ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P2.05
https://doi.org/10.26493/1855-3974.2764.fe5
(Also available at http://amc-journal.eu)
Regular dessins with moduli fields of the form
Q(ζp, p
√
q)*
Nicolas Daire
Département de Mathématiques et Applications, École Normale Supérieure,
45 rue d’Ulm, 75005 Paris, France
Fumiharu Kato , Yoshiaki Uchino
Department of Mathematics, Tokyo Institute of Technology, 2-12-1 Ookayama, Meguro,
Tokyo 152-8551, Japan
Received 10 December 2021, accepted 29 March 2023, published online 27 September 2023
Abstract
Gareth Jones asked during the 2014 SIGMAP conference for examples of regular dessins
with nonabelian fields of moduli. In this paper, we first construct dessins whose moduli
fields are nonabelian Galois extensions of the form Q(ζp, p
√
q), where p is an odd prime
and ζp is a pth root of unity and q ∈ Q is not a pth power, and we then show that their
regular closures have the same moduli fields. Finally, in the special case p = q = 3 we
give another example of a regular dessin of degree 219 ·34 and genus 14155777 with moduli
field Q(ζ3, 3
√
3).
Keywords: Dessins d’enfants, coverings.
Math. Subj. Class. (2020): 14H57, 14H30
1 Introduction
Grothendieck first coined the term Dessin d’enfant in Esquisse d’un Programme [4] to
denote a connected bicolored graph embedded on a compact connected oriented topolog-
ical surface. The study was motivated by the one to one correspondance between dessins
d’enfant, the combinatorial data of the associated cartographical group, and the geomet-
ric concept of coverings of P1 by compact Riemann surfaces ramified at most over three
*The authors are grateful to Professor Jürgen Wolfart for valuable comments.
E-mail addresses: nicolas.daire@ens.psl.eu (Nicolas Daire), bungen@math.titech.ac.jp (Fumiharu Kato),
uchino.y.ab@m.titech.ac.jp (Yoshiaki Uchino)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 24 (2024) #P2.05
points. Moreover, by Belyi’s theorem any such covering is given the structure of an alge-
braic curve defined over a number field, therefore we obtain a natural action of the absolute
Galois group Gal(Q̄/Q) on the set of isomorphism classes of dessins. A lot of the interest
for dessins stems from the fact that this action is faithful, providing a way to study the
absolute Galois group through its action on the set of dessins. A particularly interesting
family of dessins is that of regular dessins, characterized by the fact that their automor-
phism groups act transitively on their sets of edges, and the Galois action was proved to
remain faithful when restricted to the subset of isomorphism classes of regular dessins [3].
To any dessin we associate a number field called its moduli field, which is defined as the
subfield of Q̄ fixed by the subgroup of Gal(Q̄/Q) that fixes the dessin up to isomorphism.
Conder, Jones, Streit and Wolfart noted in [1] that the moduli fields of all the examples
of regular dessins known at the time were abelian Galois extensions of Q. Herradón con-
structed in [6] an explicit equation for a regular dessin whose moduli field Q( 3
√
2) is not
a Galois extension of Q, and Hidalgo later generalized his construction in [7] to produce
regular dessins whose moduli fields are of the form Q( p
√
2) where p is an odd prime num-
ber. However there is as of yet no known example of regular dessin whose moduli field
is a nonabelian Galois extension of Q. This is the starting point of this paper, in which
we will exhibit examples of regular dessins with moduli fields that are nonabelian Galois
extensions of Q.
In the present paper, we begin by recalling the main definitions and results on dessins
d’enfant. We will then expose constructions of regular dessins whose moduli fields are
nonabelian Galois extensions of Q. We first exhibit dessins whose moduli fields are of
the form Q(ζ3, 3
√
q), where ζ3 is a primitive third root of unity and q ∈ Q is not a third
power, and show that the regular closures of these dessins possess the same moduli fields.
We then generalize this construction to show that there exist regular dessins with moduli
fields Q(ζp, p
√
q), where ζp is a primitive pth root of unity and q ∈ Q>0 is not a pth power.
Finally, we give an example of a regular dessin of degree 219 ·34 and genus 14155777 with
moduli field Q(ζ3, 3
√
3).
Notations
• SE : the group of self-bijections of the set E, similarly Sn is the group of permu-
tations of a set of n elements (we favor a right action, hence we write the product
στ := τ ◦ σ)
• Gal(E/F ): the Galois group of F -automorphisms of E
• ζk: the kth primitive root of unity exp( 2iπk )
• F2: the free group of rank 2 with generators (ξ, η)
• Crit: the set of critical values of a function
2 Preliminaries on dessins d’enfant
We refer the reader to existing expositions of the theory such as [5, 8, 9] and [2] for proofs
of the presented facts and further details.
A dessin d’enfant is a connected bipartite graph embedded on a compact connected
orientable topological surface, such that the complement of the graph is a disjoint union of
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 3
2-cells. Two such dessins are equivalent if there exists an orientation preserving homeomor-
phism between the underlying surfaces that induces an isomorphism between the embedded
