Complete determination of the number of Galois points for a smooth plane curve

Complete Determination of the Number of Galois Points for a Smooth Plane Curve

SATORUFUKASAWA(*)

Dedicated to my son Atsumu and my wife Kaori

ABSTRACT- LetCbe a smooth plane curve. A pointPin the projective plane is said to be Galois with respect toCif the function field extension induced by the pro- jection fromPis Galois. We denote byd(C) (resp.d⁰(C)) the number of Galois points contained inC(resp. inP²nC). In this article, we determine the numbers d(C) andd⁰(C) in any remaining open cases. Summarizing results obtained by now, we will present a complete classification theorem of smooth plane curves by the numberd(C) ord⁰(C). In particular, we give new characterizations of Fermat curve and Klein quartic curve by the numberd⁰(C).

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

KEYWORDS. Galois point, plane curve, positive characteristic, Galois group.

1. Introduction

Let the base field K be an algebraically closed field of characteristic p0 and letCP²be a smooth plane curve of degreed4. In 1996, H.

Yoshihara introduced the notion of Galois point (see [14, 17] or survey paper [5]). If the function field extension K(C)=K(P¹), induced by the projectionpP:C!P¹from a pointP2P², is Galois, then the pointPis said to be Galois with respect toC. When a Galois pointPis contained inC (resp.P²nC), we callPan inner (resp. outer) Galois point. We denote by d(C) (resp.d⁰(C)) the number of inner (resp. outer) Galois points forC. It is

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

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

remarkable that many classification results of algebraic varieties have been given in the theory of Galois point.

Yoshihara and K. Miura determined d(C) and d⁰(C) in characteristic pˆ0 ([14, 17]). In characteristicp>0, M. Homma [13] settledd(H) and d⁰(H) for the Fermat curveHof degreep^e‡1. Recently, the present au- thor determinedd(C) whenp>2 ord 1 is not a power of 2 ([3, 4]), and d⁰(C) whendis not divisible byp,dˆp, ordˆ2^einpˆ2 ([3, 4, 7]). The following problems remain open ([4, Part III, Problem], [5, Problem 2]).

PROBLEM. (1)Let pˆ2and let e2. Find and classify smooth plane curves of degree dˆ2^e‡1withd(C)ˆd.

(2) Let p>0, e1and let dˆp^el, where l is not divisible by p. As- sume that(p^e;l)6ˆ(p;1);(2^e;1). Then, determined⁰(C).

In this article, we give a complete answer to these problems.

THEOREM1. Let pˆ2, let e2and let C be a smooth plane curve of degree dˆ2^e‡1. Then,d(C)ˆd if and only if C is projectively equivalent to the curve given by

Y

a2F₂e

(x‡ay‡a²)‡cy^2e‡1ˆ0;

(1c)

where c2Kn f0;1g.

THEOREM2. Let the characteristic p>0, let e1, let l be not divisible by p, and let C be a smooth plane curve of degree dˆp^el4. If (p^e;l)6ˆ(2^e;1), thend⁰(C)1.

Summarizing Theorems 1 and 2 and the results of Yoshihara, Miura, Homma and the present author, we obtain the following classification theorem of smooth plane curves by the numberd(C) ord⁰(C).

THEOREM3 (Yoshihara, Miura, Homma, Fukasawa). Let C be a smooth plane curve of degree d4in characteristic p0. Then:

(I) d(C)ˆ0;1;d or(d 1)³‡1. Furthermore, we have the following.

(i) d(C)ˆ(d 1)³‡1 if and only if p>0, dˆp^e‡1 and C is projectively equivalent to the Fermat curve.

(ii) d(C)ˆd5if and only if pˆ2, dˆ2^e‡1and C is projectively equivalent to the curve defined by Q

a2F2e

(x‡ay‡a²)‡cy^2e‡1ˆ0, where c2Kn f0;1g.

(iii) d(C)ˆdˆ4 if and only if p6ˆ2;3 and C is projectively equivalent to the curve defined by x³‡y⁴‡1ˆ0.

(II) d⁰(C)ˆ0;1;3;7or(d 1)⁴ (d 1)³‡(d 1)². Furthermore, we have the following.

(i) d⁰(C)ˆ(d 1)⁴ (d 1)³‡(d 1)² if and only if p>0, d 1 is a power of p and C is projectively equivalent to the Fermat curve.

(ii) d⁰(C)ˆ7 if and only if pˆ2, dˆ4 and C is projectively equivalent to Klein quartic curve.

(iii) d⁰(C)ˆ3 and three Galois points are not contained in a com- mon line if and only if d is not divisible by p, d 1 is not a power of p, and C is projectively equivalent to the Fermat curve.

(iv) d⁰(C)ˆ3 and three Galois points are contained in a common line if and only if pˆ2, dˆ4and C is projectively equivalent to the curve defined by

(x²‡x)²‡(x²‡x)(y²‡y)‡(y²‡y)²‡cˆ0;

where c2Kn f0;1g.

This is a modified and extended version of the paper [4, Part IV] (which will have been published only in arXiv).

2. Preliminaries

Let CP² be a smooth plane curve of degreed4 in characteristic p>0. For a pointP2C, we denote byT_PCP²the (projective) tangent line atP. For a projective linelP² and a pointP2C\l, we denote by I_P(C;l) the intersection multiplicity ofCandlatP. We denote byPRthe line passing through pointsPandRwhenP6ˆR, and byp_P :C!P¹;R7!PR the projection from a pointP2P². IfR2C, we denote byeRthe ramifica- tion index ofp_PatR. It is not difficult to check the following.

LEMMA1. Let P2P²and let R2C. Then forp_Pwe have the following.

(1) If RˆP, then e_RˆI_R(C;T_RC) 1.

(2) If R6ˆP, then e_RˆI_R(C;PR).

LetPbe a Galois point. We denote byGPthe group of birational maps fromCto itself corresponding to the Galois group Gal(K(C)=p_PK(P¹)). We

find easily that the group GP is isomorphic to a subgroup of the auto- morphism group Aut(C) ofC. We identifyGPwith the subgroup. When we use the symbolgfor an automorphism of the curveC, we use the symbolg for the automorphism of the function fieldK(C) corresponding tog.

If a Galois coveringu:C!C⁰between smooth curves is given, then the Galois group G acts on C naturally. We denote by G(R) the stabilizer subgroup ofR. The following fact is useful to find Galois points (see [15, III.

7.1, 7.2 and 8.2]).

FACT1. Let u:C!C⁰ be a Galois covering of degree d with Galois group G and let R;R⁰2C. Then we have the following.

(1) For anys2G, we haveu(s(R))ˆu(R).

(2) If u(R)ˆu(R⁰), then there exists an element s2G such that s(R)ˆR⁰.

(3) The order of G(R)is equal to e_R at R for any point R2C.

