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Singular plane curves with infinitely many Galois points
Journal of Algebra 323 (2010) 10–13
Contents lists available at ScienceDirect
Journal of Algebra www.elsevier.com/locate/jalgebra
Singular plane curves with infinitely many Galois points Satoru Fukasawa a,∗,1 , Takehiro Hasegawa b a b
Department of Mathematics, School of Science and Engineering, Waseda University, Ohkubo 3-4-1, Shinjuku, Tokyo 169-8555, Japan Department of Mathematics, School of Education, Waseda University, Nishi-Waseda 1-6-1, Shinjuku, Tokyo 169-8050, Japan
a r t i c l e
i n f o
Article history: Received 2 August 2007 Available online 21 October 2009 Communicated by Laurent Moret-Bailly Keywords: Galois point Singular plane curve Positive characteristic Strange curve
a b s t r a c t For a plane curve C , we call a point P ∈ P2 a Galois point with respect to C if the point projection from P induces a Galois extension of function fields. We give an example of a singular plane curve having infinitely many inner and outer Galois points. We also classify plane curves whose general points are inner Galois points. Before our results, known examples in the theory of Galois points have only finitely many Galois points, except trivial cases. © 2009 Elsevier Inc. All rights reserved.
1. Introduction Let the base field K be an algebraically closed field of characteristic p 0. A Galois point P ∈ P2 for a plane curve C ⊂ P2 of degree d 3 is a point from which the point projection π P : C P1 induces a Galois extension K (P1 ) ⊂ K (C ) of function fields. This notion was introduced by H. Yoshihara in 1996 (see [11,16] or survey paper [7]). We call the Galois point P ∈ P2 an inner (resp. outer) Galois point for C if P ∈ C (resp. P ∈ / C ). Galois points for a plane curve have been studied in characteristic zero [2,11–13,16,17], and recently also in positive characteristic [3–5,10]. However, it is difficult to investigate Galois points in the case that C is singular, even if the characteristic is zero [2,12,13,17]. In particular, there are very few results on the distribution of Galois points for a singular curve in positive characteristic. In characteristic zero, Miura [13] proved that the number of inner Galois points on a singular plane curve such that d − 1 is prime satisfies a certain inequality, especially which is finite (if d 4).
*
Corresponding author. Current address: Department of Mathematical Sciences, Faculty of Science, Yamagata University, Kojirakawa-Machi 1-4-12, Yamagata 990-8560, Japan. E-mail addresses: [email protected], [email protected] (S. Fukasawa), [email protected] (T. Hasegawa). 1 The first author was supported by Research Fellowships of the Japan Society for the Promotion of Science for Young Scientists. 0021-8693/$ – see front matter © 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2009.09.025
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Yoshihara [17] proved that rational curves have at most three outer Galois points if d = 12, 24, 60. The first purpose of this paper is to give an example of a singular curve having infinitely many inner and outer Galois points in positive characteristic p: The curve defined by X Z q−1 − Y q = 0, where q is a power of p, is such an example. Our example implies that Miura’s and Yoshihara’s assertions do not hold in positive characteristic. The second purpose of this paper is to classify plane curves whose almost all points are inner Galois points. We will prove the following: Theorem 1. Let C ⊂ P2 be a plane curve of degree d 4 and let be the set consisting of all inner Galois points on C . Then, the following conditions are equivalent: (i) p > 0, d is a power of p and C is projectively equivalent to a curve defined by X Z d−1 − Y d = 0, (ii) the set is a non-empty Zariski open set of C . 