2000 Mathematics Subject Classification: Primary 14H55; Secondary 14H30, 14H40, 20M14.Let H be a 4-semigroup, i.e., a numerical semigroup whose
minimum positive element is four. We denote by 4r(H) + 2 the minimum
element of H which is congruent to 2 modulo 4. If the genus g of H is
larger than 3r(H) − 1, then there is a cyclic covering π : C −→ P^1
of curves with degree 4 and its ramification point P such that the Weierstrass
semigroup H(P) of P is H (Komeda [1]). In this paper it is showed that we
can construct a double covering of a hyperelliptic curve and its ramification
point P such that H(P) is equal to H even if g ≤ 3r(H) − 1.* Partially supported by Grant-in-Aid for Scientific Research (15540051), Japan Society for
the Promotion of Science.
** Partially supported by Grant-in-Aid for Scientific Research (15540035), Japan Society
for the Promotion of Science

Komeda, Jiryo

Ohbuchi, Akira

Bulgarian Digital Mathematics Library at IMI-BAS

Serdica Math. J. 30 (2004), 43–54WEIERSTRASS POINTS WITH FIRST NON-GAP FOUR ONA DOUBLE COVERING OF A HYPERELLIPTIC CURVEJiryo Komeda∗, Akira Ohbuchi∗∗Communicated by V. KanevAbstract. Let H be a 4-semigroup, i.e., a numerical semigroup whoseminimum positive element is four. We denote by 4r(H) + 2 the minimumelement of H which is congruent to 2 modulo 4. If the genus g of H islarger than 3r(H) − 1, then there is a cyclic covering pi : C −→ P1 ofcurves with degree 4 and its ramification point P such that the Weierstrasssemigroup H(P ) of P is H (Komeda [1]). In this paper it is showed that wecan construct a double covering of a hyperelliptic curve and its ramificationpoint P such that H(P ) is equal to H even if g ≤ 3r(H)− 1.2000 Mathematics Subject Classification: Primary 14H55; Secondary 14H30, 14H40, 20M14.Key words: Weierstrass semigroup of a point, double covering of a hyperelliptic curve,4-semigroup.∗ Partially supported by Grant-in-Aid for Scientific Research (15540051), Japan Society forthe Promotion of Science.∗∗ Partially supported by Grant-in-Aid for Scientific Research (15540035), Japan Societyfor the Promotion of Science.44 Jiryo Komeda, Akira Ohbuchi1. Introduction. Let Z≥0 be the additive semigroup of non-negativeintegers. A subsemigroup of Z≥0 is called a numerical semigroup if the comple-ment of H in Z≥0 is finite. The cardinality of Z≥0\H is said to be the genus ofH, which is denoted by g(H). For a 4-semigroup H let S(H) = {4, s1, s2, s3}be the standard basis for H, i.e., si = Min{h ∈ H|h ≡ i mod 4} for i = 1, 2, 3.We set s2 = 4r(H) + 2. On the other hand, let C be a complete non-singularirreducible curve over an algebraically closed field k of characteristic 0, which iscalled a curve in this paper. For any point P of C we define the Weierstrasssemigroup H(P ) of P as follows:H(P ) = {n ∈ Z≥0|there exists f ∈ K(C) with (f)∞ = nP}where K(C) denotes the function field of C. It is known that H(P ) is a numericalsemigroup whose genus is equal to the genus of the curve C. In the case whereg(H) ≤ 3r(H) − 1 it was only shown that the moduli