b\in {\mathbb F}_3$ . The claim is now immediate in this remaining case.

Let $\chi \colon\ G\twoheadrightarrow ({\mathbb Z}/M\ell {\mathbb Z})^\times $ be the cyclotomic character that satisfies $\chi (\operatorname {\mathrm {Frob}}_p) \equiv p \ \pmod {M\ell }$ for all $p\nmid M\ell $ . The set is thus the union of cosets of $H'$ in H. By the claim above, there exists an with $\operatorname {\mathrm {tr}}(\alpha )^2/\det (\alpha )~\neq~0$ . By the Chebotarev density theorem, there is a prime $p\nmid M\ell $ satisfying $p\equiv a \ \pmod {M\ell }$ and . The lemma follows since $r_p/p^{k-1}$ modulo $\lambda $ agrees with $\operatorname {\mathrm {tr}}(\alpha )^2/\det (\alpha )\neq 0$ .

Case 1: Assume that k is odd or $[{\mathbb F}_\lambda :{\mathbb F}_\ell ]$ is even.

First, suppose that . By Lemma 3.2, there are primes $p_1,\ldots , p_m \nmid N\ell $ that generate the group $({\mathbb Z}/M{\mathbb Z})^\times $ and satisfy $r_{p_i}\not \equiv 0 \ \pmod {\lambda }$ for all $1\leq i\leq m$ . By Lemma 3.1 and the assumption , the image of $r_{p_i}/{p_i}^{k-1}$ in ${\mathbb F}_\lambda $ is a nonzero square for all $1\leq i \leq m$ . For each $1\leq i \leq m$ , the assumption that k is odd or $[{\mathbb F}_\lambda :{\mathbb F}_\ell ]$ is even implies that $p_i^{k-1}$ is a square in ${\mathbb F}_\lambda $ , and hence the image of $r_{p_i}$ in ${\mathbb F}_\lambda $ is a nonzero square. Since $\lambda \nmid 2$ , we deduce that each $r_{p_i}$ is a square in $K_\lambda $ . By Lemma 2.3(ii), the prime $\lambda $ splits completely in L.

Now, suppose that . There exists an element $\alpha \in \operatorname {PGL}_2({\mathbb F}_\lambda ) - \operatorname {PSL}_2({\mathbb F}_\lambda )$ with $\operatorname {\mathrm {tr}}(\alpha )^2/\det (\alpha ) \neq 0$ . By the Chebotarev density theorem, there is a prime $p\nmid N\ell $ such that . We have $r_p \equiv \operatorname {\mathrm {tr}}(\alpha )^2/\det (\alpha ) \not \equiv 0\ \pmod {\lambda }$ . Since , Lemma 3.1 implies that the image of $r_p/p^{k-1}$ in ${\mathbb F}_\lambda $ is not a square. Since k is odd or $[{\mathbb F}_\lambda :{\mathbb F}_\ell ]$ is even, the image of $r_p$ in ${\mathbb F}_\lambda $ is not a square. Therefore, $r_p$ is not a square in $K_\lambda $ . By Lemma 2.3(ii), we deduce that $\lambda $ does not split completely in L.

Case 2: Assume that k is even, $[{\mathbb F}_\lambda :{\mathbb F}_\ell ]$ is odd, and $\ell \nmid N$ .

Since $\ell \nmid N$ , there is an integer $a\in {\mathbb Z}$ such that $a\equiv 1 \ \pmod {M}$ and a is not a square modulo $\ell $ . By Lemma 3.2, there is a prime $p\equiv a \ \pmod {M\ell }$ such that $r_p\not \equiv 0 \ \pmod {\lambda }$ .

We claim that $a_p\in R$ and $\varepsilon (p)=1$ . With notation as in Section 2.1, take any $\gamma \in \Gamma $ . Since the conductor of $\chi _\gamma $ divides M and $p\equiv 1 \ \pmod {M}$ , we have $\gamma (a_p)=\chi _\gamma (p) a_p =a_p$ . Since $\gamma \in \Gamma $ was arbitrary, we have $a_p \in K$ by Lemma 2.2. Therefore, $a_p\in R$ since it is an algebraic integer. We have $\varepsilon (p)=1$ since $p\equiv 1\ \pmod {N}$ .

Since $a_p\in R$ and $r_p\not \equiv 0 \ \pmod {\lambda }$ , the image of $a_p^2$ in ${\mathbb F}_\lambda $ is a nonzero square. Since k is even, $p^k$ is a square in ${\mathbb F}_\lambda $ . Since a, and hence p, is not a square modulo $\ell $ and $[{\mathbb F}_\lambda :{\mathbb F}_\ell ]$ is odd, the prime p is not a square in ${\mathbb F}_\lambda $ . So the image of

$$\begin{align*}a_p^2/(\varepsilon(p)p^{k-1}) = p\cdot a_p^2/p^k \end{align*}$$

in ${\mathbb F}_\lambda $ is not a square. Lemma 3.1 implies that . Therefore, .

4 An effective version of Theorem 1.1

Take a newform f with notation and assumptions as in Section 1.1. Let $\lambda $ be a nonzero prime ideal of R, and let $\ell $ be the prime lying under $\lambda $ . Let $k_\lambda $ be the subfield of ${\mathbb F}_\lambda $ generated by the image of $r_p$ modulo $\lambda $ with primes $p\nmid N\ell $ . Take any prime ideal $\Lambda $ of ${\mathcal O}$ that divides $\lambda $ .

In this section, we describe how to compute an explicit finite set S of prime ideals of R as in Theorem 1.1. First, some simple definitions:

• • Let ${\mathbb F}$ be an extension of ${\mathbb F}_\Lambda $ of degree $\gcd (2,\ell )$ .

• • Let $e_0=0$ if $\ell \geq k-1$ and $\ell \nmid N$ , and $e_0=\ell -2$ otherwise.

• • Let $e_1=0$ if N is odd, and $e_1=1$ otherwise.

• • Let $e_2=0$ if $\ell \geq 2k$ , and $e_2=1$ otherwise.

• • Define ${\mathcal M}=4^{e_1} \ell ^{e_2}\prod _{p|N} p$ .

We will prove the following in Section 5.

Theorem 4.1 Suppose that all the following conditions hold:

1. (a) For every integer $0\leq j \leq e_0$ and character $\chi \colon\ ({\mathbb Z}/N{\mathbb Z})^\times \to {\mathbb F}^\times $ , there is a prime $p\nmid N\ell $ such that $\chi (p)p^j \in {\mathbb F}$ is not a root of the polynomial $x^2 - a_px + \varepsilon (p)p^{k-1} \in {\mathbb F}_\Lambda [x]$ .

