On the uniqueness of elliptic K3 surfaces with maximal singular fibre

0.23 MB·English

by  Matthias Schuett

On the uniqueness of elliptic K3 surfaces with
7 maximal singular fibre
0
0
2
Matthias Sch¨utt, Andreas Schweizer
c
e
D

2 Abstract
2

We explicitly determine the elliptic K3 surfaces with a maximal singular fibre. If
]
G the characteristic of the ground field is different from 2, for each of the two possible
A maximalfibretypes,I19 andI1∗4,thesurfaceisunique. Incharacteristic2themaximal
∗
fibre types are I and I , and there exist two (resp. one) one-parameter families of
. 18 13
h
such surfaces.
t
a
m
Keywords: elliptic surface, K3 surface, maximal singular fibre, wild ramification.
[

1 MSC(2000): 14J27; 14J28;11G05.
v
3
7 1 Introduction
8
3
2. The question of maximal singular fibres for elliptic surfaces has gained some interest
1 recently. Shiodain[11]treatedthecaseofellipticK3surfacesincharacteristic0where
7 the maximal fibres have type I resp. I∗ . Using the Artin invariant, the first author
19 14
0 proved in [6] that these are also the maximal fibres in characteristic p> 0 if p is odd.
v: Meanwhile in characteristic 2, types I and I∗ were shown to be maximal.
18 13
i
X Then we considered in [8] the maximal fibres of general elliptic surfaces over P1. We
r proved that in general the maximal fibres are strictly larger in positive characteristic
a
than in characteristic zero. Moreover,we also derived partial uniqueness results.
Inthispaper,wewanttoprovetheuniquenessintheK3case,thusansweringaquestion
of Shioda:
Theorem 1
Let p=2 andk analgebraicallyclosedfield of characteristicp. Then there are unique
elliptic6 K3 surfaces over k with singular fibre I resp. I∗ .
19 14
For both fibre types we have an obvious candidate at hand - the mod p reduction of
the corresponding elliptic K3 surface over Q (cf. [11], [9]). This approach only fails if
p = 2; in this case we will determine the families of elliptic K3 surfaces which realise
the respective maximal fibres I and I∗ (Prop.s 20, 22).
18 13
Althoughweareprimarilyinterestedinpositivecharacteristic,ourmethodatthesame
time reproves the maximality and uniqueness in characteristic 0.
1
2 The configurations in odd characteristic 2
We note that the K3 case is special not only in the sense that the maximal fibres
from characteristiczero hold. Evenover Q, the uniqueness breaks down as soon as we
consider honestly elliptic surfaces, i.e. with Euler number e 36 (cf. [12]).
≥
For characteristic different from 2 and 3 the result on maximality and uniqueness can
be reformulated in the following elementary way.
Theorem 2
Let k be an algebraically closed field of characteristic different from 2 and 3. Fix
M = 3 or 4. Let f,g k[T] with deg(f) = 2M, deg(g) = 3M. Then f3 = g2 implies
∈ 6
deg(f3 g2) M +1. Moreover, up to affine transformation of T and scaling there
− ≥
exists exactly one pair (f,g) with deg(f3 g2)=M +1.
−
Proof: If there is no polynomial a with a4 f and a6 g, then the equation
| |
Y2 =X3 3fX 2g
− −
defines an elliptic K3 surface with discriminant ∆ = 108(f3 g2). The fibre at
has type I if M = 4, and I∗ if M = 3, with n = 6−M deg−(∆). Hence the clai∞m
n n −
followsfromThm.1. Onthe otherhand,ifthereis apolynomialawitha2 f anda3 g,
then a6 (f3 g2). In particular, deg(f3 g2) 6>M +1. | |2
| − − ≥
For moreinformationonthe degreeoff3 g2 (for generalM)andits connectionwith
−
elliptic surfaces see [14] and [12] for characteristic 0 and [8] for positive characteristic.
Indeed, one can try to prove the uniqueness of the K3 surfaces in characteristic p 5
≥
by spelling out the equation deg(f3 g2) = M +1 and solving for the coefficients of
−
f and g. In Sections 3 and 4 we will use a different Weierstrass equation that makes
the calculations much less involved and that works in characteristic 3 as well. For
characteristic 2 a somewhat more structural approachapplies, based on results by the
second author in [10]. Here we also make use of a classification of wild ramification
of singular fibres. Although this might be known to the experts, we could not find a
reference for it, so we include it in Section 5 (Prop. 16).
