# Galois Group at Each Point for Some Self-Dual Curves

Authors: Hiroyuki Hayashi; Hisao Yoshihara
Journal: Geometry
 (2013)

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## Abstract

We study the Galois group defined by a point projection for plane curve. First, we present a sufficient condition that the group is primitive and then determine the structure at each point for some self-dual curves.


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## Body

## 1. Introduction

This study is a continuation of [1–4], and so forth. In general, it is not easy to determine the Galois group GP at every point P for plane curve, in particular for curve with singular point. When we determine the structure of GP, it is important to know whether it is primitive or not. However, there are not so many results which are useful for our purpose (cf. [5]). In this paper we give a geometrical criterion and then determine the group at each point for some self-dual curves.Letk be an algebraically closed field of characteristic zero. We fix it as the ground field of our discussions. Let C be an irreducible plane curve of degree d (≥2) and K=k(C) the rational function field of C. Let (X:Y:Z) be a set of homogeneous coordinates on ℙ2 and put P1=(0:0:1),P2=(0:1:0),andP3=(1:0:0). Let F(X,Y,Z) be the defining equation of C and put f(x,y)=F(X,Y,Z)/Zd, where x=X/Z,y=Y/Z.


### 1.1. Galois Group

Letr:C~→C be the resolution of singularities of C. For a point P∈ℙ2, let P^ be the dual line in the dual space ℙ2^ of ℙ2 corresponding to P. We define the morphism πP by
(1)πP:C~∋Q↦ℓPR^∈P^≅ℙ1,
where ℓPR^ is the point in ℙ2^ corresponding to the line ℓPR, which passes through P and R=r(Q) if P≠R. In case P=R, the line ℓPR is the tangent line to the branch of C at R. Clearly, we have degπP=d-mP(C) and a field extension πP*:k(ℙ1)↪K=k(C~), where mP(C) denotes the multiplicity of C at P. In case P∉C, we understand mP(C)=0. We put n(P)=d-mP(C); if there is no fear of confusion, we simply denote it by n. Since the extension depends only on P, we denote k(ℙ1) by KP, that is, we have πP*:KP↪K. Let LP be the Galois closure of K/KP and GP the Galois group Gal(LP/KP).Definition 1.
We callGP the Galois group at P for C. In case K/KP is a Galois extension, the point P is said to be a Galois point.In casek is the field of complex numbers, GP is isomorphic to the monodromy group of the covering πP:C~→ℙ1 [6, 7].


### 1.2. Self-Dual Curve

Definition 2.
A pointQ∈C is said to be a cusp of C if it is a singular point and r-1(Q) consists of a single point. Furthermore, if μ:BQ(ℙ2)→ℙ2 is a blow-up and μ-1(Q) is a nonsingular point of the proper transform of μ-1(C), the point Q is said to be a simple cusp.Denote byC^ the dual curve of C.Definition 3.
IfC^ is projectively equivalent to C, then C is said to be a self-dual curve.SupposeC is smooth. Then, C is self-dual if and only if d=2. However, if C has a singular point, the condition that C is self-dual becomes complicated. The following proposition has been known (cf. [8]).Proposition 4.
SupposeC is one of the following curves:(1)
Chas just one singular point;(2)
C is rational and has only simple cusps as singular points.Then, C is a self-dual curve if and only if C is projectively equivalent to the curve defined by y=xd.Example 5.
It seems that only a few self-dual curves have been known. Here, we present some of them(I)
C(e,d): the curve defined by YeZd-e=Xd,gcd(e,d)=1,1≤e≤d-1;(II)
C(4): the curve defined by (YZ-X2)2=X3Y (cf. [9]);(III)
C54: the curve defined by (XY-XZ+YZ)3+54X2Y2Z2=0 (cf. [10]).For the curve C(e,d), if 1<e<d-1, then P1=(0:0:1) and P2=(0:1:0) are not simple cusps and C(e,d) has no flex. The curve C(4) has two cusps P1 and P2, where P1 is not a simple cusp. The curve C54 has three cusps: P1,P2, and P3 and the normalization is an elliptic curve. It is easy to find the dual curve of C(e,d); however, in the other curves we need some consideration, for the details, see [9, 10].Remark 6.
LetΦC be the rational map ℙ2⇢ℙ2 giving the dual of C, that is,
(2)ΦC(X:Y:Z)=(∂XF:∂YF:∂ZF),
where F is the defining equation of C. In the case where C=C(e,d), the map ΦC turns out to be a quadratic transformation of ℙ2:
(3)ΦC(X:Y:Z)=(-dYZ:eZX:(d-e)XY).
We use the following notation:(i)
Zm: the cyclic group of order m;(ii)
Sd: the symmetric group of degree d;(iii)
i(X1,X2;Q): the intersection number of two curves X1 and X2 at Q;(iv)
ℓPQ: the line passing through P and Q;(v)
ℓP: a line passing through P;(vi)
TQ=TQ(C): the tangent line to C at Q.