2. Example We denote by G R the Galois group induced from the projection π R at R ∈ P2 when R is a Galois point. A plane curve C ⊂ P2 is said to be strange if there exists a point Q ∈ P2 such that the tangent line at any smooth point passes through Q (see [1]). Then Q is called the strange center. Note that the projection π Q : C P1 from a point Q ∈ P2 is not separable for a plane curve C if and only if C is strange and Q is the strange center. Example 1. (See below for the proofs.) Let p > 0 be the characteristic, q = p e with q 3 and let C ⊂ P2 be a plane curve given by X Z q−1 − Y q = 0. Let L be the line defined by Z = 0. Then, we have the followings:
/ C is a strange center. The point projec(0) P = (1 : 0 : 0) ∈ C is a singular point and Q = (0 : 1 : 0) ∈ tion π P from P (resp. π Q from Q ) is birational (resp. purely inseparable of degree q) onto the projective line. (1) All point R ∈ C \ { P } is an inner Galois point. The Galois group G R is a cyclic group of order q − 1. (2) All point R ∈ L \ { P , Q } is an outer Galois point. The Galois group G R is isomorphic to (Z/ p Z)⊕e . (3) Any point R ∈ P2 \ (C ∪ L ) is not a Galois point. We prove (1). Note that C ∩ L = { P }. Let R ∈ C \ L. We can write R = (a : b : 1) with a − bq = 0. Then, π R = ( Xˆ : Yˆ ) where Xˆ = X − a Z and Yˆ = Y − b Z , and X Z q−1 − Y q = Xˆ Z q−1 − Yˆ q . We have a field extension K ( z, y )/ K ( y ) with zq−1 − y q = 0 by π R . This is obviously a Galois extension. More precisely, which is a cyclic extension. Therefore, R is a Galois point. We prove (2). Let R ∈ L. Since P = (1 : 0 : 0) is a singular point, we may assume that R = (a : 1 : 0). Then, π R = ( Xˆ : Z ) where Xˆ = X − aY and X Z q−1 − Y q = −Y q + aY Z q−1 + Xˆ Z q−1 . We have a field extension K ( y , x)/ K (x) with y q − ay − x = 0 by π R . If a = 0 then this is a Galois extension and the Galois group is isomorphic to (Z/ p Z)⊕e . If a = 0 then this is a purely inseparable extension of degree q. We prove (3). Let R = (a : b : 1) ∈ P2 \ (C ∪ L ). Now, the singular curve C is the image of the morphism π : P1 → P2 ; (s : t ) → (sq : st q−1 : t q ). Here, π : P1 → C is birational, hence a desingularization. We have π R ◦ π (s : t ) = (sq − at q : st q−1 − bt q ), π R ( P ) = (1 : 0) and (π R ◦ π )−1 (π R ( P )) = {(b : 1), (1 : 0)}. We find easily that the ramification index of π R ◦ π at (1 : 0) is q − 1. If we assume that R is a Galois point, then π R ◦ π is a Galois covering of degree q. It is known that any ramification index of a Galois covering between smooth curves divides the degree of the field extension (see, for example, [14, III.7.2]), hence this is a contradiction. Remark 1. Example 1 is a famous example as a strange curve. However, in the authors’ best knowledge, the distribution of Galois points for the curves had never been investigated. Strange curves appear only in positive characteristic. See [1] for the geometry of strange curves.
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3. Plane curves whose general points are Galois points In this section, we prove Theorem 1. We recall some notions. Let C ⊂ P2 be a plane curve of degree d 4. For a smooth point R ∈ C , we denote by T R C the tangent line at R. There exists an integer q such that the intersection multiplicity I R (C , T R C ) q for all smooth point R ∈ C and the equality holds for a general point [8,9,15]. We call the number q the generic order of contact for C . We say that C is non-reflexive if the dual map γ of C onto its dual has inseparable degree q > 1 [8]. If C is non-reflexive, then it is also known that q = q, hence q is a power of p. Let π : Cˆ → C be a desingularization. For a point R ∈ P2 , we write the projection from R by π R and the composite π R ◦ π by πˆ R . We denote the set consisting of all inner Galois points on C by and the line