2. (b) For every nontrivial character $\chi \colon\ ({\mathbb Z}/{\mathcal M}{\mathbb Z})^\times \to \{\pm 1\}$ , there is a prime $p\nmid N\ell $ such that $\chi (p)=-1$ and $r_p \not \equiv 0 \ \pmod {\lambda }$ .

3. (c) If $\#k_\lambda \notin \{4,5\}$ , then at least one of the following holds:

   • • $\ell> 5k-4$ and $\ell \nmid N$ ,

   • • $\ell \equiv 0, \pm 1 \ \pmod {5}$ and $\#k_\lambda \neq \ell $ ,

   • • $\ell \equiv \pm 2\ \pmod {5}$ and $\#k_\lambda \neq \ell ^2$ ,

   • • there is a prime $p\nmid N\ell $ such that the image of $a_{p}^2/(\varepsilon (p) p^{k-1})$ in ${\mathbb F}_\lambda $ is not equal to $0$ , $1$ , and $4$ , and is not a root of $x^2-3x+1$ .

4. (d) If $\#k_\lambda \notin \{3,5,7\}$ , then at least one of the following holds:

   • • $\ell> 4k-3$ and $\ell \nmid N$ ,

   • • $\#k_\lambda \neq \ell $ ,

   • • there is a prime $p\nmid N\ell $ such that the image of $a_{p}^2/(\varepsilon (p) p^{k-1})$ in ${\mathbb F}_\lambda $ is not equal to $0$ , $1$ , $2$ , and $4$ .

5. (e) If $\#k_\lambda \in \{5,7\}$ , then for every nontrivial character $\chi \colon\ ({\mathbb Z}/ 4^{e_1}\ell N{\mathbb Z})^\times \to \{\pm 1\}$ , there is a prime $p\nmid N\ell $ such that $\chi (p)=1$ and $a_{p}^2/(\varepsilon (p) p^{k-1})\equiv 2 \ \pmod {\lambda }$ .

Then the group is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to $\operatorname {PSL}_2(k_\lambda )$ or $\operatorname {PGL}_2(k_\lambda )$ .

Remark 4.2 Note that the above conditions simplify greatly if one also assumes that $\ell \nmid N$ and $\ell>5k-4$ .

Although we will not prove it, Theorem 4.1 has been stated so that all the conditions (a)–(e) hold if and only if is conjugate to $\operatorname {PSL}_2(k_\lambda )$ or $\operatorname {PGL}_2(k_\lambda )$ . In particular, after considering enough primes p, one will obtain the minimal set S of Theorem 1.1 (one could use an effective version of Chebotarev density to make this a legitimate algorithm for computing this minimal set).

Let us now describe how to compute a set of exceptional primes as in Theorem 1.1. Define $M=N$ if N is odd and $M=4N$ otherwise. Set ${\mathcal M}':=4^{e_1} \prod _{p|N} p$ . We first choose some primes:

• • Let $q_1,\ldots , q_n$ be primes congruent to $1$ modulo N.

• • Let $p_1,\ldots , p_m \nmid N$ be primes with $r_{p_i}\neq 0$ such that for every nontrivial character $\chi \colon\ ({\mathbb Z}/{\mathcal M}' {\mathbb Z})^\times \to \{\pm 1\}$ , we have $\chi (p_i)=-1$ for some $1\leq i \leq m$ .

• • Let $p_0 \nmid N$ be a prime such that ${\mathbb Q}(r_{p_0})=K$ .

That such primes $p_1,\ldots , p_m$ exist is clear since the set of primes p with $r_p\neq 0$ has density $1$ . That such a prime q exists follows from Lemma 2.2 (the set of such q actually has positive density). Define the ring $R':={\mathbb Z}[a_{q}^2/\varepsilon (q)]$ ; it is an order in R.

Define S to be the set of nonzero primes $\lambda $ of R, dividing a rational prime $\ell $ , that satisfy one of the following conditions:

• • $\ell \leq 5k-4$ or $\ell \leq 7$ ,

• • $\ell | N$ ,

• • for all $1\leq i \leq n$ , we have $\ell =q_i$ or $r_{q_i} \equiv (1+q_i^{k-1})^2 \ \pmod {\lambda }$ ,

• • for some $1\leq i \leq m$ , we have $\ell =p_i$ or $r_{p_i} \equiv 0 \ \pmod {\lambda }$ ,

• • $\ell =q$ or $\ell $ divides $[R:R']$ .

Note that the set S is finite (the only part that is not immediate is that $r_{q_i} - (1+q_i^{k-1})^2\neq 0$ ; this follows from Deligne’s bound $|r_{q_i}|=|a_{q_i}| \leq 2q_i^{(k-1)/2}$ and $k>1$ ). The following is our effective version of Theorem 1.1.

Theorem 4.3 Take any nonzero prime ideal $\lambda \notin S$ of R, and let $\Lambda $ be any prime of ${\mathcal O}$ dividing $\lambda $ . Then the group is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to either $\operatorname {PSL}_2({\mathbb F}_\lambda )$ or $\operatorname {PGL}_2({\mathbb F}_\lambda )$ .

Proof Let $\ell $ be the rational prime lying under $\lambda $ . We shall verify the conditions of Theorem 4.1.

We first show that condition (a) of Theorem 4.1 holds. Take any integer $0\leq j \leq e_0$ and character $\chi \colon\ ({\mathbb Z}/N{\mathbb Z})^\times \to {\mathbb F}^\times ={\mathbb F}_\Lambda ^\times $ . We have $\ell>5k-4>k-1$ and $\ell \nmid N$ since $\lambda \notin S$ , so $e_0=0$ and hence $j=0$ . Take any $1\leq i \leq n$ . Since $q_i\equiv 1\ \pmod {N}$ and $j=0$ , we have $\chi (q_i)q_i^j=1$ and $\varepsilon (q_i)=1$ . Since $\lambda \notin S$ , we also have $q_i \nmid N \ell $ ( $q_i\nmid N$ is immediate from the congruence imposed on $q_i$ ). If $\chi (q_i)q_i^j=1$ was a root of $x^2-a_{q_i}x+\varepsilon (q_i)q_i^{k-1}$ in ${\mathbb F}_\Lambda [x]$ , then we would have $a_{q_i} \equiv 1 + q_i^{k-1} \ \pmod {\Lambda }$ ; squaring and using that $\varepsilon (q_i)=1$ , we deduce that $r_{q_i} \equiv (1+q_i^{k-1})^2 \ \pmod {\lambda }$ . Since $\lambda \notin S$ , we have $r_{q_i} \not \equiv (1+q_i^{k-1})^2 \ \pmod {\lambda }$ for some $1\leq i \leq n$ and hence $\chi (q_i)q_i^j$ is not a root of $x^2-a_{q_i}x+\varepsilon (q_i)q_i^{k-1}$ .

