Galois points for a plane curve in characteristic two

Journal of Pure and Applied Algebra 218 (2014) 343–353

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Galois points for a plane curve in characteristic two Satoru Fukasawa Department of Mathematical Sciences, Faculty of Science, Yamagata University, Kojirakawa-machi 1-4-12, Yamagata 990-8560, Japan

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Article history: Received 30 January 2012 Received in revised form 19 May 2013 Available online 26 June 2013 Communicated by S. Iyengar MSC: 14H50; 12F10; 14H05

abstract Let C be an irreducible plane curve. A point P in the projective plane is said to be Galois with respect to C if the function field extension induced by the projection from P is Galois. We denote by δ ′ (C ) the number of Galois points contained in P2 \ C . In this article we will present two results with respect to determination of δ ′ (C ) in characteristic two. First we determine δ ′ (C ) for smooth plane curves of degree a power of two. In particular, we give a new characterization of the Klein quartic in terms of δ ′ (C ). Second we determine δ ′ (C ) for a generalization of the Klein quartic, which is related to an example of Artin–Schreier curves whose automorphism group exceeds the Hurwitz bound. This curve has many Galois points. © 2013 Elsevier B.V. All rights reserved.

1. Introduction Let K be an algebraically closed field of characteristic p ≥ 0 and let V ⊂ PKn+1 be an irreducible hypersurface of degree d ≥ 4. In 1996, H. Yoshihara introduced the notion of Galois point (see [12,16] or survey paper [4]). If the function field extension K (V )/K (Pn ), induced by the projection πP : V 99K Pn from a point P ∈ Pn+1 , is Galois, then the point P is said to be Galois with respect to V . When a Galois point P is contained in Pn+1 \ V (resp. in V ), we call P an outer (resp. inner) Galois point. We denote by δ ′ (V ) the number of outer Galois points for V . It is remarkable that many classification results of algebraic varieties have been given in the theory of Galois points (see, for example, [15–18]). Yoshihara determined δ ′ (C ) for any smooth plane curve C in characteristic p = 0 [16]. In characteristic p > 0, M. Homma [10] settled δ ′ (H ) for the Fermat curve H of degree pe + 1. Recently, the present author determined δ ′ (C ) except for the case where d > p and d is divisible by p [3]. The following problem remains open (see also [4, Problem 2]). Problem 1. Let C ⊂ P2K be a smooth plane curve C of degree d ≥ 4 over an algebraically closed field K of characteristic p. If p > 0, p divides d and d > p, determine the number of outer Galois points δ ′ (C ) for C . In this article, we settle this problem if p = 2 and d = 2e . We consider an equation

(x2 + x)2 + (x2 + x)(y2 + y) + (y2 + y)2 + λ = 0,

(1)

where λ ∈ K \ {0}, in characteristic p = 2. When we refer to ‘‘the defining equation’’ of C in this article, we are only doing so on the affine patch Z ̸= 0 and that C is the closure in P2 . It is known that the curve C defined by (1) with λ = 1 is projectively equivalent to the Klein quartic (see [11, p. 18]). The first main theorem is as follows. Theorem 1. Let p = 2, e ≥ 2, and C be a smooth plane curve of degree d = 2e . Then, δ ′ (C ) = 0, 1, 3 or 7. Furthermore, we have the following. (a) δ ′ (C ) = 7 if and only if d = 4 and C is projectively equivalent to the Klein quartic.

E-mail address: [email protected]. 0022-4049/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.jpaa.2013.06.006

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(b) δ ′ (C ) = 3 if and only if d = 4 and C is projectively equivalent to the curve defined by Eq. (1) with λ ̸= 0, 1. In this case, the three Galois points are collinear. Theorem 1 is a step in the complete determination of δ ′ (C ) for positive characteristic [6]. Since curves C satisfying δ ′ (C ) > 1 are rare [19], we give a family of curves generalizing those with Eq. (1) and δ ′ (C ) arbitrarily large in Eq. (2). Let v ≥ 1, let q = 2v and let v

g (x) := x2 + αv−1 x2

v−1

+ · · · + α1 x2 + x,

where αv−1 , . . . , α1 ∈ K . We consider the curve of degree 2q defined by g 2 (x) + g (x)g (y) + λ1 g 2 (y) + λ0 = 0,

(2)

where λ0 ∈ K \ {0} and λ1 ∈ ( {i|αi ̸=0} F2i ) ∩ F2v . In particular, Eq. (2) with (v, λ1 ) = (1, 1) is the same as Eq. (1). In this article, we distinguish a particular type of Galois point.



Definition 1. A Galois point P is said to be extendable if any birational map from C to itself induced by the Galois group GP at P is the restriction of a linear transformation. We denote by δ0′ (C ) the number of extendable outer Galois points. All Galois points for a smooth plane curve are extendable Galois (see [1, Appendix A, 17 and 18] or [2]). Obviously

δ0′ (C ) ≤ δ ′ (C ). We will prove the following for δ0′ (C ) ≥ 2.

Theorem 2. Let p = 2, let e ≥ 2 and let C ⊂ P2K be an irreducible curve of degree d = 2e . We have δ0′ (C ) ≥ 2 if and only if C is projectively equivalent to an irreducible plane curve given by one of the following equations: (a)

e

i=0

i

αi x2 +

e

j=0

j

βj y2 = 0, where αi , βj ∈ K . e−1

2 2 (b) g 2 (x) + g (x)g (y) + λ1 g (y) + λ0 = 0, where g (x) = x λ1 ∈ ( {i|αi ̸=0} F2i ) ∩ F2v \ {0}.

e−2

+ αe−2 x2

+ · · · + α1 x2 + x, αi ∈ K , λ0 ∈ K \ {0} and

Furthermore, if C is smooth, then the defining equation is of the form (b) and d = 4. Remark 1. According to [5, Proposition 2], δ0′ (C ) = ∞ for the irreducible plane curve C with equation (a) in Theorem 2. The curve with Eq. (2) and λ1 ̸= 0 is characterized by the condition that d is a power of 2 and 2 ≤ δ0′ (C ) < ∞. We give more details on the curve with Eq. (2) as follows. Theorem 3. Let p = 2, let e ≥ 2 and let α ∈ K satisfy α 2 + α + λ1 = 0. Assume that λ0 ̸= 0 and (v, λ1 ) ̸= (1, 1). Let C be the plane curve given by Eq. (2). We denote by F2w the finite field ( {i|αi ̸=0} F2i ) ∩ F2v . Then, we have the following. (a) The curve C is irreducible and has two singular points. (b) The geometric genus satisfies (q − 1)2 ≤ pg (C ) < (2q − 1)(q − 1) and (q − 1)2 = pg (C ) ⇔ α ∈ F2w . (c) If α ∈ F2w (resp. α ̸∈ F2w ), then δ0′ (C ) = 2w − 1 (resp. δ0′ (C ) = 2w + 1). Moreover, the extendable outer Galois points lie on the line containing the two singular points. (d) w = v ⇔ g (x) = xq + x and in this case δ ′ (C ) = δ0′ (C ). Example 1. There exist curves C satisfying δ0′ (C ) = 2u − 1 (resp. 2u + 1) for each positive u dividing v . Indeed, let  u v α 2 +α+λ1 = 0 and take g (x) = x2 + x2 + x if u < v . Then, F2w = {i|αi ̸=0} F2i ∩ F2v = F2u . Let Φ : F2u → F2u ; X → X 2 + X ,

which is a group morphism of additive groups. Since the kernel is F2 , the order of ImΦ is 2u−1 . If we take λ1 ∈ ImΦ , for example λ1 = 0, then α ∈ F2u and we have δ0′ (C ) = 2u − 1. If we take λ1 ∈ F2u \ ImΦ , then α ̸∈ F2u and we have δ0′ (C ) = 2u + 1. Example 2. Taking g (x) = xq + x and λ1 = 0 and making the variable change Y = x + y we obtain the equation

(xq + x)(Y q + Y ) + λ0 = 0 studied by D. Subrao [14], [9, Example 11.89], and this is known as an example of an irreducible (ordinary) curve whose automorphism group exceeds the Hurwitz bound under certain assumptions.