bipartite graphs.
A dessin is determined up to isomorphism by a pair (C, β) where C is a smooth alge-
braic curve and β : C → P1 is a meromorphic mapping ramified at most over {0, 1,∞},
and by Belyi’s theorem we can further ask for C and β to both be defined over a number
field. We call (C, β) a Belyi pair and β a Belyi function. The corresponding graph embed-
ding on the underlying surface is recovered by pulling back the segment [0, 1] along β, we
define black and white vertices as the preimages of 0 and 1 respectively, and the edges as
the preimages of ]0, 1[.
By covering theory a dessin is also determined up to isomorphism by the monodromy
action of the fundamental group of the complex projective line π1(P1) on the fiber over the
point 12 which is identified to the set of edges of the dessin. The fundamental group π1(P
1)
is isomorphic to the free group of rank two F2 = ⟨ξ, η⟩ with generators ξ and η which are
two loops with base point 12 and circling counter-clockwise around 0 and 1 respectively.
The monodromy action of the generators ξ and η then corresponds to the product of the
counter-clockwise cyclic permutation of the edges around black and white vertices respec-
tively. We call monodromy map M : F2 → SE the map that associates to each element of
F2 the corresponding permutation of the set of edges, and we call cartographic group the
image of the monodromy map, which is a transitive subgroup of the group of permutations
of the set of edges.
When the automorphism group of a dessin D acts transitively on the set of edges, we say
that D is a regular dessin. When that is the case the cartographic group G acts transitively
and freely on the set of edges, the monodromy action is thus given by the canonical action of
G on itself. There is a natural bijection between regular dessins and finite groups generated
by two distinguished elements ξ and η up to isomorphism. Two regular dessins determined
by G1 = ⟨ξ1, η1⟩ and G2 = ⟨ξ2, η2⟩ respectively are isomorphic if and only if there exists
an isomorphism between G1 and G2 that preserves the distinguished generators. Given
a dessin D, there exists a unique regular dessin D̃ with a morphism ϕ : D̃ → D such
that any morphism from a regular dessin to D factors through ϕ. We call D̃ the regular
closure of D. Moreover, there exists an isomorphism Cart(D̃) ∼= Cart(D) that preserves
the distinguished generators. There exists a natural action of the absolute Galois group
Gal(Q̄/Q) on the set of isomorphism classes of dessins, we denote by Dσ the action of an
automorphism σ on a dessin D, and this Galois action commutes with regular closure, i.e.
we have (D̃)σ ∼= (̃Dσ).
Given a dessin D, we say that a number field k is a field of definition of D if D is
isomorphic to a dessin defined over k. However there does not necessarily exist a smallest
field of definition. We thus define the moduli field of a dessin D as the subfield of Q̄ fixed
by the subgroup of Gal(Q̄/Q) constituted of the elements fixing D up to isomorphism. The
moduli field of a dessin is contained in all fields of definition but is not necessarily itself a
field of definition, however it is the case in particular for regular dessins.
3 Constructions of regular dessins with nonabelian moduli fields
We are now ready to give examples of regular dessins whose moduli fields are nonabelian
Galois extensions of Q. To do so, we will first exhibit dessins with such moduli fields, and
then prove that their regular closures admit the same moduli fields.
4 Ars Math. Contemp. 24 (2024) #P2.05
Before proceeding with the examples, let us first present a classic family of Belyi poly-
nomials that we will use in the following constructions. For positive integers m,n ∈ N we
define the polynomial
Bm,n :=
(m+ n)m+n
mmnn
Xm(1−X)n ∈ Q[X].
By computing the derivative B′m,n =
(m+n)m+n
mmnn X
m−1(1 −X)n−1(m − (m + n)X) we
verify that Bm,n : P1 → P1 is a Belyi function that ramifies only at 0, 1, ∞ and mm+n with
ramification indices m, n, m + n and 2 respectively, and Bm,n(0) = 0, Bm,n(1) = 0,
Bm,n(∞) = ∞ and Bm,n( mm+n ) = 1 (see Figure 1).
0 1
m
m+n
1
2
3
m − 1
m
1
2
3
n − 1
n
Figure 1: Dessin corresponding to the Belyi pair (P1, Bm,n).
3.1 Regular dessins with moduli fields of the form Q(ζ3, 3
√
q)
Let q ∈ Q>0 be a positive rational number that is not a third power.
Let m,n ∈ N be coprime positive integers such that 2727+q2 = mm+n , and let
C : y2 = x(x− (1− ζ3))(x− 3
√
q),
β : C → P1, (x, y) 7→ 1
27mq2n
(x6 + 27)m(q2 − x6)n.
The function β is given by the composition β = β1 ◦ β0 ◦ π of the following maps.
1. π : C → P1 is the projection on the coordinate x, which is ramified over {0, 1 −
ζ3, 3
√
q,∞}.
2. β0 := X6 ∈ Q[X], Crit(β0) = {0} so β1 ◦ π ramifies over {0, (1 − ζ3)6 =
−27, q2,∞}.
3. β1 := Bm,n(X+27q2+27 ), so β = β1 ◦ β0 ◦ π ramifies over {0, 1,∞}.
The pair (C, β) is thus a Belyi pair, and we call D the corresponding dessin. The dessin
D is defined over Q(ζ3, 3√q), so its moduli field is a subfield of Q(ζ3, 3√q). By taking
the regular closure we then obtain the inclusion of moduli fields M(D̃) ⊆ M(D) ⊆
Q(ζ3, 3
√
q), and moreover D̃ is regular so it is defined over M(D̃). We shall prove that
M(D̃) is in fact exactly Q(ζ3, 3√q), which is a nonabelian Galois extension of Q with
Galois group
Gal(Q(ζ3, 3
√
q)/Q) ∼= S3.
