R ESEARCH I NSTITUTEFOR M ATHEMATICAL S CIENCESKYOTOUNIVERSITY,Kyoto,Japan ByAkiraSARASHINAMar2018 Reconstructionofone-puncturedellipticcurvesinpositivecharacteristicbytheirgeometricfundamentalgroups RIMS-1876

RIMS-1876

Reconstruction of one-punctured elliptic curves in positive characteristic by their geometric fundamental groups

By

Akira SARASHINA

Mar 2018

R _ESEARCH I NSTITUTE FOR M ATHEMATICAL S _CIENCES

KYOTO UNIVERSITY, Kyoto, Japan

Reconstruction of one-punctured elliptic curves in positive characteristic by their geometric fundamental groups

Akira Sarashina

1 Introduction

Letkbe a field,Gk the absolute Galois group ofk,U an algebraic variety overk(i.e. a geometrically connected separated scheme of finite type overk) andπ1(U) the ´etale fundamental group of U.

Whenkis a number field or, more generally, a field finitely generated over the prime field, the following philosophy of anabelian geometry, which is sometimes called the Grothendieck conjecture, was advocated by A.Grothendieck.

WhenU is an“anabelian variety”, the geometry of U is determined byπ1(U)↠Gk.

Whenk is an algebraically closed field of characteristic 0 andU is a curve (i.e. an integral separated regular scheme of finite type overkand of dimension 1), the isomorphism class ofπ1(U) as a topological group is determined by the cardinality of cusps of U and the genus of U. Therefore the isomorphism class ofU as a scheme cannot be determined only byπ1(U).

Whenkis an algebraically closed field of characteristicp >0, the isomorphism class ofπ1(U) cannot be determined by easy invariants such as the cardinality of cusps or the genus. Thus, we can even consider the following problem.

Is the isomorphism class of U as a scheme determined only byπ₁(U) ?

Regarding this problem, various results are known (cf. [4] [5] [7] [8] [11] [12] ). Among others, the following theorem is known.

Theorem 1.1 ([11]Theorem 3.5)

Let k be an algebraically closed field of characteristic p > 0, U a curve over k, F ⊂ k the algebraic closure ofFp,U0a curve defined overFandX0a smooth compactification ofU0. Assume that the genus ofX0 is 0. Then

π1(U)≃π1(U0)⇔U ≃U0×Fk(as a scheme)

■ The main result of the present paper is the following generalization of Theorem 1.1.

Theorem 1.2 (Theorem 4.9)

Let kbe an algebraically closed field of characteristicp̸= 0,2,U a curve over k,F ⊂k the algebraic closure ofFp,U0a curve defined overFandX0a smooth compactification ofU0. Assume that the genus ofX0 is 1 and that the cardinality ofX0\U0 is 1. Then

π1(U)≃π1(U0)⇔U ≃U0×Fk(as a scheme)

In the second section, we will review the reconstruction of various invariants byπ1(U), which will be used in the later sections.

In the third section,U is assumed to be an open subscheme of an elliptic or hyperelliptic curve. We will prove that linear relations of the images of cusps in P¹ are encoded in π₁(U) and a certain closed subgroupL_U ⊂π₁(U) (see the third section for the definition ofL_U).

In the fourth section, U is assumed to be a curve of (1,1)-type. At first we will prove that we can apply the main theorem of the third section to certain ´etale covers ofU. Then we will prove that the isomorphism class ofU as a scheme is determined only byπ1(U).

2 The reconstruction of various invariants ([11] § 1, § 2)

In this section, we will review the reconstruction of various invariants that was shown in [11].

The theorems in the first section are about curves of genus 0 or 1, while the theorems in this section are about curves of arbitrary genus.