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2. Preliminaries Let the characteristic of K be p ≥ 0 and let C ⊂ P2K be an irreducible plane curve of degree d ≥ 4. We introduce a system (X : Y : Z ) of homogeneous coordinates on P2 with local coordinates x = X /Z , y = Y /Z for the affine piece Z ̸= 0. Note that if F (X , Y , Z ) is the defining polynomial on P2 , then f (x, y, 1) is the defining polynomial on the open affine space Z ̸= 0, and that K (C ) is the fraction field of K [x, y]/(f (x, y)), so that K (C ) = K (x, y), where x, y are the images of x, y. We denote by Sing(C ) the singular locus of C and by π : Cˆ → C the normalization of C . For a point R ∈ C \ Sing(C ), we denote by TR C ⊂ P2 the (projective) tangent line at R. For a projective line ℓ ⊂ P2 and a point R ∈ C ∩ ℓ, we denote by IR (C , ℓ) the intersection multiplicity of C and ℓ at R. We denote by PR the line passing through points P and R when P ̸= R and by πP : C 99K P1 the projection from a point P ∈ P2 . We can identify P1 with the set of all lines passing through P, and the point πP (R) ∈ P1 with the point corresponding to the line PR if P ̸= R. We denote by πˆ P the composition map πP ◦ π : Cˆ → P1 . If Rˆ ∈ Cˆ , we denote ˆ Further, if πˆ P is separable, we denote by d ˆ the different exponent (i.e. the degree of by eRˆ the ramification index of πˆ P at R. R

the ramification divisor) of πˆ P at Rˆ (see [9, Section 7.3], [13, III. 4, III. 5]). If R ∈ C \ Sing(C ) and π (Rˆ ) = R, then we use the same symbols eR for eRˆ and dR for dRˆ , by abuse of terminology. We note the following elementary fact. Fact 1. Let P ∈ P2 \ C , let Rˆ ∈ Cˆ and let π (Rˆ ) = R. Let h = 0 be a local equation for the line PR in a neighborhood of R. Then, for πP we have the following. (a) eRˆ = ordRˆ (π ∗ h). In particular, if R is smooth, then eR = IR (C , PR).

ˆ If πˆ P is separable, then d ˆ = ord ˆ dπ h . (b) Let t be a local parameter at R. R R dt ∗

Proof. We may assume that P = (1 : 0 : 0) and R = (x0 : y0 : 1). Then, we also may assume that h = y − y0 . Since πP (x : y : 1) = (y : 1), πP (R) = (y0 : 1). Then, the function y − y0 is a local parameter at (y0 : 1) ∈ P1 . The pull-back of y − y0 for πˆ P is π ∗ πP∗ (y − y0 ) = π ∗ h. Then, the ramification index at Rˆ is equal to ordRˆ π ∗ h, and the degree of the ramification divisor at Rˆ is calculated as ordRˆ dπdt h (see [8, IV. Proposition 2.2], [9, Section 7.3]). We have the conclusion.  ∗

If a Galois covering θ : C → C ′ between smooth curves is given, then the Galois group G acts on C naturally. We denote by G(R) the stabilizer subgroup of R ∈ C . The following fact is useful in finding Galois points (see [13, III. 7.1, 7.2 and 8.2]). Fact 2. Let θ : C → C ′ be a Galois covering of degree d with Galois group G and let R, R′ ∈ C . Then we have the following. (a) (b) (c) (d) (e)

For any σ ∈ G, we have θ (σ (R)) = θ (R). If θ(R) = θ (R′ ), then there exists an element σ ∈ G such that σ (R) = R′ . The order of G(R) is equal to eR at R for any point R ∈ C . If θ(R) = θ (R′ ), then eR = eR′ . The index eR divides the degree d. Finally in this section, we note the following facts about automorphisms of P1 .

Fact 3. Let p > 0. We denote by Aut(P1 ) the automorphism group of P1 . Then, we have the following. (a) Let P1 , P2 , P3 ∈ P1 be three distinct points and let σ1 , σ2 ∈ Aut(P1 ). If σ1 (Pi ) = σ2 (Pi ) for i = 1, 2, 3, then σ1 = σ2 . (b) Each σ ∈ Aut(P1 ) of order p has a unique fixed point. Proof. Fact (a) is easily proved, if we use the classical fact that any automorphism of P1 is a linear transformation. We prove (b). We can take a matrix Aσ whose p-th power Apσ is identity representing σ . Since the eigenvalues satisfy λp = 1, hence λ = 1, and Aσ is the identity if and only if the eigenspace has dimension 2.  3. Galois points yielding extendable automorphisms In this section, we consider an irreducible (maybe singular) plane curve C of degree d with an extendable outer Galois point. The purpose of this section is to prove Theorem 2. Let P ∈ P2 be a Galois point. Note in general that there is a bijection between (a) elements of Gal(K (C )/πP∗ K (P1 ))

(b) birational maps C 99K C which commute with πP : C 99K P1 and (c) isomorphisms Cˆ → Cˆ which commute with πP ◦ π . We identify these sets and denote them by GP . When we use the symbol σ for an automorphism of the curve Cˆ , we use the symbol σ ∗ for the automorphism of the function field K (C ) corresponding to σ . When we have a Galois point and a linear transformation, we have the following. Lemma 1. Let P ∈ P2 and let φ ∈ Aut(P2 ). If P is Galois (resp. extendable Galois) for C , then the point φ(P ) is Galois (resp. extendable Galois) for φ(C ). Proof. The automorphism group φ GP φ −1 becomes the Galois group at φ(P ) with respect to φ(C ). If P is extendable Galois, then any element of φ GP φ −1 is a linear transformation. 

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Let char K = p > 0 and let C be a curve of degree d = pe ≥ 4. Assume P is extendable outer Galois, then change coordinates so that P = (1 : 0 : 0). The projection πP : C → P1 is given by (x : y : 1) → (y : 1). We have a field extension K (C )/K (P1 ) = K (x, y)/K (y) via πP . By assumption, the extension K (x, y)/K (y) is Galois and each σ ∈ GP is the restriction of a linear transformation σ˜ . Let Aσ˜ = (aij ) be a 3 × 3 matrix representing σ˜ . Since σ ∈ GP , σ ∗ (y) = y. Then, (a21 x + a22 y + a23 ) − (a31 x + a32 y + a33 )y = 0 in K (x, y). Since d ≥ 4, we have a21 = a23 = a31 = a32 = 0 and a22 = a33 . We take the representative matrix with a22 = a33 = 1. We prove the uniqueness of the matrix Aσ˜ with a22 = a33 = 1. Let A = (aij ), B = (bij ) represent σ . Then, a11 x + a12 y + a13 = b11 x + b12 y + b13 on the curve C . Since d ≥ 4, we have a11 = b11 , a12 = b12 and a13 = b13 . Therefore, the uniqueness follows. Then, we have a well-defined group homomorphism e

pe

˜ P the image. Since σ p = 1, we have a11 = 1. We have a11 = 1 since the characteristic is p. GP ↩→ Aut(P2 ). We denote by G Therefore, σ˜ is represented by a matrix  Aσ˜ =

1 0 0

a12 (σ ) 1 0

a13 (σ ) 0 1



and σ ∗ (x) = x + a12 (σ )y + a13 (σ ) for some a12 (σ ), a13 (σ ) ∈ K . Let g0 (x, y) :=



σ ∈GP (x

+ a12 (σ )y + a13 (σ )), which is

˜ P is abelian of order pe and σ˜ p = 1 for any automorphism σ˜ ∈ G˜ P , we have GP ∼ contained in K [y][x]. Since G = (Z/pZ)⊕e . Note that the set of roots {a12 (σ )y + a13 (σ )} ⊂ K (y) forms an additive subgroup of K (y). According to [7, Proposition 1.1.5 and Theorem 1.2.1], g0 (x, y) has only terms of degree a power of p in variable x. Since σ ∗ g0 (x, y) = g0 (x, y) for any σ ∈ GP , we have g0 (x, y) ∈ K (y). There exists h(y) ∈ K (y) such that g0 (x, y) + h(y) = 0 in K (x, y). Let h(y) = h1 (y)/h2 (y), where h1 , h2 ∈ K [y]. Then, g0 (x, y)h2 (y) + h1 (y) = 0 on C . Let f (x, y) be a defining polynomial. Then, there exists v(x, y) ∈ K [x, y] e such that f (x, y)v(x, y) = g0 (x, y)h2 (y) + h1 (y) as polynomials. Since (1 : 0 : 0) ̸∈ C , the xp term of f (x, y) has non-zero e coefficient. Therefore, f (x, y) has the term of degree p in variable x. Comparing the coefficient of degree pe in variable x, we have v(x, y) ∈ K [y] and v(x, y) = h2 (y) up to a constant. Then, h2 (y) divides h1 (y) and we have h(y) ∈ K [y]. Therefore, g0 (x, y) + h(y) is a defining polynomial. Proposition 1. Let C ⊂ P2K be irreducible of degree pe , where the characteristic p is positive. Assume that P = (1 : 0 : 0) is an   extendable outer Galois point with Galois group GP and let g0 (x, y) = σ˜ ∈G˜ P σ˜ ∗ x = σ ∈GP (x + a12 (σ )y + a13 (σ )). Then, we have the following. (a) GP ∼ = (Z/pZ)⊕e . (b) g0 (x, y) ∈ K [y][x] has only terms of degree a power of p in variable x. (c) The defining equation of C is of the form g0 (x, y) + h(y) = 0. While the following Lemma is a corollary of Fact 2(a)(b) for smooth curves and their points, we generalize this for singular curves if we have an extendable Galois point P.

