**AN APPLICATION OF MAEDA’S CONJECTURE TO THE**
**INVERSE GALOIS PROBLEM**

Gabor Wiese

Abstract. It is shown that Maeda’s conjecture on eigenforms of level 1 implies that
*for every positive even d and every p in a density-one set of primes, the simple group*
PSL_{2}(F_{p}d) occurs as the Galois group of a number ﬁeld ramifying only at p.

**1. Introduction**

The purpose of this paper is to support the approach to the inverse Galois problem
for certain ﬁnite groups of Lie type through automorphic forms. There have been a
number of promising results in the recent past, e.g., [2, 11] for groups of the type
PSL_{2}(F*d*), and [1, 6, 7] for more general groups. The general idea is to take varying

automorphic forms over Q and to study the images of the residual Galois repre-sentations attached to them. Currently, one only obtains positive-density or inﬁnity results. The main technical obstacle to improving the mentioned results to density 1 seems to be the lack of control on the ﬁelds of coeﬃcients of the automorphic forms involved.

In the easiest case, that of ‘classical’ modular forms, i.e., of automorphic forms for
GL_{2}overQ, there is a strong conjecture due to Maeda on the coeﬃcient ﬁelds of level 1
modular forms. In order to demonstrate the potential of the modular approach to the
inverse Galois problem, we show that the control on the coeﬃcient ﬁelds provided by
Maeda’s conjecture suﬃces to yield the following strong result on the inverse Galois
problem.

**Theorem 1.1.** *Assume the following form of Maeda’s conjecture on level 1 modular*
*forms:*

*For any k and any normalised eigenform f* *∈ Sk(1) (the space of*

*cuspidal modular forms of weight k and level 1), the coeﬃcient ﬁeld*

Q*f* :=*Q(an(f )| n ∈ N) has degree equal to dk*:= dimC*Sk(1) and the*

*Galois group of its normal closure overQ is the symmetric group Sdk.*

*(a) Let 2≤ d ∈ N be even. Then the set of primes p such that there is a number ﬁeld*

*K/Q ramiﬁed only at p with Galois group isomorphic to PSL*_{2}(F_{p}d) has density 1.

*(b) Let 1≤ d ∈ N be odd. Then the set of primes p such that there is a number ﬁeld*

*K/Q ramiﬁed only at p with Galois group isomorphic to PGL*_{2}(F_{p}d) has density 1.

Received by the editors December 7, 2012.

*1991 Mathematics Subject Classiﬁcation. MSC (2010): 11F11 (primary); 11F80, 11R32, 12F12*
(secondary).

Maeda’s conjecture was formulated as Conjecture 1.2 in [5]. It has been checked up to weight 12000 (see [4]). We also mention that a generalization of a weaker form of Maeda’s conjecture to square-free levels has recently been proposed by Tsaknias [9]. Throughout the paper the notion of density can be taken to be either natural density or Dirichlet density.

It is certainly possible to give an eﬀective version of Theorem 1.1. Suppose that
*Maeda’s conjecture has been checked for weights up to B. Then for all d≤ dim*_{C}*SB*(1)

one can work out an explicit lower bound for the density of the sets in the theorem,
*depending on B.*

The proof of Theorem 1.1 is given in the remainder of the paper. It is based on a meanwhile classical ‘big image result’ of Ribet [8], Chebotarev’s density theorem, some combinatorics in symmetric groups, and Galois theory.

**2. Proof**

In this section, the main result is proved. We use the convention that the symmetric
*group Sn* is the group of permutations of the set *{1, 2, . . . , n}.*

**2.1. Splitting of primes in extensions with symmetric Galois group.** In this
part, we give a possibly non-standard proof of the well-known fact that the splitting
*behaviour of unramiﬁed primes in a simple extension K(a)/K can be read oﬀ from*
the cycle type of the Frobenius, seen as an element of the permutation group of the
*roots of the minimal polynomial of a. (A more ‘standard’ proof would consider the*
*factorization into irreducibles of the reduction of the minimal polynomial of a, as*
in [10], p. 198).