space MH of pointedcurves (C,P ) with H(P ) = H is non-empty (Komeda [1] Corollary 4.13). Wedid not give a curve C and its point P with H(P ) = H. In this paper even ifg(H) ≤ 3r(H) − 1, it will be shown that we can find a double covering C of ahyperelliptic curve with its ramification point P such that H(P ) = H. We notethat such a curve C is not a cyclic covering of P1 with degree 4.2. On double coverings of a hyperelliptic curve. In this sec-tion we construct a double covering of a hyperelliptic curve using the method ofMumford [2] and investigate the Weierstrass semigroup of a ramification point ofthe covering. Let C be a curve. For any even number t let P1, . . . , Pt be distinctpoints of C. Let us take an invertible sheaf L and an isomorphism φ such thatφ : L⊗2 ∼= OC(−t∑i=1Pi)⊂ OC .Let S be a sheaf of OC−algebras of the form S ∼= OC ⊕ L where multiplicationis given by(a, l) · (b,m) = (a · b+ φ(l ⊗m), a ·m+ b · l).Then the canonical morphism pi : C˜ = Spec S −→ C is a double covering ofcurves whose branch locus ist∑i=1Pi (Mumford [2]). Hence, if r is the genus of C,Weierstrass Points with First Non-gap Four. . . 45then by Riemann-Hurwitz formula the genus of C˜ is 2r − 1 +t2. For any i letP˜i ∈ C˜ such that pi(P˜i) = Pi.Proposition 2.1. For any i and any positive integer n we haveh0(C, pi∗OC˜(2nP˜i)) = h0(C,OC (nPi)) + h0(C,L ⊗OC(nPi)).P r o o f. First we note that pi∗OC˜∼= OC ⊕L. Hence for any point P of Cwe getpi∗OC˜(npi∗P ) ∼= pi∗(OC˜ ⊗OC˜(npi∗P )) ∼= pi∗(OC˜ ⊗ pi∗OC(nP ))∼= pi∗OC˜ ⊗OC(nP )∼= (OC ⊕ L)⊗OC(nP ) ∼= OC(nP )⊕ (L ⊗OC(nP )).Since we have pi∗OC˜(npi∗Pi) = pi∗OC˜(2nP˜i), we get the desired equality. From now on we consider the case where C is a hyperelliptic curve ofgenus r ≥ 2.Lemma 2.2. Let Pi be a Weierstrass point on C. Then C˜ is non-hyperelliptic. Moreover, H(P˜i) is a 4-semigroup.P r o o f. Since the curve C is not rational, H(P˜i) 6∋ 2 follows from Propo-sition 2.1. Next we will show that H(P˜i) 6∋ 3. Assume that H(P˜i) ∋ 3. We knowthat H(P˜i) also contains 4, because Pi is a Weierstrass point on a hyperellipticcurve C. Hence, we obtain g(H(P˜i)) ≤ 3. But3 ≥ g(H(P˜i)) = 2r − 1 +t2≥ 2× 2− 1 + 1 = 4,which is a contradiction. Therefore, H(P˜i) is a 4-semigroup. Assume that C˜were hyperelliptic. Since H(P˜i) is a 4-semigroup, P˜i is not a Weierstrass point.Therefore, we getZ≥0\H(P˜i) = {1, . . . , n}for some n. But H(P˜i) is a 4-semigroup, which implies that n+1 = 4. Hence, weget Z≥0\H(P˜i) = {1, 2, 3}. Thus, we see that H(P˜i) is generated by 4, 5, 6 and 7,which implies that there is f ∈ K(C˜) such that (f)∞ = 6P˜i. Let us take a local46 Jiryo Komeda, Akira Ohbuchiparameter t at P˜i such that σ∗t = −t where σ is the involution on C˜ such thatC˜/ < σ >∼= C. Then f is written byf =at6+ (higher order)locally at P˜i where a is a non-zero element of k. Moreover, we haveσ∗f =at6+ (higher order)locally at P˜i. Hence f +σ∗f is a non-zero function on C˜ such that (f +σ∗f)∞ =6P˜i on C˜, which implies that (f + σ∗f)∞ = 3Pi on C. Hence we get H(Pi) ∋ 3.Since Pi is a Weierstrass