˜ P ⊂ Aut(P2 ) defined above. Lemma 2. Let P be extendable outer Galois with G ˜ P and any line ℓ passing through P, σ˜ (ℓ) = ℓ. In particular, for any σ˜ ∈ G˜ P and R ∈ P2 , πP (σ˜ (R)) = πP (R). (a) For any σ˜ ∈ G ˜ P acts transitively on C ∩ ℓ for each line P ∈ ℓ. (b) G Proof. We prove assertion (a). It follows from the form of Aσ˜ that σ˜ (ℓ) ⊂ ℓ. Since σ˜ (ℓ) is a line, we have σ˜ (ℓ) = ℓ. We prove assertion (b). Let ℓ be a line with P ∈ ℓ and let R, R′ ∈ C ∩ ℓ. We have the commutative diagram

σ

Cˆ

→

↓ Cˆ

→

C

↩→

P2

↩→

P2

↓ C

↓

σ˜

ˆ Rˆ ′ ∈ Cˆ be points with π (Rˆ ) = R and π (Rˆ ′ ) = R′ . It follows from Fact 2(b) that there exists σ ∈ GP such for any σ ∈ GP . Let R, that σ (Rˆ ) = Rˆ ′ . Since π ◦ σ = σ˜ ◦ π by the commutative diagram above, σ˜ (R) = R′ .  ˜ P \ {1}, there exists a unique line Lσ such that σ˜ (R) = R for each R ∈ Lσ , which is defined by For each element σ˜ ∈ G a12 (σ )Y + a13 (σ )Z = 0. We have P ∈ Lσ . It follows from Fact 3(b) that Lσ = PR if R ̸= P and σ˜ (R) = R. Assume that δ0′ (C ) ≥ 2. Let P1 ̸= P be an extendable outer Galois point with respect to C . By taking a suitable system of

˜ P → Aut(PP1 ) coordinates, we may assume that P1 = (0 : 1 : 0). We consider the group homomorphism rP = rP [PP1 ] : G ˜P : defined by the restriction, which is well-defined by Lemma 2(a). Then, the kernel KP = KP [PP1 ] := {σ ∈ G σ fixes each point on PP1 } of rP [PP1 ] coincides with the set {σ ∈ G˜ P : σ (P1 ) = P1 } because of Fact 3(b). In particular, 1 ̸= σ ∈ KP ⇔ Lσ = PP1 . We denote by pv the order of KP . Obviously, 0 ≤ v ≤ e. We denote by CP the image of rP . Then, CP ∼ = (Z/pZ)⊕e−v . Hereafter, we assume p = 2. We denote by L∞ the line PP1 . Proposition 2. If p = 2, then e − v ≤ 1.

S. Fukasawa / Journal of Pure and Applied Algebra 218 (2014) 343–353

347

e−v Proof. Assume that p = 2 and e −v ≥ 2. Now we have CP ∼ = (Z/2Z)⊕e−v . Let CP = {τi }i2=1 , let σ˜i ∈ G˜ P satisfy rP (σ˜ i ) = τi for 2e−v each i, and let σ˜ 1 = 1. We take Pi = σ˜ i (P1 ). Then, by Lemma 2(a), we have {Pi }i=1 ⊂ L∞ . Let R ∈ C ∩ L∞ and let Ri = σ˜ i (R). e−v Then, by Lemma 2(a)(b), we have {Ri }2i=1 = C ∩ L∞ .

We prove that Pi ̸= Pj and Ri ̸= Rj if i ̸= j. Assume that Pi = Pj . Then, (τj−1 ◦ τi )(P ) = (σ˜ j−1 ◦ σ˜ i )(P ) = P and

(τj

−1

◦ τi )(P1 ) = (σ˜ j−1 ◦ σ˜ i )(P1 ) = P1 . It follows from Fact 3(b) that τj = τi . Therefore, we have Pi ̸= Pj . Similarly, we have

Ri ̸= Rj . Since Pi ∈ L∞ is extendable Galois for each i by Lemma 1, we have at least 2e−v + 1 ≥ 5 extendable Galois points on the e−v line L∞ . We also have the set {Ri }2i=1 = C ∩ L∞ consisting of exactly 2e−v ≥ 4 points. For each i with 1 ≤ i ≤ 2e−v + 1,

˜ Pi such that η˜ i (R1 ) = R2 , by Lemma 2(b). Since the order of η˜ i is two, η˜ i (R2 ) = R1 . Since the number of there exists η˜ i ∈ G elements of {η˜ i } is 2e−v + 1 and an inequality 2e−v − 2 < 2e−v + 1 holds, there exist i, j with i ̸= j such that η˜ i (R3 ) = η˜ j (R3 ). Then, η˜ i |L∞ = η˜ j |L∞ , by Fact 3(a). Hence η˜ i (Pj ) = η˜ j (Pj ) = Pj . It follows from Fact 3(b) that η˜ i is identity on L∞ . This is a contradiction. Therefore, we have e − v ≤ 1.  Proposition 3. If e − v = 0, then C is projectively equivalent to one with the equation of Theorem 2(a) and C intersects L∞ in a single point, which is the only singularity of C . Proof. Assume that e − v = 0. Then, KP = GP and Lσ = L∞ for any σ ∈ GP ∗. SinceLσ is defined by a12 (σ )Y + a13 (σ )Z = 0 and L∞ is defined by Z = 0, we have a12 (σ ) = 0 and g0 (x, y) = ˜ x = σ ∈GP (x + a13 (σ )) ∈ K [x]. On the other σ˜ ∈G˜ P σ

ξ˜ ∗ y ∈ K [y] for P1 = (0 : 1 : 0). Then, by Proposition 1, the defining equation is of the form g0 (x) + h0 (y) + c = 0, where c ∈ K , e e 2i 2j g0 (x) = i=0 αi x and h0 (y) = j=0 βj y for some αi , βj ∈ K . We take a constant γ ∈ K such that g0 (γ ) + c = 0. Then the polynomial gˆ0 (x) := g0 (x + γ ) satisfies gˆ0 (x) = g0 (x) + g0 (γ ) = g0 (x) + c from the additive property of g0 . Making the coordinate change x → x − γ , we may assume that c = 0. e e e e Let F (X , Y , Z ) = Z 2 g0 (X /Z ) + Z 2 h0 (Y /Z ). Then, the single point Q of C ∩ L∞ is given by Z = αe X 2 + βe Y 2 = 0. e e We have differentials ∂ F /∂ X = α0 Z 2 −1 , ∂ F /∂ Y = β0 Z 2 −1 and ∂ F /∂ Z = (2e − 1)Z 2e−2 (α0 X + β0 Y ). Then, we have ∂ F /∂ X = ∂ F /∂ Y = ∂ F /∂ Z = 0 at Q and no singular points in the affine plane Z ̸= 0.  hand, by Lemma 2(b), C ∩ L∞ consists of a unique point Q . This implies that KP1 = GP1 and h0 (x, y) :=



ξ˜ ∈GP1

Proposition 4. If e − v = 1, then C is projectively equivalent to one with the equation of Theorem 2(b) and C is singular if d > 4. Proof. Assume that e − v = 1. For any automorphism σ˜ ∈ KP , a12 (σ ) = 0, since the fixed line Lσ = L∞ . We take an element τ˜ ∈ G˜ P \ KP . Then, τ˜ is represented by a matrix

