Galois points for a plane curve and its dual curve

(1)

Galois points for a plane curve and its dual curve

SATORUFUKASAWA(*) - KEIMIURA(**)

ABSTRACT- A pointPin projective plane is said to be Galois for a plane curve of degree at least three if the function field extension induced by the projection fromPis Galois. Further we say that a Galois point is extendable if any birational transformation by the Galois group can be extended to a linear transformation of the projective plane. In this article, we propose the following problem:If a plane curve has a Galois point and its dual curve has one, what is the curve?We give an answer. We show that the dual curve of a smooth plane curve does not have a Galois point. On the other hand, we settle the case where both a plane curve and its dual curve have extendable Galois points. Such a curve must be defined by X^d Y^eZ^{d e}0, which is a famous self-dual curve.

MATHEMATICSSUBJECTCLASSIFICATION(2010). 14H50, 14H05, 12F10.

KEYWORDS. Galois point, plane curve, dual curve, self-dual curve.

1. Introduction

Let the base field K be an algebraically closed field of characteristic p0 and let CP² be an irreducible plane curve of degree d3. We recall the notions ofdual curveandGalois point.

Let P² be the dual projective plane which parameterizes projective lines of P² and let (X:Y:Z), (U:V:W) be systems of homogeneous coordinates of P² and ofP² respectively. We denote by Sing(C) the singular locus ofC. IfCis defined by a homogeneous polynomialF(X;Y;Z),

(*) Indirizzo dell'A.: Department of Mathematical Sciences, Yamagata Uni- versity, Kojirakawa-machi 1-4-12, Yamagata 990-8560, Japan.

E-mail: s.fukasawa@sci.kj.yamagata-u.ac.jp

The first author was partially supported by JSPS KAKENHI Grant Numbers 22740001, 25800002.

(**) Indirizzo dell'A.: Department of Mathematics, Ube National College of Technology, Ube, Yamaguchi 755-8555, Japan.

E-mail: kmiura@ube-k.ac.jp

(2)

we have a rational map gg_C:C ^>P², which sends a smooth point Q2CnSing(C) to the point

@F

@X(Q):@F

@Y(Q):@F

@Z(Q)

2P² para- meterizing the projective tangent lineT_QCtoCatQ. This rational map is called thedual mapofCand (the closure of) the image ofCis called thedual curve, which is denoted by C. It is well-known that projective duality C Cholds (see, for example, [7, 11]).

If the function field extension K(C)=K(P¹) induced by the projection p_P:C ^>P¹ from a pointP2P² is Galois, then the pointPis said to be Galois. Moreover, denoting by GP the Galois group associated to the projectionpP, we say that a Galois pointPisextendableif any birational transformation ofCinduced by the Galois groupGPcan be extended to a linear transformation ofP².

The notion of Galois point was introduced by H. Yoshihara (see e.g.

[4, 9, 13]) and it is an interesting topic on plane curves. For instance, a well-known theorem of Noether and later results assure that ifCP²is a smooth curve of degree d, then the minimum degree of a morphism C!P¹ is d 1, and all the maps of degree d 12 and d5 are projectionsp_P:C ^>P¹ from some pointP2CandP2P²nC respectively (cf. [3, 5, 10]). Thus describing Galois points on a smooth plane curve is equivalent to detect all the Galois coverings C!P¹ having minimal degrees.

The singular curve defined by X^d Y^eZ^{d e} 0, where e1 andd;e are coprime, has lovely properties. Its dual curve is defined by the same equation (up to a projective equivalence, see Lemma 2.5). Therefore, this curve is a ``self-dual'' curve. It has an (extendable) Galois point (1:0:0) (Proposition 2.3) and its dual curve has one, by the self-duality. In the light of this fact, we propose the following problem.

PROBLEM1.1. If a plane curve has a Galois point and its dual curve has one, what is the curve? Is it a self-dual curve?

In this article, we give an answer. We show the following for smooth curves.

THEOREM1.2. Let C be a smooth plane curve of degree d3. Then, there exist no Galois points for the dual curve of C.

By projective duality, we have the following.

(3)

COROLLARY1.3. Let C be a plane curve of degree d3and let Cbe the dual curve. If both C and Chave Galois points, then they are singular.