joining two points P and Q by P Q . For basic properties of Galois covering, see [14, III. 7.2]. We often use the fact that the ramification indices at any two points contained in a common fiber, are the same for a Galois covering. We first note the following. Proposition 1. Let C ⊂ P2 be a plane curve of degree d 4. We assume that the set is a non-empty Zariski open set of C . Then, the generic order q of contact for C is strictly greater than 2. Proof. First, we prove that the dual map γ is purely inseparable onto its dual. We assume that γ has separable degree s 2. Then, for a general point R ∈ C , there exists R ∈ C such that γ ( R ) = γ ( R ), hence T R C = T R C . We may assume that I R (C , T R C ) = I R (C , T R C ) = q and R ∈ . Then, the ramification indices of πˆ R are q − 1 at R and q at R , this is a contradiction to that R is a Galois point. We assume that the generic order q for C is equal to 2. Let T be the set consisting of lines satisfying one of the following: tangent lines have two or more contact points; tangent lines have a contact with multiplicity > 2; tangent lines passing through a singular point. If C is not strange (see Section 2 for definition), then T is a finite set. The set C \ S with S = L ∈ T L is a non-empty Zariski open set. For a general point R ∈ , there are points R 1 , R 2 such that T R 1 C passes through R and R 2 but is not tangent to R nor R 2 . Therefore, πˆ R at R is ramified at R 1 and unramified at R 2 . This is a contradiction to that R is a Galois point. We may assume that C is strange with a center Q , then the tangent line T R C = R Q at a general point R does not meet a smooth point different from R and Q , because the dual map is purely inseparable. From Bézout’s theorem, a general tangent line meets the singular locus of C . Therefore, we have a strange center Q ∈ Sing C . It follows from the uniqueness of the strange center that Q = Q . We note that all singular point of C has the multiplicity < d − 1. Then πˆ R at a general Galois point R is unramified at any point. In fact, πˆ R is unramified at any smooth point by the strangeness. Also at Q because the ramification index at R is q − 1 = 1, and at (any point of the fiber of π on) each singular point Q = Q because the line R Q intersects C at another smooth point. This is a contradiction to Riemann–Hurwitz formula. 2 Proof of Theorem 1. We prove (ii) ⇒ (i). From Proposition 1, the generic order q > 2 and hence the characteristic p is positive. As in the proof of Proposition 1, the dual map γ is purely inseparable onto its dual. For a general point R, the morphism πˆ R is ramified of index q − 1 at R. We assume R ∈ . Then, a smooth point in C ∩ T R C is a ramification point of πˆ R , because R is a Galois point. Since the dual map is purely inseparable, (C ∩ T R C ) \ { R } ⊂ Sing C (for a general point R ∈ C ). If (C ∩ T R C ) \ { R } = ∅ for a general point R, then there exists a strange center Q ∈ Sing C which has the multiplicity d − q. If (C ∩ T R C ) \ { R } = ∅, then we have d = q. We consider the case that there exists a strange center Q ∈ SingC which has the multiplicity d − q. Then, the projection π Q from Q is purely inseparable onto the projective line, since π Q can be identified with the dual map γ . This implies that C is rational. Let R be a general Galois point. The morphism πˆ R is unramified at any other smooth point (by the strangeness) and at (any point of the fiber of π on) each singular point Q not Q , because the line R Q intersects C at another smooth point. Therefore, for a general Galois point R, the ramification locus of πˆ R is equal to the fiber (πˆ R )−1 (π R ( Q )). This implies that πˆ R has only tame ramification. From Riemann–Hurwitz formula, we have an equality −2 = (d − 1)(−2) + (q − 2)(d − 1)/(q − 1). This gives a contradiction.