 Aτ˜ =

1 0 0

a12 (τ ) 1 0

a13 (τ ) 0 1



for some a12 (τ ), a13 (τ ) ∈ K with a12 (τ ) ̸= 0. If we take the linear transformation φ(X : Y : Z ) = (X : a12 (τ )Y +a13 (τ )Z : Z ), then φ(P ) = (P ), φ(P1 ) = P1 and φ −1∗ (a12 (τ )y + a13 (τ )) = y. Therefore, we may assume that a12 (τ ) = 1 and a13 (τ ) = 0. ˜ P , since KP is a subgroup of index two in GP and τ represents the non-identity coset. Let g (x) := Then, KP ∪ τ˜ KP = G σ˜ ∈KP (x + a13 (σ )). It follows from [7, Proposition 1.1.5 and Theorem 1.2.1] that g (x)

v

v−1

= x2 + αv−1 x2 + · · · + α1 x2 + α0 x for some αv−1 , . . . , α0 ∈ K . Since g is separable, α0 ̸= 0. Note that g has the additive property g (x + x′ ) = g (x) + g (x′ ).  ˜ P = KP ∪ τ˜ KP , we have σ˜ ∈G˜ σ˜ ∗ x = g (x)g (x + y). By Proposition 1, there exists a polynomial h(y) of degree at most Since G P d = 2e with h(0) = 0 and a constant c ∈ K such that f (x, y) := g (x)g (x + y) + h(y) + c = 0 gives a defining equation of C . 

Then, we have f (x, y) = g 2 (x) + g (x)g (y) + h(y) + c from the additive property of g. ˜ P1 . We can define KP1 similarly. Any automorphism ξ˜ ∈ KP1 is represented by a matrix We consider the action by G

 Bξ˜ =

1 0 0

0 1 0

0



b23 (ξ ) 1

for some b23 (ξ ) ∈ K , since the fixed line Lξ = L∞ . There are exactly two points in C ∩ L∞ , by CP ∼ = Z/2Z and Fact 3(b) and Lemma 2(b). Since any automorphism of P1 fixing P and both points C ∩ L∞ is identity by Fact 3(a), we have CP1 ∼ = Z/2Z. Now we have

 1 B− Aτ˜ Bξ˜ ξ˜

=

1 0 0

1 1 0

b23 (ξ ) 0 1

 .

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Since ξ˜ −1 τ˜ ξ˜ commutes with the projection from P by this expression, we have ξ˜ −1 τ˜ ξ˜ ∈ G˜ P . Then, g (b23 (ξ )) = 0. Let g2 (y) = ξ˜ ∈KP1 (y + b23 (ξ )). Then, g2 (y) = g (y), since both are monic with the same distinct roots. We take an element

η˜ ∈ G˜ P1 \ KP1 . Then, η˜ is represented by a matrix   Bη˜ =

1 b21 (η) 0

0 1 0

0 b23 (η) 1

for some b21 (η), b23 (η) ∈ K with b21 (η) ̸= 0. Since g2 (y) = g (y) and GP1 = KP1 ∪ η˜ KP1 , by Proposition 1, the polynomial f2 (x, y) := g (y)g (y + b21 (η)x + b23 (η)) + h2 (x) + c2 , where h2 (x) is a polynomial with h2 (0) = 0 and c2 ∈ K , is a defining one of C . Then, kf (x, y) = f2 (x, y) = g 2 (y) + g (y)g (b21 (η)x) + g (b23 (η))g (y) + h2 (x) + c2 for some k ∈ K . Then, we have (i) (ii) (iii) (iv)

kg 2 (x) = h2 (x), kg (x) = g (b21 (η)x), kh(y) = g 2 (y) + g (b23 (η))g (y), and kc = c2 . i

In the relation (ii), by considering the non-zero terms of g, we have k = b21 (η)2 for any i, where 2i is the degree of a non-zero term of g. In particular, we have k = b21 (η). Therefore, we have the equation of the form f2 (x, y) = g 2 (y) + b21 (η)g (y)g (x) + g (b23 (η))g (y) + b21 (η)g 2 (x) + c2 = 0, where b21 (η) satisfies g (b21 (η)x) = b21 (η)g (x). Note that, for γ ∈ K with b21 (η)g (γ ) + g (b23 (η)) = 0, f2 (x + γ , y) = g 2 (y) + b21 (η)g (y)g (x) + b21 (η)g 2 (x) + b21 (η)g 2 (γ ) + c2 . Dividing f2 (x + γ , y) by b21 (η), taking λ1 = 1/b21 (η) and making the coordinate change x → x − γ , we have the equation of the form g 2 (x) + g (x)g (y) + λ1 g 2 (y) + λ0 = 0, where λ1 , λ0 ∈ K and λ1 satisfy λ1 ̸= 0 and g (λ1 x) = λ1 g (x). We also have λ0 ̸= 0, since the defining polynomial must √ be irreducible. By taking a linear transformation (X : Y : Z ) → (X : Y : (1/ q−1 α0 )Z ), we may assume that α0 = 1. By Lemma 3, we have equation (b) in Theorem 2. By Lemma 4 below, C is smooth only if d = 4.  v

Lemma 3. Let γ ∈ K and let g (x) = x2 + αv−1 x2 if and only if g (γ y) = γ g (y) as a polynomial.

v−1

 + · · · + α1 x2 + x, where αv−1 , . . . , α1 ∈ K . Then, γ ∈ ( {i|αi ̸=0} F2i ) ∩ F2v i

Proof. The coefficient of g (γ y) − γ g (y) of degree 2i is equal to (γ 2 − γ )αi for i = 1, . . . , v . 2v



2v−1

Lemma 4. Let v ≥ 1, let αv−1 , . . . , α1 , λ1 , λ0 ∈ K and let g (x) = x + αv−1 x + · · · + α1 x2 + x. Assume that λ0 ̸= 0 and λ1 ∈ F2v . If (v, λ1 ) ̸= (1, 1), then the plane curve C given by f (x, y) := g 2 (x) + g (x)g (y) + λ1 g 2 (y) + λ0 = 0 is singular, and Sing(C ) = {(α : 1 : 0) : α 2 + α + λ1 = 0}. v+1

v

v

Proof. Let F (X , Y , Z ) = Z 2 f (X /Z , Y /Z ) and let G(X , Z ) = Z 2 g (X /Z ). We have differentials ∂ F /∂ X = G(Y , Z )Z 2 −1 , v v v ∂ F /∂ Y = G(X , Z )Z 2 −1 and ∂ F /∂ Z = XZ 2 −2 G(Y , Z ) + G(X , Z )YZ 2 −2 . Therefore, we find that C is smooth in thev affine plane Z ̸= 0. Assume that v > 1. Then, we have ∂ F /∂ X (X , Y , 0) = ∂ F /∂ Y (X , Y , 0) = ∂ F /∂ Z (X , Y , 0) = 0. Since λ21 = λ1 , v+1

v

v

v+1

v

we have singular points given by F (X , Y , 0) = X 2 + X 2 Y 2 + λ1 Y 2 = (X 2 + XY + λ1 Y 2 )2 = 0 on the line Z = 0. We have the conclusion. Assume that v = 1. Then, ∂ F /∂ Z (X , Y , 0) = XY 2 + X 2 Y = XY (X + Y ). Note that F (X , Y , 0) = (X 2 + XY + λ1 Y 2 )2 , where λ1 = 0 or 1. By the assumption (v, λ1 ) ̸= (1, 1), we have λ1 = 0. Then, we have singular points (0 : 1 : 0) and (1 : 1 : 0).  We have only-if part of Theorem 2 by two Propositions above. It is known that δ0′ (C ) = ∞ for the curve with equation (a) of Theorem 2 (see [5, Proposition 2]). The curve with equation (b) of Theorem 2 satisfies δ0′ (C ) ≥ 2 due to the following: v

v−1

+ · · · + α1 x2 + x and let Lemma 5. Let P2 = (γ : 1 : 0), v ≥ 1, αv−1 , . . . , α1 , λ1 , λ0 ∈ K , g (x) = x2+ αv−1 x2 f (x, y) := g 2 (x) + g (x)g (y) + λ1 g 2 (y) + λ0 . Assume that λ0 ̸= 0, λ1 ∈ F2w := ( {i|αi ̸=0} F2i ) ∩ F2v and C is an irreducible

curve given by f (x, y) = 0. Then, the point P2 is extendable outer Galois if and only if γ ∈ F2w and γ 2 + γ + λ1 ̸= 0.