## 1.1. Galois Group

Letr:C~→C be the resolution of singularities of C. For a point P∈ℙ2, let P^ be the dual line in the dual space ℙ2^ of ℙ2 corresponding to P. We define the morphism πP by
(1)πP:C~∋Q↦ℓPR^∈P^≅ℙ1,
where ℓPR^ is the point in ℙ2^ corresponding to the line ℓPR, which passes through P and R=r(Q) if P≠R. In case P=R, the line ℓPR is the tangent line to the branch of C at R. Clearly, we have degπP=d-mP(C) and a field extension πP*:k(ℙ1)↪K=k(C~), where mP(C) denotes the multiplicity of C at P. In case P∉C, we understand mP(C)=0. We put n(P)=d-mP(C); if there is no fear of confusion, we simply denote it by n. Since the extension depends only on P, we denote k(ℙ1) by KP, that is, we have πP*:KP↪K. Let LP be the Galois closure of K/KP and GP the Galois group Gal(LP/KP).Definition 1.
We callGP the Galois group at P for C. In case K/KP is a Galois extension, the point P is said to be a Galois point.In casek is the field of complex numbers, GP is isomorphic to the monodromy group of the covering πP:C~→ℙ1 [6, 7].


## 1.2. Self-Dual Curve

Definition 2.
A pointQ∈C is said to be a cusp of C if it is a singular point and r-1(Q) consists of a single point. Furthermore, if μ:BQ(ℙ2)→ℙ2 is a blow-up and μ-1(Q) is a nonsingular point of the proper transform of μ-1(C), the point Q is said to be a simple cusp.Denote byC^ the dual curve of C.Definition 3.
IfC^ is projectively equivalent to C, then C is said to be a self-dual curve.SupposeC is smooth. Then, C is self-dual if and only if d=2. However, if C has a singular point, the condition that C is self-dual becomes complicated. The following proposition has been known (cf. [8]).Proposition 4.
SupposeC is one of the following curves:(1)
Chas just one singular point;(2)
C is rational and has only simple cusps as singular points.Then, C is a self-dual curve if and only if C is projectively equivalent to the curve defined by y=xd.Example 5.
It seems that only a few self-dual curves have been known. Here, we present some of them(I)
C(e,d): the curve defined by YeZd-e=Xd,gcd(e,d)=1,1≤e≤d-1;(II)
C(4): the curve defined by (YZ-X2)2=X3Y (cf. [9]);(III)
C54: the curve defined by (XY-XZ+YZ)3+54X2Y2Z2=0 (cf. [10]).For the curve C(e,d), if 1<e<d-1, then P1=(0:0:1) and P2=(0:1:0) are not simple cusps and C(e,d) has no flex. The curve C(4) has two cusps P1 and P2, where P1 is not a simple cusp. The curve C54 has three cusps: P1,P2, and P3 and the normalization is an elliptic curve. It is easy to find the dual curve of C(e,d); however, in the other curves we need some consideration, for the details, see [9, 10].Remark 6.
LetΦC be the rational map ℙ2⇢ℙ2 giving the dual of C, that is,
(2)ΦC(X:Y:Z)=(∂XF:∂YF:∂ZF),
where F is the defining equation of C. In the case where C=C(e,d), the map ΦC turns out to be a quadratic transformation of ℙ2:
(3)ΦC(X:Y:Z)=(-dYZ:eZX:(d-e)XY).
We use the following notation:(i)
Zm: the cyclic group of order m;(ii)
Sd: the symmetric group of degree d;(iii)
i(X1,X2;Q): the intersection number of two curves X1 and X2 at Q;(iv)
ℓPQ: the line passing through P and Q;(v)
ℓP: a line passing through P;(vi)
TQ=TQ(C): the tangent line to C at Q.