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We consider the case that d = q. Then, for all smooth point R, we have I R (C , T R C ) = q = d. This implies that the morphism πˆ R , for any smooth point R ∈ C , is unramified at any smooth point R = R on C . We prove that there exists a singular point P such that R P ∩ C = { R , P }. If this is proven, then we find that the projection π P is birational (since C does not have a strange center on C ). We assume that the projection for any singular point P has separable degree s > 1. Then, for a general point R ∈ C and each singular point P , R P ∩ C contains another smooth point R ∈ C . This implies that πˆ R is unramified at each point in the fiber π −1 ( P ), because πˆ R is a Galois covering and is unramified at R . Therefore, πˆ R is ramified only at R. We note that πˆ R is ramified of index q − 1 at R, which is tamely. From Riemann–Hurwitz formula, this is a contradiction. Finally, we prove that C is projectively equivalent to a curve defined by X Z q−1 − Y q = 0 in the case that d = q. By virtue of a result of Homma [1, (3.4)], [9, Theorem 3.4], C is projectively equivalent to a curve defined by G ( X , Z ) − Y q = 0 where G is a homogeneous polynomial of degree q if d = q. Now C has a singular point with multiplicity q − 1. Hence we have the form X Z q−1 − Y q = 0. 2 Remark 2. (1) Let be the set of all smooth points R of C such that the ramification indices of πˆ R at any two points contained in a common fiber, are the same. We consider the relation between (i), (ii) in Theorem 1 and the following condition: (iii) The set is a non-empty Zariski open set of C . The statement (ii) ⇒ (iii) is true from a property of Galois covering. Our proof of (ii) ⇒ (i) in Theorem 1 coincides with the proof of (iii) ⇒ (i). (2) Theorem 1 also holds for a weaker condition (ii)’: is an infinite set, by easy modification of the proof. (3) After this paper was submitted, the first author classified plane curves having infinitely many “outer” Galois points [6]. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17]
V. Bayer, A. Hefez, Strange curves, Comm. Algebra 19 (1991) 3041–3059. C. Duyaguit, K. Miura, On the number of Galois points for plane curves of prime degree, Nihonkai Math. J. 14 (2003) 55–59. S. Fukasawa, Galois points on quartic curves in characteristic 3, Nihonkai Math. J. 17 (2006) 103–110. S. Fukasawa, On the number of Galois points for a plane curve in positive characteristic, Comm. Algebra 36 (2008) 29–36. S. Fukasawa, On the number of Galois points for a plane curve in positive characteristic, II, Geom. Dedicata 127 (2007) 131–137. S. Fukasawa, Classification of plane curves with infinitely many Galois points, J. Math. Soc. Japan, submitted for publication. S. Fukasawa, Galois points for a plane curve in arbitrary characteristic, Geom. Dedicata 139 (2009) 211–218. A. Hefez, S. Kleiman, Notes on the duality of projective varieties, in: Geometry Today, in: Prog. Math., vol. 60, Birkhäuser, Boston, 1985, pp. 143–183. M. Homma, Funny plane curves in characteristic p > 0, Comm. Algebra 15 (1987) 1469–1501. M. Homma, Galois points for a Hermitian curve, Comm. Algebra 34 (2006) 4503–4511. K. Miura, H. Yoshihara, Field theory for function fields of plane quartic curves, J. Algebra 226 (2000) 283–294. K. Miura, Field theory for function fields of singular plane quartic curves, Bull. Austral. Math. Soc. 62 (2000) 193–204. K. Miura, Galois points on singular plane quartic curves, J. Algebra 287 (2005) 283–293. H. Stichtenoth, Algebraic Function Fields and Codes, Universitext, Springer-Verlag, Berlin, 1993. K.O. Stöhr, J.F. Voloch, Weierstrass points and curves over finite fields, Proc. London Math. Soc. (3) 52 (1986) 1–19. H. Yoshihara, Function field theory of plane curves by dual curves, J. Algebra 239 (2001) 340–355. H. Yoshihara, Galois points for plane rational curves, Far East J. Math. Sci. 25 (2007) 273–284.
Contents lists available at ScienceDirect
Journal of Algebra www.elsevier.com/locate/jalgebra
Singular plane curves with infinitely many Galois points Satoru Fukasawa a,∗,1 , Takehiro Hasegawa b a b
Department of Mathematics, School of Science and Engineering, Waseda University, Ohkubo 3-4-1, Shinjuku, Tokyo 169-8555, Japan Department of Mathematics, School of Education, Waseda University, Nishi-Waseda 1-6-1, Shinjuku, Tokyo 169-8050, Japan
a r t i c l e
i n f o
Article history: Received 2 August 2007 Available online 21 October 2009 Communicated by Laurent Moret-Bailly Keywords: Galois point Singular plane curve Positive characteristic Strange curve
a b s t r a c t For a plane curve C , we call a point P ∈ P2 a Galois point with respect to C if the point projection from P induces a Galois extension of function fields. We give an example of a singular plane curve having infinitely many inner and outer Galois points. We also classify plane curves whose general points are inner Galois points. Before our results, known examples in the theory of Galois points have only finitely many Galois points, except trivial cases. © 2009 Elsevier Inc. All rights reserved.