point on a hyperelliptic curve of genus r ≥ 2, we getH(Pi) ∋ 2. Thus, 2 ≤ r = g(H(Pi)) ≤ 1, which is a contradiction. Hence C˜ isnon-hyperelliptic. Lemma 2.3. Let the notation be as in Lemma 2.2. If r ≥ 3, then C˜ isnot bielliptic.P r o o f. We note thatg(C˜) = 2r − 1 +t2≥ 2× 3− 1 +22= 6.If H(P˜i) ∋ 6, then H(Pi) ∋ 3, which is a contradiction. Thus, H(P˜i) 6∋ 6. Sinceby Lemma 2.2 H(P˜i) is a 4-semigroup, C˜ is not bielliptic (Komeda [1] Lemma2.8.) Proposition 2.4. Let Pi be a Weierstrass point on a hyperelliptic curveC of genus r ≥ 5. There exists an odd number s with 1 ≤ s ≤ t− 1 such thatS(H(P˜i)) = {4, 2r + s, 2r + 2t− s, 4r + 2}.P r o o f. In view of r ≥ 5 the genus of C˜ is at least 10. By Lemmas 2.2and 2.3 we must haveS(H(P˜i)) = {4, 4r + 2, 4m+ 1, 4n + 3}(Komeda [1] Proposition 3.1). We getMin{h ∈ H(P˜i)|h is odd} ≥ 2r + 1,Weierstrass Points with First Non-gap Four. . . 47because 4r + 2 ∈ S(H(P˜i)). We setMin {h ∈ H(P˜i))|h is odd} = 2r + swith odd s ≥ 1. If s > t− 1, then we obtain2r +t2− 1 = g(H(P˜i)) ≥ r +[2r + t+ 14]+[2r + t+ 34]= 2r +t2where for any real number x the symbol [x] denotes the largest integer less thanor equal to x. This is a contradiction. Thus, s ≤ t − 1. Let S(H(P˜i)) ={4, 2r + s, h, 4r + 2}. Then we must have[h4]= r − 1 +t2−[2r + s4].Since h is an odd number such that h 6≡ 2r+s mod 4, we obtain h = 2r+2t−s. Example 2.5. Let the notaion be as in Proposition 2.4. If t = 2, thenS(H(P˜i)) = {4, 2r + 1, 2r + 3, 4r + 2}.In this case the semigroup H(P˜i) is generated by 4, 2r + 1 and 2r + 3.Combining Proposition 2.4 with Proposition 2.1 we get the following:Theorem 2.6. Let Pi be a Weierstrass point on a hyperelliptic curve Cof genus r ≥ 5. Let t ≤ 2r and s an odd number with 1 ≤ s ≤ t − 1. Then thefollowing conditions are equivalent:i) S(H(P˜i)) = {4, 2r + s, 2r + 2t− s, 4r + 2}.ii) h0(C,L ⊗OC((r +s+ 12)Pi))= 1 and h0(C,L ⊗OC((r +s− 12)Pi))= 0.P r o o f. By Proposition 2.1 we haveh0(C, pi∗OC˜((2r + s+ 1) P˜i))= h0(C, pi∗OC˜(2(r +s+ 12)P˜i))= h0(C,OC((r +s+ 12)Pi))+ h0(C,L ⊗OC((r +s+ 12)Pi)).48 Jiryo Komeda, Akira OhbuchiSince Pi is a Weierstrass point on a hyperelliptic curve and we have t ≤ 2r ands ≤ t− 1, we geth0(C, pi∗OC˜((2r+s+1)P˜i)) =[2r + s+ 14]+1+h0(C,L ⊗OC((r +s+ 12)Pi)).First we show that i) implies ii). Since 2r + s ∈ H(P˜i), we haveh0(OC˜((2r + s)P˜i)) =[2r + s4]+ 2.Hence, we geth0(OC˜((2r + s+ 1) P˜i))=[2r + s+ 14]+ 2.By the above formula we obtainh0(C,L ⊗OC((r +s+ 12)Pi))= 1.Since we haveh0(OC˜((2r + s− 1) P˜i))=[2r + s− 14]+ 1,we geth0(C,L ⊗OC((r +s− 12)Pi))= 0.Assume that ii) holds. By Proposition 2.4 there exists an odd number s′ with1 ≤ s′ ≤ t− 1 such thatS(H(P˜i)) = {4, 2r + s′, 2r + 2t− s′, 4r + 2}.If s′ ≤ s− 2, we haveh0(C˜,OC˜((2r + s′ + 1)P˜i))= h0(C,OC((r +s′ + 12)Pi))+ h0(C,L ⊗OC((r +s′ + 12)Pi))Weierstrass Points with First Non-gap Four. . . 