We now show that condition (b) of Theorem 4.1 holds. We have $e_2=0$ since $\lambda \notin S$ , and hence ${\mathcal M}'={\mathcal M}$ . Take any nontrivial character $\chi \colon\ ({\mathbb Z}/{\mathcal M}{\mathbb Z})^\times \to \{\pm 1\}$ . By our choice of primes $p_1,\ldots ,p_m$ , we have $\chi (p_i)=-1$ for some $1\leq i \leq m$ . The prime $p_i$ does not divide $N\ell $ (that $p_i\neq \ell $ follows since $\lambda \notin S$ ). Since $\lambda \notin S$ , we have $r_{p_i}\not \equiv 0 \ \pmod {\lambda }$ .

Since $\lambda \notin S$ , the prime $\ell \nmid N$ is greater that $7$ , $4k-3$ , and $5k-4$ . Conditions (c)–(e) of Theorem 4.1 all hold.

Theorem 4.1 now implies that is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to either $\operatorname {PSL}_2(k_\lambda )$ or $\operatorname {PGL}_2(k_\lambda )$ . It remains to prove that $k_\lambda ={\mathbb F}_\lambda $ . We have $q\neq \ell $ since $\lambda \notin S$ . The image of the reduction map $R' \to {\mathbb F}_\lambda $ thus lies in $k_\lambda $ . We have $\ell \nmid [R:R']$ since $\lambda \notin S$ , so the map $R'\to {\mathbb F}_\lambda $ is surjective. Therefore, $k_\lambda ={\mathbb F}_\lambda $ .

5 Proof of Theorem 4.1

5.1 Some group theory

Fix a prime $\ell $ and an integer $r\geq 1$ . A Borel subgroup of $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ is a subgroup conjugate to the subgroup of upper triangular matrices. A split Cartan subgroup of $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ is a subgroup conjugate to the subgroup of diagonal matrices. A nonsplit Cartan subgroup of $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ is a subgroup that is cyclic of order $(\ell ^r)^2-1$ . Fix a Cartan subgroup ${\mathcal C}$ of $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ . Let ${\mathcal N}$ be the normalizer of ${\mathcal C}$ in $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ . One can show that $[{\mathcal N}:{\mathcal C}] = 2$ and that $\operatorname {\mathrm {tr}}(g)=0$ and $g^2$ is scalar for all $g\in {\mathcal N}-{\mathcal C}$ .

Lemma 5.1 Fix a prime $\ell $ and an integer $r\geq 1$ . Let G be a subgroup of $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ , and let be its image in $\operatorname {PGL}_2({\mathbb F}_{\ell ^r})$ . Then at least one of the following holds:

1. (1) G is contained in a Borel subgroup of $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ .

2. (2) G is contained in the normalizer of a Cartan subgroup of $\operatorname {GL}_2({\mathbb F}_{\ell ^r})$ .

3. (3) is isomorphic to $\mathfrak {A}_4$ .

4. (4) is isomorphic to $\mathfrak {S}_4$ .

5. (5) is isomorphic to $\mathfrak {A}_5$ .

6. (6) is conjugate to $\operatorname {PSL}_2({\mathbb F}_{\ell ^s})$ or $\operatorname {PGL}_2({\mathbb F}_{\ell ^s})$ for some integer s dividing r.

Proof This can be deduced directly from a theorem of Dickson (cf. [10, Satz 8.27]), which will give the finite subgroups of $\operatorname {PSL}_2(\overline {{\mathbb F}}_\ell ) = \operatorname {PGL}_2(\overline {{\mathbb F}}_\ell )$ . The finite subgroups of $\operatorname {PGL}_2({\mathbb F}_{\ell ^r})$ have been worked out in [9].

Lemma 5.2 Fix a prime $\ell $ and an integer $r\geq 1$ . Take a matrix $A\in \operatorname {GL}_2({\mathbb F}_{\ell ^r})$ , and let m be its order in $\operatorname {PGL}_2({\mathbb F}_{\ell ^r})$ .

1. (i) Suppose that $\ell \nmid m$ . If m is $1$ , $2$ , $3$ , or $4$ , then $\operatorname {\mathrm {tr}}(A)^2/\det (A)$ is $4$ , $0$ , $1$ , or $2$ , respectively. If $m=5$ , then $\operatorname {\mathrm {tr}}(A)^2/\det (A)$ is a root of $x^2-3x+1$ .

2. (ii) If $\ell | m$ , then $\operatorname {\mathrm {tr}}(A)^2/\det (A) = 4$ .

Proof The quantity $\operatorname {\mathrm {tr}}(A)^2/\det (A)$ does not change if we replace A by a scalar multiple or by a conjugate in $\operatorname {GL}_2(\overline {{\mathbb F}}_\ell )$ . If $\ell \nmid m$ , then we may thus assume that $A=\left (\begin {smallmatrix} \zeta & 0 \\ 0 & 1 \end {smallmatrix}\right )$ where $\zeta \in \overline {{\mathbb F}}_\ell $ has order m. We have $\operatorname {\mathrm {tr}}(A)^2/\det (A) = \zeta +\zeta ^{-1} + 2$ , which is $4$ , $0$ , $1$ , or $2$ when m is $1$ , $2$ , $3$ , or $4$ , respectively. If $m=5$ , then $ \zeta +\zeta ^{-1} + 2$ is a root of $x^2-3x+1$ . If $\ell | m$ , then after conjugating and scaling, we may assume that $A=\left (\begin {smallmatrix} 1 & 1 \\ 0 & 1 \end {smallmatrix}\right )$ and hence $\operatorname {\mathrm {tr}}(A)^2/\det (A) =4$ .

5.2 Image of inertia at $\ell $

Fix an inertia subgroup ${\mathcal I}_\ell $ of $G=\operatorname {\mathrm {Gal}}({\overline {\mathbb Q}}/{\mathbb Q})$ for the prime $\ell $ ; it is uniquely defined up to conjugacy. The following gives important information concerning the representation for large $\ell $ . Let $\chi _\ell \colon\ G\twoheadrightarrow {\mathbb F}_\ell ^\times $ be the character such that for each prime $p\nmid \ell $ , $\chi _\ell $ is unramified at p and $\chi _\ell (\operatorname {\mathrm {Frob}}_p)\equiv p \ \pmod {\ell }$ .