2 The configurations in odd characteristic
The following result will be useful throughout the paper.
Theorem 3 (Pesenti-Szpiro [5, special case of Thm. 0.1])
Let S P1 be a non-isotrivial elliptic surface with conductor and Euler number
→ N
e(S). Then
e(S) 6pd(deg 2).
≤ N −
where pd is the inseparability degree of the j-invariant of S.
Remark 4
We mention thatthis inequality allowsanapproachto determining the maximalfibres
that doesn’t need the Artin invariant.
Firstofall, if S has afibre I , then necessarilyn 21since there areno non-isotrivial
n
≤
elliptic surfaces with conductor of degree less than 4. Analogously, m 16 for fibres
∗ ≤
I (at least in characteristic different from 2).
m
If there is a fibre I , I , I∗ or I∗ and the characteristic is different from 2,3,5,7,
21 20 16 15
the surface is separable. Moreover, the degree of the conductor is at most 5, which
2 The configurations in odd characteristic 3
by the inequality of Pesenti and Szpiro contradicts e(S) = 24. In small characteristic
some more fine-tuning is required.
The remainingresults ofthis sectionwouldfollowimmediately fromthe explicitdeter-
minationoftheellipticsurfacesinthenexttwosections. However,sincethecalculations
of the equations are long, in particular for a fibre of type I , we have decided to also
19
include the following structural proofs.
Lemma 5
Let S be an elliptic K3-surface in characteristic p 0 with a fibre I∗ . If p=2,7 the
≥ 14 6
configuration of S is [14*,1,1,1,1].
Proof: Since p = 2, the j-invariant has a pole of order 14, so S is non-isotrivial.
6
Moreover, S is separable, since p = 2,7. The degree of the conductor of S is at most
6
6. Actually, it has to be exactly 6, since otherwise by the bound of Pesenti-Szpiro the
surface would be rational.
So if the fibres outside I∗ are all multiplicative, the configuration must be as stated.
14
To finish the proof, it suffices to show that there is no additive fibre outside I∗ .
14
Assume on the contrary that there is another additive fibre. Then we can apply a
∗
quadratic twist that ramifies exactly at that fibre and at I . The twisted surface has
14
conductor of degree 5 while being separable, so by Thm. 3 it is rational. However,the
j-invariant still has a pole of order 14, contradiction. 2
Lemma 6
Let S be anelliptic K3-surfacein characteristicp 0 with a fibre I . If p=2,19 the
19
≥ 6
configuration of S is [19,1,1,1,1,1].
Proof: The proof is similar to that of Lemma 5. The degree of the conductor of S
can only be 6 by Thm. 3. If there was an additive fibre, we would apply a quadratic
twistthatramifiesexactlyatthisadditivefibreandI . Theconductoroftheresulting
19
surface has degree 7. By Thm. 3, the twisted surface is rational or K3. But this is
impossible with a fibre of type I∗ . 2
19
Lemma 7
AnellipticK3surfaceincharacteristic19withafibreI isnecessarilyinseparableand
19
its configuration must be [I ,II,III]. Moreover, the surface is unique. If we place
19
the fibres I , II and III at , 0 and 1, its equation is
19
∞
Y2 =X3 3T7(T 1)X +2T(T 1)11.
− − −
Proof: If the surface is separable, then by the same proof as for Lemma 6 the config-
uration must be [19,1,1,1,1,1]. The j-invariant of this surface gives a map of degree
24 from P1 to P1. By the Hurwitz formula we obtain 1 = 24+ 1deg(D) where D
− − 2
is the different. The points above 0 have ramification indices that are divisible by 3,
so they contribute at least 8(3 1) = 16 to the degree of the different. Similarly the
−
points above 1 contribute at least 12(2 1)=12. Finally the j-invarianthas a pole of
−
order 19 which, since the ramificationis wild, contributes at least 19. Summing up we
obtain a contradiction.
Wehaveseenthatthesurfaceisinseparable. Inconsequenceitsconfigurationcanonly
be [I ,II,III]. Hence the surface is the Frobenius base change of a (rational) elliptic
19
3 Fibre I∗ in characteristic =2 4
14 6
∗
surface with configuration[I ,II,III ]. Up to isomorphism, this surface is unique (in
1
any characteristic =2,3) by [8, Lem. 8.2]. It can be given in Weierstrass equation
6
Y2 =X3 3T(T 1)3 2T(T 1)5.