*Let M/K be a separable ﬁeld extension of degree n and let L/M be the Galois*
*closure of M over K. By the theorem of the primitive element there is a* *∈ M such*
*that M = K(a). Let f* *∈ K[X] be the minimal polynomial of a over K and let*

*a = a*_{1}*, a*_{2}*, . . . , an* *be the roots of f in L. The map ψ : G := Gal(L/K)* *→ Sn*,

*sending σ to the permutation ψ(σ) given by σ(ai) = aψ(σ)(i)* is an injective group

*homomorphism, which maps H := Gal(L/M ) onto StabSn*(1)*∩ ψ(G).*

**Proposition 2.1.** *Assume the preceding set-up with K a number ﬁeld. Let* *p be a*
*prime of K andP a prime of L dividing p. We suppose that P/p is unramiﬁed. Then*
*the cycle lengths in the cycle decomposition of ψ(Frob _{P/p}*)

*∈ Sn*

*are precisely the*

*residue degrees of the primes of M lying above* *p.*

*Proof. Let g* *∈ Gal(L/K). Denote by FrobgP/p* *the Frobenius element of gP/p in*

*Gal(L/K) and by f _{(gP∩M)/p}*

*the inertial degree of the prime gP ∩ M of M over p.*

*Write ϕ := Frob*for short. We have

_{P/p}*f _{(gP∩M)/p}*= min

*i∈N*(Frob

*i*

*gP/p*

*∈ H) = min*

*i∈N(ϕ*

*i*

_{∈ g}−1_{Hg).}From this we obtain the equivalences:

*∃g ∈ G : f(gP∩M)/p= d,*

*⇔ ∃g ∈ G : ψ(ϕd*_{)}_{∈ Stab}*Sn(ψ(g*
*−1*_{)(1))}
and *∀ 1 ≤ i < d : ψ(ϕi*)* ∈ StabSn(ψ(g*
*−1 _{)(1)),}*

*⇔ ∃j ∈ {1, . . . , n} : ψ(ϕd*

_{)}

_{∈ Stab}*Sn(j) and∀ 1 ≤ i < d : ψ(ϕ*

*i*

_{)}

_{ ∈ Stab}*Sn(j)*

*⇔ ψ(ϕ) contains a d-cycle.*

This proves the proposition.

**2.2. Combinatorics in symmetric groups.** We will eventually be interested in
*primes of a ﬁxed residue degree d in an extension with symmetric Galois group. The*
results of the previous part hence lead us to consider elements in symmetric groups
*having a d-cycle, which we do in this part.*

The contents of this part is presumably also well-known. Since the techniques are
very simple and straight forward, I decided to include the proofs rather than to look
*for suitable references. Let d≥ 1 be a ﬁxed integer. Deﬁne recursively for i ≥ 1 and*
1*≤ j ≤ i*
*a(0) := 0,* *b(i, j) :=* 1
*j!dj*(1*− a(i − j)), a(i) :=*
*i*
*k*=1
*b(i, k).*

**Lemma 2.2.** *With the preceding deﬁnitions we have*
*a(i) =*
*i*
*j*=1
(*−1)j*+1
*j!dj* = 1*− exp*
*−1*
*d*
+
*∞*
*j=i+1*
(*−1)j*
*j!dj* *.*

*Proof. This is a simple induction. For the convenience of the reader, we include the*

inductive step:
*a(i + 1) =*
*i*+1
*k*=1
*b(i + 1, k) =*
*i*+1
*k*=1
1
*k!dk*(1*− a(i + 1 − k))*
=
*i*+1
*k*=1
1
*k!dk*
⎛
*⎝1 −i+1−k*
*j*=1
(*−1)j*+1
*j!dj*
⎞
⎠ =*i*+1
*k*=1
⎛
⎝ 1
*k!dk* *−*
*i+1−k*
*j*=1
(*−1)j*+1
*k!j!dj+k*
⎞
⎠
=
*i*+1
*m*=1
1
*m!dm* +
*i*+1
*m*=2
1
*m!dm*
*m−1*_{}
*j*=1
*m*
*j*
(*−1)j* =
*i*+1
*m*=1
(*−1)m*+1
*m!dm* *.*
*For i→ ∞ the convergence a(i) → 1 − exp(−1 _{d}* ) is very quick because of the simple
estimate of the error term

_{ }

*∞*

_{j}_{=i+1}_{j}_{!}1

_{d}_{j}

_{ ≤}*2*

_{(i+1)!d}*i+1*.