Definition

Letkbe an algebraically closed field of characteristicp >0,U a curve overk(i.e. an integral separated regular scheme of finite type over k and of dimension 1), π₁(U) the ´etale fundamental group of U, U_H the ´etale cover of U that corresponds to an open subgroup H ⊂ π₁(U), X = U^cpt the smooth compactification ofU,g(X) the genus of X, SU =X\U the complement of U inX,nU the cardinality ofSU, K the function field ofU, K^sep a separable closure ofK, ˜K the maximal Galois extension ofK inK^sep that is unramified over U, ˜X the integral closure ofX in ˜K, ˜SU the inverse image ofSU under X˜ → X, IP˜ the inertia subgroup in π1(U) associated to ˜P ∈ S˜U, I^wild_˜

P the Sylow p-subgroup of IP˜, I^tame_˜

P ^def=IP˜/I^wild_˜

P ,Sub(π1(U)) =^def{H ⊂π1(U)|H is a closed subgroup},F the algebraic closure ofFp in k, (Q/Z)^′def={a∈Q/Z| the order ofais prime top}andFP˜^def=(I_P^tame_˜ ⊗Z(Q/Z)^′)⨿

{∗}({∗}means one point set, ˜P ∈S˜U).

Theorem 2.1 ([11]§1,§2) Fromπ₁(U)

• (g(X), n_U) can be recovered group-theoretically

• When (g(X), n_U)̸= (0,0),pcan be recovered group-theoretically

• π1(X) can be recovered group-theoretically as a quotient group ofπ1(U)

• S˜U can be recovered group-theoretically as a subset of Sub(π1(U)). More precisely, ˜SU can be identified with a subset of Sub(π1(U)) via ˜SU → Sub(π1(U)), ˜P → IP˜, and this subset can be recovered group-theoretically.

• SU can be recovered group-theoretically as a quotient set of ˜SU

• The field structure of FP˜ obtained by identifyingFP˜ withF can be recovered group-thoretically

■

Definition

SetI=IP˜. Letdbe any positive integer. We defineχI,das follows χI,d : I↠I^tame/(p^d−1) =I^tame⊗Z 1

p^d−1Z/Z,→F^×_˜

P

Corollary 2.2 ([11]Corollary 2.11)

LetM be anFp[π₁(U)]-module that can be recovered group-theoretically fromπ₁(U). LetI=IP˜,d≥1 andi∈Z. Then

M(χⁱ_I,d) =^def{x∈M ⊗_FpFP˜ |γx=χⁱ_I,d(γ)x(γ∈I)} can be recovered group-theoretically.

■

3 Linear relations of the images of cusps in P

In this section, we will use the same symbols as in the previous sections, and we assume thatp̸= 0,2 and thatX is an elliptic or hyperelliptic curve.

We will prove that linear relations of the images of cusps in P¹ are encoded in π₁(U) and a certain closed subgroupL_U ⊂π₁(U).

Definition

Let x : X → P¹ be a finite morphism of degree 2, S =^defx(SU), λ0, λ_∞, λ1, λ2,· · ·, λm ∈ X ramified points of xand Pi the image of λi in P¹ (i = 0,∞,1,2,· · ·, m). By Hurwitz’s formula, m is an even number. In this section, we assume thatλ0, λ_∞, λ1, λ2,· · · , λm∈SU, SU\{λ0, λ_∞, λ1, λ2,· · ·, λm} ̸=∅ andx⁻¹(S) =SU. Letµ_(1,1), µ_(1,2), µ_(2,1),· · ·, µ_(l,1), µ_(l,2)∈SU be unramified points (µ_(i,1)is conjugate withµ_(i,2)),R1, R2,· · · , Rlthe images ofµ_(1,1), µ_(1,2), µ_(2,1),· · · , µ_(l,1), µ_(l,2)∈SU in P¹.

SetSU,unr^def={µ(1,1), µ(1,2), µ(2,1),· · ·, µ(l,1), µ(l,2)}, SU,ram^def={λ0, λ_∞, λ1, λ2,· · ·, λm}, Sunr^def={R1, R2,· · ·, Rl},Sram^def={P0, P_∞, P1, P2,· · ·, Pm}.

Let Iλ˜ ⊂π₁(U) be the inertia group corresponding to ˜λ∈ X,˜ Iλ,˜P¹ ⊂π₁(P¹\S) be the inertia group corresponding to ˜λ∈P˜¹ (Here, ˜P¹ stands for the integral closure ofP¹ in ˜K. By definition, ˜X= ˜P¹).

SetQ^def=π₁(P¹\S)^ab,p′ (the maximal pro-prime-to-pabelian quotient of π₁(P¹\S)), L_U ^def=ker(π₁(U)→ π₁(P¹\S)→Q) andQ_U^def=π₁(U)/L_U.