## 2. Statement of Results

We need some preparations before stating the results. A curve means a nonsingular projective algebraic curve. LetX1 and X2 be curves and f:X1→X2 a surjective morphism, which we call a covering for short. We denote by e(R,f) the ramification index of f at R∈X1. If there is no fear of confusion, we simply denote it by e(R).Definition 7.
Letf:X1→X2 be the covering above. If there exists a curve X3 and coverings α:X1→X3 and β:X3→X2 such that f=βα, degα≥2 and degβ≥2, then f is said to be decomposable and X3 an intermediate covering. If such a curve X3 does not exist, then f is said to be indecomposable (cf. [11]).Definition 8.
Letf:X1→X2 be the covering above and R1,…,Rr all the ramification points for f. Put e(Ri)=ei(1≤i≤r). The covering f is said to be an s-covering overf(Ri) if there exists no ramification point in f-1f(Ri) except Ri. The f is said to be an s-covering if it is an s-covering over each f(Ri)(1≤i≤r).Definition 9.
With the same notation as in Definition8, we call {(R1,…,Rr),(e1,…,er)} (or, simply (e1,…,er)) the ramification data for f.We give several sufficient conditions thatf is indecomposable. Some of them will not be used later in this paper.Proposition 10.
Letf:X1→X2 be the covering above and n=degf. If one of the following conditions is satisfied, then f is indecomposable.(1)
For somei(1≤i≤r), ei is prime and n<2ei.(2)
e1=n-1.(3)
X2 is a rational curve, f is an s-covering except over f(R1) and ei is prime for each i≥s+1, where f-1f(R1)={R1,…,Rs}.Proposition 11.
With the same notation as in Proposition10, if f is an s-covering and satisfies one of the following conditions, then f is indecomposable.(1)
X1 is a rational curve, e1≥e2, n-1≥e2, and ei is prime for each i≥3.(2)
X1 is a rational curve and ei is prime for each i≥2.(3)
X2 is a rational curve and ei is prime for each i.
Hereafter, we follow the notation in Section1. By taking a suitable projective change of coordinates, we can assume the projection center is P1 without changing the structure of GP. Putting y=tx, we have KP=k(t) and K=k(x,y)=k(t,x). Put g(x)=f(x,tx)/xm∈k(t)[x], where m=mP(C) and let {x1,…,xn}  (n=n(P)) be the roots of g(x)=0. Then, we can consider GP as a permutation subgroup of Sn. Note that GP is a transitive subgroup of Sn. Hence, GP is a primitive group if and only if the isotropy subgroup of an element of {x1,…,xn} is a maximal subgroup of Sn.Theorem 12.
The groupGP is primitive if and only if πP is indecomposable. In particular, if n(P) is a prime number, then GP is primitive for P∈ℙ2.Definition 13.
AssumeQ∈C is a smooth point or a cusp. A line ℓ=ℓPQ is said to be a simple e-tangent line to C if the following conditions are satisfied:(1)
ifQ≠P (resp., Q=P), then i(C,ℓ;Q)=e (resp., e+m), where e≥2 and m=mP(C);(2)
the curvesC and ℓ have normal crossings except at Q.Sometimes we call ℓ a simple e-tangent for short.Note that a simplee-tangent ℓPQ yields an s-covering over πP(Q).Lemma 14.
We have the following assertions forGP.(1)
If each lineℓP has normal crossings with C or is a simple e-tangent line to C such that e is a prime number, then GP is primitive (cf. [5, Lemma 4.4.4]).(2)
If there exists a simple2-tangent line ℓP, then GP contains a transposition.The following lemma is well known.Lemma 15.
If a permutation groupG⊂Sn is primitive and contains a transposition, then it is a full symmetric group.Combining the results above, we get the following corollary.Corollary 16.
If the coveringπP:C~→ℙ1 is one of the coverings in Propositions 10 or 11 and πP is an s-covering over πP(Ri) with ei=2 for some i(1≤i≤r), then GP is a full symmetric group. In particular, if each line ℓP has normal crossings with C or is a simple 2-tangent, then GP is a full symmetric group.Corollary16 implies [2, Theorem 1 and 1′]. Now we can state the structure of GP as follows.Theorem 17.
For the curvesC in Example 5, the Galois groups GP are as follows, where Z1 indicates the trivial group(I)
the caseC=C(e,d) (see Table 1);(II)
the caseC=C(4) (see Table 2);(III)
the caseC=C54 (see Table 3).Table 1
P
P
1
P
2
P
3
P
∈
C
∖
{
P
1
,
P
2
}
P
∈
ℙ
2
∖
C
∪
{
P
3
}
G
P
Z
d
-
e
Z
e
Z
d
S
d
-
1
S
dTable 2
P
P
1
,
P
2
P
∈
C
∖
{
P
1
,
P
2
}
P
∈
ℙ
2
∖
C
G
P
Z
2
S
3
S
4Table 3
P
P
1
,
P
2
,
P
3
P
∈
C
∖
{
P
1
,
P
2
,
P
3
}
P
∈
ℙ
2
∖
C
G
P
S
3
S
5
S
6Remark 18.
For the curves in Theorem17, P is a Galois point if and only if GP is a cyclic group. However, the same assertion does not hold true in general, see, for example, [3].


