Galois Group at Each Point for Some Self-Dual Curves

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


Introduction
is study is a continuation of [1][2][3][4], and so forth. In general, it is not easy to determine the Galois group at every point for plane curve, in particular for curve with singular point. When we determine the structure of , 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.
Let be an algebraically closed �eld of characteristic �ero. We �x it as the ground �eld of our discussions. Let be an irreducible plane curve of degree (≥ 2) and the rational function �eld of . Let be a set of homogeneous coordinates on ℙ 2 and put 1 0 0 1 , 2 0 1 0 , and 3 1 0 0 . Let , , be the de�ning equation of and put , , , , where , .

Galois Group.
Let be the resolution of singularities of . For a point ℙ 2 , let be the dual line in the dual space ℙ 2 of ℙ 2 corresponding to . We de�ne the morphism by where ℓ is the point in ℙ 2 corresponding to the line ℓ , which passes through and if . In case , the line ℓ is the tangent line to the branch of at .
Clearly, we have deg and a �eld extension * ℙ 1 ↪ , where denotes the multiplicity of at . In case , we understand 0. We put ; if there is no fear of confusion, we simply denote it by . Since the extension depends only on , we denote ℙ 1 by , that is, we have * ↪ . Let be the Galois closure of and the Galois group Gal .
De�nition 1. We call the Galois group at for . In case is a Galois extension, the point is said to be a Galois point.
In case is the �eld of complex numbers, is isomorphic to the monodromy group of the covering ℙ 1 [6,7].

Self-Dual Curve
De�nition 2. A point is said to be a cusp of if it is a singular point and 1 consists of a single point. Furthermore, if ℙ 2 ℙ 2 is a blow-up and 1 is a nonsingular point of the proper transform of 1 , the point is said to be a simple cusp.
Denote by the dual curve of .
De�nition �. If is projectively equivalent to , then is said to be a self-dual curve.
Suppose is smooth. en, is self-dual if and only if . However, if has a singular point, the condition that is self-dual becomes complicated. e following proposition has been known (cf. [8]).

Proposition 4. Suppose is one of the following curves:
(1) has just one singular point; (2) is rational and has only simple cusps as singular points.  54 0 (cf. [10]).
For the curve ( , if 1 < < 1, then 1 (0 ∶ 0 ∶ 1 and (0 ∶ 1 ∶ 0 are not simple cusps and ( has no �ex. e curve (4 has two cusps 1 and , where 1 is not a simple cusp. e curve 54 has three cusps: 1 , and Remark 6. Let Φ be the rational map ℙ ⇢ℙ giving the dual of , that is, where is the de�ning equation of . In the case where ( , the map Φ turns out to be a quadratic transformation of ℙ : We use the following notation:

Statement of Results
We need some preparations before stating the results. A curve means a nonsingular projective algebraic curve. Let 1 and be curves and ∶ 1 → a surjective morphism, which we call a covering for short. We denote by ( the rami�cation index of at 1  We give several sufficient conditions that is indecomposable. Some of them will not be used later in this paper.

Proposition 10.
Let ∶ 1 → be the covering above and deg . If one of the following conditions is satis�ed, then is indecomposable.

(3) is a rational curve, is an s-covering except over
Proposition 11. With the same notation as in Proposition 10, if is an s-covering and satis�es one of the following conditions, then is indecomposable.
(2) 1 is a rational curve and is prime for each .
(3) is a rational curve and is prime for each .
Hereaer, we follow the notation in Section 1. By taking a suitable projective change of coordinates, we can assume the projection center is 1  (1) if (resp., ), then , (resp., ), where 2 and ; (2) the curves and have normal crossings except at .
Sometimes we call a simple -tangent for short.
Note that a simple -tangent yields an s-covering over .
Lemma 14. We have the following assertions for .
(1) If each line has normal crossings with or is a simple -tangent line to such that is a prime number, then is primitive (cf. [5,Lemma 4.4.4]). e following lemma is well known.

Lemma 15. If a permutation group is primitive and contains a transposition, then it is a full symmetric group.
Combining the results above, we get the following corollary.  Table 1); (II) the case 4 (see Table 2); (III) the case 54 (see Table 3).
Remark 18. For the curves in eorem 17, is a Galois point if and only if is a cyclic group. However, the same assertion does not hold true in general, see, for example, [3].