To that end we must show that an automorphism σ ∈ Gal(Q̄/Q) fixes D̃ if and only if it
fixes ζ3 and 3
√
q, or equivalently that Gal(Q(ζ3, 3
√
q)/Q) acts freely on the orbit of D̃.
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 5
Let σ ∈ Gal(Q̄/Q), the Galois conjugate Dσ is given by the Belyi pair (Cσ, βσ),
where
Cσ : y2 = x(x− (1− σ(ζ3)))(x− σ( 3
√
q)),
and βσ has the same expression as β because all of its coefficients are rational. The orbit
of the pair (ζ3, 3
√
q) by Gal(Q̄/Q) is {ζi3, ζj3 3
√
q}1≤i≤2,0≤j≤2. Elliptic curves given by
equations of the form y2 = (x− a)(x− b)(x− c) are isomorphic if and only if the cross-
ratios of the tuples (a, b, c,∞) coincide. We verify that the cross-ratios are all distinct, so
the orbit of D is given by the six dessins Dσ for σ ∈ Gal(Q(ζ3, 3√q)/Q). As a consequence
M(D) = Q(ζ3, 3√q). To prove that the regular closures D̃σ constituting the orbit of D̃ are
also non isomorphic, we must first draw the dessins Dσ to compute their cartographic
groups.
Let us first draw the dessin D0 corresponding to the Belyi pair (P1, β1 ◦ β0) (see Fig-
ure 3). The dessin D0 is defined over Q, so the dessins Dσ in the orbit are then obtained
by lifting D0 to the curves Cσ . To simplify the graphical representations of the dessins, we
will use the notation in Figure 2 for consectutive edges incident to a vertex.
eξ
eξ
2
eξ
n−2
eξ
n−1
eeξ
n
=
n
eeξ
n
Figure 2: Notation for consecutive edges.
(P1, id)0 1
(P1, β1)
−27 0 q2
m n
(P1, β1 ◦ β0)
0
n
m
n
m
n
m
n
m
n
m
n
m
3
√
q
ζ3 3
√
q
ζ23 3
√
q
1− ζ23
1− ζ3
Figure 3: Construction of D0.
The dessins D1, . . . ,D6 conjugate to D are embedded on a torus, so in the representa-
tions in Figure 4 we will identify the outermost edges on opposite sides.
6 Ars Math. Contemp. 24 (2024) #P2.05
?
∞
0 0
00
3
√
q 3
√
q
1− ζ3
1− ζ3
n
4
m
3
n
16
m
15
n
6
m
5
n
18
m
17
n
8
m
7
n
20
m
19
n
10
m
9
n
22
m
21
n
12
m
11
n
24
m
23
n n
m
m
1 13
14
2
131
2
14
(a) D1 := D
?
∞
0 0
00
ζ3 3
√
q ζ3 3
√
q
1− ζ23
1− ζ23
n
2
m
3
n 14
m
15
n
6
m
5
n
18
m
17
n
8
m
7
n
20
m
19
n
10
m
9
n
22
m
21
n
12
m
11
n
24
m
23
n n
m
m
1 13
16
4
131
4
16
(b) D2 := Dσ, σ : (ζ3, 3√q) 7→ (ζ23 , ζ3 3
√
q)
?
∞
0 0
00
ζ3 3
√
q ζ3 3
√
q
1− ζ3
1− ζ3
n2
m
3
n 14
m 1
5
n
4
m
5
n
16
m
17
n
8
m
7
n
20
m
19
n
10
m
9
n
22
m
21
n
12
m
11
n
24
m
23
n n
m
m
1 13
18
6
131
6
18
(c) D3 := Dσ, σ : (ζ3, 3√q) 7→ (ζ3, ζ3 3√q)
?
∞
0 0
00
ζ23 3
√
q ζ23 3
√
q
1− ζ23
1− ζ23
n2
m
3
n 14
m 15
n
4
m
5
n
16
m
17
n
6
m
7
n
18
m
19
n
10
m
9
n
22
m
21
n
12
m
11
n
24
m
23
n n
m
m
1 13
20
8
131
8
20
(d) D4 := Dσ, σ : (ζ3, 3√q) 7→ (ζ23 , ζ23 3
√
q)
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 7
?
∞
0 0
00
ζ23 3
√
q ζ23 3
√
q
1− ζ3
1− ζ3
n2
m3
n 14
m 15
n
4
m
5
n 1
6
m
17
n
6
m
7
n
18
m
19
n
8
m
9
n
20
m
21
n
12
m
11
n
24
m
23
n n
m
m
1 13
22
10
131
10
22
(e) D5 := Dσ, σ : (ζ3, 3√q) 7→ (ζ3, ζ23 3
√
q)
?
∞
0 0
00
3
√
q 3
√
q
1− ζ23
1− ζ23
n2
m3
n 14
m 15
n
4
m
5
n 16
m 1
7
n
6
m
7
n
18
m
19
n
8
m
9
n
20
m
21
n
10
m
11
n
22
m
23
n n
m
m
1 13
24
12
131
12
24
(f) D6 := Dσ, σ : (ζ3, 3√q) 7→ (ζ23 , 3
√
q)
Figure 4: Dessins D1, . . . ,D6 in the Galois orbit of D.
We will now establish that D̃1 is not isomorphic to D̃2, . . . , D̃6. To that end it suffices
to show that there is no isomorphism between the cartographic groups fixing the canonical
generators. We shall therefore exhibit an element ω ∈ F2 = ⟨ξ, η⟩ such that Mk(ω)
commutes with Mk(η2) only when k = 1, where Mk is the monodromy map of Dk.
We have defined m and n to be positive coprime integers such that 2727+q2 =
m
m+n , so
we cannot have m = n = 1. We will treat the case where m ̸= 1 does not divide n, the
other case being treated similarly. Let
ω := ξnη−1ξm−nηξn.
We shall show that Mk(ω) commutes with Mk(η2) only when k = 1.
Let Ek := {1, 2, . . . , 24} be the set of edges of Dk incident to 0. The action of η fixes
the set Ek on which it induces the cyclic permutation (1, 2, . . . , 24), and every white vertex
except 0 has degree one so the action of η is trivial on the complement of Ek.
We can write Ek = Eoddk ⊔ Eevenk as the disjoint union of the sets of respectively
odd and even numbered edges incident to 0, such that η sends one to the other. The black
vertices of Eoddk are of degree m except for the two black vertices of the edges 1 and 13 that
are of degree 2m. Therefore if m does not divide some integer l then ξl sends every edge of
Eoddk to the complement of Ek, and otherwise the action of ξ
m on Eoddk corresponds to the
sole transposition (1, 13). Similarly if n does not divide l then ξl sends every edge of Eevenk
to the complement of Ek, and the action of ξn on Eevenk is the transposition (2k, 2k + 12).
In particular, by hypothesis n is not a multiple of m, so m − n is not a multiple of m
either, hence both ξn and ξm−n send the edges of Eoddk to the complement of Ek. However
η acts trivially on the latter, so ξnη−1ξm−n and ξm−nηξn both fix the set Eoddk on which
they induce the same action as ξm, i.e. the transposition (1, 13). Therefore the action
of ω = ξnη−1ξm−nηξn is the same as that of ξmηξn on Eoddk and the same as that of
ξnη−1ξm on Eevenk . See Figure 5.
The action of ω fixes the set Ek on which it induces the permutation
Mk(ω)|Ek = (1, 13)(2k, 2k + 12) · (1, 2)(3, 4) · · · (23, 24) · (1, 13)(2k, 2k + 12)
8 Ars Math. Contemp. 24 (2024) #P2.05