49=[2r + s′ + 14]+ 1because ofh0(C,L ⊗OC((r +s− 12)Pi))= 0.Hence 2r + s′ 6∈ H(P˜i), which is a contradiction. Assume that s′ ≥ s + 2. Sincewe haveh0(C,L ⊗OC((r +s+ 12)Pi)) = 1,we know thath0(C˜,OC˜((2r + s+ 1)P˜i)) =[2r + s+ 14]+ 2.Therefore there exists an odd number h with h ≤ 2r + s such that h ∈ H(P˜i).Then h < 2r + s′, which is a contradiction. Hence s′ = s. Since for a 4-semigroup H with g(H) ≥ 3r(H) there exist a cyclic coveringof the projective line P1 with degree 4 and its total ramification point P suchthat H(P ) = H (Komeda [1] §4), we want to investigate 4-semigroups H withg(H) ≤ 3r(H)− 1.Proposition 2.7. Let H be a 4-semigroup with g(H) ≤ 3r(H)−1. Thenthere exist 2 ≤ t ≤ 2r and an odd number s with 1 ≤ s ≤ t − 1 such thatS(H) = {4, 2r + s, 2r + 2t− s, 4r + 2}.P r o o f. If a 4-semigroup H satisfies g(H) ≤ 3r(H)− 1, by Komeda [1] itis one of the semigroups with the following standard basis:i) {4, 4n + 1, 4m+ 3, 4 · 2n+ 2}, 1 ≤ n ≤ m ≤ 3n− 1,ii) {4, 4n + 3, 4m+ 1, 4(2n + 1) + 2}, 2 ≤ n+ 1 ≤ m ≤ 3n+ 1,iii) {4, 4n+1, 4m+2, 4l+3}, 1 ≤ n ≤ m ≤ 2n−1,m ≤ l ≤ n+m−1, n+l ≤ 2m−1,vi) {4, 4n + 1, 4m+ 3, 4l + 2}, 2 ≤ n ≤ m ≤ 2n− 2,m+ 1 ≤ l ≤ 2n− 1,v) {4, 4n + 3, 4m + 1, 4l + 2}, 2 ≤ n+ 1 ≤ m ≤ 2n,m ≤ l ≤ 2n,vi) {4, 4n+3, 4m+2, 4l+1}, 2 ≤ n+1 ≤ m ≤ 2n,m+1 ≤ l ≤ n+m,n+l ≤ 2m−1.In the case i) let r = 2n, s = 1 and t = 2m− 2n+2. Then the set {4, 2r+ s, 2r+2t−s, 4r+2} coincides with the set {4, 4n+1, 4m+3, 4·2n+2}. In the case ii) letr = 2n+1, s = 1 and t = 2m−2n. In the case iii) let r = m, s = 4n+1−2m andt = 2n−2m+2l+2. In the case vi) let r = l, s = 4n+1−2l and t = 2n+2m−2l+2.50 Jiryo Komeda, Akira OhbuchiIn the case v) let r = l, s = 4n+3− 2l and t = 2n+2m− 2l+2. In the case vi)let r = m, s = 4n+ 3− 2m and t = 2n− 2m+ 2l + 2. 3. Construction of a point on a double covering of a hy-perelliptic curve with a given semigroup. In this section we construct apoint P˜i satisfying the conditions in Theorem 2.6 ii). For that purpose we needa hyperelliptic curve C which is a covering of degree n of another hyperellipticcurve. First we build a hyperelliptic curve C ′ which is the base of the covering.For a homogeneous polynomial F ∈ C[x, z] of degree 2b+2 which has no multiplefactor we setC1(F ) = {(s, x)|s2 = F (x, 1)}, (C1(F ))0 = {(s, x)|s2 = F (x, 1), x 6= 0},C2(F ) = {(t, z)|t2 = F (1, z)}, (C2(F ))0 = {(t, z)|t2 = F (1, z), z 6= 0}.Through the isomorphism between (C1(F ))0 and (C2(F ))0 sending (s, x) to(sxb+1,1x)we can construct the nonsingular curve C ′ = HC(F ) by patchingC1(F ) and C2(F ). We can define a morphism h : C′ = HC(F ) −→ P1 sending anelement (s, x) of C1(F ) (resp. (t, z) of C2(F )) to (x : 1) (resp. (1 : z)). Since thedegree of h is two, HC(F ) is a hyperelliptic curve of genus b. On the other hand,let ρ : P1 −→ P1 be the morphism defined by sending (u : v) to (x(u, v) : z(u, v))where z(u, v) = vn and x(u, v) = uλ(u − τ1v)(u − τ2v) · · · (u − τn−λv) with dis-tinct non-zero elements τ1, . . . , τn−λ of k. Then (0 : 1) and (1 : 0) are ramificationpoints with indices λ and n respectively. Let q1 = (0 : 1), q2, . . . , qα−1, qα = (1 : 0)be the branch points of ρ. We set (ρ∗F )(u, v) = F (x(u, v), z(u, v)). We considerthe curve HC(ρ∗F ) in the following cases:i) The case where the