Proof. We consider the projection πP2 (x : y : 1) = (x + γ y : 1) from P2 . Let t := x + γ y. By the additive property of g (y), we have fˆ (t , y) := f (t + γ y, y) = (g (t ) + g (γ y))2 + (g (t ) + g (γ y))g (y) + λ1 g 2 (y) + λ0

= g 2 (t ) + g 2 (γ y) + (g (t ) + g (γ y) + λ1 g (y))g (y) + λ0 .

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v i If we consider fˆ as an element K [t ][y], then the coefficient of y2 +2 of fˆ (t , y) is equal to that of g (γ y)g (y) for 0 ≤ i < v , v

v

i

which is γ 2 + γ or (γ 2 + γ 2 )αi . Note also that K (C ) = K (t , y) ∼ = K (t )[y]/(fˆ (t , y)). Assume that P2 is extendable outer Galois. Note that 2v + 2i = 2i (2v−i + 1) is not a power of two for 0 ≤ i < v . Since

v i v v i the coefficient of y2 +2 of fˆ (t , y) must be zero by Proposition 1, γ 2 + γ = 0 and (γ 2 + γ 2 )αi = 0 for 1 ≤ i < v . We have v

v+1

γ ∈ F2v and γ 2 = γ 2 = γ . This implies γ ∈ F2w . Since the degree of K (t , y)/K (t ) is 2v+1 and the coefficient of y2 of fˆ (t , y) is γ 2 + γ + λ1 , we have γ 2 + γ + λ1 ̸= 0. Conversely, assume that γ ∈ F2w and γ 2 +γ +λ1 ̸= 0. By Lemma 3, we have g (γ y) = γ g (y). Using the additive property of g (y) and the properties g (γ y) = γ g (y) and g (λ1 y) = λ1 g (y), we have i

fˆ (t , y) = g 2 (t ) + γ 2 g (y) + (g (t ) + γ g (y) + λ1 g (y))g (y) + λ0

= g 2 (t ) + {(γ 2 + γ + λ1 )g (y) + g (t )}g (y) + λ0 = β{g (y + (1/β)t )}g (y) + g 2 (t ) + λ0 , where β = γ 2 + γ + λ1 ̸= 0. Then, fˆ (t , y + b) = fˆ (t , y + (1/β)t + b) = fˆ (t , y) for any element b ∈ K with g (b) = 0. Therefore, the set {y + b : g (b) = 0} ∪ {y + (1/β)t + b : g (b) = 0} ⊂ K (t , y) consists of all roots of fˆ (t , y) and K (t , y)/K (t ) with the relation fˆ (t , y) = 0 is Galois. Since any automorphism induced by the Galois extension is given by (t , y) → (t , t + b) or (t , y) → (t , y + (1/β)t + b) for some b with g (b) = 0, it is the restriction of a linear transformation. Therefore, the point P2 = (γ : 1 : 0) is extendable Galois.  4. Smooth curves In this section, we consider a smooth plane curve C of degree 2e for some e ≥ 2. Assume that δ ′ (C ) ≥ 2 and P = (1 : 0 : 0) is outer Galois. We note that each automorphism σ ∈ GP can be extended to a linear transformation of P2 (see [1, Appendix A, 17 and 18] or [2]). Therefore, δ0′ (C ) = δ ′ (C ) ≥ 2. By Theorem 2, d = 4 and the defining equation is of the form

(x2 + x)2 + (x2 + x)(y2 + y) + (y2 + y)2 + λ = 0, where λ ∈ K \ {0}. Let Q = (γ : 1 : 0) ∈ L∞ . By the smoothness of C , Q is outer Galois if and only if Q is extendable outer Galois. It follows from Lemma 5 that Q is extendable outer Galois if and only if γ = 0 or 1. Therefore, we have exactly three outer Galois points P, (0 : 1 : 0) and (1 : 1 : 0) on the line L∞ . We denote the Galois point (0 : 1 : 0) by P1 . Let Q ∈ P2 \ (C ∪ L∞ ). Assume that Q is Galois. Then, Q is extendable Galois by the smoothness of C . By Proposition 2, the order of the group KQ [QP ] is 2 or 4. The latter case implies that C is in the case (a) of Theorem 2 by Proposition 3. Since C ˜ P \ KP is smooth, the former case holds. Similarly, the order of the group KQ [QP1 ] is 2. Therefore, Q ∈ Lτ ∩ Lη for some τ˜ ∈ G

˜ P1 \ KP1 . Since γ˜ ∈G˜ γ˜ ∗ x = x(x + 1)(x + y)(x + y + 1), Lτ is given by Y = 0 or Y + Z = 0. Similarly, Lη is and some η˜ ∈ G P given by X = 0 or X + Z = 0. Then, Q = (0 : 0 : 1), (1 : 0 : 1), (0 : 1 : 1) or (1 : 1 : 1). Therefore, we have δ ′ (C ) ≤ 7. Let 

φ1 (X : Y : Z ) = (X + Z : Y : Z ),

φ2 (X : Y : Z ) = (X : Y + Z : Z ).

Then, φ1 (C ) = φ2 (C ) = C and (0 : 0 : 1) = φ1 (1 : 0 : 1) = φ2 (0 : 1 : 0) = φ2 ◦ φ1 (1 : 1 : 1). Since φ1 and φ2 are linear transformations, we have δ ′ (C ) = 7 and we may assume that Q = (0 : 0 : 1). We consider the projection πQ : C → P1 from Q , which is given by (X : Y : Z ) → (X : Y ). We take x = X /Y and z = Z /Y . Then we have a field extension K (x, z )/K (x) via πQ with the relation f0 (x, z ) := (x2 + xz )2 + (x2 + xz )(1 + z ) + (1 + z )2 + λz 4

= λz 4 + (x2 + x + 1)z 2 + (x2 + x)z + x4 + x2 + 1 = 0. Note that dx/dz = (x2 + x)/(z 2 + z ). Therefore, points with x2 + x = 0 are ramification points of πQ . Let R be such a ˜ Q with σ (R) = R. Since the fixed line is given ramification point with x = 0. It follows from Fact 2(c) that there exists σ˜ ∈ G by Lσ = QR, which is given by X = 0, the linear transformation σ˜ is represented by a matrix

 Aσ˜ =

1 0

a(σ )

0 1 0

0 0 1

 ,

where a(σ ) ∈ K . Then, the polynomial

σ ∗ f0 = f0 (x, z + a(σ )x) = f0 (x, z ) + f0 (x, a(σ )x) + (x4 + x2 + 1) = f0 (x, z ) + (λa(σ )4 + a(σ )2 )x4 + (a(σ )2 + a(σ ))x3 + (a(σ )2 + a(σ ))x2 is equal to f0 (x, z ) up to a constant. We have a(σ ) = 1 and hence, λ = 1. According to [3, Part III, Proposition 6], δ ′ (C ) ≥ 7 and hence, δ ′ (C ) = 7 if C is defined by (x2 + x)2 + (x2 + x)(y2 + y) + 2 (y + y)2 + 1 = 0.