(4) Ifu(R)ˆu(R⁰), then e_Rˆe_R⁰. (5) The index e_Rdivides the degree d.

We recall a theorem on the structure of the Galois group at a Galois point (see [4, Part II]). Let d 1ˆp^el(resp.dˆp^el), wherelis not di- visible byp, letzbe a primitivel-th root of unity, and letkˆ[Fp(z):Fp].

Let Pˆ(1:0:0) be an inner (resp. outer) Galois point forC. The pro- jection p_P:C!P¹ is given by (x:y:1)7!(y:1). We have a field ex- tension K(x;y)=K(y) viapP. Let g2GP. Then, the automorphism g2GP

can be extended to a linear transformation ofP² (see [1, Appendix A, 17 and 18] or [2]). Let Agˆ(a_ij) be a 33 matrix representing g. Since g2G_P, g(y)ˆy. Then, (a₂₁x‡a₂₂y‡a₂₃) (a₃₁x‡a₃₂y‡a₃₃)yˆ0 in K(x;y). Sinced4, we havea₂₁ˆa₂₃ˆa₃₁ˆa₃₂ˆ0 anda₂₂ˆa₃₃. W e may assume thata₂₂ˆa₃₃ ˆ1. Sinceg^pelˆ1, we havea^l₁₁ˆ1. We take a group homomorphism GP!Kn0;g 7!a11(g), where a11(g) is the (1;1)- element ofAg. Then, we have the splitting exact sequence of groups

0!(Z=pZ)^e!GP! hzi !1;

and the following theorem.

THEOREM4. Let CP²be a smooth curve and let P be an inner (resp.

outer) Galois point. Then, k divides e and GP(Z=pZ)^ejhzi.

REMARK1. The condition thatkdivideseis equivalent to thatldivides p^e 1. We give a proof here. Ifkdividese,F_p(z)ˆF_pkis a subfield ofF_p^e.

Since z2Fp^e,z^{pe 1}ˆ1. Since the order of z in the multiplicative group Fp^en0 isl,ldividesp^e 1. The converse also holds.

We denote the kernel (resp. quotient) byK_P(resp. byQ_P). An element s2 K_P(resp. a generatort2 Q_P) is represented by a matrix

Asˆ

1 a12(s) a13(s)

0 1 0

0 0 1

0 B@

1 CA

resp.Atˆ

z a12(t) a13(t)

0 1 0

0 0 1

0 B@

1 CA

; where a₁₂(s);a₁₃(s);a₁₂(t);a₁₃(t)2K. For each non-identity element g2G_P, there exist s2 K_P and i such that gˆstⁱ. Then, there exists a unique line Lg, which is defined by (zⁱ 1)X‡(a12(s)‡a12(tⁱ))Y‡ (a13(s)‡a13(tⁱ))Zˆ0, such thatg(R)ˆRfor anyR2Lg. Note thatP2Lg

if and only ifg2 K_P. Furthermore, fors2 K_PandR6ˆP,L_sˆRPif and only if s(R)ˆR. For a suitable system of coordinates, we can take a₁₂(t)ˆa₁₃(t)ˆ0.

Finally in this section, we note the following facts on automorphisms ofP¹.

LEMMA2. We denote byAut(P¹)the automorphism group ofP¹. (1) Let P1;P2;P32P¹be three distinct points and letg₁;g₂2Aut(P¹).

Ifg₁(P_i)ˆg₂(P_i)for iˆ1;2;3, theng₁ˆg₂.

(2) Let P₁;P₂2P¹ be distinct points and let GAut(P¹)be a finite subgroup. Ifg(P₁)ˆP₁ andg(P₂)ˆP₂ for anyg2G, then G is a cyclic group whose order is not divisible by p if p>0.

(3) Let l be not divisible by p, let P2P¹, and let GAut(P¹) b e a subgroup of order l. Assume that G is cyclic andt(P)ˆP for any t2G. Then, there exists a unique point Q such that Q6ˆP and t(Q)ˆQ for anyt2G.

PROOF. The fact (1) is easily proved, if we use the classical fact that any automorphism ofP¹is a linear transformation. We prove (2). We may as- sume that P1ˆ(1:0) and P2ˆ(0:1). Let g2G. Since g(P1)ˆP1 and g(P₂)ˆP₂,gis represented by a matrix

A_gˆ a(g) 0

0 1

!

;

where a(g)2K. Then, the homomorphism c:G!Kn0:g7!a(g) is in- jective andc(G) is cyclic. Letmbe the order ofc(G). Then,c(G) is con-

tained in the set fx2Kn0jx^m 1ˆ0g. If m is divisible by p, the set consists of at most m=pelements. Therefore,mis not divisible byp. W e have the conclusion.

We prove (3). We may assume thatPˆ(1:0). Lettbe a generator of G. Sincet(P)ˆPandtis an automorphism of orderlnot divisible byp,tis represented by a matrix

Atˆ z b 0 1

!

;

wherezis a primitivel-th root of unity andb2K. Then,tⁱis represented by the matrix

A_tiˆ zⁱ zⁱ 1 z 1b

0 1

0 B@

1 CA:

Let Qˆ(x:1). Then, tⁱ(Q)ˆQ if and only if (z 1)x‡bˆ0. We have

the conclusion. p

3. Only-if-part of the proof of Theorem 1

Letpˆ2, letqˆ2^e4 and letCbe a plane curve of degreedˆq‡1.

Assume thatd(C)ˆd. LetP₁;. . .;P_dbe inner Galois points forC. By the results of the previous paper [4, Part III, Lemma 1, Propositions 1, 3 and 4], we have the following.

PROPOSITION1. Assume thatd(C)ˆd. Then, we have the following.

(1) Galois points P₁;. . .;P_dare contained in a common line.

(2) For any i and any elements2G_Pin f1g, the order ofsis two.

(3) For any i and any elementss;t2GPin f1gwiths6ˆt, Ls6ˆTPiC and Ls6ˆLt. In particular, the set fTP1C\TPiCj2idg con- sists of exactly d 1points.

By the condition(1)and Fact1(2), for each i with3id, there exists t_i2G_Pi such that t_i(P₁)ˆP₂. Let fQg ˆT_P1C\T_P2C. In addition, we have the following by the condition(2).

(4) For any i with3id,ti(P2)ˆP1andti(Q)ˆQ.

(5) For any i;j with 3i;jd, t_it_j(P1)ˆP1, t_it_j(P2)ˆP2 and t_it_j(Q)ˆQ.

LEMMA 3. For a suitable system of coordinates, we may assume that P1ˆ(1:0:0), P2ˆ(0:0:1)and Qˆ(0:1:0).