zeros of F (x, y) in P1 are different from q1, . . . , qα.Then HC(ρ∗F ) is a non-singular curve of genus nb+ n− 1.ii) The case where one of the zeros of F (x, y) is equal to q1 and the otherzeros are different from q2, . . . , qα. Then HC(ρ∗F ) is a singular curve with onlyone singular point. The singular point is analytically isomorphic to the point(0, 0) on the curve defined by the equation y2 = uλ. Since the singularity isresolved by[λ2]blowing-ups where [x] means the largest integer less than orequal to x, the genus of HC(ρ∗F ) is nb+ n− 1−[λ2].Weierstrass Points with First Non-gap Four. . . 51iii) The case where q1 and qα are zeros of F (x, y) and the other zeros ofF (x, y) are different from q2, . . . , qα−1. By the similar method to the case ii) thegenus of of HC(ρ∗F ) is nb+ n− 1−[λ2]−[n2].Let η : C −→ HC(ρ∗F ) be the normalization. Then we get a commutativediagramC = ˜HC(ρ∗F )η→ HC(ρ∗F )φ→ HC(F )↓ ↓P1ρ→ P1Thus C is a hyperelliptic curve whose genus takes any value of nb, nb+1, . . . , nb+n − 1. Moreover, the morphism φ˜ = φ ◦ η : C −→ HC(F ) is of degree n, whichimplies thatφ˜∗g12(HC(F )) = φ˜∗h∗OP1(1) = h∗Cρ∗OP1(1) = h∗COP1(n) = ng12(C),where hC is the composite map of η and the morphismHC(ρ∗F ) −→ P1 of degree2.Lemma 3.1. Let the notation be as in the above. We denote the genusof C by r. We set t = 2n with a positive integer n ≤ r. Let s be an odd integerwith 1 ≤ s ≤ t−1. Then there exist points P1, . . . , Pt, Q1, . . . , Q s+1−t2+r of C suchthatP1 + P2 + · · · + Pt + (r − t+s+ 12)g12(C) ∼ 2(Q1 + · · ·+Q s+1−t2+r)where P1, . . . , Pn are Weierstrass points and Q1, . . . , Q s+1−t2+r are different fromP1. Moreover, we get h0(OC(Q1 + · · ·+Q s+1−t2+r)) = 1.P r o o f. Let p be a point on C ′ = HC(F ). First we show that there arepoints q, q1, . . . , qb−1 of C′ such thatp+ q + (b− 2)g12(C′) ∼ 2(q1 + · · ·+ qb−1).Let 2A : Picb−1(C ′) −→ Pic2b−2(C ′) be the morphism defined by 2A(L) = 2L.We setΘ = {O(T1 + · · · + Tb−1)|T1, . . . , Tb−1 ∈ C′},which is a theta divisor on the abelian variety Pic b−1(C ′). Hence Θ is an ampledivisor, which implies that the divisor 2A(Θ) ⊂ Pic2b−2(C ′) is ample. By Nakai’s52 Jiryo Komeda, Akira Ohbuchicriterion for any 1-dimensional subvariety Σ ⊂ Pic2b−2(C ′) we have (Σ.2A(Θ)) >0, that is to say, Σ ∩ 2A(Θ) 6= ∅. Now we setΣ = {p+ q + (b− 2)g12(C′)|q ∈ C ′},which is a 1-dimensional locus in Pic2b−2(C ′). Therefore we get Σ ∩ 2A(Θ) 6= ∅,which implies thatp+ q + (b− 2)g12(C′) ∼ 2(q1 + · · · + qb−1)for some points q, q1, . . . , qb−1 of C′. Here let p be a Weierstrass point on thehyperelliptic curve C ′. We may assume that q1, . . . , qb−1 are distinct from p. Infact, let q1 = · · · = ql = p and let ql+1, . . . , qb−1 be distinct from p. Then we getp+ q + (b− 2− l)g12(C′) ∼ 2(ql+1 + · · ·+ qb−1)because of 2p ∼ g12(C′). Take Weierstrass points q′1, . . . , q′l on C′ which are distinctfrom p. Then we obtainp+ q + (b− 2)g12(C′) ∼ 2(q′1 + · · ·+ q′l + ql+1 + · · · + qb−1).Let φ˜∗p = P1 + · · · + Pn and φ˜∗q = Pn+1 + · · · + P2n. Since p is a Weierstrasspoint