1. Introduction Let the base field K be an algebraically closed field of characteristic p 0. A Galois point P ∈ P2 for a plane curve C ⊂ P2 of degree d 3 is a point from which the point projection π P : C P1 induces a Galois extension K (P1 ) ⊂ K (C ) of function fields. This notion was introduced by H. Yoshihara in 1996 (see [11,16] or survey paper [7]). We call the Galois point P ∈ P2 an inner (resp. outer) Galois point for C if P ∈ C (resp. P ∈ / C ). Galois points for a plane curve have been studied in characteristic zero [2,11–13,16,17], and recently also in positive characteristic [3–5,10]. However, it is difficult to investigate Galois points in the case that C is singular, even if the characteristic is zero [2,12,13,17]. In particular, there are very few results on the distribution of Galois points for a singular curve in positive characteristic. In characteristic zero, Miura [13] proved that the number of inner Galois points on a singular plane curve such that d − 1 is prime satisfies a certain inequality, especially which is finite (if d 4).
*
Corresponding author. Current address: Department of Mathematical Sciences, Faculty of Science, Yamagata University, Kojirakawa-Machi 1-4-12, Yamagata 990-8560, Japan. E-mail addresses: [email protected], [email protected] (S. Fukasawa), [email protected] (T. Hasegawa). 1 The first author was supported by Research Fellowships of the Japan Society for the Promotion of Science for Young Scientists. 0021-8693/$ – see front matter © 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2009.09.025
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Yoshihara [17] proved that rational curves have at most three outer Galois points if d = 12, 24, 60. The first purpose of this paper is to give an example of a singular curve having infinitely many inner and outer Galois points in positive characteristic p: The curve defined by X Z q−1 − Y q = 0, where q is a power of p, is such an example. Our example implies that Miura’s and Yoshihara’s assertions do not hold in positive characteristic. The second purpose of this paper is to classify plane curves whose almost all points are inner Galois points. We will prove the following: Theorem 1. Let C ⊂ P2 be a plane curve of degree d 4 and let be the set consisting of all inner Galois points on C . Then, the following conditions are equivalent: (i) p > 0, d is a power of p and C is projectively equivalent to a curve defined by X Z d−1 − Y d = 0, (ii) the set is a non-empty Zariski open set of C . 2. Example We denote by G R the Galois group induced from the projection π R at R ∈ P2 when R is a Galois point. A plane curve C ⊂ P2 is said to be strange if there exists a point Q ∈ P2 such that the tangent line at any smooth point passes through Q (see [1]). Then Q is called the strange center. Note that the projection π Q : C P1 from a point Q ∈ P2 is not separable for a plane curve C if and only if C is strange and Q is the strange center. Example 1. (See below for the proofs.) Let p > 0 be the characteristic, q = p e with q 3 and let C ⊂ P2 be a plane curve given by X Z q−1 − Y q = 0. Let L be the line defined by Z = 0. Then, we have the followings:
/ C is a strange center. The point projec(0) P = (1 : 0 : 0) ∈ C is a singular point and Q = (0 : 1 : 0) ∈ tion π P from P (resp. π Q from Q ) is birational (resp. purely inseparable of degree q) onto the projective line. (1) All point R ∈ C \ { P } is an inner Galois point. The Galois group G R is a cyclic group of order q − 1. (2) All point R ∈ L \ { P , Q } is an outer Galois point. The Galois group G R is isomorphic to (Z/ p Z)⊕e . (3) Any point R ∈ P2 \ (C ∪ L ) is not a Galois point. We prove (1). Note that C ∩ L = { P }. Let R ∈ C \ L. We can write R = (a : b : 1) with a − bq = 0. Then, π R = ( Xˆ : Yˆ ) where Xˆ = X − a Z and Yˆ = Y − b Z , and X Z q−1 − Y q = Xˆ Z q−1 − Yˆ q . We have a field extension K ( z, y )/ K ( y ) with zq−1 − y q = 0 by π R . This is obviously a Galois extension. More precisely, which is a cyclic extension. Therefore, R is a Galois point. We prove (2). Let R ∈ L. Since P = (1 : 0 : 0) is a singular point, we