We say that a Galois pointP isinner(resp. outer) if P2CnSing(C) (resp.P2P²nC). In the case where Galois points are extendable, we show the following characterization theorems.

THEOREM1.4. Let C be a plane curve of degree d3and let Cbe the dual curve of C.

(I) The following conditions are equivalent.

(1) C and C have extendable outer Galois points.

(2) C is projectively equivalent to the plane curve defined by X^d Y^eZ^{d e}0for some e1.

(II) The following conditions are equivalent.

(1) C has an extendable outer Galois point and Chas an extendable inner Galois point.

(2) C and C have extendable inner Galois points.

(3) C is projectively equivalent to the plane curve defined by X^d Y^d ¹Z0.

To prove our result we will connect Galois points for a plane curveCand its dual curve. In particular, we will show that - given an extendable Galois pointPforC- the Galois groupG_Phas a natural action on the dual curve C, and such an action preserves the fibers of a certain projection p_P:C!P¹. Namely,

PROPOSITION1.5. Let C be a plane curve with an extendable Galois point P2P² and let G_P be the Galois group. Any s2G_P induces a natural linear transformations:P² !P² (see Lemma2.2for details).

Then, there exists a unique point P2P²such that the maps7!sinduces an injective homomorphism

GP,!G[P]: ft2Bir(C)jt(C\`n fPg)` for a general line`3Pg;

where Bir(C) is the group of all birational transformations of C. In particular, the degree of the projectionp_P:C ^>P¹from P is at least the order of G_P.

In the next Section we will recall some preliminary facts about projective duality and Galois group, and we will achieve Proposition 1.5. Section 3 and Section 4 will be devoted to prove Theorems 1.2 and 1.4 respectively.

(4)

2. Preliminaries

LetCbe an irreducible plane curve of degreed3. If a pointP2P²is not inC, we define the multiplicity ofCatPas zero. We denote bydthe degree of the dual curveC.

We recall duality principle betweenP² andP². For a projective line

`P², we denote by [`] the point in P² corresponding to`. If` is defined byuXvYwZ0 for someu;v;w2K, then the point [`]2P² is given by (u:v:w). For a pointP2P², we denote by [P]P²the line corresponding to P, which parameterizes the star of lines through P.

IfP is defined by u₁Xv₁Yw₁Zu₂Xv₂Yw₂Z0 for some u1;u2;v1;v2;w1;w22K, then [P] is the line passing through the two points (u1:v1:w1);(u2:v2:w2)2P². Then, we have the following ele- mentary facts.

LEMMA 2.1. Let P2P² be a point and let `P² be a line. Then, [[P]]P and[[`]]`hold. Furthermore,

P2`,[`]2[P]

holds.

LEMMA 2.2. Let f be a linear transformation of P² and let A_f be a matrix representing f, i.e.f(X:Y:Z)(X;Y;Z)A_f. Then the induced map f:P² !P²;[`]7![f(`)] is represented by the transpose of the matrix A_f¹, that is

f(U:V :W)(U;V;W)^tA_f¹: Moreover, the diagram

commutes, where the horizontal arrows are the dual maps of C andf(C).

We often use the standard formof the defining equation for a plane curve with an extendable Galois point, which is given by the following.

PROPOSITION2.3 (see [8, 13, 15]). Let P2P² be a point with multiplicity m0. The point P is extendable Galois for C if and only if there

(5)

exists a linear transformationfonP²such thatf(P)(1:0:0)andf(C) is given by

X^{d m}Gm(Y;Z)Gd(Y;Z)0;

where G_i(Y;Z)is a homogeneous polynomial of degree i in variables Y;Z.

In this case, the Galois group G_f(P) is cyclic and there exists a primitive (d m)-th rootzof unity such that a generators2G_f(P)is represented by s(X:Y :Z)(zX:Y:Z).

Firstly, we prove the first assertion of Proposition 1.5 for the curves with the standard form as in Proposition 2.3.

LEMMA2.4. Let P(1:0:0)2P², let C be defined by X^{d m}Gm(Y;Z)Gd(Y;Z)0;

and let a generator s2G_P be represented bys(X:Y :Z)(zX:Y:Z).

Then, there exists a unique point P2P² such that the map G_P! G[P];sⁱ7!sⁱis a well-defined and injective homomorphism.