## 3. Proofs

First, we prove Propositions10 and 11.Claim 1.
Supposef and a ramification point R∈X1 satisfy the following conditions:(1)
f is an s-covering over f(R);(2)
e(R) is prime.If there exists an intermediate covering β:X3→X2, then β is unramified at R′=α(R).Proof.
Supposeβ is ramified at R′. Then, since e(R) is prime, we have e(R′,β)=e(R,f), hence R′ is not a branch point for α. Then, there will appear another ramification point for f in f-1(f(R)). This is a contradiction.The proof of Proposition10 is as follows. Suppose f is decomposable and there exists a covering β:X3→X2 as in Definition 7. First, we prove the assertion (1). Since ei is prime, β is unramified at Ri′ by Claim 1. Hence, we have e(Ri,α)=e(Ri,f). Since there exists at least two points in β-1(f(Ri)), we have n=degf≥2e(Ri,f), which contradicts the assumption. Next we prove (2). Clearly α and β are ramified at R1 and R1′, respectively. Put B1=f(R1). Then, since e1=n-1, β-1(B1) consists of one or two points. In the former case, α-1(β-1(B1)) consists of two points, on the other hand in the latter case α-1(B1i)(i=1,2) consists of one point, where β-1(B1)={B11,B12}. In each case we infer the inequality n=degf≥(n-1)+2, which is a contradiction. We go to the proof of (3). Then, by Claim 1, Bi(i≥2) is not a branch point for β. Thus, B1 is the only branch point for β. Then, by Hurwitz’s formula, we have 2g(X3)-2=-2b+c, where g(X3) is the genus of X3, b is the degree of β, and c≤b-1. Since g(X3)≥0, this inequality implies b=1, which is a contradiction.Next we prove Proposition11. In each case we use the reduction to absurdity, that is, suppose f is decomposable. So we use the notation Ri′=α(Ri)(1≤i≤r). In the case (I), by Claim 3, β is unramified at Ri′(i≥3). Since X2 and X3 are rational, from Hurwitz’s formula, we infer that β is ramified with the index e(R1′,β)=e(R2′,β)=degβ. Then, since there exists no ramification points in f-1(f(Ri)) except Ri(i=1,2), α must branch at R1′ and R2′. However, there exists an unramified point in f-1(f(R2)), this is a contradiction. Therefore, f is indecomposable. In the case (II), by Claim 1, β is unramified at Ri′ for i≥2. Since X3 is rational, by Hurwitz’s formula, we have a contradiction. In the case (III) similarly, by Claim 1, β is unramified at every point; however, since X2 is rational, β must be an identity, which is a contradiction. This completes the proof of Proposition 11.The proof of Theorem12 is as follows: suppose GP is not primitive and let Gx be the isotropy group of x=x1 in GP. Then, there exists a subgroup H of GP such that Gx⊊H⊊GP. Let CH be the nonsingular model of the intermediate field which corresponds to H by the Galois correspondence. Then, there exist the coverings α:C~→CH and β:CH→ℙ1 such that πP=βα. Thus, πP is decomposable. The converse assertion is clear from the Galois correspondence.The proof of Lemma14 is simple. In view of Definition 13, we see that the assertion (1) is another expression of (3) in Proposition 11. The assertion (2) may be well known (cf. [7]).Now we proceed to the proof of Theorem17. The structure of GP depends on the covering πP and πP depends on the position of P. We prove by examining the cases where P lies on the tangent line to C at the cusp or at the flex. Hereafter, we assume C is the curve in Theorem 17. Since C is a self-dual curve and has only cusps as the singularity, the following remark is clear.Remark 19.