WhenX is a hyperelliptic curve,xis the unique finite morphism of degree 2 (up to isomorphism ofP¹, see [2]IV Propotition 5.3). WhenXis an elliptic curve,xis not unique (therefore,S, Q, LU, QU, SU,unr, Sunr, λ0, P0, µ_(1,1), R1, etc., depend on the choice ofx). In this section, we assume thatxis fixed.

Proposition 3.1

SU,ram, SU,unr, S, Sram, Sunr, Q and the natural injective map QU ,→ Q can be recovered group- theoretically fromπ1(U) andLU.

Proof

For eachλ∈SU, we fix ˜λ∈S˜U above λ. We define an equivalence relation∼onSU by sayingν ∼λ if Iν˜/(I˜ν∩LU) = Iλ˜/(Iλ˜∩LU) (as subsets of QU). We can identify S with SU/ ∼ (see the proof of [11]Lemma 2.1). SU,unr={λ∈SU|there exists ν ∈SU\{λ} such that λ∼ν }, SU,unr andSU,ram are recovered from π1(U) andLU. AsSram (resp. Sunr) is the image of SU,ram (resp. SU,unr),Sram and S_unr are recovered fromπ₁(U) andL_U.

Via the exact sequence 0 → Q_U → Q → Z/2Z, we can regard Q as subset of ¹₂Q_U. By G.A.G.A theorems ([1]Expos´e 12 , Expos´e 13)

Q≃(⊕P∈SI_{P ,}^tame_˜

P¹)/∆, I_{P ,}^tame_˜

P¹ ≃Zˆ^p′ , ∆≃Zˆ^p′

I_λ,^tame_˜P1 /I_λ˜^tame ≃Z/2Z(λ∈SU,ram), I_λ,˜^tame_P1 /I_˜λ^tame= 0 (λ∈SU,unr) and

QU ≃((⊕P∈S_ramI_{P ,}^tame_˜P1)^∗+ ( ∑

P∈Sunr

I_{P ,}^tame_˜P1))

( (⊕λ∈S_ramI_{P ,}^tame_˜P1)^∗def=ker((⊕λ∈S_ramI_{P ,}^tame_˜P1)↠⊕Z/2Z↠^sumZ/2Z) ) therefore

Q≃( ∑

P∈S_ram

1

2I_P^tame_˜ ) + ( ∑

P∈S_unr

I_P^tame_˜ )⊂ 1 2QU

By identifyingQwith the right-hand side of this isomorphism, we obtain Q_U ,→Q.

■

We will use the following lemma in the proof of Theorem 3.3.

Lemma 3.2

Letpbe an odd prime number. For anya1,· · ·, am, b1,· · · , bl∈ {0,1,· · · , p−1}(m∈2Z≥0,l∈Z≥0and (m, l)̸= (0,0)),e1,· · ·, em, f1,· · · , fl∈Z>0with p∤(∏m

i=1ei)(∏l

j=1fj) andα1,· · ·, αm, β1,· · · , βl∈Z, there existd0,˜a1,· · · ,a˜m,˜b1,· · · ,˜bl∈Z>0such that, for anyd∈Zsuch thatd≥d0, we have (i)∼(iii)

(i) ˜c≡c mod p (c=a₁,· · · , a_m, b₁,· · ·, b_l)

(ii) ˜ai≡αi mod ei , ˜bj≡βj mod fj (1≤i≤m , 1≤j≤l) (iii) For allq,t, δ₁,· · ·, δ_m, ϵ₁,· · · , ϵ_l ∈Zs.t. 0≤q≤ m

2 ,0≤t ≤m 2 , 0≤δ_i≤a˜_i+p^d−1

2 , 0≤ϵ_j ≤˜b_j and ∑

i

δ_i+∑

j

ϵ_j =p^d−1

2 +s−q+tp^d , we have ∏

i,j

(˜ai+^pd₂⁻¹ δi

)(b˜j

ϵj

)

≡0 mod p

In particular, whenl̸= 0 and (m, l)̸= (0,1), for anya1,· · ·, am, b1,· · · , bl∈ {0,1,· · · , p−1}, there exist

d,a˜1,· · ·,˜am,˜b1,· · · ,˜bl∈Z>0 which satisfy (i),(iii),(iv),(v),(vi).