PROOF. By Lemma 2.2, s is represented by s(U:V:W) (z ¹U:V:W) and s(C)C. Let P(1:0:0)2P². By the re- presentation ofs, we have

s2G[P] ft2Bir(C)jt(C\`n fPg)` for a general line`3Pg:

We prove the uniqueness of P. By contradiction we assume that s2G[P]\G[Q], whereQ6P. LetR2Cbe a point not contained in the line passing throughPandQsuch thats(R)6R. Bys2G[P]\G[Q],s(R) is contained in the lines passing throughRandP,RandQrespectively.

This implies thats(R)R. This is a contradiction. p PROOF OF PROPOSITION 1.5. By Proposition 2.3, there exists a linear transformationfsuch thatf(P)(1:0:0),f(C) is given byX^{d m}Gm(Y;Z) Gd(Y;Z)0 and a generator s2G_f(P) is represented by s(X:Y:Z) (zX:Y :Z). By Lemma 2.4, we have an injectionf :G_f(P),!G[f(P)];sⁱ7!sⁱ. Let Pf ¹(f(P)). We also have isomorphisms g:GP!G_f(P);t7!ftf ¹ and h:G[f(P)]!G[P];h7!f ¹hf. Since (^tA_f¹)^t(A_f¹AtA_f) ¹(^tA_f¹) ¹ (^tA_f¹)^tA_f^tA_t¹ ^tA_f¹(^tA_f)^t A_t¹, we have (hfg)(t)t. Therefore, GP!G[P];t7!t is a well-defined and injective homomorphism. The uniqueness of the pointPfollows, similarly to the proof of Lemma 2.4.

(6)

We prove the second assertion. Letmbe the multiplicity ofC atP.

Then, the order ofG[P] is at mostd m, which is equal to the degree of the projection p_P:C ^>P¹. By the first assertion, we have

d md m. p

Here, we summarize the properties of the curve defined by X^d Y^eZ^{d e}0.

LEMMA 2.5. Let C be defined by X^d Y^eZ^{d e}0. Then, we have the following.

(1) The dual curve Cis given by( 1)^de^e(d e)^{d e}U^d d^dV^eW^{d e}0.

(2) Points (1:0:0)2P² and (1:0:0)2P² are extendable outer Galois for C and C respectively.

(3) If ed 1, then points (0:1:0)2P² and (0:1:0)2P² are extendable inner Galois for C and C respectively.

PROOF. Since the dual mapgis given by

(dX^d ¹: eY^e ¹Z^{d e}: (d e)Y^eZ^{d e} ¹)

(dX^d: eXY^e ¹Z^{d e}: (d e)XY^eZ^{d e} ¹)

(dYêZ^{d e}: eXYê ¹Z^{d e}: (d e)XYêZ^{d e} ¹)

(dYZ: eXZ: (d e)XY);

we have (1). We have assertions (2) and (3) by Proposition 2.3. p

3. Galois points for the dual curve of a smooth curve

LetCP² be a smooth curve of degreed3 and letC be the dual curve ofC. The dual mapg:C!Cis birational ([7], [11, Theorem 1.5.3]) andCis of degreedd(d 1), since the linear system

@F

@X;@F

@Y;@F

@Z has no base point. Note that any birational transformation of C is extendable whend4 (see [1, Appendix A, 17 and 18] or [2]). Therefore any Galois point is extendable in this case.

LEMMA3.1. Let C be a smooth plane curve of degree d4. Then, any birational map C ^>C can be extended to a linear transformation of P². Therefore, any Galois point for Cis extendable.

(7)

PROOF. Lett:C ^>C be a birational map. Then the rational map t0:g ¹tg:C ^>Cis a birational transformation ofC. SinceCis smooth of degree d4, t0 can be extended to a linear transformation of P² as observed above. Since the diagram

commutes, we havet₀j_Ct. Hencet can be extended to a linear trans-

formation ofP². p

PROOF OFTHEOREM1.2. LetCbe a smooth plane curve of degreed3.

Throughout we treat the casesd3 andd4 separately.