*We now relate the quantities a(i) and b(i, j) to proportions in the symmetric group.*
*Let n, j* *∈ N. Deﬁne*

*An(d) :={g ∈ Sn| g contains at least one d-cycle},*

*Bn(d, j) :={g ∈ Sn| g contains precisely j d-cycles}.*
**Lemma 2.3.** *For all n≥ 2d the following formulae hold, where i := n _{d}
:*

*(a) n!· a(i) = #An(d),*

*(c) n!·* *2n−d−1 _{n}_{(n−1)}*(1

*− a(i − 1)) = #{g ∈ Bn(d, 1)| the unique d-cycle contains 1 or 2},*

*(d) n!·* * _{n}_{(n−1)}*1 (1

*− a(i − 2)) = #{g ∈ Bn(d, 2)| one d-cycle contains 1, the other 2}.*

*Proof. (a) and (b) are proved by induction for n* *≥ 1. For n < d (i.e., i = 0), the*

equalities are trivially true. Now we describe the induction step:
#*Bn(d, j) =*
1
*j!·*
*n*
*d*
*· (d − 1)!*
*·*
*n− d*
*d*
*· (d − 1)!*
*· · ·*
*×*
*n− (j − 1)d*
*d*
*· (d − 1)!*
*× (n − jd)! · (1 − a(i − j)) =* *n!*
*j!dj*(1*− a(i − j)).*

*The ﬁrst equality can be seen as follows: there are j! ways of ordering the j d-cycles.*
*The number of choices for the ﬁrst d-cycle is given by (n _{d}*)

*· (d − 1)!, the one for the*second is

*n−d*

_{d}*· (d − 1)!, and so on. After having chosen j d-cycles, n − jd elements*remain. Among these remaining elements we may only take those that do not contain

*any d-cycle; their number is (n− jd)! · (1 − a(i − j)) by induction hypothesis.*

(c) The number of elements in the set in question is
2
*n− 1*
*d− 1*
*−*
*n− 2*
*d− 2*
*(d− 1)! · (n − d)! · (1 − a(i − 1)) = n!2n− d − 1*
*n(n− 1)* (1*− a(i − 1))*

because *n _{d−1}−1*

*· (d − 1)! is the number of choices for a d-cycle with one previously*chosen element (i.e., 1 or 2) and

*n−2*

_{d−2}*· (d − 1)! is the number of choices for a d-cycle*containing 1 and 2.

(d) The number of elements in the set in question is
*n− 2*
*d− 1*
*(d− 1)! ·*
*n− 2 − (d − 1)*
*d− 1*
*(d− 1)! · (n − 2d)! · (1 − a(i − 2))*
*= n!* 1
*n(n− 1)*(1*− a(i − 2))*

because *n _{d−1}−2*

*· (d − 1)! is the number of choices for a d-cycle containing 1 and*not containing 2 and

*n−2−(d−1)*
*d−1*

*· (d − 1)! is the number of choices for a d-cycle*

*containing 2 among the elements remaining after the ﬁrst choice, and again (n− 2d)! ·*
(1*− a(i − 2)) is the number of elements remaining after the two choices such that*

*they do not contain any d-cycle.*

We write*A± _{n}(d) for the subsets ofAn(d) consisting of the elements having positive*

or negative signs.