(iv)p^d>4s (s^def= ∑

c=a₁,···,a_m,b₁,···,b_l

˜ c)

(v) 2| p^d−1

(p^d−1,˜c) , 2| p^d−1

(p^d−1, s−1) (c=a₁,· · · , a_m, b₁,· · ·, b_l) (vi) (p^d−1,b˜₁) = 1

Proof

We take anyu∈Zsuch that p^u>2(∑

i

ai+∑

j

bj+m

2p+ (∑

i

ei+∑

j

fj)p)−1

(⇐⇒ p^u>(∑

i

ai+∑

j

bj+m

2p+ (∑

i

ei+∑

j

fj)p) + (

u−1

∑

h=0

p−1 2 p^h) )

and set d0^def=u+ 3. We define ˜a1,· · ·,a˜m,˜b1,· · ·,˜bl to be the unique integers that satisfy (ii) and the following condition.

˜

ai=ai+p+ 1 2 p+

∑u

h=2

p−1

2 p^h+Aip (1≤Ai≤ei)

˜bj=bj+Bjp (1≤Bj≤fj) Then for anyd≥d0, we have

s=∑

i

ai+∑

j

bj+m 2p+m

2 p^u+1+D ((m+l)p≤D^def=(∑

i

Aip) + (∑

j

Bjp)≤(∑

i

ei+∑

j

fj)p)

˜

a_i+p^d−1

2 =a_i+p−1

2 +p+ 1

2 p^u+1+ (

d−1

∑

h=u+2

p−1

2 p^h) +A_ip p^d−1

2 +s−q+tp^d= (∑

i

ai+∑

j

bj+m

2 p+D−q) + (

d−1

∑

h=0

p−1

2 p^h) +m

2p^u+1+tp^d Let∑

ga_(i,g)p^g , ∑

gb_(j,g)p^g , ∑

gδ_(i,g)p^g , ∑

gϵ_(j,g)p^g (a_(i,g), b_(j,g), δ_(i,g), ϵ_(j,g)∈ {0,1,· · ·, p−1}) be thep-adic expansions of ˜ai+^pd₂⁻¹, ˜bj, δi, ϵj, respectively.

At first, suppose either that there existi∈ {1,2,· · ·, m}, g∈ {0,1,· · ·, u−1}such thatb(j,g)< ϵ(j,g), or that there exist j ∈ {1,2,· · ·, l}, g ∈ {0,1,· · ·, u−1} such that b(j,g) < ϵ(j,g). By Lucas’ theorem ([3]),

(˜ai+^pd₂⁻¹ δ_i

)

≡0mod p or (˜bj

ϵ_j )

≡0 mod p therefore we have (iii).

Next, suppose thata_(i,g)≥δ_(i,g) andb_(j,g)≥ϵ_(j,g) hold for anyi∈ {1,2,· · ·, m}, j∈ {1,2,· · · , l}, g∈ {0,1,· · ·, u−1}. Then we have

p^u>∑

i

a_i+∑

j

b_j+m

2(p−1) +D≥(∑

i u∑−1

g=0

δ_(i,g)p^g) + (∑

j u∑−1

g=0

ϵ_(j,g)p^g)

Letη be the uth coefficient of thep-adic expansion of (∑

iδi) + (∑

jϵj) = ^pd₂⁻¹ +s−q+tp^d. Thenη satisfiesη≡∑

iδ_(i,u)+∑

jϵ_(j,u)mod p. And we have

p^u>(∑

i

ai+∑

j

bj+m

2p+D−q) + (

u−1

∑

h=0

p−1 2 p^h)

Then we have η = ^p−₂¹. Therefore there exists i ∈ {1,2,· · ·, m} such that δ_(i,u) ̸= 0 or there exists j∈ {1,2,· · ·, l}such thatϵ(j,u)̸= 0.