Firstly, we assume thatd4. Aiming for a contradiction, letR2P²be a Galois point forC, and letmbe the multiplicity ofCatR. It follows from Lemma 3.1 and Proposition 1.5 that the degree of the projection fromRis at mostd. Therefore, we haved md. SinceR2C, there exists a point R02C with g(R0)R. Let `P² be a general line with R2`. Then, C\`n fRg consists of exactly d m smooth points of C. Since C\`n fRgconsists of only smooth points, there are exactlyd mpoints in g ¹(C\`n fRg) and such points are not flexes (i.e. the intersection multiplicity ofCand the tangent line at the point is two). It follows from Lemma 2.1 that [`]2[g(Q0)]T_Q₀CP² for any point Q02g ¹(C\`).

This implies that, for a general pointP2[R][g(R₀)]T_R₀C, there exist exactlyd mpoints not ing ¹(R) which are not flexes and the tangent lines containP. Then, by Hurwitz formula for the projectionp_P:C!P¹from a general pointP2TR0CnC, we have 2g(C) 22deg (pP)degB, where g(C)(d 1)(d 2)

2 is the genus of C andB is the ramification divisor.

Since the sum of the degrees ofBgiven by points inC\T_R₀C is at most d 1, we have degB(d 1)(d m). Thus Hurwitz formula leads to the following inequality

d² 3d2d(d 1)(d m)2d 1:

This implies thatd3. This is a contradiction to the assumptiond4.

So we assume that d3. Then, we have d6. We recall the well- known fact that there are nine flexes forC. Let Flex(C)Cbe the set of all flexes of C. Then g(Flex(C))Sing(C) as C does not admit bitangent lines. Firstly we would like to determine lines `P² such that

(8)

`\Sing(C)6 ;. Let Q2Flex(C) and let P2TQC, which is maybe Q.

Using Hurwitz formula for the projectionpP, the ramification divisor has degree 6 (resp. 4) if P6Q (resp. PQ. Hence one of the following conditions holds.

(1) There exist three flexes whose tangent lines containP.

(2) There exist exactly two flexes and two other points whose four tangent lines containP.

(3) There exist exactly one flex and four other points whose five tangent lines containP.

(4) PQand there exist three points which are not flexes and tangent lines at them containP.

It follows from Lemma 2.1 that one of the following conditions holds if

`\Sing(C)6 ;for a line`P².

(1) C\`consists of three singular points.

(2) C\`consists of two singular points and two smooth points.

(3) C\`consists of one singular point and four smooth points.

(4) C\`consists of exactly one singular point and three smooth points.

LetR2P²be a point and let^pR:pRg:C!P¹. We denote byeP

the ramification index at P2C for ^pR. Note that if Q2Flex(C) and g(Q)6R, then^p_Ris ramified atQ, since the differential ofgatQis zero.

The above conditions represent types of ramification indices of flexes for

^

p_R. For example, condition (1) implies that e_Q₁e_Q₂e_Q₃2, where g(Q1);g(Q2);g(Q3)2C\`ifR2`andR62C.

Assume by contradiction thatR2P²is a Galois point forC. Then, the field extension induced by ^pR is Galois. We often use the property of Galois extensions that the ramification indexe_Q₁is equal to the onee_Q₂ if

^

p_R(Q₁)^p_R(Q₂) ([12, Corollary 3.7.2]). Moreover, we separate the cases R2CnSing(C),R2P²nCandR2Sing(C).

Firstly we suppose that R is an inner Galois point, that is R2CnSing(C). Then, the projection ^p_R is of degree five. Then, any ramification index is five, by the property mentioned above. Therefore, the set ^p_R¹(^p_R(Q)) fQg if Q2Flex(C). Since this condition does not coincide with (1),. . ., (4) we have a contradiction.

Then we assume thatRis an outer Galois point, that isR2P²nC, so that deg^pR6. Note that there exist no tangent lines with two contact points forC, by the smoothness ofCand projective duality. By the property

(9)

of Galois extensions, any line containingRand a singular point must satisfy (1). Since the number of singular points is nine, there exists at most three lines containingRand satisfying (1). Applying Hurwitz formula to the projection p^_R, we have that the ramification divisor must have degree 12. Thus there exists another ramification point which is not a flex of C, but this contradicts the property of Galois extensions.