Lemma 5.3 Fix a prime $\ell \geq k-1$ for which $\ell \nmid 2N$ . Let $\Lambda $ be a prime ideal of ${\mathcal O}$ dividing $\ell $ , and set $\lambda =\Lambda \cap R$ .

1. (i) Suppose that $r_\ell \not \equiv 0 \ \pmod {\lambda }$ . After conjugating by a matrix in $\operatorname {GL}_2({\mathbb F}_\Lambda )$ , we have

   
   In particular, contains a cyclic group of order $(\ell -1)/\gcd (\ell -1,k-1)$ .

2. (ii) Suppose that $r_\ell \equiv 0 \ \pmod {\lambda }$ . Then is absolutely irreducible and is cyclic. Furthermore, the group is cyclic of order $(\ell +1)/\gcd (\ell +1,k-1)$ .

Proof Parts (i) and (ii) follow from Theorems 2.5 and 2.6, respectively, of [8]; they are theorems of Deligne and Fontaine, respectively. We have used that $r_\ell =a_\ell ^2/\varepsilon (\ell ) \in R$ is congruent to $0$ modulo $\lambda $ if and only if $a_\ell \in {\mathcal O}$ is congruent to $0$ modulo $\Lambda $ .

5.3 Borel case

Suppose that

is a reducible subgroup of $\operatorname {GL}_2({\mathbb F})$ . There are, thus, characters $\psi _1,\psi _2 \colon\ G \to {\mathbb F}^\times $ such that after conjugating the ${\mathbb F}$ -representation

, we have

The characters $\psi _1$ and $\psi _2$ are unramified at each prime $p\nmid N\ell $ since

is unramified at such primes.

Lemma 5.4 For each $i\in \{1,2\}$ , there is a unique integer $0\leq m_i < \ell -1$ such that $\psi _i\chi _\ell ^{-m_i} \colon\ G \to {\mathbb F}^\times $ is unramified at all primes $p\nmid N$ . If $\ell \geq k-1$ and $\ell \nmid N$ , then $m_1$ or $m_2$ is $0$ .

Proof The existence and uniqueness of $m_i$ is an easy consequence of class field theory for ${\mathbb Q}_\ell $ . A choice of embedding ${\overline {\mathbb Q}}\subseteq {\overline {\mathbb Q}}_\ell $ induces an injective homomorphism $G_{{\mathbb Q}_\ell }:=\operatorname {\mathrm {Gal}}({\overline {\mathbb Q}}_\ell /{\mathbb Q}_\ell ) \hookrightarrow G$ . Let ${\mathbb Q}_\ell ^{\operatorname {ab}}$ be the maximal abelian extension of ${\mathbb Q}_\ell $ in ${\overline {\mathbb Q}}_\ell $ . Restricting $\psi _i$ to $G_{{\mathbb Q}_\ell }$ , we obtain a representation $\psi _i\colon\ G_{{\mathbb Q}_\ell }^{\operatorname {ab}}:=\operatorname {\mathrm {Gal}}({\mathbb Q}_\ell ^{\operatorname {ab}}/{\mathbb Q}_\ell ) \to {\mathbb F}^\times $ . By local class field, the inertia subgroup ${\mathcal I}$ of $G_{{\mathbb Q}_\ell }^{\operatorname {ab}}$ is isomorphic to ${\mathbb Z}_\ell ^\times $ . Since $\ell $ does not divide the cardinality of ${\mathbb F}^\times $ , we find that $\psi _i|_{{\mathcal I}}$ factors through a group isomorphic to ${\mathbb F}_\ell ^\times $ . The character $\psi _i|_{{\mathcal I}}$ must agree with a power of $\chi _\ell |_{{\mathcal I}}$ since $\chi _\ell \colon\ G_{{\mathbb Q}_\ell } \to {\mathbb F}_\ell ^\times $ satisfies $\chi _\ell ({\mathcal I})={\mathbb F}_\ell ^\times $ and ${\mathbb F}_\ell ^\times $ is cyclic.

The second part of the lemma follows immediately from Lemma 5.3.

Take any $i\in \{1,2\}$ . By Lemma 5.4, there is a unique $0\leq m_i < \ell -1$ such that the character

$$\begin{align*}\tilde{\psi}_i:=\psi_i\chi_\ell^{-m_i}\colon\ G\to {\mathbb F}^\times \end{align*}$$

is unramified at $\ell $ and at all primes $p\nmid N$ . There is a character $\chi _i \colon\ ({\mathbb Z}/N_i{\mathbb Z})^\times \to {\mathbb F}^\times $ with $N_i\geq 1$ dividing some power of N and $\ell \nmid N_i$ such that $\tilde \psi _i(\operatorname {\mathrm {Frob}}_p) = \chi _i(p)$ for all $p\nmid N\ell $ . We may assume that $\chi _i$ is taken so that $N_i$ is minimal.

Lemma 5.5 The integer $N_i$ divides N.

Proof We first recall the notion of an Artin conductor. Consider a representation $\rho \colon\ G\to \operatorname {\mathrm {Aut}}_{{\mathbb F}}(V)$ , where V is a finite-dimensional ${\mathbb F}$ -vector space. Take any prime $p\neq \ell $ . A choice of embedding ${\overline {\mathbb Q}}\subseteq {\overline {\mathbb Q}}_p$ induces an injective homomorphism $\operatorname {\mathrm {Gal}}({\overline {\mathbb Q}}_p/{\mathbb Q}_p) \hookrightarrow G$ . Choose any finite Galois extension $L/{\mathbb Q}_p$ for which $\rho (\operatorname {\mathrm {Gal}}({\overline {\mathbb Q}}_p/L))=\{I\}$ . For each $i\geq 0$ , let $H_i$ be the ith ramification subgroup of $\operatorname {\mathrm {Gal}}(L/{\mathbb Q}_p)$ with respect to the lower numbering. Define the integer

$$\begin{align*}f_p(\rho)= \sum_{i\geq 0} [H_0:H_i]^{-1}\cdot \dim_{{\mathbb F}} V/V^{H_i}. \end{align*}$$

The Artin conductor of $\rho $ is the integer $N(\rho ):=\prod _{p\neq \ell } p^{f_p(\rho )}$ .