Suppose a lineℓ satisfies the following conditions:(1)
ℓ does not pass through any cusp;(2)
ℓ is not the tangent line to C at the flex.Then, ℓ is a simple 2-tangent line to C or ℓ and C have normal crossings.Proof of the Case (I). Assume C=C(e,d). It has the following property.Claim 2.
The tangent lineTP1(resp.,TP2) is Y=0(resp.,Z=0) and TP1∩TP2={P3}, which does not lie on C. In case e=1(resp.,d-1), C has one flex at P1(resp.,P2). On the other hand, in case 1<e<d-1, C has no flex.Proof.
Calculating the Hessian ofXd-YeZd-e (cf. [12]), we infer readily the assertions.IfP=P1,P2, or P3, then GP can be determined directly. In fact, if P=P1, then consider the affine part Z≠0 of C, that is, the affine defining equation is ye-xd=0. Then, putting y=tx, we get te-xd-e=0, hence GP≅Zd-e. The other case P=P2 is similarly determined. If P=P3, then consider the affine part X≠0, we get yezd-e=1. Putting z=ty, we get td-eyd=1, hence GP≅Zd. As we have seen above, these points are Galois ones.Next we treat the caseP∈C∖{P1,P2}. First, we prove the subcase 1<e<d-1. Since C is a self-dual curve and has no flex, we see that, if a line ℓP passes through neither P1 nor P2, then it has normal crossings with C or it is a simple 2-tangent line to C. Furthermore, by Hurwitz’s formula, we see there exists a simple 2-tangent. Then, by (1) in Proposition 11 and Lemma 15, we have GP≅Sd-1. Next we prove the subcase e=1. Then, P1(resp.,P2) is a flex (resp., cusp) and the tangent line at P1(resp.,P2) does not meet C except at P1(resp.,P2). If a line ℓP does not pass through P2, then it has normal crossings with C or it is a simple 2-tangent line to C. By (2) in Proposition 11 and Lemma 15, we have GP≅Sd-1. The proof of the case e=d-1 is the same.Now we prove the case whereP∈ℙ2∖C and P≠P3. If P∈ℓP1P2 and 1<e<d-1, then πP has two ramification points R1 and R2 such that e(R1)=e, e(R2)=d-e and πP(R1)=πP(R2). Thus, πP is not an s-covering. If ℓP passes through neither P1 nor P2, then ℓP is a simple 2-tangent to C or has normal crossings with C. By (3) in Proposition 10, πP is indecomposable. Since there exists a simple 2-tangent ℓP, we conclude GP≅Sd. In case P∈ℓP1P2 and e=1 or d-1, πP is an s-covering and e1=d-1 and e2=2, hence by (2) in Proposition 10, GP is primitive and there exists a simple 2-tangent line ℓP, thus we conclude GP≅Sd. In view of Remark 19, we conclude easily from the similar argument that GP≅Sd when P∈ℙ2∖{C∪ℓP1P2}.Proof of the Case (II). Assume C=C(4). It has the following property.Claim 3.
TheTP1(resp.,TP2) is Y=0(resp.,Z=0) and TP1∩TP2={P3}, which does not lie on C. Furthermore, TP1∩C={P1} and TP2∩C={P2,(1:1:0)}. The C has one flex F of order 1, that is, i(C,TF;F)=3 and TF does not pass though P3.Proof.
The last assertion is checked by Hurwitz’s formula and the others are simple.Remark 20.
The coordinates of the flexF are computed as (-576:-4096:135).Clearly, ifP=P1 or P2, then GP≅Z2. If P∈C∖{P1,P2}, then n=3, hence GP is primitive. We divide the proof into three cases:(1)
P=F;(2)
P=(1:1:0);(3)