By Lemma 3 and Proposition 1(2)(4),t_iis given by a matrix

A_ti ˆ

0 0 1

0 a_i 0 a²_i 0 0 0

B@

1 CA;

for somea_i2K. Then,t_it_jis given by the matrix

A_titj ˆ

a²_j 0 0 0 a_ia_j 0 0 0 a²_i 0

BB

@

1 CC A:

Let H(C) be the subgroup of Aut(P²) consisting of any g2Aut(P²) satisfying

(h1) g(P1)ˆP1,g(P2)ˆP2 andg(Q)ˆQ, (h2) fg(P_i)j3idg ˆ fP_ij3idg, and (h3) g(C)ˆC.

LEMMA 4. The group H(C) is a cyclic group whose order is at most d 2ˆq 1.

PROOF. By the condition (h1) of H(C), for any g2H(C), g is re- presented by a matrix

Agˆ

a 0 0 0 b 0 0 0 1 0

B@

1 CA

for somea;b2K. We prove thatgdepends only on the image ofP₃. Pre- cisely, we show that for g₁;g₂2H(C), if g₁(P₃)ˆg₂(P₃), then g₁ˆg₂. To prove this, it suffices to show that gˆ1 if g(P3)ˆP3. Assume that g(P3)ˆP3. Sincegfixes three distinct pointsP1;P2;P3on the lineP1P2,g is identity on the line P1P2, by Lemma 2(1) in Section 2. We have aˆ1, since P₁P₂ is given by Yˆ0. On the other hand, by the condition (h3), g(T_P3C)ˆT_P3C. Then, the point Q₀ given by T_P1C\T_P3C is fixed by g.

Note that Q₀6ˆQ;P₁ by Proposition 1(3)(1) and Fact 1(3). Since g fixes

three distinct pointsP1;Q;Q0on the lineP1QandP1Qis given byZˆ0, we havebˆ1.

By the above discussion and the condition (h2), the order of H(C) is at most d 2ˆq 1. We consider the group homomorphism H(C)!P₁P₂P¹ given by restrictions, which is well-defined by the condition (h1) of H(C). Then, this is injective by the above discussion.

It follows from Lemma 2(2) that H(C) is cyclic. p We consider the setSˆ ft₃t_ij3idg. Then,SH(C) by Proposi- tion 1(5)(1). Since the cardinality ofSisq 1,SˆH(C) by Lemma 4. Since H(C) is cyclic, there existsisuch thatt₃t_iis a generator ofH(C). There- fore,t3tiis given by the matrix

At3tiˆ

z² 0 0

0 z 0

0 0 1 0

B@

1 CA;

wherezis a primitive (q 1)-th root of unity. We denotet₃t_ibyg.

By Proposition 1(3), there exists an elements2G_P1n f1gsuch that the (1;2)-elementa₁₂(s) and (1;3)-elementa₁₃(s) of a matrixA_srepresentings are not zero (see also Section 2). If we take a linear transformationfwith Y7!(1=a12(s))Y and Z7!(1=a13(s))Z, then f(P_i)ˆP_i for iˆ1;2, f(Q)ˆQ,fgf ¹ˆgandfsf ¹is represented by the matrix

A₀ˆ

1 1 1 0 1 0 0 0 1 0

B@

1 CA:

Therefore, we may assume thatsis represented by the matrixAsˆA0. The automorphismg^jsg ^jis represented by the matrix

1 z^j z^2j

0 1 0

0 0 1

0 B@

1 CA:

In particular, g^jsg ^j2GP1 for any j with 1jq 1. Since the cardinality of the set fg^jsg ^jj1jq 1g GP1 is q 1, GP1ˆ fgQ^jsg ^jj1jq 1g [ f1g. Then, the rational function g(x;y):ˆ

a2Fq

(x‡ay‡a²)2K(x;y) is fixed by any element of G_P1. Therefore, g(x;y)2K(y). There exists h(y)2K(y) such that g(x;y)‡h(y)ˆ0 in K(x;y). Let h(y)ˆh1(y)=h2(y), where h1;h22K[y]. Then, g(x;y)h2(y)‡

h1(y)ˆ0 on C. Let f(x;y) be a defining polynomial. Then, there exists v(x;y)2K[x;y] such that f(x;y)v(x;y)ˆ g(x;y)h2(y)‡h1(y) as poly- nomials. Since P12C is smooth and the tangent line T_P1CˆP1Q is given by Zˆ0, the coefficient of x^2e for f(x;y) as in (K[y])[x] is a con- stant. Comparing the coefficient of degree 2^e in variable x, we have v(x;y)2K[y] and v(x;y)ˆh₂(y) up to a constant. Then, h₂(y) divides h1(y) and we have h(y)2K[y]. Therefore, we may assume that f(x;y)ˆg(x;y)‡h(y), where h(y)2K[y]. By the condition that the tangent line T_P2CˆP₂Q is given by Xˆ0, g(x;y)‡cy^q‡1ˆ0 for some c2Kn0. Therefore, we have a defining equation f(x;y)ˆ g(x;y)‡cy^q‡1ˆ0.

Finally in this section, we investigate conditions for the smooth- ness of C. Let G(X;Y;Z):ˆZ^q‡1g(X=Z;Y=Z) and let F(X;Y;Z):ˆ

Z^q‡1f(X=Z;Y=Z). Then, by direct computations, we have F(Z;Y;X)ˆ F(X;Y;Z). Since there exist exactly d points contained in C and the line defined by Y ˆ0, such points are smooth. Therefore, singular points should lie on Y 6ˆ0. Leth(x;z)ˆG(x;1;z). We consider has an element of K(z)[x]. Then, the set fa‡a²zja2Fqg K(z), which con- sists of all roots of h(x;z)ˆ0, forms an additive subgroup of K(z).

According to [8, Proposition 1.1.5 and Theorem 1.2.1], we have the following.

LEMMA 5. The polynomial h(x;z)2K(z)[x] has only terms of degree equal to some power of p. In particular, h_x(x;z)ˆz^q‡z, where h_x is a partial derivative by x.

Assume that (x;z)2C is a singular point, i.e. hx(x;z)ˆhz(x;z)ˆ0.

Then, (x;z) isFq-rational by Lemma 5. We havec6ˆ1 by the following.

LEMMA6. The equalityfh(x;z)jx;z2F_qg ˆ f0;1gholds.

PROOF. If zˆ0, then h(x;z)ˆ0. We fix z₀2F_qn0. We consider h(x;z₀)ˆz₀ Q

a2Fq

(x‡a‡a²z₀)2F_q[x]. For eacha2F_q, there exists a un- iqueb2F_q withb6ˆa such thata‡a²z₀ˆb‡b²z₀. Therefore, the car- dinality of the setS₀:ˆ fa‡a²z₀ja2F_qgisq=2ˆ2^{e 1}. By direct compu- tations, we find that any element ofS0is a root of the separable polynomial h₀(x)ˆ^eP¹

iˆ0z²₀ⁱx²ⁱ, which is of degreeq=2. Then,h(x;z₀)ˆh₀(x)²as elements ofF_q[x]. Then, by direct computations, we haveh₀(x)(h₀(x)‡1)ˆz₀(x^q‡x)

as elements ofFq[x]. Assumex2Fq. Then,h0(x)(h0(x)‡1)ˆ0. Therefore, h(x;z0)ˆ0 or 1. If we take x2FqnS0, then h0(x)6ˆ0 and hence,

h(x;z0)ˆ1. p

4. If-part of the proof of Theorem 1

We use the same notation as in the previous section,g;f;F, and so on.