0
m
ξn
η−1
ξm−n
n ξn
η
e
eω
0
m
ξm−n
η
ξn
n ξn
η−1
eω
e
Figure 5: Action of ω on Eoddk \ {1, 2k − 1} and on Eevenk \ {2, 2k}.
Therefore for k = 1,
M1(ω)|E1 = (1, 13)(2, 14) · (1, 2)(3, 4) · · · (23, 24) · (1, 13)(2, 14)
= (1, 2)(3, 4) · · · (23, 24)
so ω and η2 commute on E1. Moreover η acts trivially on the complement of E1 so
M1(ω)|D1\E1 and M1(η2)|D1\E1 automatically commute. Finally, we obtain that M1(ω)
and M1(η2) commute.
For k = 2, we observe that 4ωη
2
= 15η
2
= 17 but 4η
2ω = 6ω = 5. Similarly, for
3 ≤ k ≤ 6, we observe that 1ωη2 = 14η2 = 16 but 1η2ω = 3ω = 4. We have thus shown
that Mk(ω) and Mk(η2) commute only for k = 1.
This concludes the proof that D̃ is a regular dessin with moduli field Q(ζ3, 3√q).
3.2 Regular dessins with moduli fields of the form Q(ζp, p
√
q)
Let p be an odd prime, and q ∈ Q>0 a positive rational number that is not a pth power. In
this example we will need an additional parameter γ ∈ Q \ {0}. Let
C : y2 = x(x− (1− ζp))(x− γ p
√
q).
We construct the Belyi function β : C → P1 as the composition β = β2 ◦ β1 ◦ β0 ◦ π
of the following maps.
1. π : C → P1 is the projection on the coordinate x, which ramifies over {0, 1 −
ζp, γ p
√
q,∞}.
2. β0 := X2p ∈ Q[X], and Crit(β0) = {0,∞} so β0 ◦ π ramifies over {0, (1 −
ζp)
2p, γ2pq2,∞}.
3. β1 ∈ Q[X] is chosen independently of γ such that Crit(β1) ∪ {β1((1 − ζp)2p)} =
{0, 1,∞}, β1((1−ζp)2p) = 0 < β1(0) < 1 and β′1(0) > 0. The existence of β1 veri-
fying those conditions is assured by Proposition 3.2 below. Under those assumptions
β1 ◦ β0 ◦ π ramifies over {0, 1, β1(0), β1(γ2pq2),∞}.
4. γ ∈ Q>0 is then chosen small enough so that β′1 > 0 on [0, γ2pq2]. This guarantees
us that we have 0 < β1(0) < β1(γ2pq2) < 1.
5. β2 := Br,s ◦ Bm,n, where (m,n) and (r, s) are pairs of coprime positive integers
such that β1(γ2pq2) = mm+n and Bm,n(β1(0)) =
r
r+s . Finally, β = β2 ◦ β1 ◦ β0 ◦ π
ramifies over {0, 1,∞}.
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 9
The pair (C, β) is thus a Belyi pair, and we call D the corresponding dessin. With
the same arguments as before, the moduli field of D is Q(ζp, p√q), which is a nonabelian
Galois extension of Q with Galois group
Gal(Q(ζp, p
√
q)/Q) ∼= Z/pZ ⋊ (Z/pZ)×
generated by σ : ζip p
√
q 7→ ζi+1p p
√
q and τ : ζip p
√
q 7→ ζgip p
√
q where g generates (Z/pZ)×.
We shall show that there exists γ ∈ Q \ {0} such that the regular closure of the dessin D
thus obtained also has moduli field Q(ζp, p
√
q).
Remark 3.1. In the previous subsection we treated the case p = 3. In that specific case
we gave a simpler expression for β, mainly due to the fact that β0 ◦ π already had all of its
critical values in Q∪{∞}. However in the general case we must use the intermediate map
β1 as well as the parameter γ to conclude the proof.
Let us first prove the existence of β1.
Proposition 3.2. Let E ⊂ Q̄ ∩ R \ {0} be a finite set. Then there exists P ∈ Q[X] such
that P (E) ⊆ {0}, Crit(P ) ⊆ {0, 1}, 0 < P (0) < 1 and P ′(0) > 0.
Remark 3.3. In the context of this proposition we only deal with polynomials so for P ∈
Q[X] we define Crit(P ) := {P (z)| z ∈ C, P ′(z) = 0}, which does not include the point
at infinity to simplify notations.
Proof. To show this we will proceed similarly as in the proof of the only if part of Belyi’s
theorem, by applying additional transformations to ensure that 0 < P (0) < 1. Let us first
prove that we can reduce to the case where E is a subset of rational numbers.
Lemma 3.4. Let E ⊂ Q̄ ∩ R \ {0} be a finite set fixed by Gal(Q̄/Q). Then there exists
P ∈ Q[X] such that P (0) = 0 and Crit(P ) ∪ P (E) ⊂ Q \ {0}.
Proof. Let {a1, . . . , am} = E∩Q and {b1, . . . , bn} = E\Q. We construct P by induction
on the number n of non rational elements of E.
For α ∈ Q, define Fα, Gα ∈ Q[X] by
Fα :=
n∏
j=1
(X − (bj − α)2) and Gα := Fα((X − α)2) =
n∏
j=1
(X − bj)(X + bj − 2α).
Let us first assume that there exists α ∈ Q such that Gα(0) ̸∈ Crit(Gα) ∪ Gα(E).
Define P1(X) := Gα(X)−Gα(0) ∈ Q[X], then P1(0) = 0 ̸∈ E′ := Crit(P1)∪P1(E) ⊂
Q̄ ∩ R \ {0}. Note that E′ is stable under the action of Gal(Q̄/Q), and |E′ \ Q| =
|Crit(Fα) ∪ Fα(0) \ Q| = |Crit(Fα) \ Q| < degFα = n. By induction, there exists
P2 ∈ Q[X] such that P2(0) = 0 and Crit(P2)∪P2(E′) ⊂ Q\{0}. Now P := P2 ◦P1 has
the desired properties, since P (0) = 0 and Crit(P )∪P (E) = Crit(P2)∪P2(Crit(P1))∪
P2(P1(E)) = Crit(P2) ∪ P2(E′) ⊂ Q \ {0}.
Let us now prove that there exists α ∈ Q such that Gα(0) ̸∈ Crit(Gα) ∪ Gα(E).
Let us first treat the case where 0 < b1 < b2, . . . , bn. When α approaches 12 , Gα(0) =∏n
j=1 −bj(bj−2α) approaches 0 but the critical values of Gα do not. Indeed, Crit(Gα) =
CritFα∪Fα(Crit((X−α)2)) = Crit(Fα)∪{Fα(0)}; Fα(0) approaches F b1
2
(0) ̸= 0, and
since F b1
2
does not have multiple roots, the critical values of Fα approach the critical values
10 Ars Math. Contemp. 24 (2024) #P2.05
of F b1
2
which are all non zero. Therefore for α ̸= b12 in the neighborhood of b12 we have
Gα(0) ̸∈ Crit(Gα). Moreover Gα(0), Gα(a1), . . . , Gα(am) are all distinct polynomials in
the indeterminate α, so they coincide at only finitely many points. In particular for α ̸= b12
in the neighborhood of b12 we have Gα(0) ̸∈ {Gα(a1), . . . , Gα(am)}. Since α ∈ Q we
also have Gα(0) ̸= 0 = Gα(b1) = · · · = Gα(bn) hence Gα(0) ̸∈ Gα(E), proving the
existence of α as desired.