on C ′, P1, · · · , Pn are also Weierstrass points on C. We obtainP1 + · · ·+ Pt +(r − t+s+ 12)g12(C) ∼φ˜∗(p+ q + (b− 2)g12(C′)) +((r − t+s+ 12)− (nb− 2n))g12(C)because of φ˜∗g12(C′) = ng12(C). Since r − t+s+ 12≥ nb− 2n, we getP1 + · · · + Pt +(r − t+s+ 12)g12(C) ∼ 2(Q1 + · · ·+Q s+1−t2+r)for some points Q1, . . . , Q s+1−t2+r of C distinct from P1. Lastly we may assumethat h0(OC(Q1 + · · ·+Q s+1−t2+r))= 1. In fact, if h0(OC(Q1 + · · ·+Q s+1−t2+r))≥ 2, then we must have (upon renumbering of the points Qi)Q1 + · · ·+Q s+1−t2+r ∼ lg12(C) +Q1 + · · · +Q s+1−t2+r−2l.Weierstrass Points with First Non-gap Four. . . 53Hence we getP1 + · · · + Pt +(r − t+s+ 12− 2l)g12(C) ∼ 2(Q1 + · · ·+Q s+1−t2+r−2l).Let us take distinct Weierstrass points Q s+1−t2+r−2l+1, . . . , Q s+1−t2+r on C whichare different from P1, Q1, . . . , Q s+1−t2+r−2l. Then we getP1 + · · · + Pt +(r − t+s+ 12)g12(C) ∼ 2(Q1 + · · ·+Q s+1−t2+r)again where h0(OC(Q1 + · · · +Q s+1−t2+r))= 1 and Q1, . . . , Q s+1−r2+r are dif-ferent from P1. We setL = OC(Q1 + · · · +Q s+1−t2+r −(r +s+ 12)P1).Then by Lemma 3.1 we getL⊗2 ∼= OC(P1 + P2 + · · ·+ Pt − tg12(C))∼= OC(−ι(P1)− · · · − ι(Pt))where ι is the hyperelliptic involution on C.Theorem 3.2. Let the notation be as in the above. Let pi : C˜ =Spec(OC ⊕ L) −→ C be the canonical morphism. We set pi−1(P1) = {P˜1}. Ifr ≥ 5, then we getS(H(P˜1)) = {4, 2r + s, 2r + 2t− s, 4r + 2}P r o o f. By Lemma 3.1 we geth0(C,L ⊗OC((r +s+ 12)P1))= h0(OC(Q1 + · · · +Q s+1−t2+r))= 1andh0(C,L ⊗OC((r +s− 12)P1))= h0(OC(Q1 + · · ·+Q s+1−t2+r)− P1)= 0By Theorem 2.6 we get our desired result. Combining Theorem 3.2 with Proposition 2.7 we get the following:54 Jiryo Komeda, Akira OhbuchiMain Theorem 3.3. Let H be a 4-semigroup of genus g(H) ≥ 10 withS(H) = {4, 4r1 + 1, 4r2 + 2, 4r3 + 3}.Assume that g(H) ≤ 3r2 − 1. Then there exist a double covering pi : C˜ −→ C ofa hyperelliptic curve and its ramification point P˜ ∈ C˜ such that H(P˜ ) = H.Considering the result of the case where H is a 4-semigroup with 4r2+2 ∈S(H) and g(H) ≥ 3r2, the following statement holds:Corollary 3.4. Let H be a 4-semigroup of genus ≥ 10. Then there exista double covering of a hyperelliptic curve and its ramification point P˜ such thatH(P˜ ) = H.REFERENCES[1] J. Komeda. On Weierstrass points whose first non-gaps are four. J. ReineAngew. Math. 341 (1983), 68–86.[2] D. Mumford. Prym varieties I. Contributions to Analysis, Academic Press,New York, 1974, 325–350.Jiryo KomedaDepartment of MathematicsKanagawa Institute of TechnologyAtsugi, 243-0292, Japane-mail: komeda@gen.kanagawa-it.ac.jpAkira OhbuchiDepartment of MathematicsFaculty of Integrated Arts and SciencesTokushima UniversityTokushima, 770-8502, Japane-mail: ohbuchi@ias.tokushima-u.ac.jp Received July 26, 2003

Weierstrass Points with First Non-Gap Four on a Double Covering of a Hyperelliptic Curve

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Weierstrass Points with First Non-Gap Four on a Double Covering of a Hyperelliptic Curve

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