Using that the character $\tilde \psi _i\colon\ G \to {\mathbb F}^\times $ is unramified at $\ell $ , one can verify that $N(\tilde \psi _i)=N_i$ . Consider our representation . For a fixed prime $p\neq \ell $ , take L and $H_i$ as above. The semisimplification of is $V_1\oplus V_2$ , where $V_i$ is the one-dimensional representation given by $\psi _i$ . We have since $\dim _{{\mathbb F}} V^{H_i} \leq \dim _{{\mathbb F}} V_1^{H_i} + \dim _{{\mathbb F}} V_2^{H_i}$ . By using this for all $p\neq \ell $ , we deduce that $N(\psi _1) N(\psi _2)=N_1 N_2$ divides . The lemma follows since divides N (cf. [11, Proposition 0.1]).

Fix an $i\in \{1,2\}$ ; if $\ell \geq k-1$ and $\ell \nmid N$ , then we may suppose that $m_i=0$ by Lemma 5.4. Since the conductor of $\chi _i$ divides N by Lemma 5.5, assumption (a) implies that there is a prime $p\nmid N\ell $ for which $\chi _i(p) p^{m_i} \in {\mathbb F}$ is not a root of $x^2-a_p x + \varepsilon (p)p^{k-1} \in {\mathbb F}[x]$ . However, this is a contradiction since

$$\begin{align*}\chi_i(p) p^{m_i} = \tilde\psi_i(\operatorname{\mathrm{Frob}}_p) \chi_\ell(\operatorname{\mathrm{Frob}}_p)^{m_i} = \psi_i(\operatorname{\mathrm{Frob}}_p) \end{align*}$$

is a root of $x^2-a_p x + \varepsilon (p)p^{k-1}$ .

Therefore, the ${\mathbb F}$ -representation is irreducible. In particular, is not contained in a Borel subgroup of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ .

5.4 Cartan case

Lemma 5.6 The group is not contained in a Cartan subgroup of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ .

Proof Suppose that is contained in a Cartan subgroup ${\mathcal C}$ of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ . If $\ell =2$ , then ${\mathcal C}$ is reducible as a subgroup of $\operatorname {GL}_2({\mathbb F})$ since ${\mathbb F}/{\mathbb F}_\Lambda $ is a quadratic extension. However, we saw in Section 5.3 that is an irreducible subgroup of $\operatorname {GL}_2({\mathbb F})$ . Therefore, $\ell $ is odd. If ${\mathcal C}$ is split, then is a reducible subgroup of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ . This was ruled out in Section 5.3, so ${\mathcal C}$ must be a nonsplit Cartan subgroup with $\ell $ odd.

Recall that the representation is odd, i.e., if $c\in G$ is an element corresponding to complex conjugation under some embedding ${\overline {\mathbb Q}}\hookrightarrow {\mathbb C}$ , then . Therefore, has order $2$ and determinant $-1\neq 1$ (this last inequality uses that $\ell $ is odd). A nonsplit Cartan subgroup ${\mathcal C}$ of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ is cyclic and hence $-I$ is the unique element of ${\mathcal C}$ of order $2$ . Since $\det (-I)= 1$ , we find that does not belong to ${\mathcal C}$ ; this gives the desired contradiction.

5.5 Normalizer of a Cartan case

Suppose that

is contained in the normalizer ${\mathcal N}$ of a Cartan subgroup ${\mathcal C}$ of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ . The group ${\mathcal C}$ has index $2$ in ${\mathcal N}$ , so we obtain a character

The character $\beta _\Lambda $ is nontrivial since

by Lemma 5.6.

Lemma 5.7 The character $\beta _\Lambda $ is unramified at all primes $p\nmid N\ell $ . If $\ell \geq 2k$ and $\ell \nmid N$ , then the character $\beta _\Lambda $ is also unramified at $\ell $ .

Proof The character $\beta _\Lambda $ is unramified at each prime $p\nmid N\ell $ since is unramified at such primes. Now, suppose that $\ell \geq 2k$ and $\ell \nmid N$ . We have $\ell>2$ , so $\ell \nmid |{\mathcal N}|$ and hence Lemma 5.3 implies that is cyclic. Moreover, Lemma 5.3 implies that is cyclic of order $d\geq (\ell -1)/(k-1)$ . Our assumption $\ell \geq 2k$ ensures that $d>2$ .

Now, take a generator g of . Suppose that $\beta _\Lambda $ is ramified at $\ell $ and hence g belongs to ${\mathcal N}-{\mathcal C}$ . The condition $g\in {\mathcal N}-{\mathcal C}$ implies that $g^2$ is a scalar matrix and hence is a group of order $1$ or $2$ . This contradicts $d>2$ , so $\beta _\Lambda $ is unramified at $\ell $ .

Let $\chi $ be the primitive Dirichlet character that satisfies $\beta _\Lambda (\operatorname {\mathrm {Frob}}_p)=\chi (p)$ for all primes $p\nmid N\ell $ . Since $\beta _\Lambda $ is a quadratic character, Lemma 5.7 implies that the conductor of $\chi $ divides ${\mathcal M}$ . The character $\chi $ is nontrivial since $\beta _\Lambda $ is nontrivial. Assumption (b) implies that there is a prime $p\nmid N\ell $ satisfying $\chi (p)=-1$ and $r_p \not \equiv 0 \ \pmod {\lambda }$ . We thus have $g\in {\mathcal N}-{\mathcal C}$ and $\operatorname {\mathrm {tr}}(g)\neq 0$ , where . However, this contradicts that $\operatorname {\mathrm {tr}}(A)=0$ for all $A\in {\mathcal N}-{\mathcal C}$ .

Therefore, the image of does not lie in the normalizer of a Cartan subgroup of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ .

5.6 $\mathfrak {A}_5$ case

Assume that is isomorphic to $\mathfrak {A}_5$ with $\#k_\lambda \notin \{4,5\}$ .

The image of $r_p/p^{k-1}= a_p^2/(\varepsilon (p)p^{k-1})$ in ${\mathbb F}_\lambda $ is equal to $\operatorname {\mathrm {tr}}(A)^2/\det (A)$ with . Every element of $\mathfrak {A}_5$ has order $1$ , $2$ , $3$ , or $5$ , so Lemma 5.2 implies that the image of $r_p/p^{k-1}$ in ${\mathbb F}_\lambda $ is $0$ , $1$ , $4$ , or is a root of $x^2-3x+1$ for all $p\nmid N\ell $ . If $\lambda | 5$ , then Lemma 5.2 implies that $k_\lambda ={\mathbb F}_5$ , which is excluded by our assumption on $k_\lambda $ . So $\lambda \nmid 5$ and Lemma 5.2 ensures that $k_\lambda $ is the splitting field of $x^2-3x+1$ over ${\mathbb F}_\ell $ . So $k_\lambda $ is ${\mathbb F}_\ell $ if $\ell \equiv \pm 1 \ \pmod {5}$ and ${\mathbb F}_{\ell ^2}$ if $\ell \equiv \pm 2 \ \pmod {5}$ .