**Corollary 2.4.** *Let d, n∈ N, n ≥ 2d ≥ 2 and put i := n _{d}
. Then the estimates*

*and*
#*A*+*n(d)− #A−n(d)*
#*An(d)*
* ≤* * _{n}_{− 1}*1

*·*2

1*− exp(−*1* _{d}*)

*−*2

_{(i+1)!d}*i+1*

*hold.*

*Proof. Consider the bijection φ : Sn*

*g→g◦(12)*

*−−−−−−→ Sn. For j > 2 the image of* *A*+*n(d)∩*

*Bn(d, j) under φ lands in* *A−n(d) because the multiplication with (1 2) can at most*

*remove two d-cycles. Now consider g* *∈ A*+_{n}(d)∩ Bn(d, 2). Clearly φ(g)∈ A−n(d) unless

*one of the d-cycles contains 1 and the other one contains 2. For g∈ A*+_{n}(d)∩ Bn(d, 1)

*we ﬁnd that φ(g)* *∈ A− _{n}(d) unless the single d-cycle of g contains 1 or 2. In view of*
Lemma 2.3, we thus obtain the inequality

#*A*+_{n}(d)− #A−_{n}(d)≤ n! ·*2n− d − 1*
*n(n− 1)* (1*− a(i − 1)) +*
1
*n(n− 1)*(1*− a(i − 2))*
*≤ n! ·* 2
*n− 1.*

By exchanging the roles of + and*− we obtain the ﬁrst estimate. The second estimate*
then is an immediate consequence of Lemma 2.2 and the trivial estimate of the error

term mentioned after it.

**2.3. Density of primes with prescribed residue degree in composites of**
**ﬁeld extensions with symmetric Galois groups.**

**Lemma 2.5.** *Let 1* *≤ d ∈ N, K be a ﬁeld and L/K, F/K be two ﬁnite Galois*
*extensions such that Gal(L/K) ∼= Snwith n≥ max(5, 2d) and L is not a subﬁeld of F .*

*Let C* *⊆ G := Gal(F/K) be a subset and put c :=* *#C _{#G}*

*and a :=*

*#An(d)*

_{#Sn}*= a(n*

_{d}).*Let X := Gal(LF/K) and Y be the subset of X consisting of those elements that*
*project to an element inAn(d)⊆ Sn∼= Gal(L/K) or to an element in C* *⊆ Gal(F/K)*

*under the natural projections. Then*

*#Y*
*#X* *= a + c− (1 + δ)ac,*
*where* _{⎧}
⎨
⎩
*δ = 0,* *if L∩ F = K,*
*|δ| ≤* 1
*n−1* *·* *1−exp(−*1 2
*d)−* 2
*(1+ n _{d )!d}1+ nd *

*,*

*otherwise.*

*Proof. The intersection L∩ F is a Galois extension of K which is contained in L.*

*The group structure of Sn* *(more precisely, the fact that the alternating group An*

*is the only non-trivial normal subgroup of Sn) hence implies that [L∩ F : K] ≤ 2;*

*for, if L∩ F were equal to L, then L would be a subﬁeld of F , which is excluded by*
assumption.

*Assume ﬁrst L∩ F = K, then Gal(LF/K) ∼= Gal(L/K)× Gal(F/K) and thus*
*#Y = #An(d)· #G + #Sn· #C − #An(d)· #C,*

*from which the claimed formula follows by dividing by #X = #G· #Sn*.

*An(d) that project to the identity of Gal(N/K) are precisely those in* *A*+*n(d). In*

*a similar way, we denote by C*+ *those elements of C projecting to the identity of*
*Gal(N/K), and by C−*the others. Then we have

*#Y = #An(d)·*
*#G*
2 +
*#Sn*
2 *· #C − #A*
+
*n(d)· #C*+*− #A−n(d)· #C−.*
*Dividing by #X =* *#Sn*_{2}*·#G* we obtain
*#Y*
*#X* *= a + c− (1 + δ)ac, where δ =*
*#C*+*− #C−*
*#C* *·*
#*A*+* _{n}(d)− #A−_{n}(d)*
#

*An(d)*

*.*

The claim is now a consequence of Corollary 2.4.