Finally, we assume that R2Sing(C). Let R02Flex(C) with g(R0)R. By the property of Galois extensions, any line containing R and another singular point must satisfy (1). Let points P2P² and Q₁;Q₂;Q₃2Flex(C) satisfy P2T_Q_iC for any i, as in condition (1). We may assume that P(1:0:0);Q₁(0:1:0);Q₂(0:0:1) and C is defined by

X³G₁(Y;Z)X²G₂(Y;Z)XG₃(Y;Z)0;

whereG_iis a homogeneous polynomial of degreei. SinceQ1;Q22Flex(C) andT_Q₁C(resp.T_Q₂C) is defined byZ0 (resp.Y 0),G10 andG2;G3

is divisible by YZ. Since T_Q₃C is defined by aYbZ0 for some a;b2Kn0, we also haveG₂0. We have an equationX³G₃(Y;Z)0.

We can take a linear transformation fofP² such thatf(P)P and the three points given by XG3(Y;Z)0 move to (0:1: 1), (0:1:v), (0:1:v²) byf, where vis a cubic root of 1 different from 1. Then, f(C) is given byX³a(Y³Z³)0 for somea2K. We may assume that Cis defined byX³Y³Z³0. Note that the nine flexes are contained in the union of lines X0,Y 0 and Z0. For this Fermat curve, if P1;P22Flex(C), then there exists a linear transformation c such that c(P1)P2 and c(C)C. Therefore, we may assume that R0 (0:1: 1). The dual map is given by (X²:Y²:Z²). We find that g(Flex(C)\ fX0g)C\ fU0g, and analogous equalities holds for the pairsY;V andZ;W. Here, we haveR(0:1:1). LetR⁰(1:0:1) and let`be the line passing throughR;R⁰2Sing(C), which is defined by UV W0. Since R2C\ fU0g, R⁰2C\ fV0g and ` sat- isfies condition (1), we have that ` must meet C at another singular point lying on the line fW 0g. Thus we have a contradiction as ` in- tersects fW0g at (1: 1:0)62Sing(C). p

4. Curves with extendable Galois points

Before proving Theorem 1.4, we present a preliminary lemma involved in the proof.

(10)

LEMMA4.1. Let G(Y;Z);H(V;W)be homogeneous polynomials of degree d. Assume that d^d( G)^d ¹ H(GY;GZ) holds and m1 is the maximal number such that Y^mdivides G. Then, W^{d m}divides H.

PROOF. Let GY^mF. Then F is not divisible by Y and, by Euler formula (d m)FYF_YZF_Z, F_Z also is not. Then, G_YmY^m ¹F

Y^mFY Y^m ¹(mFYFY) and GZY^mFZ. We denote by H P^d

i0aiVⁱW^{d i}. Then, by the assumption,

d^d( 1)^d ¹Y^m(d ¹⁾F^d ¹ H(GY;GZ)

X

i

a_iY^i(m ^{1)m(d i)}(mFYF_Y)ⁱF^{d i}_Z :

Since (mFYF_Y)ⁱF^{d i}_Z is not divisible byYfor anyi, we havea_i0 for any i such that m(d 1)>i(m 1)m(d i), i.e. i>m. Therefore, HP^m

i0a_iVⁱW^{d i}W^{d m}P^m

i0a_iVⁱW^{m i}. p

PROOF OF THEOREM 1.4(I). The assertion (2) ) (1) is nothing but Lemma 2.5(2). We prove (1) ) (2). By using projective duality, we may assume thatdd. By Proposition 2.3, we may assume thatP(1:0:0) be an extendable outer Galois point,Cis given by

X^dG(Y;Z)0;

whereG(Y;Z) is a homogeneous polynomial of degreed, and any birational transformation s2GP is represented by s(X:Y:Z)(zⁱX:Y:Z) for some i. By Lemma 2.4, we have an injection G_P,!G[P], where P(1:0:0). Then,ddandG_PG[P]. ThereforeP2P² is extendable outer Galois forC. Furthermore,Cis given by

U^dH(V;W)0;

where His a homogeneous polynomial of degree d. Since the dual map g:C ^>C is given by g(dX^d ¹ :GY:GZ), we have (dX^d ¹)^d H(GY;GZ)0. SinceX^d GonC, we have an equation

d^d( G)^d ¹H(G_Y;G_Z)0

onC. Since the variableX does not appear, this equation holds as polynomials. For a suitable system of coordinates, by Lemma 2.2, G(Y;Z) is divisible byYZ. Letm1 (resp.n1) be the maximal number such that Y^m (resp.Zⁿ) dividesG. Note thatmnd. Then, by Lemma 4.1,His

(11)

divisible by V^{d n}W^{d m}. Since (d n)(d m)2d (nm)d degH, we haveHaV^{d n}W^{d m} for somea2K. Therefore,Cis projectively equivalent to the curve given byU^d V^eW^{d e}0 for somee1.