On the other hand, anyi∈ {1,2· · · , m} satisfies p^u> a_i+p−1

2 +A_ip

Therefore we havea_(i,u)= 0. It is clear that anyj satisfiesb_(j,u)= 0. By Lucas’s theorem ([3]), (˜ai+^pd₂⁻¹

δ_i )

≡0 mod p or (˜bj

ϵ_j )

≡0 mod p

Thus, in both cases, we have (iii). By definition of ˜a1,· · ·,˜am,˜b1,· · ·,˜bl, we have (i),(ii). This proves the first half of the lemma.

Next, we will prove the second half of the lemma.

•Supposeb₁= 0

We setf₁= 1 and take anye₁,· · · , e_m, f₂,· · ·, f_l, α₁,· · ·, α_m, β₁,· · ·, β_lthat satisfyp/(| ∏m

i=1e_i)(∏l j=1f_j).

We apply the first half of the lemma to them. By the proof of the first half of the lemma, we can take

˜b₁=p. We can take a sufficiently largedthat satisfies (v), becausep̸= 2. Therefore we can takedthat satisfies (iv),(v) and (vi).

•Supposeb1̸= 0 andl≡0mod 2.

By Dirichlet’s theorem on arithmetic progressions, there exists N ∈ Z>0 such that b1+p+N p² is a prime number. We take f1 = 1 +N p, β1 = b1, e1 = e2 = · · · = em = f2 = · · · = fl = 2, α1 = · · · = αm = β2 = · · · = βl = 1. We apply the first half of the lemma to them. By the proof of the first half of the lemma, we can take ˜b1 =b1+p+N p². Then ˜b1 is a prime number and

˜b1≥1 +p+p², in particular (p²−p,˜b1) = 1. Anyd≥d0satisfies (v), because ˜a1,· · · ,a˜m,˜b1,· · ·,˜bl, s−1 are odd numbers. Thus, if we take sufficiently largedthat satisfies (iv), d,˜a₁,· · · ,a˜_m,˜b₁,· · ·,˜b_lsatisfy (i),(iii)∼ (v). If ˜b₁ ∤ p^d−1, then we also have (vi). If ˜b₁|p^d−1 (i.e. (vi) is not satisfied), we have p^d+1−1 = (p−1)(p^d+ (p^d−1+· · ·+p+ 1)) ≡(p−1)p^d mod b˜₁. Hence d+ 1,˜a₁,· · · ,a˜_m,˜b₁,· · ·,˜b_l satisfy (i),(iii)∼(vi).

•Supposeb1̸= 0 andl≡1mod 2.

By assumption, we have m ̸= 0 or l ≥ 3. Suppose l ≥ 3 (resp. m ̸= 0). By Dirichlet’s theorem on arithmetic progressions, there exists N ∈ Z>0 such that b1+p+N p² is a prime number. We take

f1 = 1 +N p, β1 = b1 f2 = 4, β2 = 2, e1 = · · · =em =f3 = · · · =fl = 2, α1 =· · · = αm = β3 =

· · ·=βl= 1 (resp. f1= 1 +N p, β1=b1, e1 = 4, α1 = 2, e2 =· · ·=em =f2 =· · · =fl = 2, α2=

· · ·=αm=β2=· · ·=βl= 1). We apply the first half of the lemma to them. By the proof of the first half of the lemma, we can take ˜b1 = b1+p+N p². Then ˜b1 is a prime number and ˜b1 ≥ 1 +p+p², in particular (p³−p,˜b1) = 1. ˜a1,· · ·,˜am,˜b1,· · · ,˜bl, s−1 are odd numbers except ˜b2 (resp. ˜a1), and ˜b2

(resp. ˜a1)≡2 mod4. Hence alld∈2Z>0 satisfy (v). Thus, if we take sufficiently largedthat satisfies (iv),d,˜a1,· · · ,a˜m,˜b1,· · ·,˜bl satisfy (i),(iii)∼(v). If ˜b1 ∤ p^d−1, then we also have (vi). If ˜b1|p^d−1, then we have p^d+2−1 = (p−1)(p^d(p+ 1) + (p^d−1+· · ·+p+ 1))≡(p−1)(p^d(p+ 1)) mod˜b1. Hence d+ 2,˜a₁,· · · ,a˜_m,˜b₁,· · ·,˜b_l satisfy (i),(iii)∼(vi).