may assume that R = (a : 1 : 0). Then, π R = ( Xˆ : Z ) where Xˆ = X − aY and X Z q−1 − Y q = −Y q + aY Z q−1 + Xˆ Z q−1 . We have a field extension K ( y , x)/ K (x) with y q − ay − x = 0 by π R . If a = 0 then this is a Galois extension and the Galois group is isomorphic to (Z/ p Z)⊕e . If a = 0 then this is a purely inseparable extension of degree q. We prove (3). Let R = (a : b : 1) ∈ P2 \ (C ∪ L ). Now, the singular curve C is the image of the morphism π : P1 → P2 ; (s : t ) → (sq : st q−1 : t q ). Here, π : P1 → C is birational, hence a desingularization. We have π R ◦ π (s : t ) = (sq − at q : st q−1 − bt q ), π R ( P ) = (1 : 0) and (π R ◦ π )−1 (π R ( P )) = {(b : 1), (1 : 0)}. We find easily that the ramification index of π R ◦ π at (1 : 0) is q − 1. If we assume that R is a Galois point, then π R ◦ π is a Galois covering of degree q. It is known that any ramification index of a Galois covering between smooth curves divides the degree of the field extension (see, for example, [14, III.7.2]), hence this is a contradiction. Remark 1. Example 1 is a famous example as a strange curve. However, in the authors’ best knowledge, the distribution of Galois points for the curves had never been investigated. Strange curves appear only in positive characteristic. See [1] for the geometry of strange curves.
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3. Plane curves whose general points are Galois points In this section, we prove Theorem 1. We recall some notions. Let C ⊂ P2 be a plane curve of degree d 4. For a smooth point R ∈ C , we denote by T R C the tangent line at R. There exists an integer q such that the intersection multiplicity I R (C , T R C ) q for all smooth point R ∈ C and the equality holds for a general point [8,9,15]. We call the number q the generic order of contact for C . We say that C is non-reflexive if the dual map γ of C onto its dual has inseparable degree q > 1 [8]. If C is non-reflexive, then it is also known that q = q, hence q is a power of p. Let π : Cˆ → C be a desingularization. For a point R ∈ P2 , we write the projection from R by π R and the composite π R ◦ π by πˆ R . We denote the set consisting of all inner Galois points on C by and the line joining two points P and Q by P Q . For basic properties of Galois covering, see [14, III. 7.2]. We often use the fact that the ramification indices at any two points contained in a common fiber, are the same for a Galois covering. We first note the following. Proposition 1. Let C ⊂ P2 be a plane curve of degree d 4. We assume that the set is a non-empty Zariski open set of C . Then, the generic order q of contact for C is strictly greater than 2. Proof. First, we prove that the dual map γ is purely inseparable onto its dual. We assume that γ has separable degree s 2. Then, for a general point R ∈ C , there exists R ∈ C such that γ ( R ) = γ ( R ), hence T R C = T R C . We may assume that I R (C , T R C ) = I R (C , T R C ) = q and R ∈ . Then, the ramification indices of πˆ R are q − 1 at R and q at R , this is a contradiction to that R is a Galois point. We assume that the generic order q for C is equal to 2. Let T be the set consisting of lines satisfying one of the following: tangent lines have two or more contact points; tangent lines have a contact with multiplicity > 2; tangent lines passing through a singular point. If C is not strange (see Section 2 for definition), then T is a finite set. The set C \ S with S = L ∈ T L is a non-empty Zariski open set. For a general point R ∈ , there are points R 1 , R 2 such that T R 1 C passes through R and R 2 but is not tangent to R nor R 2 . Therefore, πˆ R at R is ramified at R 1 and unramified at R 2 . This is a contradiction to that R is a Galois point. We may assume that C is strange with a center Q , then the tangent line T R C = R Q at a general point R does not meet a smooth point different from R and Q , because the dual map is purely inseparable. From Bézout’s theorem, a general tangent line meets the singular locus of C . Therefore, we have a strange center Q ∈ Sing C . It follows from the uniqueness of the strange center that Q = Q . We note that all singular point of C has the multiplicity < d − 1. Then πˆ R at a general Galois point R is unramified at any point. In fact, πˆ R is unramified at any smooth point by the strangeness. Also at Q because the ramification index at R is q − 1 = 1, and at (any point of the fiber of π on) each singular point Q = Q because the line R Q intersects C at another smooth point. This is a contradiction to Riemann–Hurwitz formula. 