− − − −
Frobenius base change gives the claimed equation after minimalising. 2
Remark 8
Byasimilarargumentonecaneasilyshowthatincharacteristic7anellipticK3surface
with a fibre I∗ is inseparable and must have configuration [I∗ ,II,II]. Placing these
14 14
fibres at , 1 and 1 one obtains the equation
∞ −
Y2 =X3 3(T +1)3(T 1)3X 2T7(T +1)(T 1),
− − − −
which is exactly the surface from [11, Thm. 3.2].
3 Fibre I∗ in characteristic = 2
14
6
In this section, we proveThm. 1 for the maximalfibre type I∗ . In characteristiczero,
14
the uniqueness was proven in [14].
Let k be an algebraically closed field of characteristic different from 2. Let S be an
ellipticK3surfaceoverkwithasection. Byassumption,wecanworkwithanextended
Weierstrass form
S : Y2 =X3+A(T)X2+B(T)X +C(T). (1)
In this setting, the discriminant is given by
∆= 27C2+18CAB+A2B2 4A3C 4B3.
− − −
Note that we have four normalisations available: Mo¨bius transformation in T (e.g. to
fixthecusps)andrescalingbyX α2X,Y α3Y. Inthefollowing,wewillusethese
7→ 7→
normalisations in a convenient way.
Proposition 9
∗
There is a unique elliptic K3-surfaceoverk with a fibre I . Its equation canbe given
14
as
Y2 =X3+(T3+2T)X2 2(T2+1)X +T.
−
This surface has discriminant ∆=4T4+13T2+32.
Proof: Assume that S has a fibre of type I∗ . We locate the cusp of the special fibre
14
at . If the fibre type is I∗ with n 8, then we can assume after a translation
∞ n ≥
X X +α(T) that
7→
A(T)=a T3+a T2+a T +a , B(T)=b T2+b T +b , C(T)=c T +c .
3 2 1 0 2 1 0 1 0
Herea =0,sinceotherwiseS wouldberational. Hencewecanscalesuchthata =1.
3 3
6
By construction, ∆ has degree at most 10:
10
∆= d Tl.
l
l=0
X
3 Fibre I∗ in characteristic =2 5
14 6
We ask for all solutions to the system of equations
d =...=d =0.
5 10
In the first instance,
d = 4c +b2, d = 4c +2b b +2b2a 12a c
10 − 1 2 9 − 0 2 1 2 2− 2 1
give c ,c . Then we claim that b = 0. Otherwise, d implies b = 0. Thus b = 0
0 1 2 8 1 0
6
by d , so B C 0 gives a contradiction. Hence we can rescale to set b = 2.
6 2
≡ ≡ −
By a translation in T, we achieve b = 0. This uses up our final normalisation. In
1
consequence,
d = 4(a +b a2), d = 4(a +a a +2b a 2a3)
8 − 1 0− 2 7 − 0 1 2 0 2− 2
give a and a . It follows that
1 0
d = 2a (b 2a2), d =(b 2a2+2)(b 2a2 2).
5 − 2 0− 2 6 0− 2 0− 2−
The two choices of b give rise to isomorphic elliptic curves. To see this, apply the
0
scaling a ia ,T iT,x ix,y ζy with ζ2 = i = √ 1. Let a = µ,b =
2 2 2 0
7→ 7→ 7→ − 7→ −
2µ2 2. We obtain a one dimensional family parametrisingelliptic K3 surfaces with a
fibre−of type at least I∗ :
13
Y2 =X3+((2 µ2)(µ+T)+µT2+T3)X2+2(µ2 1 T2)X µ+T. (2)
− − − −
This family has discriminant
∆ = 8µT5+(8µ2+4)T4+(32µ 16µ3)T3+(24µ2+13 16µ4)T2
− − (3)
+(86µ+8µ5 32µ3)T +32 35µ2+8µ6 28µ4.
− − −
Hence there is a unique specialisation with a fibre of type I∗ at µ=0. This gives the
14
claimed equation and discriminant for S in any characteristic p=2. 2
6
Remark 10
It is immediate fromthe shape ofA,B,C, that S arisesfroma rationalelliptic surface
by quadratic base change and twisting. It follows that this fibration is uniquely deter-
mined by its configuration of singular fibres. The configuration is [7,1,1,III], if p = 7,
6
and [7,II,III], if p=7.