From assumption (c), we find that $\ell> 5k-4$ and $\ell \nmid N$ . By Lemma 5.3, the group contains an element of order at least $(\ell -1)/(k-1)> ((5k-4)-1)/(k-1) = 5$ . This is a contradiction since $\mathfrak {A}_5$ has no elements with order greater than $5$ .

5.7 $\mathfrak {A}_4$ and $\mathfrak {S}_4$ cases

Suppose that is isomorphic to $\mathfrak {A}_4$ or $\mathfrak {S}_4$ with $\#k_\lambda \neq 3$ .

First suppose that $\#k_\lambda \notin \{5,7\}$ . The image of $r_p/p^{k-1}= a_p^2/(\varepsilon (p)p^{k-1})$ in ${\mathbb F}_\lambda $ is equal to $\operatorname {\mathrm {tr}}(A)^2/\det (A)$ with . Since every element of $\mathfrak {S}_4$ has order at most $4$ , Lemma 5.2 implies that $r_p/p^{k-1}$ is congruent to $0$ , $1$ , $2$ , or $4$ modulo $\lambda $ for all primes $p\nmid N\ell $ . In particular, $k_\lambda ={\mathbb F}_\ell $ . By assumption (d), we must have $\ell>4k-3$ and $\ell \nmid N$ . By Lemma 5.3, the group contains an element of order at least $(\ell -1)/(k-1)> ((4k-3)-1)/(k-1) = 4$ . This is a contradiction since $\mathfrak {S}_4$ has no elements with order greater than $4$ .

Now, suppose that $\#k_\lambda \in \{5,7\}$ . By assumption (e), with any $\chi $ , there is a prime $p\nmid N\ell $ such that $a_p^2/(\varepsilon (p)p^{k-1}) \equiv 2 \ \pmod {\lambda }$ . The element

has order $1$ , $2$ , $3$ , or $4$ . By Lemma 5.2, we deduce that g has order $4$ . Since $\mathfrak {A}_4$ has no elements of order $4$ , we deduce that

is isomorphic to $\mathfrak {S}_4$ . Let $H'$ be the unique index $2$ subgroup of H; it is isomorphic to $\mathfrak {A}_4$ . Define the character

The quadratic character $\beta $ corresponds to a Dirichlet character $\chi $ whose conductor divides $4^e \ell N$ . By assumption (e), there is a prime $p\nmid N\ell $ such that $\chi (p)=1$ and $a_p^2/(\varepsilon (p)p^{k-1})\equiv 2 \ \pmod {\lambda }$ . Since $\beta (\operatorname {\mathrm {Frob}}_p)=\chi (p)=1$ , we have

. Since $H'\cong \mathfrak {A}_4$ , Lemma 5.2 implies that the image of $a_p^2/(\varepsilon (p)p^{k-1})$ in ${\mathbb F}_\lambda $ is $0$ , $1$ , or $4$ . This contradicts $a_p^2/(\varepsilon (p)p^{k-1})\equiv 2 \ \pmod {\lambda }$ .

Therefore, the image of is not isomorphic to either of the groups $\mathfrak {A}_4$ or $\mathfrak {S}_4$ .

5.8 End of proof

In Section 5.3, we saw that is not contained in a Borel subgroup of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ . In Section 5.5, we saw that is not contained in the normalizer of a Cartan subgroup of $\operatorname {GL}_2({\mathbb F}_\Lambda )$ .

In Section 5.6, we showed that if $\#k_\lambda \notin \{4,5\}$ , then is not isomorphic to $\mathfrak {A}_5$ . We want to exclude the cases $\#k_\lambda \in \{4,5\}$ since $\operatorname {PSL}_2({\mathbb F}_4)$ and $\operatorname {PSL}_2({\mathbb F}_5)$ are both isomorphic to $\mathfrak {A}_5$ .

In Section 5.7, we showed that if $\#k_\lambda \neq 3$ , then is not isomorphic to $\mathfrak {A}_4$ and not isomorphic to $\mathfrak {S}_4$ . We want to exclude the case $\#k_\lambda =3$ since $\operatorname {PSL}_2({\mathbb F}_3)$ and $\operatorname {PGL}_2({\mathbb F}_3)$ are isomorphic to $\mathfrak {A}_4$ and $\mathfrak {S}_4$ , respectively.

By Lemma 5.1, the group must be conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to $\operatorname {PSL}_2({\mathbb F}')$ or $\operatorname {PGL}_2({\mathbb F}')$ , where ${\mathbb F}'$ is a subfield of ${\mathbb F}_\Lambda $ . One can then show that ${\mathbb F}'$ is the subfield of ${\mathbb F}_\Lambda $ generated by the set . By the Chebotarev density theorem, the field ${\mathbb F}'$ is the subfield of ${\mathbb F}_\Lambda $ generated by the images of $a_p^2/(\varepsilon (p) p^{k-1})=r_p/p^{k-1}$ with $p\nmid N\ell $ . Therefore, ${\mathbb F}'=k_\lambda $ and hence is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to $\operatorname {PSL}_2(k_\lambda )$ or $\operatorname {PGL}_2(k_\lambda )$ .

6 Examples

6.1 Example from Section 1.2

Let f be the newform from Section 1.2. We have $E={\mathbb Q}(i)$ . We know that $\Gamma \neq 1$ since $\varepsilon $ is nontrivial. Therefore, $\Gamma =\operatorname {\mathrm {Gal}}({\mathbb Q}(i)/{\mathbb Q})$ and $K=E^{\Gamma }$ equals ${\mathbb Q}$ . So $\Gamma $ is generated by complex conjugation and we have for $p\nmid N$ . As noted in Section 1.2, this implies that $r_p$ is a square in ${\mathbb Z}$ for all $p\nmid N$ and hence L equals $K={\mathbb Q}$ . Fix a prime $\ell =\lambda $ and a prime ideal $\Lambda | \ell $ of ${\mathcal O}={\mathbb Z}[i]$ .