■ Definition ([11]§3)

Letγbe an integer such thatγ≥1 ,p/γ| and 2|γ. We define

H˜(Z/γZ) =^def{(cP)P∈S, cP ∈Z/γZ|(< cP >P∈S = Z/γZ) and (∑

P∈S

cP = 0)} H(Z/γZ) =^defH˜(Z/γZ)/(Z/γZ)^×

The natural identification Surj(Q,Z/γZ)≃H˜(Z/γZ) and the restriction map Hom(Q,Z/γZ)→Hom(QU,Z/γZ) yield the following map

H(Z/γZ)≃ {H^′ ⊂π1(P¹\S)：open subgroup |π1(P¹\S)/H^′≃Z/γZ}

→ {H ⊂π₁(U) : open subgroup | (π₁(U)/H ≃Z/γZorπ₁(U)/H ≃Z/1

2γZ) andL_U ⊂H}

Fix closed pointsρ0 ̸=ρ_∞ ∈P¹. For each isomorphism ϕ:P¹≃P¹ with ϕ(ρ0) = 0, ϕ(ρ_∞) =∞, we obtain a bijectionP¹(k)\{ρ_∞} ≃P¹(k)\{∞}=k. This bijection does not depend on the choice ofϕup to scalar multiplication. Hence the additive structure on P¹(k)\{ρ_∞} that is induced by this bijection does not depend on the choice ofϕ, and only depends on the choice ofρ₀ andρ_∞.

Theorem 3.3

For anyaP ∈Fp (P ∈S \{P0, P_∞}), consider the following condition

∑

P∈S\{P0,P_∞}

a_PP =P₀ (with respect to the additive structure associated withP₀andP_∞) Then whether this condition holds or not can be determined group-theoretically byπ₁(U) andL_U. Proof  

We define a1,· · ·, am, b1,· · ·, bl ∈ {0,1,· · ·, p −1} by ai mod p = aP_i, bj mod p = aR_j (i = 1,· · ·, m , j = 1,· · · , l), and apply Lemma 3.2 to them. Then we obtain ˜aP_i^def=a˜i, ˜aR_j^def=˜bj, dthat satisfy (i),(iii),(iv),(v),(vi). LetH (resp. H^′) be the open subgroup ofπ1(U) (resp. π1(P¹\S)) associated with (cP)P∈S ∈H(Z/(p^d−1)Z), wherecP_∞ = 1, cP₀ =s−1 =^def∑

P∈S\{P₀,P_∞}˜aP−1, cP =−˜aP(P ̸=P0, P_∞).

SetXH^def=(UH)^cpt, (P¹)H′^def=((P¹\S)H′)^cpt, ϕ:XH→X andψ:XH→(P¹)H′.

P¹ oo ^x X

(P¹)H^′

OO

XH ϕ

OO

oo ψ

By Lemma3.2 (vi), we have (p^d−1,˜aR₁) = 1. Then R1 is totally ramified in (P¹)H′ → P¹. On the other hand, by definition, R1 is unramified in X → P¹. Hence the above commutative diagram is a cartesian product on generic points. In particular, the degree of XH →X is p^d−1, that is the degree of (P¹)H^′ →P¹. Then by Theorem 2.1 and Corollary 2.2, whether (π1(XH)^ab/p)(χ⁻_µ˜^a˜R1

(1,1),d) = 0 holds or not can be determined group-theoretically (here, ˜µ(1,1)∈X˜ is a point above µ(1,1)). By Artin-Schreier theory,

(π1(XH)^ab/p)^∗def=Hom(π1(XH)^ab/p,Fp) =Homcont(π1(XH),Fp) =H_et¹(XH,Fp) =H¹(XH,OX_H)[F−1]