**Lemma 2.6.** *Let (an*)*n≥1* *be a sequence of non-negative real numbers such that*

_{∞}

*n*=1*an* *diverges.*

*(a) Let γ > 0, b*_{0} *∈ R. Assume that an* *<* 1_{γ}*for all n* *≥ 1. We deﬁne a sequence*

*(bn*)*n≥0* *by the rule*

*bn:= bn−1+ an− γbn−1an*

*for all n≥ 1. Then the sequence (bn*)*n≥1* *tends to 1/γ for n→ ∞.*

*(b) Let (δn*)*n≥1* *be a sequence of real numbers tending to 0 and let c*0 *∈ R. Assume*

*that lim sup _{n→∞}an< 1. We deﬁne the (modiﬁed) inclusion–exclusion sequence as*

*cn:= cn−1+ an− (1 + δn)cn−1an* *for n≥ 1.*

*Then the sequence (cn*)*n≥1* *tends to 1.*

*Proof. (a) We let*

*rn*:= 1*− γbn*= 1*− γ(bn−1+ an− γbn−1an*) = (1*− γbn−1*)(1*− γan*)

= (1*− γb*_{0})(1*− γa*_{1})(1*− γa*_{2})*· · · (1 − γan).*

*To see that the limit of (γbn*)*n≥0* is 1, we take the logarithm of (1*− γa*1)(1*− γa*2)*· · ·*

(1*− γan*):
*n*
*i*=1
log(1*− γai*) =*−γ*
*n*
*i*=1
*ai−*
*n*
*i*=1
*∞*
*j*=2
*(γai*)*j*
*j* *≤ −γ*
*n*
*i*=1
*ai.*

By our assumption this diverges to *−∞ for n → ∞, so that limn→∞rn*= 0, proving

the lemma.
(b) Let min
*1,*_{lim supn→∞}1 _{a}*n* *− 1*

*> > 0. There is N such that* *|δn| < and*

*an* *<* * _{1+}*1

*for all n*

*≥ N. By enlarging N if necessary we may also assume that*

*cN* *≥ 0. The reason for the latter is that cN+n* *> cN* +

*n*

*i*=1*aN+i* *if cN+i< 0 for all*

0*≤ i ≤ n.*

We consider the two sequences

*bN* *:= cN* *and bn= bn−1+ an− (1 + )bn−1an, for n > N*

and

*dN* *:= cN* *and dn= dn−1+ an− (1 − )dn−1an, for n > N.*

By (a) we know lim*n→∞bn*= * _{1+}*1 and lim

*n→∞dn*=

*1*

_{1−}*. For n≥ N by induction we*

obtain the estimate:

*Thus, there is M such that* * _{1+}*1

*− ≤ cn*

*≤*

*1*

_{1−}*+ for all n≥ M. As is arbitrary,*

we ﬁnd lim*n→∞cn*= 1.

**Proposition 2.7.** *Let 1* *≤ d ∈ N, K be a ﬁeld and let Ln* *for n* *∈ N be Galois*

*extensions of K with Galois group Gal(Ln/K) ∼= SNn* *such that Nn* *< Nn*+1 *for all*

*n≥ 1. Denote by Gn* *the Galois group of the composite ﬁeld L*1*L*2*· · · Ln* *over K and*

*for 1≤ i ≤ n denote by πi* *: Gn→ Gal(Li/K) the natural projection. Consider*

*cn*:=

#*{g ∈ Gn| ∃i ∈ {1, . . . , n} : πi(g)∈ Gal(Li/K) ∼= SNi* *contains a d-cycle}*

*#Gn*

*.*
*Then the sequence cn* *tends to 1 for n→ ∞.*

*Proof. Without loss of generality we can assume that max(5, 2d)≤ N*_{1}*. Let c*_{0} := 0
*and an* *:= a(N _{d}n
) =*

*#A*

_{#S}Nn(d)*Nn* . By Lemmas 2.2 and 2.3 it is clear that

_{∞}

*n*=1*an*

diverges.