Set $q_1=109$ and $q_2=379$ ; they are primes that are congruent to $1$ modulo $27$ . Set $p_1=5$ , and we have $\chi (p_1)=-1$ , where $\chi $ is the unique nontrivial character $({\mathbb Z}/3{\mathbb Z})^\times \to \{\pm 1\}$ . Set $q=5$ ; the field ${\mathbb Q}(r_q)$ equals $K={\mathbb Q}$ and hence ${\mathbb Z}[r_{q}]={\mathbb Z}$ .

One can verify that $a_{109}=164$ , $a_{379}=704$ , and $a_5=-3i$ , so $r_{109}=164^2$ , $r_{379}=704^2$ , and $r_5=3^2$ . We have $r_{109} - (1+109^2)^2 = -2^2\times 3^3\times 7\times 19\times 31\times 317$ and $r_{379} - (1+379^2)^2 = -2^2 \times 3^3\times 2,647 \times 72,173$ . So if $\ell \geq 11$ , then there is an $i\in \{1,2\}$ such that $\ell \neq q_i$ and $r_{q_i}\not \equiv (1+q_i^2)^2\ \pmod {\ell }$ .

Let S be the set from Section 4 with the above choice of $q_1$ , $q_2$ , $p_1$ , and q. We find that $S = \{2,3,5,7,11\}$ . Theorem 4.3 implies that is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to $\operatorname {PSL}_2({\mathbb F}_\ell )$ when $\ell>11$ .

Now, take $\ell \in \{7,11\}$ . Choose a prime ideal $\Lambda $ of ${\mathcal O}$ dividing $\ell $ . We have $e_0=e_1=e_2=0$ and ${\mathcal M}=3$ . The subfield $k_\ell $ generated over ${\mathbb F}_\ell $ by the images of $r_p$ modulo $\ell $ with $p\nmid N\ell $ is of course ${\mathbb F}_\ell $ (since the $r_p$ are rational integers). We now verify the conditions of Theorem 4.1.

We first check condition (a). Suppose there is a character $\chi \colon\ ({\mathbb Z}/27{\mathbb Z})^\times \to {\mathbb F}_\ell ^\times $ such that $\chi (q_2)$ is a root of $x^2-a_{q_2} x + \varepsilon (q_2) q_2^2$ modulo $\ell $ . Since $q_2\equiv 1\ \pmod {27}$ and $a_{q_2}=704$ , we find that $1$ is a root of $x^2-a_{q_2} x + q_2^2 \in {\mathbb F}_\ell [x]$ . Therefore, $a_{q_2} \equiv 1+q_2^2 \ \pmod {\ell }$ and hence $r_{q_2}^2 = a_{q_2}^2 \equiv (1+ q_2^2)^2 \ \pmod {\ell }$ . However, $r_{379} - (1+379^2)^2 \not \equiv 0 \ \pmod {\ell }$ from the explicit value of it computed above. So the character $\chi $ does not exist and we have verified condition (a).

We now check condition (b). Let $\chi \colon\ ({\mathbb Z}/3{\mathbb Z})^\times \to \{\pm 1\}$ be the nontrivial character. We have $\chi (5)=-1$ and $r_5 = 9 \not \equiv 0 \ \pmod {\ell }$ . Therefore, condition (b) holds.

We now check condition (c). If $\ell =7$ , we have $\ell \equiv 2 \ \pmod {5}$ and $\#k_\ell = \ell \neq \ell ^2$ , so condition (c) holds. Take $\ell =11$ . We have $a_5^2/(\varepsilon (5) 5^2) = 9/5^2 \equiv 3 \ \pmod {11}$ , which verifies condition (c).

Condition (d) holds since $\#k_\ell = 5$ if $\ell = 7$ , and $\ell>4k-3=9$ and $\ell \nmid N$ if $\ell =11$ .

Finally, we explain why condition (e) holds when $\ell =7$ . Let $\chi \colon\ ({\mathbb Z}/7\cdot 27 {\mathbb Z})^\times \to \{\pm 1\}$ be any nontrivial character. A quick computation shows that there is a prime $p\in \{13,37,41\}$ such that $\chi (p)=1$ and that $a_p^2/(\varepsilon (p) p^2) \equiv 2 \ \pmod {7}$ .

Theorem 4.1 implies that is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to $\operatorname {PSL}_2({\mathbb F}_\ell )$ or $\operatorname {PGL}_2({\mathbb F}_\ell )$ . Since $L=K$ , the group isomorphic to $\operatorname {PSL}_2({\mathbb F}_{\ell })$ by Theorem 1.2(i).

6.2 Example from §1.3

Let f be a newform as in Section 1.3; we have $k=3$ and $N=160$ . The Magma code below verifies that f is uniquely determined up to replacing by a quadratic twist and then a Galois conjugate. So the group , up to isomorphism, does not depend on the choice of f.

Define $b = \zeta _{13}^1 + \zeta _{13}^5+\zeta _{13}^8 +\zeta _{13}^{12}$ , where $\zeta _{13}$ is a primitive $13$ th root of unity in ${\overline {\mathbb Q}}$ (note that $\{1,5,8,12\}$ is the unique index $3$ subgroup of ${\mathbb F}_{13}^\times $ ). The characteristic polynomial of b is $x^3 + x^2 - 4x + 1$ , and hence ${\mathbb Q}(b)$ is the unique cubic extension of ${\mathbb Q}$ in ${\mathbb Q}(\zeta _{13})$ . The code below shows that the coefficient field E is equal to ${\mathbb Q}(b,i)$ (it is a degree $6$ extension of ${\mathbb Q}$ that contains roots of $x^3 + x^2 - 4x + 1$ and $x^2+1$ ).

Fix notation as in Section 2.1. We have $\Gamma \neq 1$ since $\varepsilon $ is nontrivial. The character $\chi _\gamma ^2$ is trivial for $\gamma \in \Gamma $ (since $\chi _\gamma $ is always a quadratic character times some power of $\varepsilon $ ). Therefore, $\Gamma $ is a $2$ -group. The field $K=E^{\Gamma} $ is thus ${\mathbb Q}(b)$ , which is the unique cubic extension of ${\mathbb Q}$ in E. Therefore, $r_p = a_p^2/\varepsilon (p)$ lies in $K={\mathbb Q}(b)$ for all $p\nmid N$ .

The code below verifies that $r_3$ , $r_7$ , and $r_{11}$ are squares in K that do not lie in ${\mathbb Q}$ (and, in particular, are nonzero). Since $3$ , $7$ , and $11$ generate the group $({\mathbb Z}/40{\mathbb Z})^\times $ , we deduce from Lemma 2.3 that the field $L=K(\{\sqrt {r_p}:p\nmid N\})$ is equal to K.