This, together with [9]Proposition 9, implies (π₁(X_H)^ab/p)(χ⁻_µ˜^˜aR1

(1,1),d) = 0

⇔(π1(XH)^ab/p)^∗((χ⁻_µ˜^˜aR1

(1,1),d)⁻¹) = 0

⇔The FrobeniusF on∑

r

H¹(XH,OX_H)((χ⁻_µ˜^˜aR1

(1,1),d)^−pr) is nilpotent

⇔The Cartier operatorC on∑

r

H⁰(XH,ΩX_H)((χ⁻_µ˜^˜aR1

(1,1),d)^pr) is nilpotent

By fixing a suitable coordinate choice ofP¹, setB^def=k[x, x⁻¹,(x−P1)⁻¹,(x−P2)⁻¹,· · · ,(x−Pm)⁻¹,(x− R1)⁻¹,· · · ,(x−Rl)⁻¹][z]/ < z² −x(x−P1)· · ·(x−Pm) >, then we can write U = SpecB. Set BH =^defB[y]/ < y^pd−1−x^s−1∏

P∈S\{P₀,P_∞}(x−P)^−˜aP >, then we can writeUH = SpecBH. Because Ω_P1\S =O_P¹_\S(dx) =O_P¹_\S(dx/x) and P¹\S ←UH is ´etale, we have ΩU_H =OU_H(dx/x). By Lemma 3.2(vi), we have (p^d−1,˜aR₁) = 1, which implies that we have Γ(UH,ΩU_H)(χ⁻_µ˜^˜aR1

(1,1),d) =By(dx/x).

Letf ∈B and set ω = f y(^dx_x)∈ Γ(U_H,Ω_UH)(χ⁻_µ˜^a˜R1

(1,1),d). We will consider a necessary and sufficient condition forω∈Γ(X_H,Ω_XH)(χ⁻_µ˜^˜aR1

(1,1),d). This can be checked at each ν ∈X_H\U_H. Lett_ν be a prime element ofOX_H,ν.

•Supposeϕ(ν) =λ_∞

The ramification index ofψ(ν) over P_∞ is p^d−1. The ramification index ofϕ(ν) =λ_∞ over P_∞ is 2.

By Abhyankar’s lemma, the ramification index ofν overλ_∞is (p^d−1)/2 andν is unramified overψ(ν).

By (dx/dtν) =−x²(dx⁻¹/dtν) andordν(dx⁻¹/dtν) =p^d−2, we haveordν(dx/dtν) =−p^d, and ordν(f ydx

dtν

x⁻¹) = p^d−1

2 ordλ_∞(f) + 1−p^d+ (p^d−1)

= p^d−1

2 ordλ_∞(f)

therefore

ω= (f ydx dtν

x⁻¹)dtν ∈ΩX_H,ν

⇔ordν(f ydx dtν

x⁻¹)≥0

⇔ord_λ∞(f)≥0

•Supposeϕ(ν) =λ0

SeteP₀^def=(p^d−1)/(p^d−1, s−1) which is the ramification index ofψ(ν) overP0. By Lemma 3.2(v), we have 2|eP₀. By the same argument as above, the ramification index of ν over λ0 is eP₀/2 and thatν is unramified overψ(ν). Then

ordν(f ydx dtν

x⁻¹) =eP0

2 ordλ₀(f) +(s−1)eP0

p^d−1 + (eP₀−1)−eP₀

=e_P0

2 (ordλ0(f) +2((s−1)−(p^d−1, s−1))

p^d−1 )

By Lemma 3.2(iv), we havep^d−1≥2s >2((s−1)−(s−1, p^d−1))≥0. Therefore ω= (f ydx

dtν

x⁻¹)dtν ∈ΩX_H,ν

⇔ordν(f ydx dtν

x⁻¹)≥0

⇔ordλ₀(f)≥ −2((s−1)−(p^d−1, s−1)) p^d−1

⇔ord_λ0(f)≥0

•Supposeϕ(ν) =λi (i= 1,2,· · · , m)

Set eP_i =^def((p^d−1)/(p^d −1,˜aP_i)), this is the ramification index of ψ(ν) over Pi. By Lemma 3.2(v), we have 2|eP_i. By the same argument as above, the ramification index of ν over λi is eP_i/2 and ν is unramified overψ(ν). Then

ordν(f ydx dtν

x⁻¹) =eP_i

2 ordλ_i(f)−˜aP_ieP_i

p^d−1 + (eP_i−1)

=eP_i

2 (ordλ_i(f) + 2(p^d−1)−(˜aP_i+ (p^d−1,˜aP_i))

p^d−1 )