LetCbe the plane curve given by Equation (1c) withc2Kn f0;1g. As in the previous section, C is smooth. We prove d(C)ˆd. We consider the projection p_P1 from P₁ ˆ(1:0:0). Then, we have the field extension K(x;y)=K(y) with f(x;y)ˆg(x;y)‡cy^q‡1ˆ0. Since (x‡ay‡a²)‡

by‡b²ˆx‡(a‡b)y‡(a‡b)², we have f(x‡ay‡a²;y)ˆf(x;y) for any a2F_q. Therefore, P₁ is Galois. By the symmetric property of F(X;Y;Z) forX;Z, we find that a point (0:0:1) is also inner Galois forC.

We also find that there exist a tangent line Tsuch that I_Q(C;T)ˆ2 for someQ2C\T. Therefore,Cis not projectively equivalent to the Fermat curve of degree q‡1 (see, for example, [13]). According to [4, Part III, Lemma 1 and Proposition 1], we haved(C)ˆd.

REMARK2. The projective equivalence class of the plane curve given by Equation (1c) is uniquely determined by a constant c2Kn0. There- fore, infinitely many classes exist. Precisely, we have Lemma 7 below.

LEMMA7. Let a;b2Kn0and let C_a(resp. C_b) be the plane curve given by Equation (1c) with cˆa (resp. cˆb). If there exists a projective transformationfsuch thatf(Ca)ˆCb, then aˆb.

PROOF. Let P1;. . .;P_d be inner Galois points for Ca, which are con- tained in the line defined byY ˆ0. Then, P₁;. . .P_d are also inner Galois for C_b. Since the tangent lines T_PC_a and T_PC_b at Pˆ(a²:0:1) with a2Fq are given by the same equationX‡aY‡a²Zˆ0,TPiCaˆTPiCb

for iˆ1;. . .;d. We may assume that P1ˆ(1:0:0), P2ˆ(0:0:1) and P3ˆ(1:0:1). Let Q2ˆ(0:1:0) and let Q3ˆ(1:1:0). Then, T_P1Ca\ T_P2C_aˆT_P1C_b\T_P2C_bˆ fQ₂g and T_P1C_a\T_P3C_aˆT_P1C_b\T_P3C_bˆ fQ₃g.

Letfbe a linear transformation such thatf(C_a)ˆC_b.

Iff(P₁)ˆP_ifor somei6ˆ1, then we takes2G_Pj(C_b) for somejsuch thats(Pi)ˆP1, by Fact 1(2). Then,sf(P1)ˆP1. Therefore, there exists a linear transformation f such that f(Ca)ˆC_b and f(P1)ˆP1. If f(P₂)ˆP_i for some 3id, then we take t2G_P1(C_b) such that

t(P3)ˆP2, by Fact 1(2). Then, tf(P2)ˆP2. Therefore, there exists a linear transformationfsuch thatf(Ca)ˆCb,f(P1)ˆP1andf(P2)ˆP2. If f(P3)ˆP_ifor some 4id, then we takeg2H(C_b) such thatg(P_i)ˆP3, whereH(C_b) is the group forC_bdiscussed in the previous section. Then, gf(P₃)ˆP₃. Therefore, there exists a linear transformationfsuch that f(C_a)ˆC_b, f(P_i)ˆP_i for iˆ1;2;3 and f(Q_j)ˆQ_j for jˆ2;3. Since f(P1)ˆP1;f(P2)ˆP2andf(Q2)ˆQ2,fis represented a matrix

A_fˆ

l1 0 0 0 l2 0 0 0 l₃ 0

B@

1 CA

for some l₁;l₂;l₃2Kn0. Since f(P₃)ˆP₃ and f(Q₃)ˆQ₃, we have l₁ˆl₃andl₁ˆl₂. Then,fis identity. Therefore, by considering the de- fining equations ofCaandCb, we should haveaˆb. p REMARK3. Ifcˆ1, then the plane curveCdefined by Equation (1c) is parameterized asP¹!P²:(s:1)7!(s^q‡1:s^q‡s:1):The distribution of Galois points for this curve has been settled in [6].

5. Proof of Theorem 2 (The case wherel3)

If we have two outer Galois points, then we note the following (see Section 2).

LEMMA 8. Let P;P2 be outer Galois points for C. Then, any element g2G_P can be extended to a linear transformation of P², and hence g(P₂)2P²is also outer Galois for C.

Let dˆp^el, where e1, l3 and l divides p^e 1, and let Pˆ(1:0:0) be an outer Galois point. It follows from a generalization of Pardini's theorem by Hefez [9, (5.10) and (5.16)] and Homma [12] that the generic order of contact forCis equal to 2, i.e.I_R(C;T_RC)ˆ2 for a general pointR2C(see also [10, 11]).

Let MP²be a projective line withP2M. Note thatg(M)ˆMfor any g2GP, by the forms of the matricesAs andAt as in Section 2. The homomorphismr_P[M]:G_P!Aut(M), which is induced by the restriction, is well-defined. Then, the kernel Kerr_P[M] is a subgroup of K_P and the cardinality of Ker r_P[M] is a power ofp, sinceg2Kerr_P[M] if and only if

LgˆM. We denote it byp^v[M]. Since the kernel KerrP[M] is a subspace of GPasFp-vector spaces, we have the following diagram.

Using lower splitting exact sequence as groups, we have the following.

LEMMA9. The integer l divides p^{e v[M]} 1for any line M with P2M.

Hereafter in this section, we assume that P22P²n fPg is an outer Galois point forC.

PROPOSITION2. Assume thatl3. Then:

(1) v[PP₂]ˆe, and there exists a unique point Q2P²with Q6ˆP such thatg(Q)ˆQ for anyg2GP.

Let Q be the point as in(1). Furthermore, we have the following.

(2) If l5, then P₂ˆQ.

(3) If lˆ4 and P₂6ˆQ, then Q2C or there exist two outer Galois point P3;P4such thatg(P4)ˆP4 for anyg2GP3.

(4) If lˆ3and P26ˆQ, then Q2C.

PROOF. Letg2G_Pn K_P and letL_gbe the line, which is a fixed locus, defined as in Section 2. The set C\L_g consists of d points, because T_RCˆPR6ˆL_gifR2C\L_gby Fact 1(3) and Lemma 1(2). Lett2 Q_Pbe a generator and letLtbe the line, defined as in Section 2. We denotev[PP2] byvand assume thatv5e.