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5. Plane curves with many extendable Galois points Let v ≥ 1, let q = 2v and let v

g (x) := x2 + αv−1 x2

v−1

+ · · · + α1 x2 + x,

where αv−1 , . . . , α1 ∈ K . We consider the curve of degree 2q defined by f (x, y) := g 2 (x) + g (x)g (y) + λ1 g 2 (y) + λ0 = 0, where λ0 ∈ K \ {0} and λ1 ∈ (



{i|αi ̸=0}

F2i ) ∩ F2v . In this section, we assume that λ0 ̸= 0 and (v, λ1 ) ̸= (1, 1). Let

F (X , Y , Z ) = Z f (X /Z , Y /Z ) and let G(X , Z ) = Z q g (X /Z ). We denote by F2w the finite field {i|αi ̸=0} F2i ∩ F2v . For a constant a ∈ K with g (a) = 0, we denote by σa the linear transformation defined by (X : Y : Z ) → (X + aZ : Y : Z ). Let P = (1 : 0 : 0), let KP := {σa : g (a) = 0} and let τ be the linear transformation given by (X : Y : Z ) → (X + Y : Y : Z ). Then, KP is a subgroup of Aut(P2 ) of order q. Let G := KP ⟨τ ⟩. Since ⟨τ ⟩KP = KP ⟨τ ⟩, G is a subgroup of Aut(P2 ) of order 2q. Let ℓ be a line passing through P. Then, ℓ is given by cY + dZ = 0 for some c , d ∈ K . Since σa (x : −d : c ) = (x + ac : −d : c ) and τ (x : −d : c ) = (x − d : −d : c ), we have ξ (ℓ) = ℓ for any ξ ∈ G. Since g (x + a) = g (x) + g (a) = g (x) and g (x + y + a) = g (x) + g (y) + g (a) = g (x) + g (y) for any a with g (a) = 0, ξ (C ) = C for any ξ ∈ G. The group G will be ˜ P at P if one is able to prove that C is irreducible, because G gives 2q automorphisms of C which commute the Galois group G with the projection πP : C → P1 . Similarly to Section 3, for each ξ ∈ G \ {1}, we define the line Lξ as the line satisfying that ξ (R) = R for all R ∈ Lξ . 2q



Lemma 6. The curve C with Eq. (2) is irreducible. Furthermore, the geometric genus of C satisfies (q − 1)2 ≤ pg (C ) < (2q − 1)(q − 1). Proof. Let C0 be an irreducible component of C with degree d0 and geometric genus pg (C0 ). Since C is smooth in the affine plane Z ̸= 0 by Lemma 4, in the affine plane, C0 is smooth and C0 does not intersect another irreducible component. Let ξ ∈ G \ KP . For some a with g (a) = 0, ξ (X : Y : Z ) = (X + Y + aZ √ : Y : Z ). Then, the fixed locus L√ξ is given by Y + aZ = 0 and C ∩ Lξ is given by F (X , aZ , Z ) = G(X , Z )2 + λ0 Z 2q = (G(X , Z ) + λ0 Z q )2 = 0. Since G(X , Z ) + λ0 Z q has exactly q roots by that g (x) is a separable polynomial of degree q, the multiplicity of the set C ∩ Lξ is two at each point and the set C ∩ Lξ consists of exactly q points. Therefore, the multiplicity of the set C0 ∩ Lξ is two at each point and the set C0 ∩ Lξ consists of exactly d0 /2 points. We consider the projection πP : C0 → P1 from P. Since points of C0 ∩ Lξ are wild ramifications with index 2, we have 2pg (C0 ) − 2 ≥ d0 (−2) + q × (d0 /2)2 by the Riemann–Hurwitz formula. Then, 2pg (C0 ) − 2 ≥ (q − 2)d0 . On the other hand, by genus formula, we have 2pg (C0 ) − 2 ≤ (d0 − 3)d0 . Therefore, d0 ≥ q + 1. We find that any irreducible component of C is of degree at least q + 1. Since the degree of C is 2q, C is irreducible. By Lemma 4, C is singular. Using formulas as above, we have an inequality{(q − 2)d + 2}/2 ≤ pg (C ) < (d − 1)(d − 2)/2. Since d = 2q, we have (q − 1)2 ≤ pg (C ) < (2q − 1)(q − 1).  Lemma 7. Let α 2 + α + λ1 = 0, let Q = (α : 1 : 0) (i.e. Q is a singular point) and let π : Cˆ → C be the normalization. We consider the projection πˆ Q = πQ ◦ π from Q . (a) There exists no ramification point of πˆ Q contained in π −1 (C \ L∞ ). (b) Assume that α ∈ F2w . Then, we have the following. (1) The degree of the projection πˆ Q is q and Q is extendable inner Galois. (2) The fiber π −1 (Q ) consists of exactly q points and there are q distinct tangent directions at Q . (3) pg (C ) = (q − 1)2 . (c) Assume that α ̸∈ F2w . Then, we have the following. (1) The degree of the projection πˆ Q is at least q + 1. (2) The fiber π −1 (Q ) consists of at most q/2 points. (3) pg (C ) ≥ q2 − q + 1. Proof. We prove (a). Let t := x + α y and let fˆ (t , y) := f (t + α y, y). Then, fˆ (t , y) = (g 2 (α y) + g (α y)g (y) + λ1 g 2 (y)) + g (t )g (y) + g 2 (t ) + λ0 . v i If we consider fˆ as an element K [t ][y], then the coefficient of y2 +2 of fˆ (t , y) is equal to that of g (α y)g (y) for 0 ≤ i < v , v

v

i

v

i

v

which is α 2 + α or (α 2 + α 2 )αi . Note that α 2 = α and α 2 = α 2 = α for αi ̸= 0, if g 2 (α y) + g (α y)g (y) + λ1 g 2 (y) = 0 in K [y]. Therefore, g (α y) = α g (y) if and only if g 2 (α y) + g (α y)g (y) + λ1 g 2 (y) = 0 in K [y]. Note that K (C ) = K (t , y) ∼ = K (t )[y]/(fˆ (t , y)). The extension K (t , y)/K (t ) is of degree q (resp. at least q + 1) if g (α y) = α g (y) (resp. g (α y) ̸= α g (y)). Note that dt /dy = (α g (y) + g (α y) + g (t ))/g (y). Since the projection πQ is given by πQ (x : y : 1) = (t : 1), the ramification

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locus in the affine plane Z ̸= 0 is given by α g (y) + g (α y) + g (t ) = 0. By using the condition fˆ (t , y) = 0, we have (α 2 + α + λ1 )g 2 (y) + λ0 = 0. Since α 2 + α + λ1 = 0, there exists no ramification point in the affine plane Z ̸= 0. We prove (b). Assume that α ∈ F2w . Then, g (α y) = α g (y) and we have the field extension K (t , y)/K (t ) of degree q with the relation g (y) + g (t ) + (λ0 /g (t )) = 0. Since g (y + a) = g (y) + g (a) = g (y) for any a ∈ K with g (a) = 0, the set {y + a : g (a) = 0} ⊂ K (t , y) consists of all solutions of g (y) + g (t ) + (λ0 /g (t )) = 0 and K (t , y)/K (t ) with this relation is Galois. We denote by GQ the subgroup of Aut(Cˆ ) corresponding to the Galois group. Let σ ∈ GQ . Since the set {y + a : g (a) = 0} ⊂ K (t , y) consists of all solutions of g (y) + g (t ) + (λ0 /g (t )) = 0, σ ∗ t = t and σ ∗ y = y + a for some a ∈ K with g (a) = 0. Therefore, σ is the restriction of the linear transformation given by (t , y) → (t , y + a). We have (1). According to [13, III. 1.14], for each solution t0 of g (t ) = 0, πˆ Q−1 ((t0 : 1)) consists of a single point Qˆ t0 with eQˆ t = q. 0

Therefore, π −1 (Q ) consists of exactly q points and there are exactly q distinct tangent directions. We have (2). We consider the projection πP from P = (1 : 0 : 0). The set πˆ P−1 (πP (Q )) = π −1 (C ∩ L∞ ) consists of exactly 2q points, by (2). Therefore,