The curveCis also by Lemma 2.5(1). p

PROOF OF THEOREM 1.4(II). The assertion (3) ) (1) is nothing but Lemma 2.5. We prove (1))(2). LetPbe an extendable outer Galois point forCand letQbe an extendable inner Galois point forC. By Proposition 1.5,ddanddd 1. Then we treat the casesdd 1 anddd separately.

Firstly, we assume thatdd 1. By Proposition 2.3, we may assume that P(1:0:0), a generator s2G_P is represented by s(X:Y:Z) (zX:Y :Z) andCis given by

X^dG(Y;Z)0:

By Lemma 2.4, we have an injection G_P,!G[P], where P(1:0:0).

Considering the orders ofGPandG[P], we haveGPG[P]. HenceP2Cis smooth and extendable inner Galois. Considering the action,Cis given by

U^dH1(V;W)Hd1(V;W)0;

whereH_iis a homogeneous polynomial of degreei. For a suitable system of coordinates, by Lemma 2.2, we may assume thatH₁W. We denote by H_d1H. ThenHis not divisible byW asC cannot contain lines, and alsoHVis not by Euler formula. Since the dual mapg_C:C ^>CCis given by

g_C(dU^d ¹W :H_V :U^dH_W)(dU^d ¹W² :WH_V :U^dWWH_W)

(dU^d ¹W²:WHV : HWHW) by usingU^dW H, we have

(dU^d ¹W²)^dG(WH_V; HWH_W)0:

SinceU^dW HonC, we have an equation

d^d( H)^d ¹W^d1 G(WH_V; HWH_W)

onC. Since the variableU does not appear, this equation holds as polynomials. LetG(Y;Z)P^d

i0a_iYⁱZ^{d i}. Then, we have d^d( H)^d ¹W^d1 X

i

aiWⁱHⁱ_V( HWHW)^{d i}:

(12)

SinceHV and HWHWare not divisible byW,ai0 for anyi. This is a contradiction.

Then we assume thatdd. By Proposition 1.5, we have an injection G_Q,!G[Q]. Considering the orders, we haveG_QG[Q]. HenceQ2Cis smooth and extendable inner Galois. We have assertion (2).

Then it remains to prove (2))(3). LetPbe an extendable inner Galois point forCand letQbe an extendable inner Galois point forC. By projective duality, we may assume thatdd, and Proposition 1.5 assures thatd d 1. Hence we deal with the casesdd 1 andddsep- arately.

Assume thatdd 1. By Proposition 1.5 and considering the orders, we haveGPG[P]. Then,Pis extendable outer Galois forC. Similarly to the above discussion to prove (1))(2), we have a contradiction.

Therefore, we have dd. By Proposition 2.3, we may assume that P(1:0:0) andCis given by

X^d ¹ZG(Y;Z)0:

By Lemma 2.4, we have an injection G_P,!G[P], where P(1:0:0).

Considering the orders of GP andG[P], we have GPG[P]. Therefore, P2P² is extendable inner Galois forC. Furthermore,C is given by

U^d ¹(aVbW)H(V;W)0;

wherea;b2K. Since the dual mapg:C ^>C is given by

g((d 1)X^d ²Z:GY :X^d ¹GZ)((d 1)X^d ²Z²:ZGY :X^d ¹ZZGZ)

((d 1)X^d ²Z²:ZGY: GZGZ) by usingX^d ¹Z G, we have

((d 1)X^d ²Z²)^d ¹(aZG_Yb( GZG_Z))H(ZG_Y; GZG_Z)0:

SinceX^d ¹Z GonC, we have an equation

(d 1)^d ¹( G)^d ²Z^d(aZGYb( GZGZ)) H(ZGY; GZGZ) onC. Since the variable X does not appear, this equation holds as polynomials. LetH(V;W)P^d

i0a_iVⁱW^{d i}. Then, we have (d 1)^d ¹( G)^d ²Z^d(aZG_Yb( GZG_Z)) X

i

a_iZⁱGⁱ_Y( GZG_Z)^{d i}: Since G_Y and GZG_Z are not divisible by Z, a_i0 for any i<d.