Let us now treat the general case where b1, . . . , bn are not assumed to be positive by
reducing it to the previous case. For α′ ∈ Q, define Hα′ ∈ Q[X] by
Hα′ := (X − α′)2 − α′2 ∈ Q[X].
Note that Crit(Hα′) = {−α′}. For α′ > 0 sufficiently small we have −α′2 < Hα′(0) =
0 < Hα′(a1), . . . ,Hα′(am), Hα′(b1), . . . ,Hα′(bn). Let E′′ := Crit(Hα′) ∪ Hα′(E).
The set E′′ is a finite subset of Q̄ ∩ R \ {0} fixed by Gal(Q̄/Q), and E′′ has at most
n non rational elements, which are all positive. By the above, there exists P3 ∈ Q[X]
such that Crit(P3) ∪ P3(E′′) ⊂ Q \ {0} and P3(0) = 0. Then P := P3 ◦ Hα′ has the
desired properties, since P (0) = 0 and Crit(P ) ∪ P (E) = Crit(P3) ∪ P3(Crit(Hα′)) ∪
P3(Hα′(E)) = Crit(P3) ∪ P3(E′′) ⊂ Q \ {0}.
Let us denote by P1 the polynomial obtained using this lemma, which verifies P1(0) =
0 and E′ := Crit(P1) ∪ P1(E) ⊂ Q \ {0}. We can further assume that P ′1(0) > 0 by
taking (−P1) if necessary. We now send the points E′ to {0, 1}.
Lemma 3.5. Let E ⊂ Q \ {0} a finite set. Then there exists P ∈ Q[X] such that P (E) ⊆
{0}, Crit(P ) ⊆ {0, 1}, 0 < P (0) < 1 and P ′(0) > 0.
Proof. For α ∈ Q, let Fα := (X − α)2 ∈ Q[X], and note that Crit(Fα) = {0}. There
exists α < 0 sufficiently small such that 0 < Fα(0) < Fα(a) for all a ∈ E. We take
F :=
Fα
maxa∈E Fα(a)
.
Let {a1, . . . , al} = F (E) such that 0 < F (0) < a1 < · · · < al = 1. We also add a
rational point a0 ∈ Q such that F (0) < a0 < a1.
Let m and n be the coprime positive integers such that al−1 = mm+n . We recall that
Bm,n verifies Crit(Bm,n) = {0, 1}, Bm,n(0) = Bm,n(1) = 0, Bm,n( mm+n ) = 1, and
Bm,n is strictly increasing between 0 and mm+n . Let P1 := Bm,n, then Crit(P1) = {0, 1}
and 0 < P1 ◦F (0) < P1(a0) < · · · < P1(al−1) = 1. There is one point fewer than before,
so we can iteratively construct P2, . . . , Pl in the same way, so that P := Pl ◦ · · · ◦ P1
verifies Crit(P ) ⊆ {0, 1}, P (a1) = · · · = P (al) = 0 < P (F (0)) < 1 = P (a0) and
P ′(F (0)) > 0. Therefore P ◦ F has the desired properties.
Let us denote by P2 the polynomial obtained using this lemma with the finite set E′
obtained previously. Then the polynomial P := P2 ◦ P1 verifies P (E) ⊆ {0}, Crit(P ) ⊆
{0, 1}, 0 < P (0) < 1 and P ′(0) > 0, thus concluding the proof of Proposition 3.2.
We can now use Proposition 3.2 with the finite set
E := {(1− ζkp )2p}1≤k≤ p−12
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 11
to obtain the map β1 as desired. For 1 ≤ k ≤ p−12 we have (1 − ζkp )2p = (|1 −
ζkp |ζ2k−12p )2p = |1 − ζkp |2p ∈ R so E ⊂ Q̄ ∩ R, and the set E is the Galois orbit of
(1− ζp)2p so it is fixed by Gal(Q̄/Q), hence E verifies the conditions of Proposition 3.2.
Let us denote by D(β1) the dessin corresponding to the Belyi pair (P1, β1). The Belyi
pair (P1, β1) is fixed by the action of the complex conjugation, so the embedding of D(β1)
on P1 admits a symmetry along the real line. Moreover the Belyi function β1 is a polyno-
mial, so D(β1) ∩ R is a (graph theoretic) path. Let vl < · · · < v1 be the negative vertices
on the path, and let ek denote the edge (vk−1, vk). By hypothesis β′1(0) > 0 so v1 is a
black vertex, and for k < l, the vertex vk is of even degree 2dk. We then have e
ξdk
k = ek+1
and eξ
dk
k+1 = ek if k is odd, or e
ηdk
k = ek+1 and e
ηdk
k+1 = ek if k is even. See Figure 6.
As remarked earlier, the Galois orbit of (1− ζp)2p is {(1− ζkp )2p}1≤k≤ p−12 ⊂ R−, and
(1− ζ
p−1
2
p )2p < · · · < (1− ζp)2p < 0. By construction β1((1− ζp)2p) = 0, so (1− ζp)2p
and all its Galois conjugates are black vertices of D(β1) lying on the path (v1, · · · , vl). Let
t > 0 be the index such that vt = (1− ζp)2p, and vt is a black vertex so t is odd. Then
µ0 := ξ
d1ηd2 · · · ηdt−1ξ2dtηdt−1 · · · ηd2ξd1
fixes the edge e1 (Figure 6).
vt v2 v1 ×0
e3 e2 e1
µ0
ξd1
ξd1
ηd2
ηd2
ξdt
ξdt
Figure 6: Dessin D(β1) corresponding to (P1, β1).
Let γ > 0 small enough so that β′1 > 0 on [0, γ
2pq2]. Let us next draw the dessin
D(β2) corresponding to the Belyi pair (P1, β2 = Br,s ◦Bm,n). See Figure 7.
(P1, id)0 1
(P1, Br,s)
0
r
r+s 1
r s
(P1, β2 = Br,s ◦Bm,n)0
m
m+nβ1(0) 1
s
s
mr nr
Figure 7: Dessin D(β2) corresponding to (P1, β2).
By lifting the dessin D(β2) along β1 we obtain the dessin D(β2 ◦ β1) corresponding to
the Belyi pair (P1, β2 ◦ β1). This amounts to replacing each edge of D(β1) by a copy of
D(β2). Note that the degrees of the black and white vertices are thus multiplied by mr and
12 Ars Math. Contemp. 24 (2024) #P2.05
nr, respectively. Analogously to µ0 we define
µ :=(ξmrd1ηξsη)(ξnrd2ηξsη) · · · (ξmrdt−2ηξsη)(ξnrdt−1ηξsη)
· (ξ2mrdtηξsη)(ξnrdt−1ηξsη)(ξmrdt−2ηξsη) · · · (ξnrd2ηξsη)ξmrd1
and we verify again that µ fixes the edge (0, v1). Note also that ξ2s fixes the edge (0, γ p
√
q).
See Figure 8.
vt v2 v1
γ2pq20
s
s
s
s
mr
mr
mr
mr
mr
mr
mr
mr
µ
nr
nr
nr
nr
η
η
η
η
η
η
ξs
ξs
ξs
ξs
ξmrd1
ξmrd1
ξnrd2
ξnrd2
ξmrdt
ξmrdt
Figure 8: Dessin D(β2 ◦ β1) corresponding to (P1, β2 ◦ β1).
Let D0 be the dessin corresponding to the Belyi pair (P1, β2 ◦ β1 ◦ β0). To simplify
the representations of the dessins we only show the vertices 0, ζk2p(1 − ζp), 1 − ζkp , and
ζk2pγ p
√
q. We decorate the vertices ζk2p(1 − ζp) (which map to (1 − ζp)2p ∈ R− by β0)
and ζk2pγ p
√
q (which map to γ2pq2 ∈ R+ by β0) respectively with the symbols ⊖ and ⊕ to
distinguish them. See Figure 9.
0
⊕
γ p
√
q
	