πˆ P is not ramified at any point in π −1 (C ∩ L∞ ). Let Rˆ be a ramification point for πˆ P with R = π (Rˆ ) = (x0 : y0 : 1). Since πP is given by (x : y : 1) → (y : 1), the function x − x0 is a local parameter at R and dy/dx = g (y)/g (x). Then g (y0 ) = 0 by Fact 1(b), g 2 (x0 ) + λ0 = 0 by the equation for f and √ g (y) (g (x) + λ0 )2 dy = = . dx g (x) g (x)(g (x) + λ1 g (y)) √ √ Note that ordR (g (x)) = ordR (g (x) + λ1 g (y)) = 0. We have ordR (g (x) + λ0 ) = 1, since g (x) + λ0 is separable. It follows from Fact 1(b) that the different exponent for πˆ P at Rˆ is equal to two. By the Riemann–Hurwitz formula, we have pg (C ) = (q − 1)2 . We complete the proof of (b). We prove (c). Assume that α ̸∈ F2w . Then, g (α y) ̸= α g (y) and we have the field extension K (t , y)/K (t ) of degree at least q + 1. We have (1). Furthermore, the multiplicity at Q is at most q − 1, and the set π −1 (C ∩ L∞ ) consists of at most 2q − 2 points. If all points in π −1 (C ∩ L∞ ) are not ramification points, then, by that P = (1 : 0 : 0) gives Galois covering and Fact 2(d), the sum of ramification indices along π −1 (C ∩ L∞ ) is at most 2q − 2 < 2q. This is a contradiction. By Fact 2(d) again, all points in π −1 (C ∩ L∞ ) are ramification points for πˆ P . Therefore, π −1 (Q ) consists of at most q/2 < q − 1 points. We have (2). Let e be the ramification index at each point of π −1 (C ∩ L∞ ). Since e divides the degree 2q, where 2q is a power of two, e is divisible by 2. By the Riemann–Hurwitz formula, we have 2pg (C ) − 2 ≥ 2q(−2) + q × (2q/2) × 2 + (2q/e) × e. Then, we have pg (C ) ≥ q2 − q + 1.  We consider the case where g (x) = xq + x. Lemma 8. Let g (x) = xq + x. Then, we have the following. (a) Let P2 = (γ : 1 : 0) with γ ∈ K . If P2 is outer Galois, then γ ∈ Fq . (b) All outer Galois points lie on L∞ . Proof. We prove (a). Let t := x + γ y and let fˆ (t , y) := f (t + γ y, y). Then, fˆ (t , y) = (γ 2q + γ q + λ1 )y2q + (γ q + γ )yq+1 + (γ 2 + γ + λ1 )y2 + (t q + t )(yq + y) + (t q + t )2 + λ0 . The projection πP2 is given by πP2 (x : y : 1) = (t : 1). Note that dt /dy = ((γ q + γ )yq + (t q + t ))/(yq + y). Assume that γ q ̸= γ . For each t there is at most one ramification point mapping to (t : 1), since one can solve uniquely for y and then for x. Substituting t q + t = (γ q + γ )yq into fˆ (t , y), we have a polynomial of the form

(γ 2 + γ + λ1 )y2q + (γ 2 + γ + λ1 )y2 + λ0 in the variable y. Therefore, fˆ (t , y) = dt /dy = 0 has a solution. This implies that there exists a smooth point R in the affine plane Z ̸= 0 such that R is the unique ramification point in πP−21 (πP2 (R)). Assume that P2 is outer Galois. It follows from Fact 2(d) that there is a ramification point R of index d. Then, IR (C , TR C ) = d, by Fact 1(a). This is a contradiction to Lemma 9(a). We prove (b). Let α 2 + α + λ1 = 0. We consider the projection from the singular point Q = (α : 1 : 0) again. Let t := x + α y and let fˆ (t , y) := (t + α y, y). Then, fˆ (t , y) = (α q + α)yq+1 + (t q + t )(yq + y) + (t q + t )2 + λ0 . Assume that α ̸∈ F2v and P ′ ∈ P2 \ L∞ . Then, the multiplicity at Q is q − 1, and for each t ∈ K , there exist a solution y of fˆ (t , y) = 0. The former implies that πˆ P ′ is ramified at some points in π −1 (Q ), by Lemma 7(c)(2). The latter implies that (C ∩ P ′ Q ) \ L∞ ̸= ∅. By Lemma 7(a), πˆ P ′ is not ramified at points in (C ∩ P ′ Q ) \ L∞ . Using Fact 2(d), P ′ is not Galois. Assume that α ∈ F2v . Let P3 = (0 : 0 : 1) and let Q = (α : 1 : 0) with α 2 + α + λ1 = 0. Then, the line P3 Q is given by x + α y = 0. Then, we have f (α y, y) = (α 2 + α + λ1 )(yq + y)2 + λ0 = λ0 .

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Therefore, C ∩ P3 Q = {Q }. Note that this line corresponds to a tangent direction at Q . By Lemma 7(b)(2), πˆ P3 is ramified at exactly one of the q points in π −1 (Q ). It follows from Fact 2(d) that P3 is not Galois. If P4 = (β1 : β2 : 1) is Fq -rational, then the linear transformation φ given by (X : Y : Z ) → (X + β1 Z : Y + β2 Z : Z ) satisfies φ(C ) = C and φ(P4 ) = P3 , since g (x + β) = g (x) for any β ∈ Fq . Therefore, there exist both ramified and unramified points in π −1 (Q ) for πˆ P4 . It follows from Fact 2(d) that P4 is not outer Galois. More strongly, any point in the line P4 Q is not outer Galois. Assume that P ′ is outer Galois in the affine plane Z ̸= 0 and is not contained in any line which is given as a tangent direction at a singular point. Then, P ′ is not Fq -rational. We have only to prove the existence of two distinct tangent lines containing P ′ and at least two points where the lines are tangent to the curve, by Lemma 9(b). Note that the degree of the ramification divisor of the projection πˆ P ′ is 2(q − 1)2 − 2 − 2q(−2) = 2q2 . In particular, there is a tangent line at a smooth point containing P ′ . It follows from Lemma 9(a) and Fact 2(d) that such a tangent line contains at least two points where the line is tangent to C . Therefore, P ′ is contained in an Fq -line, by Lemma 9(b). Let P ′ = (a : b : 1) be contained in the line L defined by β1 X + β2 Y + β3 Z = 0 with β1 , β2 , β3 ∈ Fq . Then (β1 , β2 ) ̸= (0, 0) and aβ1 + bβ2 + β3 = 0. If β1 ̸= 0 (resp. β2 ̸= 0), then we take the linear transformation φ1 given by (X : Y : Z ) → (X + (β3 /β1 )Z : Y : Z ) (resp. (X : Y : Z ) → (X : Y + (β3 /β2 )Z : Z )). Then, we have φ1 (C ) = C because g (x + β) = g (x) for any β ∈ Fq , and φ1 (L) is given by X + β2 Y = 0 (resp. β1 X + Y = 0). Therefore, we may assume that β3 = 0. Assume that β1 ̸= 0. We take a linear transformation φ2 with (X : Y : Z ) → (X + (β2 /β1 )Y : Y : Z ). Then φ2 (C ) is given by

(xq + x)2 + (xq + x)(yq + y) + ((β2 /β1 )2 + (β2 /β1 ) + λ1 )(yq + y)2 + λ0 = 0, and φ2 (L) is given by X = 0. Therefore, we may assume that β2 = 0 and β1 = 1. Then, a = 0. Since the line L defined by X = 0 is not any line given by a tangent direction at a singular point, we have λ1 ̸= 0. If q = 2, then λ1 = 1 since λ1 ∈ F2 . Therefore, we have q > 2, by the assumption (v, λ1 ) ̸= (1, 1). Note that bq + b ̸= 0, since P ′ is not Fq -rational. We take a linear transformation φ3 (X : Y : Z ) = (X : Y + bZ : Z ). Then, φ3 (P ′ ) = (0 : 0 : 1) and φ3 (C ) is given by F1 (X , Y , Z ) := F (X , Y + bZ , Z ) = 0. Let f1 (x, z ) := F1 (x, 1, z ). We consider the projection πφ3 (P ′ ) (x : 1 : z ) = (x : 1). Then, we have dx dz

=

x(xq−1 + (bq + b)z q + 1) {(xq + xz q−1 )((1 + bz )q + (1 + bz )z q−1 )}z = . {(xq + xz q−1 )((1 + bz )q + (1 + bz )z q−1 )}x z ((1 + bz )q + (1 + bz )z q−1 )

Let R ∈ C ∩ L, which is a smooth point of C . Since F1 (0, Y , Z ) has q solutions and L is the tangent line at R, we have ordR (x) = 2 and ordR (xq−1 ) = 2(q−1). Note that ordR ((bq +b)z q +1) = q or 0. Then, dR = ordR (dx/dz ) = 2+ordR (xq−1 +(bq +b)z q +1) = 2 + q or 2, by Fact 1. If (xq−1 + (bq + b)z q + 1)(R) = (bq + b)z q + 1 = 0, then R is uniquely determined for x = 0. This implies that dR = 2 for any R ∈ C ∩ L, or dR = q + 2 for one point R ∈ C ∩ L and dR′ = 2 for any point R′ ∈ C ∩ L \ {R}. Therefore, the sum of the different exponents at points in the line X = 0 is at most 2(q − 1) + (2 + q) = 3q < 2q2 . Then, there exists another tangent line containing P ′ . Therefore, P ′ is given as the intersection point by two distinct tangent lines containing at least two points where the lines are tangent to C . Assume β1 = 0. Then, L is given by Y = 0 and b = 0. We take a linear transformation φ4 (X : Y : Z ) = (X + aZ : Y : Z ). Then, φ4 (P ′ ) = (0 : 0 : 1) and φ4 (C ) is given by F2 (X , Y , Z ) := F (X + aZ , Y , Z ) = 0. Let f2 (y, z ) := F2 (1, y, z ). We consider the projection πφ4 (P ′ ) (1 : y : z ) = (1 : y). Then, we have dy dz