We consider the case where g(P2)ˆP2 for some g2G_Pn K_P. Let s2 K_P2. Then,s(R)2L_g ands(R)6ˆR ifR2C\L_g, by Fact 1(1)(3) and that L_g consists of exactly d points. Furthermore, s(P)2T_s(R1)C\ Ts(R2)Cˆ fPg, ifR1;R22C\LgwithR16ˆR2. Therefore, we should have s(P)ˆP. This is a contradiction tov5e.

We consider the case where g(P₂)6ˆP₂ for anyg2G_Pn K_P. Assume that g₁(P₂)ˆg₂(P₂) forg₁;g₂2G_P. Note that any element g2G_P is re- presented as gˆstⁱ for some s2 K_P and some i (see Section 2). Let

g₁ ˆs1tⁱandg₂ˆs2t^j, wheres1;s22 KP. Since (t ^js₂¹s1t^j)t^{i j}(P2)ˆP2

and t ^js₂¹s1t^j 2 KP, we have iˆj and g₂¹g₁2 KP by the assumption.

Furthermore, we have g₂¹g₁2KerrP[PP2], since g₂¹g₁(P)ˆP and g₂¹g₁(P2)ˆP2. Therefore, we havep^{e v}l‡1 outer Galois points on the line PP2, by Lemma 8 and that the group ImrP[PP2] is isomorphic to (Z=pZ)^{e vj}hzi.

LetR2C\L_t. We consider points on the linePR. Let]G_P(R)ˆp^bl, whereG_P(R) is the stabilizer subgroup atR. Then we havep^{e b}flexes of order (]GP(R) 2) by Fact 1(3). We note that (p^bl 2)p^{e b}p^e(l 2).

Furthermore, for each outer Galois points, we spent at least degree (d 1)(p^e(l 2)) as the degree of the Wronskian divisor. Therefore, it follows from the degree of Wronskian divisor ([16, Theorem 1.5]) that

(p^{e v}l‡1)(d 1)p^e(l 2)3d(d 2):

Then, we have

(p^{e v}l‡1)p^e(l 2)53dˆ3p^el:

Therefore, (p^{e v}l‡1)(l 2) 3l50. Note thatp^{e v} 1lby Lemma 9.

Then, (l²‡l‡1)(l 2) 3l50. This is a contradiction. Therefore,vˆe.

In particular, the group ImrP[PP2] is a cyclic group of order l. By Lemma 2(3) in Section 2, a fixed point by the group Imr_P[PP2] which is different fromPis uniquely determined. We denote it byQ. Then,g(Q)ˆQ for anyg2G_P, sincegˆstⁱfor somes2 K_Pand somei. We have (1).

We prove (2). Assume that P26ˆQ. Since the group ImrP[PP2] is a cyclic group of orderl, we havel‡1 outer Galois points on the linePP2, by Lemma 8. Furthermore, for each outer Galois point, we spent at least degree (d 1)p^e(l 2) as the degree of the Wronskian divisor, similarly to the proof of (1). Therefore, it follows from the degree of Wronskian divisor ([16, Theorem 1.5]) that

(l‡1)(d 1)p^e(l 2)3d(d 2):

Then, we have

(l‡1)p^e(l 2)53dˆ3p^el:

Therefore, (l‡1)(l 2) 3l50. Then,l² 4l 250. We havel4.

We prove (3). Assume that P₂6ˆQandQ62C. Since Imr_P[PP₂] is a cyclic group of orderl, the cardinality of C\PP₂ is equal toland there exists l‡1 outer Galois points on PP2, by Fact 1(3), Lemma 8 and the assumption. Let C\PP2ˆ fR1;. . .;Rlg and let P;P2;. . .;Pl‡1 be outer Galois points.

Letlˆ4. The restrictionrP[PP2](t) of the generatort2 QP is a gen- erator of ImrP[PP2]. We may assume thatt(Ri)ˆRi‡1 foriˆ1;2;3;4, where R5ˆR1. We can take h_j2Imr_Pj[PP2] such thath_j(R1)ˆR2 for jˆ2;3;4;5 by Fact 1(2). We consider the case where at least three ele- ments offh_jgare of order 4. We may assume thath₂,h₃,h₄are of order 4.

Assume that h_j(R₂)ˆR₄ for any j with 2j4. Then, we have h_j(R4)ˆR3. Since three points on the linePP2has the same images under h₂;h₃;h₄, these are the same automorphism of the linePP2by Lemma 2(1).

Then,h_jfixesP₂;P₃;P₄forjˆ2;3;4, becauseh_j(P_j)ˆP_j. This implies that h_j is identity on PP₂, by Lemma 2(1). This is a contradiction. Therefore, there existsjsuch thath_j(R₂)ˆR₃. Then, we haveh_j(R₃)ˆR₄. Therefore, t coincides with h_j on the line PP2. Then, t(Pj)ˆh_j(Pj)ˆPj 6ˆQ. This implies thattfixesP1;P_jandQ. This is a contradiction.

We consider the case where there exist distinctj;ksuch thath_jandh_kis of order 2. Then,h_j(R₂)ˆR₁,h_j(R₃)ˆR₄andh_j(R₄)ˆR₃. This holds also for h_k. Then h_j ˆh_k on the line PP₂ by Lemma 2(1). Then, h_j(P_k)ˆ h_k(Pk)ˆPk. Since the group ImrPj[PP2] is cyclic, h(Pk)ˆPk for any h2ImrPj[PP2], by Lemma 2(3). If we takejˆ3 andkˆ4, then we have the conclusion, since any element ofG_Pjis a product of elements ofK_Pjand ofQ_Pj.

We prove (4). Letlˆ3. Assume thatP₂6ˆQandQ62C. We may assume thatt2GPsatisfies thatt(Ri)ˆRi‡1foriˆ1;2;3, whereR4ˆR1. We can takeh2GP2such thath(R1)ˆR2, by Fact 1(2). Then, we haveh(R2)ˆR3

andh(R3)ˆR1. This implies thattcoincides withhonPP2, by Lemma 2(1).

Therefore,t(P₂)ˆh(P₂)ˆP₂6ˆQ. This is a contradiction. p LetQ2P²n fPgbe the point such thatg(Q)ˆQfor anyg2GP, as in Proposition 2. We may assume thatQˆ(0:1:0) for a suitable system of coordinates. Then, the linePQˆPP2 is defined byZˆ0. Using Propo- sition 2(1), we can determine the defining equation ofC, as follows.

PROPOSITION 3. The curve C is projectively equivalent to a plane curve whose defining equation is of the form f(x; y)ˆ P

0meamx^pm_l

‡ h(y)ˆ0;wherea_e;. . .;a₀2Kandh(y)2K[y]is a polynomial. Further- more,a_ea₀6ˆ0, the derivativeh⁰(y)is of degreed 2, and polynomialsh(y) andh⁰(y)do not have a common root.