Proofs
First, we prove Propositions 10 and 11. (1) is an s-covering over .
If there exists an intermediate covering ∶ 3 2 , then is unrami�ed at ′ .
Proof. Suppose is rami�ed at ′. en, since is prime, we have ′, , , hence ′ is not a branch point for . en, there will appear another rami�cation point for in 1 . is is a contradiction.
e proof of Proposition 10 is as follows. Suppose is decomposable and there exists a covering ∶ 3 2 as in �e�nition 7. First, we prove the assertion (1). Since is prime, is unrami�ed at ′ by Claim 1. Hence, we have , , . Since there exists at least two points in 1 , we have 2 , , which contradicts the assumption. Next we prove (2). Clearly and are rami�ed at 1 and ′ 1 , respectively. Put 1 1 . en, since 1 1, 1 1 consists of one or two points. In the former case, 1 1 1 consists of two points, on the other hand in the latter case 1 1 1, 2 consists of one point, where 1 1 11 , 12 }. In each case we infer the inequality 1 2, which is a contradiction. We go to the proof of (3). en, by Claim 1, 2 is not a branch point for . us, 1 is the only branch point for . en, by Hurwitz's formula, we have 2 3 is the genus of 3 , is the degree of , and 1. Since 3 0, this inequality implies 1, which is a contradiction.
Next we prove Proposition 11. In each case we use the reduction to absurdity, that is, suppose is decomposable. So we use the notation ′ 1 . In the case (I), by Claim 3, is unrami�ed at ′ 3 . Since 2 and 3 are rational, from Hurwitz's formula, we infer that is rami�ed with the index . en, since there exists no rami�cation points in 1 except 1, 2 , must branch at ′ 1 and ′ 2 . However, there exists an unrami�ed point in 1 2 , this is a contradiction. erefore, is indecomposable. In the case (II), by Claim 1, is unrami�ed at ′ for 2. Since 3 is rational, by Hurwitz's formula, we have a contradiction. In the case (III) similarly, by Claim 1, is unrami�ed at every point; however, since 2 is rational, must be an identity, which is a contradiction. is completes the proof of Proposition 11. e proof of eorem 12 is as follows: suppose is not primitive and let be the isotropy group of 1 in . en, there exists a subgroup of such that ⊊ ⊊ e proof of Lemma 14 is simple. In view of �e�nition 13, we see that the assertion (1) is another expression of (3) in Proposition 11. e assertion (2) may be well known (cf. [7]). Now we proceed to the proof of eorem 17. e structure of depends on the covering and depends on the position of . We prove by examining the cases where lies on the tangent line to at the cusp or at the �ex. Hereaer, we assume is the curve in eorem 17. Since is a self-dual curve and has only cusps as the singularity, the following remark is clear.
Remark 19. Suppose a line ℓ satis�es the following conditions: (1) ℓ does not pass through any cusp; (2) ℓ is not the tangent line to at the �ex.
en, ℓ is a simple 2-tangent line to or ℓ and have normal crossings.
Proof of the Case (I). Assume = ( . It has the following property. 1 , has one �ex at 1 (resp. 2 . On the other hand, in case 1 < < 1, has no �ex. Proof. Calculating the Hessian of (cf. [12]), we infer readily the assertions. If = 1 2 , or 3 , then can be determined directly. In fact, if = 1 , then consider the affine part 0 of , that is, the affine de�ning e�uation is = 0. en, putting = , we get = 0, hence ≅ . e other case = 2 is similarly determined. If = 3 , then consider the affine part 0, we get = 1. Putting = , we get = 1, hence ≅ . As we have seen above, these points are Galois ones.
Next we treat the case { 1 2 }. First, we prove the subcase 1 < < 1. Since is a self-dual curve and has no �ex, we see that, if a line ℓ passes through neither 1 nor 2 , then it has normal crossings with or it is a simple 2tangent line to . Furthermore, by Hurwitz's formula, we see there exists a simple 2-tangent. en, by (1) in Proposition 11 and Lemma 15, we have ≅ 1 . Next we prove the subcase = 1. en, 1 (resp. 2 is a �ex (resp., cusp) and the tangent line at 1 (resp. 2 does not meet except at 1 (resp. 2 . If a line ℓ does not pass through 2 , then it has normal crossings with or it is a simple 2-tangent line to . By (2) in Proposition 11 and Lemma 15, we have ≅ 1 . e proof of the case = 1 is the same.