1− ζp
	
1− ζ̄p
⊕
⊕
	
1−
ζ 2p
	 1−
ζ
−
2
p
⊕
ζ p
γ
p√
q
⊕
ζ −
1
p
γ
p √
q
	
1
−
ζ
3 p
	
1−
ζ −
3
p
D0 = D(β2 ◦ β1 ◦ β0)
D(β2 ◦ β1) 0
γ2pq2
⊕
(1− ζp)2p
	(1− ζ2p)2p
(1− ζ3p)2p
Figure 9: Dessin D0 corresponding to (P1, β2 ◦ β1 ◦ β0).
We may now draw the Galois conjugates Dσ for σ ∈ Gal(Q̄/Q) by lifting the dessin
D0 along the projection π, by treating separately the cases σ(ζp) ∈ {ζp, ζ̄p} and σ(ζp) ∈
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 13
{ζ2p , . . . , ζp−2p }. We call the dessins respectively Dk and Djk, see Figure 10. We identify
the outermost edges on opposite sides in the representations.
?
∞
A
A
C
C
B B
D D
0 0
00
ζ•p p
√
q ζ•p p
√
q
⊕
⊕
1− ζ±1p
1− ζ±1p
	
	
⊕
	
×(2p− k)
⊕
	
×(2p− k)
⊕
	
×(k − 1)
⊕
	
×(k − 1)
µ
µ
ξ2s
ξ2s
(a) Dk := Dσ with σ(ζp) = ζ±1p .
?
∞
A
A
C
C
B B
D D
0 0
00
ζ•p p
√
q ζ•p p
√
q
⊕
⊕
1− ζjp
1− ζjp
	 	