(13)

Therefore,H(V;W)cV^dfor somec2K. We have an equation U^d ¹(aVbW)cV^d0

forC. The dual curveC is projectively equivalent to the curve given by U^d V^d ¹W0. Therefore,Cis also by Lemma 2.5(1). p REMARK 4.2. In order to deal with Problem 1.1, the condition P62Sing(C) in the definition of inner Galois point is crucial. In particular, Theorem 1.4 fails to hold true as we extend the assertion to (non-extendable) outer Galois point forCand Galois point lying on Sing(C).

For example, let CP² be curve defined by (XYZ)³ 27XYZ0. Then (1:0:0) is an outer Galois point forC([14, Example 2]).

SinceCis a rational nodal curve with three flexes, the dual curveCis a quartic plane curve with three singular double points. ThusCis not a self- dual curve. However, any singular pointQ2Sing(C) is a Galois point for Cas the projectionpQ:C ^>P¹ has degree two.

REMARK4.3. According to [6], the set of all Galois points for the plane curve defined byX^d Y^eZ^{d e}0 is equal to f(1:0:0);(0:1:0);(0:0:1)g (ifd4.

Acknowledgments. The authors are grateful to Professor Yoshiaki Fukuma for helpful comments. The authors thank Professor Hisao Yosh- ihara for informing a new result of him with Hayashi [6]. The authors also thank the referee for helpful advice.

REFERENCES

[1] E. ARBARELLO, M. CORNALBA, P. A. GRIFFITHSand J. HARRIS,Geometry of algebraic curves, Vol. I. Grundlehren der Mathematischen Wissenschaften 267, Springer-Verlag, New York, 1985.

[2] H. C. CHANG,On plane algebraic curves, Chinese J. Math.6(1978), 185-189.

[3] C. CILIBERTO, Alcune applicazioni di un classico procedimento di Castel- nuovo, Seminari di geometria, 1982-1983 (Bologna, 1982/1983), Univ. Stud.

Bologna, Bologna, 1984, 17-43.

[4] S. FUKASAWA, Galois points for a plane curve in arbitrary characteristic, Proceedings of the IV Iberoamerican conference on complex geometry, Geom.

Dedicata139(2009), 211-218.

[5] R. HARTSHORNE,Generalized divisors on Gorenstein curves and a theorem of Noether, J. Math. Kyoto Univ.26(1986), 375-386.

(14)

[6] H. HAYASHIand H. YOSHIHARA,Galois group at each point for some self-dual curves, Geometry2013(2013), Article ID 369420, 6 pages.

[7] S. L. KLEIMAN,Tangency and duality, In:Proceedings of the 1984 Vancouver conference in algebraic geometry, CMS Conference Proceedings, 6, Amer.

Math. Soc., Providence, RI, 1986, pp. 163-226.

[8] K. MIURA,Galois points for plane curves and Cremona transformations, J.

Algebra320(2008), 987-995.

[9] K. MIURAand H. YOSHIHARA,Field theory for function fields of plane quartic curves, J. Algebra226(2000), 283-294.

[10] M. NAMBA, Families of meromorphic functions on compact Riemann surfaces, Lecture Notes in Math.767, Springer-Verlag, Berlin, 1979.

[11] M. NAMBA,Geometry of projective algebraic curves, Marcel Dekker, Inc., New York, 1984.

[12] H. STICHTENOTH,Algebraic function fields and codes, Universitext, Springer- Verlag, Berlin, 1993.

[13] H. YOSHIHARA, Function field theory of plane curves by dual curves, J.

Algebra239(2001), 340-355.

[14] H. YOSHIHARA,Galois points for plane rational curves, Far East J. Math. Sci.

25(2007), 273-284.

[15] H. YOSHIHARA, Rational curve with Galois point and extendable Galois automorphism, J. Algebra321(2009), 1463-1472.

Manoscritto pervenuto in redazione il 29 Novembre 2012.