2 Proof of Theorem 1. We prove (ii) ⇒ (i). From Proposition 1, the generic order q > 2 and hence the characteristic p is positive. As in the proof of Proposition 1, the dual map γ is purely inseparable onto its dual. For a general point R, the morphism πˆ R is ramified of index q − 1 at R. We assume R ∈ . Then, a smooth point in C ∩ T R C is a ramification point of πˆ R , because R is a Galois point. Since the dual map is purely inseparable, (C ∩ T R C ) \ { R } ⊂ Sing C (for a general point R ∈ C ). If (C ∩ T R C ) \ { R } = ∅ for a general point R, then there exists a strange center Q ∈ Sing C which has the multiplicity d − q. If (C ∩ T R C ) \ { R } = ∅, then we have d = q. We consider the case that there exists a strange center Q ∈ SingC which has the multiplicity d − q. Then, the projection π Q from Q is purely inseparable onto the projective line, since π Q can be identified with the dual map γ . This implies that C is rational. Let R be a general Galois point. The morphism πˆ R is unramified at any other smooth point (by the strangeness) and at (any point of the fiber of π on) each singular point Q not Q , because the line R Q intersects C at another smooth point. Therefore, for a general Galois point R, the ramification locus of πˆ R is equal to the fiber (πˆ R )−1 (π R ( Q )). This implies that πˆ R has only tame ramification. From Riemann–Hurwitz formula, we have an equality −2 = (d − 1)(−2) + (q − 2)(d − 1)/(q − 1). This gives a contradiction.
S. Fukasawa, T. Hasegawa / Journal of Algebra 323 (2010) 10–13
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We consider the case that d = q. Then, for all smooth point R, we have I R (C , T R C ) = q = d. This implies that the morphism πˆ R , for any smooth point R ∈ C , is unramified at any smooth point R = R on C . We prove that there exists a singular point P such that R P ∩ C = { R , P }. If this is proven, then we find that the projection π P is birational (since C does not have a strange center on C ). We assume that the projection for any singular point P has separable degree s > 1. Then, for a general point R ∈ C and each singular point P , R P ∩ C contains another smooth point R ∈ C . This implies that πˆ R is unramified at each point in the fiber π −1 ( P ), because πˆ R is a Galois covering and is unramified at R . Therefore, πˆ R is ramified only at R. We note that πˆ R is ramified of index q − 1 at R, which is tamely. From Riemann–Hurwitz formula, this is a contradiction. Finally, we prove that C is projectively equivalent to a curve defined by X Z q−1 − Y q = 0 in the case that d = q. By virtue of a result of Homma [1, (3.4)], [9, Theorem 3.4], C is projectively equivalent to a curve defined by G ( X , Z ) − Y q = 0 where G is a homogeneous polynomial of degree q if d = q. Now C has a singular point with multiplicity q − 1. Hence we have the form X Z q−1 − Y q = 0. 2 Remark 2. (1) Let be the set of all smooth points R of C such that the ramification indices of πˆ R at any two points contained in a common fiber, are the same. We consider the relation between (i), (ii) in Theorem 1 and the following condition: (iii) The set is a non-empty Zariski open set of C . The statement (ii) ⇒ (iii) is true from a property of Galois covering. Our proof of (ii) ⇒ (i) in Theorem 1 coincides with the proof of (iii) ⇒ (i). (2) Theorem 1 also holds for a weaker condition (ii)’: is an infinite set, by easy modification of the proof. (3) After this paper was submitted, the first author classified plane curves having infinitely many “outer” Galois points [6]. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17]
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