PROOF. Lets2 K_Pand lett2 Q_Pbe a generator, as in Section 2. We may assume thatt(x)ˆzxandtyˆyfort:K(C)!K(C), wherezis a

primitivel-th root of unity. LetAsbe a matrix representings2 KPas in Section 2. SinceLsis defined byZˆ0, the (1;2)-element ofAsis zero. Since the groupK_Pis aFp(z)-vector space, we have a system of basisb1;. . .;bm, where kmˆe. For any s2 K_P, the (1;3)-element of A_s is given by a₁bQ₁‡ ‡a_mb_m for some (a₁;. . .;a_m)2 ^mF_p(z). We define g₀(x)ˆ

(a1;...;am)(x‡S_ia_ib_i), where the subscript (a₁;. . .;a_m)2 ^mF_p(z) is taken over all elements. Let gˆg^l₀. Then, we find easily thatgg(x)ˆg(x) for any element g2GP. Therefore, there exists an elementh(y)2K(y) such thatg(x)‡h(y)ˆ0 inK(C). Then,h(y) is a polynomial of degree at most d by considering the degree of C. On the other hand, the set n P

i a_ib_ija_i2F_p(z)o

K, which consists of all roots ofg₀(x)ˆ0, forms an additive subgroup of K. According to [8, Proposition 1.1.5 and Theorem 1.2.1], the polynomialg₀has only terms of degree equal to some power of p, i.e. g0ˆaex^pe‡ ‡a1x^p‡a0x for some ae;. . .;a02K. Since g0 is separable and has p^eroots, we haveaea06ˆ0.

Finally, we prove that the degree ofh⁰(y) isd 2, andh(y) andh⁰(y) do not have a common root. Sinceh(y) is of degree at mostdˆp^el,h⁰(y) is of degree at most d 2. Let F(X;Y;Z)ˆf(X=Z;Y=Z)Z^d, G₀(X;Z)ˆg₀(X=Z)Z^pe andH(Y;Z)ˆh(Y=Z)Z^d. Then, FX ˆlG^l₀ ¹(a0Z^{pe 1}),FY ˆHY andFZˆ lG^l₀ ¹(a0XZ^{pe 2})‡HZ. We haveFX(X;Y;0)ˆ0. Sincedˆp^el,FY(X;Y;0)ˆ H_Y(Y;0)ˆ0. Assume that h⁰(y) is of degree at most d 3. Then, F_Z(X;Y;0)ˆ0‡H_Z(Y;0)ˆ0. Therefore,Chas singular points on the line defined byZˆ0. This is a contradiction to the smoothness ofC. On the other hand, if there existb2K such thath(b)ˆh⁰(b)ˆ0, then a point (a:b:1)

withg0(a)ˆ0 is a singular point. p

LEMMA 10. Let C be the plane curve given by the equation as in Proposition 3.Then, Q2P²nC and Q6ˆP₂.

PROOF. It follows from Lemma 2(1) and Fact 1(4) thatL_sˆPP₂ for anys2 K_P2. Therefore, any ramified pointR2Cofp_P2withZ6ˆ0 is tame.

Let pQbe the projection from Q. Note that pQ(x:y:1)ˆ(x:1). By the form ofp_Q, ifx x0is a local parameter at (x0;y0)2C, then (x0;y0) is not a ramification point. For a point (x₀;y₀) withf_x(x₀;y₀)ˆlg₀(x₀)^{l 1}6ˆ0,y y₀ is a local parameter. Therefore, ramification points of p_Qin Z6ˆ0 is con- tained in the locus dx

dyˆ h⁰(y)

fx ˆ0, which is equivalent to h⁰(y)ˆ0.

Therefore, there existd 2 linesl₁;. . .;l_{d 2}which containPanddrami-

fication points ofpQ, by Proposition 3. SinceP26ˆP, for any ramification pointRofp_Q, the cardinality of the setP2R\ fR⁰2CjQ2TR⁰Cg P2R\

S

d 2

iˆ1liis at mostd 2 .

Assume that Q2C. It follows from Fact 1(3) that I_Q(C;PQ)ˆd. By Fact 1(3) again,g(Q)ˆQfor anyg2G_P2. LetR2Cbe a ramification point ofp_QinZ6ˆ0. It follows from Lemma 1(2) thatQ2T_RC. Sinceg(Q)ˆQ for any g2G_P2, Q2T_g(R)C for any g2G_P2. Then, the cardinality of C\P2RisdandQ2TR⁰Cfor anyR⁰2C\P2R. This is a contradiction to that the cardinality of P2R\ fR⁰2CjQ2TR⁰Cgis at most d 2. There- fore,Q2P²nC.

Assume thatQ2P²nCandQˆP₂. Then, the setC\PP₂ containsl points, since the group Imr_P[PP₂] is cyclic of orderl. Lett₂2 Q_P2 be a generator and letLt2 be the line defined as in Section 2. Then, the locus Sˆ S

s2KP2

s(Lt2) consists of p^elines. By considering the order of GP2, the ramification locus ofp_Qin the affine planeZ6ˆ0 is contained in the locusS.

Note that the set T

s2KP2

s(Lt2) consists of a unique point, which is not con- tained in C, by Fact 1(3) and that the setC\PP₂ contains two or more distinct points. Since the setC\s(L_t2) consists of exactlydpoints for any s2 KP2, the number of ramification points inZ6ˆ0 is exactlyp^ed. On the other hand, for eachb2K withh⁰(b)ˆ0, there exist exactlydpoints (a;b) such that f(a;b)ˆ0, sinceaea06ˆ0 and h(b)6ˆ0 by Proposition 3.

Therefore,h⁰(y) has exactlyp^eroots. LetRbe a ramification point ofp_Q which is contained in Z6ˆ0. Since R is tame (stated above),e_R is com- puted as the order of dx

dyˆ h⁰(y)

f_x at R plus one. Since eRˆl for any ramification point R2C withZ6ˆ0, the polynomialh⁰(y) is divisible by (y b)^{l 1} if h⁰(b)ˆ0. Therefore, h⁰(y) should be of the form c(y b1)^{l 1} (y bp^e)^{l 1}, which is of degree p^e(l 1). Since h⁰(y) is of degree p^el 2, by Proposition 3, we have p^eˆ2. Since l3 divides

p^e 1ˆ1, this is a contradiction. p

PROOF OF THEOREM 2 (when l3). It follows from Lemma 10 that P26ˆQandQ62C. Ifl5 orlˆ3, then this is a contradiction to Propo- sition 2(2)(4). Assume thatlˆ4. Then, by Proposition 2(3), there exists two distinct outer Galois points P₃;P₄ such that g(P₄)ˆP₄ for any g2G_P3. Then, the pointP₄satisfies the condition of ``Q'' as in Proposition 2(1) for P₃. Then, this is a contradiction to Lemma 10. p