P is the other point.In any case, by Hurwitz’s formula, we infer that there exists at least one simple2-tangent line passing through P, hence GP≅S3. Then consider the case P∈ℙ2∖C. If P∈ℓP1P2, then πP has ramification points R1 and R2 such that e(R1)=e(R2)=2 and πP(R1)=πP(R2). Thus, πP is not an s-covering. Consider πP for the most special case ℓP1P2∩TF={P}. We infer from Hurwitz’s formula that the ramification data is (3,24):=(3,2,2,2,2). By (3) in Proposition 10, we have GP≅S4. There are several cases of position of P which yield different ramification data; however, it is easy to see that there exists i such that ei=2. Then from Propositions 10 or 11, we conclude GP≅S4.Proof of the Case (III). Assume C=C54. It has the following property. There exists a projective transformation σ such that σ(C)=C and σ(X,Y,Z)=(Y,X,-Z), (-X,Z,Y) or (Z,Y,X) so that σ interchanges Pi(i=1,2,3).Claim 4.
The flexes ofC are F1=(4:-1:4),F2=(1:-4:4), and F3=(4:-4:1), hence the tangent lines to C at them are L1:X+8Y+Z=0,L2:8X+Y-Z=0, and L3:-X+Y+8Z=0, respectively. On the other hand, the tangent lines to C at P1,P2, and P3 are L4:X=Y,L5:X=Z, and L6:Y=Z, respectively. There exist just three points Qi(i=1,2,3) satisfying the following conditions:(1)
Qi∉C;(2)
ifℓ=ℓQi does not pass through any cusp, then ℓ and C have normal crossings or there exist two points Q′∈C satisfying i(C,ℓ;Q′)≥3.Such Qi is an intersection Lj∩Lk, where {i,j,k}={1,2,3}, indeed Q1=(1:-7:1),Q2=(7:-1:1), and Q3=(1:-1:7). Therefore, if P∈ℙ2∖{C,Q1,Q2,Q3}, then there exists a line ℓ passing through P such that ℓ is a simple 2-tangent line to C.Proof.
Making use of the results in [10] and observing the self-duality of C, we can check the assertions by direct computations.Now let us begin the proof. IfP=P1, then n=3, hence GP is primitive. The lines ℓP1P2 and ℓP1P3 yield the ramification points of order three of πP, hence we infer from Hurwitz’s formula that there exists i such that ei=2. Thus, we get GP1≅S3. For P=P2 or P3, using the projective transformation σ above, we see GPi≅S3(i=2,3).Next consider the caseP∈C∖{P1,P2,P3}. Then we have n=5, hence GP is primitive. Using Hurwitz’s formula or the self-duality of C, we see that there exists a simple 2-tangent line to C, thus we have GP≅S5.Finally, we consider the remaining caseP∈ℙ2∖C.Claim 5.
Letni be the number of ramification points with index i. Then we have n2+2n3+3n4=12, where n4≤3. In particular, if n4=3(resp.,2), then P=(1:1:1)(resp.,Qi), furthermore; n3=0(resp.,3) and n2=3(resp.,0).Proof.
The former assertion is clear from Claim4 and Hurwitz’s formula. The proof of the latter assertion is as follows: observing Claim 4, we infer that, if n4=3, then P is unique (1:1:1), which is the intersections of the three lines L4,L5, and L6 (Figure 1). Similarly observing Claim 4, we infer that if n4=2, then P=Q1,Q2 or Q3. In this case, we have i(C,ℓPPi;Pi)=3, hence n3=3.Figure 1Claim 6.
IfπP is an s-covering, then πP is indecomposable.Proof.