=

y(yq−1 + (aq + a)z q + 1) {(yq + yz q−1 )((1 + az )q + (1 + az )z q−1 )}z = . q q − 1 q q − 1 {(y + yz )((1 + az ) + (1 + az )z )}y z ((1 + az )q + (1 + az )z q−1 )

Similarly to the case where β1 ̸= 0, P ′ is given as the intersection point by two distinct tangent lines containing at least two points where the lines are tangent to C , if q > 2. Assume q = 2. Let R ∈ C ∩ {Y = 0}. Then, by Fact 2(c), there exists σ ∈ GP ′ \ {1} of order two such that σ (R) = R. We take double coverings ϵ : Cˆ → Cˆ /⟨σ ⟩ and θ : Cˆ /⟨σ ⟩ → P1 , and consider πP ′ : Cˆ → P1 as the composition map of ϵ and θ . Since θ is a double covering over P1 , there exist a ramification point. Let S be such a point. Then, ϵ(R) ̸= S, since the ramification index for πˆ P ′ at R is two. Further, we have θ (ϵ(R)) ̸= θ (S ). Then, we have two ramification points of πP ′ which are contained in different fibers. We have two tangent lines containing P ′ .  Lemma 9. Let g (x) = xq + x. Then, we have the following. (a) There exists no smooth point R ∈ C such that IR (C , TR C ) = d. (b) If TR1 C = TR2 C for distinct smooth points R1 , R2 ∈ C , then TR1 C is defined over Fq . Proof. We prove (a). Let y = ax + b be the equation of the tangent line at a smooth point of C intersecting with multiplicity d = deg C . Then f (x, ax + b) = c2 (x − c1 )d for some c1 , c2 ∈ K with c2 ̸= 0, but then 2q

c2 (x2q − c1 ) = f (x, ax + b)

= x2q (1 + aq + λ1 a2q ) + xq+1 (aq + a) + xq g (b) + x2 (1 + a + λ1 a2 ) + xg (b) + λ1 g 2 (b) + λ0 so 1 + aq + λ1 a2q ̸= 0, aq + a = 0 and 1 + a + λ1 a2 = 0. This is impossible because aq = a implies the other two are equal.

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We prove (b). The Gauss map γ : C 99K (P2 )∗ ∼ = P2 , which sends a smooth point R to the tangent line TR C at R, is given by

(∂ F /∂ X : ∂ F /∂ Y : ∂ F /∂ Z ) = (yq + y : xq + x : x(yq + y) + y(xq + x)). q

q

Let R1 = (x1 : y1 : 1) and R2 = (x2 : y2 : 1) with TR1 C = TR2 C . Assume that g (x1 ) = x1 + x1 ̸= 0. Then, x2 + x2 ̸= 0. Since q q q q TR1 C = TR2 C , (y1 + y1 )/(x1 + x1 ) = (y2 + y2 )/(x2 + x2 ) and q λ0 y1 + y1 = 1 + + λ1 q (xq1 + x1 )2 x1 + x1



q

q

y1 + y1 q

x1 + x1 q

2 =

λ0 (xq2 + x2 )2 q

q

by the defining equation, we have x1 + x1 = x2 + x2 and y1 + y1 = y2 + y2 . Then, there exist βx , βy ∈ Fq such that x2 = x1 +βx q q q q and y2 = y1 + βy . If βx = 0, then βy = 0, by the condition x1 (y1 + y1 )/(x1 + x1 ) + y1 = x2 (y2 + y2 )/(x2 + x2 ) + y2 . Therefore, q q q q q βx ̸= 0. By using the condition x1 (y1 + y1 )/(x1 + x1 ) + y1 = x2 (y2 + y2 )/(x2 + x2 ) + y2 , we have (y1 + y1 )/(xq1 + x1 ) = βy /βx . q q Note that (y1 + x1 (βy /βx ))q = (y1 + x1 (βy /βx )), since (y1 + y1 )/(x1 + x1 ) = βy /βx . Then, TR1 C is defined by

  βy βy X + Y + y1 + x1 Z = 0, βx βx which is defined over Fq , since (y1 + x1 βy /βx ) ∈ Fq . q If g (y1 ) = y1 + y1 ̸= 0, then we have the same assertion, similarly to the above discussion.  Proof of Theorem 3. Assertion (a) is derived from Lemmas 4 and 6. Assertion (b) is given by Lemmas 6 and 7(b)(3) and (c). We prove (c). Assume that P ′ is extendable outer Galois. Since σ˜ (Sing(C )) = Sing(C ) for any σ˜ ∈ GP ′ , P ′ lie on the line L∞ . If one singular point Q = (α : 1 : 0) is F2w -rational, then the other singular point (α + 1 : 1 : 0) is also F2w -rational. By Lemma 5, δ0′ (C ) = 2w − 1 if α ∈ F2w , and 2w + 1 if α ̸∈ F2w . If w = v , then αi = 0 for 1 ≤ i ≤ v − 1. Assertion (d) is derived from Lemmas 7 and 8.  Acknowledgments The author thanks the referee for very careful reading and helpful comments. In particular, the author could improve the proof of Proposition 2. The author was partially supported by JSPS KAKENHI Grant Numbers 22740001 and 25800002. References [1] E. Arbarello, M. Cornalba, P.A. Griffiths, J. Harris, Geometry of Algebraic Curves, Vol. I, in: Grundlehren der Mathematischen Wissenschaften, vol. 267, Springer-Verlag, New York, 1985. [2] H.C. Chang, On plane algebraic curves, Chinese J. Math. 6 (1978) 185–189. [3] S. Fukasawa, On the number of Galois points for a plane curve in positive characteristic, Comm. Algebra 36 (2008) 29–36; Part II, Geom. Dedicata 127 (2007) 131–137; Part III, Geom. Dedicata 146 (2010) 9–20. [4] S. Fukasawa, Galois points for a plane curve in arbitrary characteristic, in: Proceedings of the IV Iberoamerican Conference on Complex Geometry, in: Geom. Dedicata, vol. 139, 2009, pp. 211–218. [5] S. Fukasawa, Classification of plane curves with infinitely many Galois points, J. Math. Soc. Japan 63 (2011) 195–209. [6] S. Fukasawa, Complete determination of the number of Galois points for a smooth plane curve, Rend. Sem. Mat. Univ. Padova (in press). [7] D. Goss, Basic Structures of Function Field Arithmetic, Springer-Verlag, Berlin, 1996. [8] R. Hartshorne, Algebraic Geometry, in: GTM, vol. 52, Springer-Verlag, 1977. [9] J.W.P. Hirschfeld, G. Korchmáros, F. Torres, Algebraic Curves Over a Finite Field, in: Princeton Ser. Appl. Math., Princeton Univ. Press, Princeton, 2008. [10] M. Homma, Galois points for a Hermitian curve, Comm. Algebra 34 (2006) 4503–4511. [11] N. Hurt, Many Rational Points, Kluwer Academic Publishers, Dordrecht, 2003. [12] K. Miura, H. Yoshihara, Field theory for function fields of plane quartic curves, J. Algebra 226 (2000) 283–294. [13] H. Stichtenoth, Algebraic Function Fields and Codes, in: Universitext, Springer-Verlag, Berlin, 1993. [14] D. Subrao, The p-rank of Artin–Schreier curves, Manuscripta Math. 16 (1975) 169–193. [15] T. Takahashi, Galois points on normal quartic surfaces, Osaka Math. J. 16 (2005) 57–66. [16] H. Yoshihara, Function field theory of plane curves by dual curves, J. Algebra 239 (2001) 340–355. [17] H. Yoshihara, Galois points on quartic surfaces, J. Math. Soc. Japan 53 (2001) 731–743. [18] H. Yoshihara, Galois points for smooth hypersurfaces, J. Algebra 264 (2003) 520–534. [19] H. Yoshihara, S. Fukasawa, List of problems, available at: http://hyoshihara.web.fc2.com/openquestion.html.