Let $N_{K/{\mathbb Q}}\colon\ K\to {\mathbb Q}$ be the norm map. The following code verifies that $N_{K/{\mathbb Q}}(r_3) = 2^6$ , $N_{K/{\mathbb Q}}(r_7) = 2^6$ , $N_{K/{\mathbb Q}}(r_{11}) = 2^{12}5^4$ , $N_{K/{\mathbb Q}}(r_{13})=2^{12} 13^2$ , $N_{K/{\mathbb Q}}(r_{17}) = 2^{18}5^2$ , and that

(6.1) $$ \begin{align} \gcd\Big( 641\cdot N_{K/{\mathbb Q}}(r_{641}-(1+642^2)^2), \, 1,061\cdot N_{K/{\mathbb Q}}(r_{1061}-(1+1,061^2)^2)\Big) = 2^{12}.\\[-25pt] \nonumber \end{align} $$

Set $q_1=641$ and $q_2=1,061$ ; they are primes congruent to $1$ modulo $160$ . Let $\lambda $ be a prime ideal of R dividing a rational prime $\ell>3$ . From (6.1), we find that $\ell \neq q_i$ and $r_{q_i} \not \equiv (1+q_i^2)^2 \ \pmod {\lambda }$ for some $i\in \{1,2\}$ (otherwise, $\lambda $ would divide $2$ ).

Set $p_1=3$ , $p_2= 7$ , and $p_3=11$ . For each nontrivial quadratic character $\chi \colon\ ({\mathbb Z}/40 {\mathbb Z})^\times \to \{\pm 1\}$ , we have $\chi (p_i)=-1$ for some prime $i\in \{1,2,3\}$ (since $3$ , $7$ , and $11$ generate the group $({\mathbb Z}/40{\mathbb Z})^\times $ ). From the computed values of $N_{K/{\mathbb Q}}(r_p)$ , we find that $r_{p_i} \not \equiv 0 \ \pmod {\lambda }$ for all $i\in \{1,2,3\}$ and all nonzero prime ideals $\lambda \nmid N$ of R.

Set $q=3$ . We have noted that $r_{q}\in K-{\mathbb Q}$ , so $K={\mathbb Q}(r_{q})$ . The index of the order ${\mathbb Z}[r_{q}]$ in R is a power of $2$ since $N_{K/{\mathbb Q}}(q)$ is a power of $2$ .

Let S be the set from Section 4 with the above choice of $q_1$ , $q_2$ , $p_1$ , $p_2$ , $p_3$ , and q. The above computations show that S consists of the prime ideals $\lambda $ of R that divide a prime $\ell \leq 11$ .

Now, let $\ell $ be an odd prime congruent to $\pm 2$ , $\pm 3$ , $\pm 4$ , or $\pm 6$ modulo $13$ . Since K is the unique cubic extension in ${\mathbb Q}(\zeta _{13})$ , we find that the ideal $\lambda := \ell R$ is prime in R and that ${\mathbb F}_\lambda \cong {\mathbb F}_{\ell ^3}$ . The above computations show that $\lambda \notin S$ when $\ell \notin \{3,7,11\}$ . Theorem 4.3 implies that if $\ell \notin \{3,7,11\}$ , then is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to $\operatorname {PSL}_2({\mathbb F}_\lambda )$ or $\operatorname {PGL}_2({\mathbb F}_\lambda )$ , where $\Lambda $ is a prime ideal of ${\mathcal O}$ dividing $\lambda $ . So if $\ell \notin \{3,7,11\}$ , the group isomorphic to $\operatorname {PSL}_2({\mathbb F}_\lambda )\cong \operatorname {PSL}_2({\mathbb F}_{\ell ^3})$ by Theorem 1.2(i) and the equality $L=K$ .

Now, take $\lambda = \ell R$ with $\ell \in \{3,7,11\}$ ; it is a prime ideal. Choose a prime ideal $\Lambda $ of ${\mathcal O}$ dividing $\lambda $ . We noted above that ${\mathbb Z}[r_3]$ is an order in R with index a power of $2$ ; the same argument shows that this also holds for the order ${\mathbb Z}[r_7]$ . Therefore, the field $k_\lambda $ generated over ${\mathbb F}_\ell $ by the images of $r_p$ modulo $\lambda $ with $p\nmid N\ell $ is equal to ${\mathbb F}_\lambda $ . Since $\#{\mathbb F}_\lambda =\ell ^3$ , we find that conditions (c)–(e) of Theorem 4.1 hold.

We now show that condition (a) of Theorem 4.1 holds for our fixed $\Lambda $ . We have $e_0=0$ , so take any character $\chi \colon\ ({\mathbb Z}/N{\mathbb Z})^\times \to {\mathbb F}_\Lambda ^\times $ . We claim that $\chi (q_i)\in {\mathbb F}_\Lambda $ is not a root of $x^2-a_{q_i} x + \varepsilon (q_i) q_i^2$ for some $i\in \{1,2\}$ . Since $q_i\equiv 1\ \pmod {N}$ , the claim is equivalent to showing that $a_{q_i} \not \equiv 1 + q_i^2 \ \pmod {\Lambda }$ for some $i\in \{1,2\}$ . So we need to prove that $r_{q_i} \equiv (1 + q_i^2)^2 \ \pmod {\lambda }$ for some $i\in \{1,2\}$ ; this is clear since otherwise $\ell $ divides the quantity (6.1). This completes our verification of condition (a).

We now show that condition (b) of Theorem 4.1 holds. We have $r_p \not \equiv 0 \ \pmod {\lambda }$ for all primes $p\in \{3,7,11,13,17\}$ ; this is a consequence of $N_{K/{\mathbb Q}}(r_p)\not \equiv 0 \ \pmod {\ell }$ . We have ${\mathcal M}= 120$ if $\ell =3$ and ${\mathcal M}=40$ otherwise. Condition (b) holds since $({\mathbb Z}/{\mathcal M}{\mathbb Z})^\times $ is generated by the primes $p\in \{3,7,11,13,17\}$ for which $p\nmid {\mathcal M}\ell $ .

Theorem 4.1 implies that is conjugate in $\operatorname {PGL}_2({\mathbb F}_\Lambda )$ to $\operatorname {PSL}_2({\mathbb F}_\lambda )$ or $\operatorname {PGL}_2({\mathbb F}_\lambda )$ . Since $L=K$ , the group is isomorphic to $\operatorname {PSL}_2({\mathbb F}_\lambda )\cong \operatorname {PSL}_2({\mathbb F}_{\ell ^3})$ by Theorem 1.2(i).