	 	
⊕
	
×(2p− k)
⊕
	
×(2p− k)
⊕
	
×(k − 1)
⊕
	
×(k − 1)
µ
µ
ξ2s
ξ2s
(b) Djk := Dσ with σ(ζp) = ζjp where j ∈
{2, . . . , p− 2}.
Figure 10: Dessins (a) Dk and (b) Djk in the Galois orbit of D.
For all k, we have in fact D2k−1 = Dσ where σ : (ζp, p√q) 7→ (ζp, ζk−1p p
√
q), and
D2k = Dσ where σ : (ζp, p√q) 7→ (ζ̄p, ζkp p
√
q). We have similar expressions for the dessins
Djk.
Let k be fixed, and let us consider the dessin Dk. Let A denote one of the two
edges incident to 0 and on the path to the ramification point 1 − ζ±1p . We also call
B := Aη
2k−1
, C := A4p, D := Cη
2k−1
(See Figure 10a). Let E denote the set of edges inci-
dent to 0. The action of η induces the cyclic permutation of the edges of E = {Aηi}0≤i<8p.
Furthermore by construction every white vertex aside from 0 has degree 1 or 2, so η2 fixes
every edge in the complement of E. We can write E = E⊖ ⊔ E⊕ as the disjoint union
of E⊖ := {A2i}0≤i<4p and E⊕ := {B2i}0≤i<4p, such that η sends one to the other. The
action of µ on E⊖ is the transposition (A,C), and similarly the action of ξ2s on E⊕ is the
transposition (B,D).
We do the same for the dessins of the form Djk, with the only difference that this time
the action of µ on E⊖ is trivial, including on the edges A and C (see Figure 10b).
We are almost in the same configuration as in the first example. We define analogously
ω := µηµ−1ξ2sη−1µ,
and we shall prove that for some choices of γ, the actions of ω and of η2 commute only for
D1. To reproduce the proof in the first example we need only show that for some choice of
γ the actions of µηµ−1ξ2s and µ−1ξ2sη−1µ on the set E⊕ is the same as that of ξ2s.
Note that for any edge e ∈ E⊕, the edge eξi is fixed by η if i is not a multiple of s.
To that end we shall show that for some choice of γ the action of µ on E⊕ is the same as
that of ξδ , where δ is the number of occurences of ξ in the word µ, and then that δ is not a
multiple of s.
14 Ars Math. Contemp. 24 (2024) #P2.05
We define the words ρ1, ρ′1, ρ2, ρ
′
2, . . . , ρ2t−2, ρ
′
2t−2 ∈ F2 to be the increasing subse-
quence of the prefixes ending in η of the word µ defined above, such that ρ1 := ξmrd1η,
ρ′1 := ρ1ξ
sη, ρ2 := ρ′1ξ
nrd2η, ρ′2 := ρ2ξ
sη, etc., and µ = ρ′2t−2ξ
mrd1 . We shall show by
induction that for some choice of γ the action of ρi (resp. ρ′i) is the same as the action of
ξδi (resp. ξδ
′
i ), where δi (resp. δ′i) is the number of occurences of ξ in the word ρi (resp.
ρ′i). By induction it suffices to show that δi, δ
′
i are not multiples of s. Modulo s we have
δi ≡ δ′i equal to the non empty partial sum of
mrd1+nrd2+ · · ·+mrdt−2+nrdt−1+2mrdt+nrdt−1+mrdt−2+ · · ·+nrd2+mrd1
consisting of the first i terms.
To proceed we shall use the following result, but let us first introduce some notations.
Let P =
∑d
i=0 ciX
i ∈ Z[X] and c ∈ Z>0 such that β1 = Pc . Note that P and c do not
depend on the choice of γ, and 0 < β1(0) =
P (0)
c < 1 so 0 < c0, c− c0. We define
α := v2(γ
2pq2), ν := v2(c0) + v2(c− c0),
where v2 denotes the 2-valuation.
Lemma 3.6. If α > ν, then there exists e ∈ Z such that em ≡ c0 mod 2α and en ≡ c−c0
mod 2α, and v2(s) ≥ α− ν.
Proof. Let a, b ∈ Z coprime such that γ2pq2 = ab 2α.
Firstly,
m
m+ n
= β1(
a
b
2α) =
P (ab 2
α)
c
=
∑d
i=0 cia
i2αibd−i
bdc
,
so there exists f ∈ Z such that fm = ∑di=0 ciai2αibd−i and f(m+ n) = bdc, so
em ≡ c0 mod 2α, en ≡ c− c0 mod 2α
for e ∈ Z such that ebd ≡ f mod 2α.
Secondly,
r
r + s
= Bm,n(β1(0)) =
β1(0)
m(1− β1(0))n
β1(
a
b 2
α)m(1− β1(ab 2α))n
=
bd(m+n)cm0 (c− c0)n
(bdP (ab 2
α))m(bdc− bdP (ab 2α))n
,
so there exists g ∈ Z such that gr = bd(m+n)cm0 (c−c0)n and g(r+s) = (bdP (ab 2α))m(bdc−
bdP (ab 2
α))n. In the expansion of (bdP (ab 2
α))m, aside from the constant term bdmcm0 , ev-
ery other term is a multiple of an integer of the form ci02
αj with i ≤ m− 1 and j ≥ m− i.
By hypothesis α > ν ≥ v2(c0), so those other terms are all multiples of 2α+(m−1)v2(c0),
hence there exists A ∈ Z such that (bdP (ab 2α))m = bdmcm0 +A2α+(m−1)v2(c0). Similarly
there exists B ∈ Z such that (bdc − bdP (ab 2α))n = bdn(c − c0)n + B2α+(n−1)v2(c−c0).
Then g(r+ s) = bd(m+n)cm0 (c− c0)n +C2α+(m−1)v2(c0)+(n−1)v2(c−c0) for some C ∈ Z,
so gr = bd(m+n)cm0 (c − c0)n and gs = C2α+(m−1)v2(c0)+(n−1)v2(c−c0). The integers r
and s are coprime, so after dividing gr and gs by their greatest common dividor we obtain
that
v2(s) ≥ α− ν > 0.
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 15
Using this lemma, we know that if α > ν, then there exists e ∈ Z such that em ≡ c0
mod 2α and en ≡ c − c0 mod 2α, v2(s) ≥ α − ν where ν does not depend on γ, and r
is coprime to s so is not a multiple of 2. Therefore there exists e′ ∈ Z such that e′mr ≡ c0
mod 2α and e′nr ≡ c − c0 mod 2α. Moreover 2α−ν is a common divisor of 2α and s,
so by the above modulo 2α−ν we have e′δi ≡ e′δ′i equal to the non empty partial sum δ̃i
consisting of the first i terms of the sum
c0d1 + (c− c0)d2 + · · ·+ c0dt−2 + (c− c0)dt−1 + 2c0dt
+(c− c0)dt−1 + c0dt−2 + · · ·+ (c− c0)d2 + c0d1.
Similarly e′δ is equal modulo 2α−ν to the whole sum
δ̃ := 2(c0d1 + (c− c0)d2 + · · ·+ c0dt−2 + (c− c0)dt−1 + c0dt).
By construction c0, c − c0, di are positive and do not depend on the choice of γ, so
0 < c0d1 ≤ δ̃i ≤ δ̃, thus for any choice of γ such that α > ν and δ̃ < 2α−ν (for instance
γ = 2
u
2v+1 with 1 ≪ u ≪ v), we obtain δ̃i, δ̃ ̸≡ 0 mod 2α−ν , and in consequence δi, δ′i
and δ are not multiples of s. Therefore we can now conclude by induction that the actions
of ρi and ρ′i are the same as that of ξ
δi and ξδ
′
i , respectively. Indeed, δ1 is not a multiple
of s so ρ1 = ξδ1η and ξδ1 have the same action on E⊕. If ρi has the same action as ξδi
on E⊕, then ρ′i = ρiξ
sη has the same action as ξδiξsη = ξδ
′
iη on E⊕, and also the same
action as ξδ
′
i because δ′i is not a multiple of s. Similarly, if ρ
′
i has the same action as ξ
δ′i
on E⊕, then ρi+1 has the same action as ξδi+1η on E⊕, and also the same action as ξδi+1
because δi+1 is not a multiple of s.
We have thus proved that µ has the same action as ξδ on E⊕, and by symmetry µ−1
has the same action as ξ−δ on E⊕. And δ and 2s− δ are not multiples of s, so µηµ−1ξ2s
and µ−1ξ2sη−1µ have the same action as ξ2s on E⊕, as announced. We shall now observe