By definition, 2 > 2((p^d−1)−(˜aP_i + (p^d −1,˜aP_i)))/(p^d−1) is clear. By Lemma 3.2(iv), we have p^d−1≥4s >2(˜aP_i+ (p^d−1,˜aP_i)), hence 2((p^d−1)−(˜aP_i+ (p^d−1,˜aP_i)))/(p^d−1)>1. Therefore

ω= (f ydx dtν

x⁻¹)dtν ∈ΩX_H,ν

⇔ordν(f ydx dtν

x⁻¹)≥0

⇔ordλ_i(f)≥ −2(p^d−1)−(˜aP_i+ (p^d−1,a˜P_i))

p^d−1 )

⇔ord_λi(f)≥ −1

•Supposeϕ(ν) =µ_(i,j)(i= 1,2,· · · , l , j= 1,2)

SeteR_i^def=((p^d−1)/(p^d−1,˜aR_i)), which is the ramification index of ψ(ν) over Ri. µ_(i,j) is unramified

overRi. Thus the ramification index ofν overµ_(i,j) iseR_i andν is unramified overψ(ν). Then ordν(f ydx

dtν

x⁻¹) =eR_iordµ_(i,j)(f)−a˜R_ieR_i

p^d−1 + (eR_i−1)

=eR_i(ordµ_(i,j)(f)−a˜Ri+ (p^d−1,˜aR_i) p^d−1 + 1) By Lemma 3.2(iv), we have p^d−1>˜aRi+ (p^d−1,˜aR_i)>0. Therefore

ω= (f ydx dtν

x⁻¹)dtν ∈ΩX_H,ν

⇔ordν(f ydx dtν

x⁻¹)≥0

⇔ordµ_(i,j)(f)≥ ˜aRi+ (p^d−1,˜aR_i) p^d−1 −1

⇔ord_µ(i,j)(f)≥0

SetD^def=λ₁+λ₂+· · ·+λ_m∈Div(X). By the above computation, ω∈ΩX_H ⇔f ∈Γ(X,L(D))

LetKXbe the canonical divisor ofX. By Hurwitz’s formula, we haveg^def=g(X) =m/2, anddeg(KX) = m−2. Thus by the Riemann-Roch theorem, we have dimkΓ(X,L(D)) = g + 1. The valuations of 1,(x/z),(x²/z),· · · ,(x^g/z) ∈ Γ(X,L(D)) at λ0 are mutually different, hence these functions are linearly independent overk. Then we have Γ(X,L(D)) =<1,(x/z),(x²/z),· · ·,(x^g/z)>. By Lemma 3.2(iv) (which impliesp^d−1> s−1) and the following formula

C^d(x^jy^αpdz^βpddx x) =

{

x^j/pdy^αz^{β dx}_x (j ∈p^dZ) 0 (j ∈Z\p^dZ) we have

C^d(ydx

x ) =C^d(x^1−s(x−P1)^˜aP1· · ·(x−Pm)^a˜m(x−R1)^˜aR1· · ·(x−Rl)^˜aRly^pddx x)

=−(aP₁P1+· · ·+aP_mPm+aR₁R1+· · ·+aR_lRl)ydx x On the other hand, for anyq∈ {1,2· · · , g}, we have

C^d(x^q z ydx

x)

=C^d(x^{(q+1−s+pd−12 )}(x−P1)^{(˜aP1+pd−12 )}· · ·(x−Pm)^{(˜aPm+pd−12 )}(x−R1)^˜aR1· · ·(x−Rl)^a˜Rl (y

z )p^d dx

x)

=∑

t

∑

(δ₁,···,δ_m,ϵ₁,···,ϵ_l)

(∏

i

(˜a_P1+^pd₂⁻¹ δi

)

(−P_i)^{(˜aPi+pd−12 −δi)})(∏

j

(a˜_Rj ϵj

)

(−R_j)^{(˜aRj−ϵj)})x^t+1y z

dx x

In this formula,t runs over all the integers that satisfy q+ 1−s+ ((p^d−1)/2)≤(t+ 1)p^d ≤q+ 1 + (m+ 1)((p^d−1)/2) (hence (m/2)−1 ≥t ≥ 0 by Lemma 3.2(iv)). δ1,· · · , δm, ϵ1,· · ·, ϵl run over all