6. Proof of Theorem 2 (The case wherel2)

Letp3, lete1, letl2 and letCbe a smooth plane curve of de- gree dˆp^el4. We denote by L₁P² the line defined byZˆ0. Let P2P²nC be Galois with respect to C. Assume that Pˆ(1:0:0). Let g2G_Pand letA_gbe a 33 matrix representingg. Then,

Agˆ

a₁₁(g) a₁₂(g) a₁₃(g)

0 1 0

0 0 1

0 B@

1 CA;

where a₁₁(g)ˆ 1 and a₁₂(g);a₁₃(g)2K. Then, g(x)ˆa₁₁(g)x‡ aQ12(g)y ‡a13(g). Note that KPˆ fg2GPja11(g)ˆ1g. Let g(x;y):ˆ

s2KP

(x‡a12(s)y‡a13(s)). Note that the set of roots fa12(s)y‡ a₁₃(s)js2 K_Pg K(y) forms an additive subgroup of K(y). Ac- cording to [8, Proposition 1.1.5 and Theorem 1.2.1], g(x;y)2 K[y][x] has only terms of degree equal to some power ofpin variablex.

Therefore, g(x;y)ˆae(y)x^pe ‡ae 1(y)x^{pe 1}‡ ‡a1(y)x^p‡a0(y)x for somea_e(y);. . .;a₀(y)2K[y] with dega_i(y)p^e pⁱforiˆ0;. . .;e. Then, a_e(y)ˆ1 anda₀(y)ˆ Q

s2KPn0(a₁₂(s)y‡a₁₃(s)).

Assume thatlˆ1. Then,K_PˆG_P andg2K(y), sincesgˆgfor any s2G_P. There existsh(y)2K(y) such thatg(x;y)‡h(y)ˆ0 inK(x;y). Let h(y)ˆh1(y)=h2(y), where h1;h22K[y]. Then,g(x;y)h2(y)‡h1(y)ˆ0 on C. Letf(x;y) be a defining polynomial. Then, there existsv(x;y)2K[x;y]

such that f(x;y)v(x;y)ˆg(x;y)h2(y)‡h1(y) as polynomials. Since (1:0:0)62C,f(x;y) has the term of degreep^ein variablex. Comparing the coefficient of degreep^ein variablex, we havev(x;y)2K[y] andv(x;y)ˆ h2(y) up to a constant. Then,h2(y) dividesh1(y) and we haveh(y)2K[y].

Therefore,g(x;y)‡h(y) is a defining polynomial.

LEMMA11. Assume that lˆ1. Then, the defining equation of C is of the form g(x;y)‡h(y)ˆ0, where g(x;y)2K[y][x] has only terms of degree equal to some power of p in variable x.

Assume thatd⁰(C)2. LetP22P²n(C[ fPg) be Galois with respect to C. By taking a suitable system of coordinates, we may assume that P₂ˆ(0:1:0). Then, PP₂ˆL₁. Similar to the previous section, we con- sider the group homomorphismrP:GP !Aut(PP2), which is induced by the restriction. The cardinality of the kernel Ker rPis a power ofp. We denote

it byp^v. Obviously, 0ve. Then, KerrPˆ fs2GPjs(P2)ˆP2g, since P22Lsif and only ifs(P2)ˆP2fors2 KP. Sincea12(s)ˆ0 if and only if s2 K_P,a0(y) is of degreep^e p^v in variabley.

LEMMA12. If lˆ1, then vˆe.

PROOF. W e assume that v5e. Then, the defining polynomial g(x;y)‡h(y) has the term a₀(y)x, which is of degree p^e p^v>0 in variable y. Since P2ˆ(0:1:0) is Galois, the defining polynomial g(x;y)‡h(y) has only terms of degree equal to some power of p in variabley, by Lemma 11. Therefore,p^e p^vˆp^v(p^{e v} 1) is a power of p. Then, p^{e v} 1ˆp^b for some integerb. This impliesbˆ0 and pˆ2.

This is a contradiction. p

By Lemmas 11 and 12, we have a defining equation g(x)‡h(y)ˆ0, whereg;hhave only terms of degree equal to some power ofp. It is not difficult to check thatCis singular. This is a contradiction.

Assume that lˆ2. Let t2GPn KP. Then, t(x;y)ˆ( x‡a12(t)y‡ a13(t);y) for some a12(t);a13(t)2K. Then, GPˆ fstⁱjs2 KP;iˆ0;1g.

Note thatst(x;y)ˆ( x‡(a12(s)‡a12(t))y‡(a13(s)‡a13(t));y). There- fore,^g(x;y):ˆ Q

g2GP

g(x)ˆg(x;y)g( x‡a₁₂(t)y‡a₁₃(t);y)ˆ g²(x;y) g(x;y)g( a12(t)y a13(t);y), since g(x;y) is linear in variable x. Since g^g(x;y)ˆ^g(x;y) for any g2GP, there exists h(y)2K(y) such that f(x;y):ˆ^g(x;y)‡h(y)ˆ0 inK(x;y). Then,h(y) is a polynomial andf(x;y) is a defining polynomial (similarly to the caselˆ1).

LEMMA13. Assume that lˆ2. Then, the defining equation of C is of the form g²(x;y)‡g(x;y)g(ay‡b;y)‡h(y)ˆ0, where a;b2K and g2K[y][x] has only terms of degree equal to some power of p in variable x.

We considerP₂ˆ(0:1:0). It follows from Lemma 13 that there exist polynomials g₁(x;y)2K[x][y] and h₁(x)2K[x] such that g₁(x;y) has only terms of degree equal to some power of p in variable y and f1(x;y):ˆg²₁(x;y)‡g1(x;y)g1(x;cx‡d)‡h1(x) is a defining polynomial of C for some c;d2K. Let g1(x;y)ˆb_e(x)y^pe‡b_{e 1}(x)y^{pe 1}‡ ‡b₀(x)y, whereb_e(x)ˆ1 andb_{e 1}(x);. . .;b₀(x)2K[x]. Sincef(x;y) andf₁(x;y) are defining polynomials ofC, we havecf(x;y)ˆf₁(x;y) for somec2K.

LEMMA14. If lˆ2, then vˆe.

PROOF. Assume thatv5e. Firstly we prove thatpˆ3 andvˆe 1.

Now,a0(y) is of degreep^e p^v>0. Considering the polynomialsg²(x;y), g(x;y)g(ay‡b;y) andh(y),f(x;y) has the terma²₀(y)x², which is of degree 2(p^e p^v) in variable y. We consider this term for f1(x;y). Since g₁(x;y)g₁(x;cx‡d)ˆP

i b_i(x)g₁(x;cx‡d)y^pi has only terms of degree equal to some power ofpin variableyand 2(p^e p^v) is not a power ofpin p>2, the term of the highest degree of a²₀(y)x² does not appear here.