*If we call Ki* *the unique quadratic extension of K inside Li*, then Lemma 18.3.9

*of [3] shows that Gal(L*_{1}*. . . Ln/K*1*· · · Kn) ∼= AN*1 *× · · · × ANn, for all n* *≥ 1. This*

*implies that Ln* *cannot be a subﬁeld of L*1*· · · Ln−1* *for any n≥ 2.*

*Lemma 2.5 inductively gives the formula cn= an+cn−1−(1+δn)ancn−1for n≥ 1,*

*where δn*is bounded by
*|δn| ≤*
1
*Nn− 1·*
2
1*− exp(−*1* _{d}*)

*−*2

*(1+ Nn*

*d*

*)!d*1+

*Nn*

*d*

*,*

*which clearly tends to 0 for n→ ∞. Lemma 2.6 yields the claim on the limit.*
By applying Chebotarev’s density theorem and noting that the set in the
proposi-tion is conjugaproposi-tion invariant, we obtain the following corollary.

**Corollary 2.8.** *Let 1≤ d ∈ N, K be a number ﬁeld and let Ln* *for n∈ N be Galois*

*extensions of K with Galois group Gal(Ln/K) ∼= SNn* *such that Nn* *< Nn*+1 *for all*

*n≥ 1.*

*Then the set of primes of K*

*{p | ∃i ∈ {1, . . . , n} : πi*(Frobp)*∈ Gal(Li/K) ∼= SNi* *contains at least one d-cycle}*

*has a density, and the density is equal to cnfrom Proposition 2.7 and hence tends to 1*

*for n→ ∞. Here Frob*_{p}= Frob_{P/p}*for any primeP of the composite ﬁeld L*_{1}*L*_{2}*· · · Ln*

*above* *p.*

The following is the main theorem of this paper concerning the density of primes with prescribed residue degree in a composite of ﬁeld extensions with symmetric Galois groups.

**Theorem 2.9.** *Let 1* *≤ d ∈ N, K be a number ﬁeld and let Mn* *for n* *∈ N be ﬁeld*

*extensions of K with splitting ﬁeld Lnover K having Galois group Gal(Ln/K) ∼= SNn*

*such that Nn< Nn*+1 *for all n≥ 1.*

*Then the set of primes of K*

*{p | ∃i ∈ {1, . . . , n}, ∃P/p prime of Mi* *of residue degree d}*

*has a density, and the density is equal to cn* *from Proposition 2.7 and hence tends*

*Proof. Because of Proposition 2.1 the set of primes in the theorem is the same as the*

set in Corollary 2.8.

**2.4. End of the proof.**

*Proof of Theorem 1.1. Since dim*_{C}*Sk*(1) tends to*∞ for k → ∞ (for even k), Maeda’s*

*conjecture implies the existence of newforms fn* of level one and increasing weight

(automatically without complex multiplication because of level 1) such that their
*coeﬃcient ﬁelds Mn*:=Q*fn* satisfy the assumptions of Theorem 2.9.

*For each n and each prime* *P of Mn* *consider the Galois representation ρ*proj_{f}*n,*P :

Gal(*Q/Q) → PGL*_{2}(F*p) attached to fn*. Theorem 3.1 of Ribet [8] implies that for

*each fn* and all but possibly ﬁnitely many P, its image is equal to PGL2(FP), if the

residue ﬁeld F_{P} ofP has odd degree over its prime ﬁeld, and equal to PSL_{2}(F_{P}) if
the residue degree is even. We will abbreviate this by PXL_{2}(F_{P}).

Consequently, the set of primes (of Q)

*{p | ∃i ∈ {1, . . . , n}, ∃P/p prime of Mi* *s.t. ρ*proj*fi,*P *∼*= PXL2(F*pd*)*}*

has the same density as the corresponding set in Theorem 2.9, implying

Theorem 1.1.

**Acknowledgments**

I wish to thank Sara Arias-de-Reyna for reading an earlier version of this paper and her corrections and valuable remarks. I also thank J¨urgen Kl¨uners for the reference to van der Waerden. I acknowledge partial support by the Priority Program 1489 of the Deutsche Forschungsgemeinschaft (DFG) and by the Fonds National de la Recherche Luxembourg (INTER/DFG/12/10).

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