By Claim5 the ramification index is 2,3, or 4. Suppose πP is decomposable. Then, degβ=2 or 3. By Claim 1, β is unramified at Ri′=α(Ri), where ei=2 or 3. By Claim 5, we have n4≤3. As we have seen in the proof of Proposition 10, β cannot be ramified at only one point. Thus, we have n4≠1. If n4=0, then the proof is clear by (3) in Proposition 11. If n4=2, then P=Qi(i=1,2,3). In case degβ=2, β is ramified at R1′ and R2′. Since degα=3, this cannot occur. In case degβ=3, β is ramified at R1′ and R2′ with e(R1′,β)=e(R2′,β)=2; however, these do not satisfy Hurwitz’s formula. If n4=3, then P=(1:1:1) and from Claim 4 and Hurwitz’s formula we infer that the ramification data is (43,23):=(4,4,4,2,2,2). Suppose πP is decomposable. Then, degβ=2 or 3. If degβ=2, then β is ramified at Ri′,(i=1,2,3). However, since degα=3, this case cannot occur. Then, we have degβ=3. We see easily that β is ramified at Ri′ with e(Ri′,β)=2(i=1,2,3). However, this does not satisfy Hurwitz’s formula. Therefore, πP is indecomposable.Now we resume the proof. We prove by examining the cases:(i)
P=(1:1:1);(ii)
P=Qi(i=1,2,3);(iii)
P∈ℓPiPj(1≤i,j≤3), P≠(1:1:1), and P≠Qi(i=1,2,3);(iv)
P is the point not appearing in the above case.By Claims5 and 6, the proof is complete for (i) and (iv). So let us treat the case (ii). By Claim 6, GP is primitive. However, there exists no simple 2-tangent line. Take Q1=(1:-7:1) and consider the affine part Z≠0. The defining equation is (xy-x+y)3+54x2y2=0. Putting u=x-1,v=y+7 and v=tu, we get h(t,u):=(tu2-8u+2tu-15)3+54(u+1)(tu-7)2=0. Here, we consider the Galois group obtained by the special value t=2. By the aid of a software, for example, PARI, we see that the polynomial h(2,u)=(2u2-4u-15)3+54(u+1)(2u-7)2 in ℚ[u] is irreducible and the Galois group of this polynomial is S6. Let u1(t),…,u6(t) be the roots of h(t,u)=0 with respect to u. Note that ui(t)   (1≤i≤6) is regular near t=2 and {u1(2),…,u6(2)} are the roots of h(2,u)=0. We can find ci∈ℚ(1≤i≤6) satisfying the conditions: u~(t)=c1u1(t)+⋯+c6u6(t) (resp., u~(2)=c1u1(2)+⋯+c6u6(2)) is a generator of the minimal splitting field of h(t,u) (resp., h(2,u)) over k(t) (resp., ℚ). Suppose the degree of u~(t) is less than 6!. Then, so is u~(2), which is a contradiction. Hence we have [k(t,u):k(t)]=6!, thus we conclude GP≅S6. The proofs of the other two cases Q2 and Q3 are almost the same.The proof of the case (iii) is as follows: here we notice that ifP∈ℓPiPj,i≠j,(i,j=1,2,3), then πP is not an s-covering. First we consider the special case where P is in some TFi, for example, ℓP1P2∩TF1={P}.Then the ramification data is {(F1,P1,P2,P3,R5,R6,R7),(4,33,23)} and πP(P1)=πP(P2). Suppose πP is decomposable. Then, by Claim 1, β:X3→ℙ1 is unramified at α(P3) and Ri′,(i≥5). Namely, β is ramified at just two points. Then, the ramification data of β is {(α(F1),α(P1)),(2,2)} or {(α(F1),α(P1)),(3,3)}, where degβ=2 or 3, respectively. However, it is easy to see that this is impossible considering α and πP, so πP is indecomposable. Since there exist ei=2(i=5,6,7), we conclude GP≅S6. On the other hand, if P is not in TFi for each i(i=1,2,3), then, by (3) in Proposition 10, f is indecomposable. Since there exists a simple 2-tangent, we have GP≅S6.Thus, we complete all the proofs.Remark 21.
In the list of Theorem17 only two kinds of group appear. Of course, other kinds will appear in other examples, for example, let us take the Fermat quartic X4+Y4+Z4=0. Then, there exist 12 points such that GP is the dihedral group of order 8 (cf. [13]).Problem. Concerning the Galois groups for C(e,d)(1<e<d-1), full symmetric group Sd degenerates into the cyclic group. How does the symmetric group degenerate for various curves?



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