Therefore, this term should appear ing²₁(x;y) (up to a constant). Since the polynomialg²₁(x;y)ˆP

i;jb_i(x)b_j(x)y^pi‡pjhas only terms of degreepⁱ‡p^j ˆ pⁱ(1‡p^{j i}) withijand 0i;jein variabley, we have 2p^v(p^{e v} 1)ˆ pⁱ(1‡p^{j i}) for some i;j with ij. Then, we should have iˆv and 2(p^{e v} 1)ˆ1‡p^{j i}. This implies that 2p^{e v} p^{j i}ˆ3. If jˆi, then pˆ2. This is a contradiction. If j6ˆi, then p^{j i}(2p^{e v j‡i} 1)ˆ3. We should havepˆ3,j iˆ1 andp^{e v 1}ˆ1. Sinceiˆvandi5jas above, iˆvˆe 1 and jˆe.

Secondly we prove thatpˆ3, eˆ1 andvˆ0. Note thatp^e p^{e 1} ˆ 2p^{e 1} in pˆ3. Since the polynomialg²(x;y)ˆP

i;ja_i(y)a_j(y)x^pi‡pj has the term 2a₀(y)x^pe‡1, which is of degree 2p^{e 1}in variabley, and the polynomial g₁(x;y)g₁(x;cx‡d) has only terms of degree equal to some power ofpin variable y, the term of the highest degree 2p^{e 1}‡p^e‡1 of 2a0(y)x^pe‡1 appears in g²₁(x;y) (up to a constant). Since g²₁(x;y)ˆP

i;jb_i(x)b_j(x)y^pi‡pj, and pⁱ‡p^jˆ2p^{e 1} implies that iˆjˆe 1,b²_{e 1}(x)y^{2pe 1} has the term of the highest degree 2p^{e 1}‡p^e‡1 of 2a₀(y)x^pe‡1. Letkbe the degree of b_{e 1}(x). Sinceb_{e 1}(x) has the term of degree at least (p^e‡1)=2, we have (p^e‡1)=2kp^e p^{e 1}ˆ2p^{e 1}. Then, b_{e 1}(x)b₀(x)y^{pe 1‡1} is of degree k‡(p^e p^{e 1})ˆk‡2p^{e 1} in variable x. Since (p^e‡1)=2‡(p^e p^{e 1})ˆ p^e‡(p^e 2p^{e 1}‡1)=2ˆp^e‡(p^{e 1}‡1)=2, the term of the highest degree ofb_{e 1}(x)b₀(x)y^{pe 1‡1}appears ing²(x;y). Sinceg²(x;y)ˆP

i;ja_i(y)a_j(y)x^pi‡pj, k‡(p^e p^{e 1})ˆpⁱ¹‡p^j1 for some i1j1. Since p^e‡(p^{e 1}‡1)=2 k‡(p^e p^{e 1})ˆpⁱ¹‡p^j1, we have j1ˆe and i1ˆe 1. Therefore, kˆ2p^{e 1}ˆp^e p^{e 1}ˆp^e p^v. W e have e 1ˆ0, since degb_i(x)ˆ p^e p^vif and only ifiˆ0.

Finally, we consider the remaining case where pˆ3, eˆ1 and vˆ0. Then, g(x;y)ˆx³ (ay‡b)²x and g1(x;y)ˆy³ (a1x‡b₁)²y for some a;b;a1;b₁2K with a;a16ˆ0. Note that g(ay‡b)ˆ (ay‡b)((a‡a)y‡(b‡b))((a a)y‡(b b)). Since f(x;y)ˆg²(x;y)‡

g(ay‡b;x)g(x;y)‡h(y), the coefficient ofxy⁵off(x;y) is equal to the one of g(ay‡b;y), which is equal toa²a(a‡a)(a a). Ifa(a‡a)(a a)6ˆ0, then the coefficient of xy⁵ of f1(x;y) is not zero. However, by considering g²₁(x;y)‡g₁(x;cx‡d)g₁(x;y), that is zero. Therefore, we should have a(a‡a)(a a)ˆ0. Then,g(ay‡b) is of degree at most two. Ifg(ay‡b;y) is of degree at most one, then two of the three conditionsaˆ0,a‡aˆ0 and a aˆ0 hold. Then, we haveaˆaˆ0. This is a contradiction toa6ˆ0.

Thereforeg(ay‡b) is of degree two. Then the coefficient ofx³y²off(x;y) is not zero. Sincey²appears only ing²₁(x;y) orh₁(y) forf₁(x;y) andg²₁(x;y)ˆ y⁶ 2(a1x‡b₁)y⁴‡(a1x‡b₁)²y², the coefficient ofx³y²off1(x;y) is zero.

This is a contradiction. p

By Lemma 14, we have p^vˆp^e. Then, g(x;y)2K[x] and g₁(x;y)2 K[y]. We denote g(x;y) by g(x) and g₁(x;y) by g₁(y). We have f(x;y)ˆg²(x)‡ g(x)(g(ay)‡g(b))‡l₁g²₁(y)‡l₂ for somel₁;l₂2K. Let G(X;Z)ˆ Z^peg(X=Z) and let G1(Y;Z)ˆZ^peg1(Y=Z). Then, F(X;Y;Z)ˆ Z^2pef(X=Z;Y=Z) ˆG²(X;Z)‡G(X;Z)(G(aY;Z)‡g(b)Z^pe)‡l1G²₁(Y;Z)‡ l₂Z^2pe. Let a (resp. b) be the coefficient of XZ^{pe 1} (resp. YZ^{pe 1}) for G(X;Z) (resp. G₁(Y;Z)). Then, F_Xˆ2G(X;Z)aZ^{pe 1}‡aZ^{pe 1}(G(aY;Z)‡

g(b)Z^pe), F_YˆaaZ^{pe 1}G(X;Z)‡2l₁G₁(Y;Z)bZ^{pe 1} and F_Zˆ 2G(X;Z) aXZ^{pe 2} aXZ^{pe 2}(G(aY;Z)‡g(b)Z^pe)‡G(X;Z)( abYZ^{pe 2}) 2l1G1(Y;Z) bYZ^{pe 2}. Therefore, FX(X;Y;0)ˆFY(X;Y;0)ˆFZ(X;Y;0)ˆ0 and we have singular points on the lineL1.

We have the assertion of Theorem 2.

Acknowledgements. The author is grateful to Professor Masaaki Homma for helpful discussions. The parameterization as in Remark 3 appeared during discussions with Homma. The author thanks Professor Hisao Yoshihara for helpful comments. The author also thanks the referee for helpful comments, by which the author could improve several proofs.

The author was partially supported by Grant-in-Aid for Young Scientists (B) (22740001), MEXT and JSPS, Japan.

REFERENCES

[1] E. ARBARELLO- M. CORNALBA- P. A. GRIFFITHS- 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), pp. 185- 189.