the action of ω = µηµ−1ξ2sη−1µ on E. Let Mk and M
j
k denote the monodromy maps of
the dessins Dk and Djk.
For the dessins Dk for 1 ≤ k ≤ 2p, the action of µ on E⊖ is the transposition (A,C),
and the action of ξ2s on E⊕ is the transposition (B,D), therefore the action of ω fixes the
set E on which it induces the permutation
Mk(ω)|E = (A,C)(B,D) ·
4p−1∏
i=0
(Aη
2i
, Aη
2i+1
) · (A,C)(B,D).
Hence for k = 1,
M1(ω)|E = (A,Aη
4p
)(Aη, Aη
4p+1
) ·
4p−1∏
i=0
(Aη
2i
, Aη
2i+1
) · (A,Aη4p)(Aη, Aη4p+1)
=
4p−1∏
i=0
(Aη
2i
, Aη
2i+1
)
so ω and η2 commute on E. Moreover η2 acts trivially on the complement of E, so finally
M1(ω) and M1(η2) commute.
16 Ars Math. Contemp. 24 (2024) #P2.05
For k = 2, we observe that Bωη
2
= Dη
−1η2 = Dη but Bη
2ω = Bη
2η−1 = Bη .
Similarly, for 3 ≤ k ≤ 2p, we observe that Aωη2 = Cηη2 = Cη3 but Aη2ω = Aη2η = Aη3 .
Therefore Mk(ω) and Mk(η2) do not commute for 2 ≤ k ≤ 2p.
For the dessins Djk for 1 ≤ k ≤ 2p and 2 ≤ j ≤ p−12 , ξ2s on E⊕ is the transposition
(B,D), and µ acts trivially on E⊖, therefore the action of ω fixes the set E on which it
induces the permutation
M jk(ω)|E = (B,D) ·
4p−1∏
i=0
(Aη
2i
, Aη
2i+1
) · (B,D).
Hence we observe that Bωη
2
= Dη
−1η2 = Dη but Bη
2ω = Bη
2η−1 = Bη , so M jk(ω) and
M jk(η
2) do not commute.
We have thus shown that the actions of ω and η2 commute only for D1, this concludes
the proof that D̃ is a regular dessin with moduli field Q(ζp, p√q).
3.3 Regular dessin with moduli field Q(ζ3, 3
√
3)
Finally, let us exhibit a regular dessin with moduli field Q(ζ3, 3
√
3) of smaller degree by
choosing a Belyi map that is a rational function instead of a polynomial as was done in the
previous subsections. Let
C : y2 = x(x− (1− ζ3))(x− 3
√
3),
β : C → P1, (x, y) 7→ (x+ 3
3)3
35(x− 32)2 .
The function β is given by the composition of the following maps β = β1 ◦ β0 ◦ π.
1. π : C → P1 is the projection on the coordinate x, which ramifies over {0, 1 −
ζ3,
3
√
3,∞}.
2. β0 := X6 ∈ Q[X], Crit(β0) = {0} so β0 ◦ π ramifies over {0, (1 − ζ3)6 =
−33, 32,∞}.
3. β1 :=
(X+33)3
35(X−32)2 , Crit(β1) = {0, 1} so β = β1 ◦ β0 ◦ π ramifies over {0, 1,∞}.
The pair (C, β) is thus a Belyi pair, and we call D the dessin corresponding to (C, β).
Similarly as in 3.1, D has moduli field Q(ζ3, 3
√
3). We will proceed analogously to show
that the regular closure D̃ has the same field of moduli. Let us first draw the dessin D0
corresponding to the Belyi pair (P1, β1 ◦ β0) (see Figure 11), and lift it to the conjugate
curves Cσ to obtain the conjugate dessins Dσ for σ ∈ Gal(Q(ζ3, 3
√
3)/Q) (see Figure 12).
As usual we identify the outermost edges on opposite sides.
We can now compute the cartographic groups of the dessins. Let Mk denote the mon-
odromy map of Dk. Then
Mk(ξ) =
(1,13,14,7,25,26)(2,15,16)(3,17,18)(4,19,20)(5,21,22)
(6,23,24)(8,27,28)(9,29,30)(10,31,32)(11,33,34)(12,35,36)
for all 1 ≤ k ≤ 6, and
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 17
(P1, id)0 1
(P1, β1)?
−33 0 1 32
(P1, β1 ◦ β0)
0
?
??
?
? ?
1− ζ23
1− ζ3
3
√
3
ζ3
3
√
3
ζ23
3
√
3
Figure 11: Construction of D0.
? ??
2
8
3
9
4
10
5
11
6
12
15
27
16
28
17
29
18
30
19
31
20
32
21
33
22
34
23
35
24
36
13
14
25
26
0 0
00
∞3√3 3
√
3
1− ζ3
1− ζ3
1 7
1 7
(a) D1 := D
? ??
2
8
3
9
4
10
5
11
6
12
15
27
16
28
17
29
18
30
19
31
20
32
21
33
22
34
23
35
24
36
13
14
25
26
0 0
00
∞
ζ3
3
√
3 ζ3
3
√
3
1− ζ23
1− ζ23
1 7
1 7
(b) D2 := Dσ, σ : (ζ3, 3√q) 7→ (ζ23 , ζ3 3
√
q)
18 Ars Math. Contemp. 24 (2024) #P2.05
? ??
2
8
3
9
4
10
5
11
6
12
15
27
16
28
17
29
18
30
19
31
20
32
21
33
22
34
23
35
24
36
13 14
25
26
0 0
00
∞
ζ3
3
√
3 ζ3
3
√
3
1− ζ3
1− ζ3
1 7
1 7
(c) D3 := Dσ, σ : (ζ3, 3√q) 7→ (ζ3, ζ3 3√q)
? ??
2
8
3
9
4
10
5
11
6
12
15
27
16
28
17
29
18
30
19
31
20
32
21
33
22
34
23
35
24
36
13
14
2526
0 0
00
∞
ζ23
3
√
3 ζ23
3
√
3
1− ζ23
1− ζ23
1 7
1 7
(d) D4 := Dσ, σ : (ζ3, 3√q) 7→ (ζ23 , ζ23 3
√
q)
? ??
2
8
3
9
4
10
5
11
6
12
15
27
16
28
17
29
18
30
19
31
20
32
21
33
22
34
23
35
24
36
13
14
25
26
0 0
00
∞
ζ23
3
√
3 ζ23
3
√
3
1− ζ3
1− ζ3
1 7
1 7
(e) D5 := Dσ, σ : (ζ3, 3√q) 7→ (ζ3, ζ23 3
√
q)
? ??
2
8
3
9
4
10
5
11
6
12
15
27
16
28
17
29
18
30
19
31
20
32
21
33
22
34
23
35
24
36
13
14
25
26
0 0
00
∞3√3 3
√
3
1− ζ23
1− ζ23
1 7
1 7
(f) D6 := Dσ, σ : (ζ3, 3√q) 7→ (ζ23 , 3
√
q)
Figure 12: Dessins D1, . . . ,D6 in the Galois orbit of D.
N. Daire et al.: Regular dessins with moduli fields of the form Q(ζp, p
√
q) 19
• M1(η) =
(1,2,3,4,5,6,7,8,9,10,11,12)(13,36)(14,15)(16,17)(18,19)
(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35) ,
• M2(η) =
(1,2,3,4,5,6,7,8,9,10,11,12)(13,36)(14,27)(15,26)(16,29)
(17,28)(18,19)(20,21)(22,23)(24,25)(30,31)(32,33)(34,35) ,
• M3(η) =
(1,2,3,4,5,6,7,8,9,10,11,12)(13,36)(14,27)(15,26)(16,17)
(18,31)(19,30)(20,21)(22,23)(24,25)(28,29)(32,33)(34,35) ,
• M4(η) =
(1,2,3,4,5,6,7,8,9,10,11,12)(13,36)(14,27)(15,26)(16,17)
(18,19)(20,33)(21,32)(22,23)(24,25)(28,29)(30,31)(34,35) ,
• M5(η) =
(1,2,3,4,5,6,7,8,9,10,11,12)(13,36)(14,27)(15,26)(16,17)
(18,19)(20,21)(22,35)(23,34)(24,25)(28,29)(30,31)(32,33) ,
• M6(η) =
(1,2,3,4,5,6,7,8,9,10,11,12)(13,24)(14,27)(15,26)(16,17)
(18,19)(20,21)(22,23)(25,36)(28,29)(30,31)(32,33)(34,35) .
Using the computer algebra system SageMath [10], we determined that
|⟨M1(ξ),M1(η)⟩| = 42467328 = 219 · 34.
Moreover, M1(ξ), M1(η) and M1(ξη) respectively have orders 6, 12 and 12, so the Euler
characteristic of the underlying surface of D̃1 is
χ = |⟨M1(ξ),M1(η)⟩| · (
1
ordM1(ξ)
+
1
ordM1(η)
+
1
ordM1(ξη)
− 1)
= −28311552 = −220 · 33,
and its genus is g = 1− χ2 = 14155777.
We will now show that D̃1 is not isomorphic to D̃2, . . . , D̃6. We claim that ω :=
[ξ−1η2ξ, ξη] ∈ kerM1 \
⋃
2≤k≤6 kerMk, thus concluding the proof. Indeed, we obtain:
• M1(ω) = id;
• M2(ω) = (13, 25)(15, 27)(21, 33)(23, 35);
• M3(ω) = (17, 29)(21, 33);
• M4(ω) = (13, 25)(15, 27)(19, 31)(21, 33);
• M5(ω) = (13, 25)(17, 29);
• M6(ω) = (13, 25)(19, 31)(21, 33)(23, 35).
We have thus constructed a regular dessin D̃ of degree 219 · 34 and genus 14155777
with moduli field Q(ζ3, 3
√
3).
ORCID iDs
Fumiharu Kato https://orcid.org/0009-0002-4800-0029
20 Ars Math. Contemp. 24 (2024) #P2.05
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