Remark 11
ThemodelinProp.9relatestotheWeierstrassequationin[11]asfollows: Thefirstau-
thorexhibitedin[6, 6]aquadratictwistofthe lattermodelwhichhasgoodreduction
§
at 3. Then use the coordinate T = 1 and twist over Q(√ 1).
s −
Corollary 12
LetS be anelliptic K3surface overk witha fibre oftype I∗ . ThenS is supersingular
14
if and only if p 3 mod 4.
≡
Proof: By Thm. 1, S arises from the corresponding elliptic K3 surface S over Q by
0
reduction. Since S is a singular K3 surface (i.e. ρ(S ) = 20), it is modular by a
0 0
resultofLivn´e[3,Rem.1.6]. Theassociatednewformofweight3hasCMbyQ(√ 1).
−
(In fact, it is easily shown to have level 16 (cf. [7, Tab. 1]).) This implies that the
characteristic polynomial of Frobenius on H2(S,Q ) has all zeroes of the shape (root
e´t ℓ
of unity) p-power if and only if p 3 mod 4. Hence the Corollary follows from the
Tate Conj×ecture [15] which is known≡for elliptic K3 surfaces. 2
4 Fibre I in characteristic =2 6
19
6
4 Fibre I in characteristic = 2
19
6
In this section, we sketch the proof of Thm. 1 for the maximal fibre type I in char-
19
acteristics p=2. In characteristic zero, this was again included in [14].
6
We employ the same approachandnotationas in Sect. 3. In particular,the prooffirst
determines the elliptic K3 surfaces with a fibre of type I and then specialises. Since
18
the proof heavily relies on the help of a machine to factor polynomials, we omit some
of the details. They can be obtained from the authors upon request.
Proposition 13
The elliptic K3-surface over k with a fibre I is unique. Its equation can be given as
19
Y2 =X3+(T4+T3+3T2+1)X2+2(T3+T2+2T)X+T2+T +1.
This surface has discriminant ∆=4T5+5T4+18T3+3T2+14T 31.
−
Proof: Let S be an elliptic K3 surface with a fibre of type I . Let S be given in
19
extendedWeierstrassequation(1). Locatingthe cuspofthespecialfibreat , wecan
∞
assume that
A(T) = a T4+a T3+a T2+a T +a ,
4 3 2 1 0
B(T) = b T3+b T2+b T +b ,
3 2 1 0
C(T) = c T2+c T +c .
2 1 0
Since the special fibre is multiplicative, a = 0. Hence we can scale such that a = 1.
4 4
6
In this setting, ∆ has degree at most 14:
14
∆= d Tl.
l
l=0
X
We ask for the solutions to the system of equations
d =...=d =0.
6 14
In the first instance, we shall ignore d , thus investigating the special fibre type I .
6 18
The vanishing of the polynomials d ,d ,d determines c ,c ,c . Then
14 13 12 2 1 0
d = b2a +..., d = b2a +...,
11 − 3 1 10 − 3 0
so we have to distinguish whether b =0.
3
1st case: b = 0: In this case, d = b (2b b a ). The choice b = 0 successively
3 11 2 1 2 3 2
−
implies b = b = 0. In consequence, B C 0, and equation (1) becomes singular.
1 0
≡ ≡
Hencewecanassumeb =1afterrescalingandobtainb fromd =0. Thesuccessive
2 1 11
factorisationsofd ,...directlygiveb ,a anda . Thenwenormalisebyatranslation
10 0 1 0
in T to assume a = 0. We obtain the following family of elliptic K3 surfaces with a
3
fibre of type I :
18
1 1 1
Y2 =X3+(T4+a T2+ a2)X2+(T2+ a )X + . (4)
2 4 2 2 2 4
There are severalnotable properties of this family:
4 Fibre I in characteristic =2 7
19
6
It arises from the (unique) rational elliptic surface a fibre of type I ,
9
•
E : Y2+TXY +Y =X3 (5)
by the family of quadratic base changes
T 2T2+a .
2
7→
To see this, apply the variable change Y Y +(T2+ a2)X + 1 to eq. (4). In
7→ 2 2
particular, every member has two 3-torsion sections (0, 1).
±2
There is no specialisation with a fibre of type I . Another way to deduce this
19
•
is to consider the discriminant
∆=(2T2+a 3)(4T4+(6+4a )T2+9+3a +a2).
2− 2 2 2
Ithasamodelwithgoodreductionat2: InsteadofthebasechangeT 2T2+a
2
• 7→
simply apply a base change which is equivalent up to Mo¨bius transformation,
e.g. T T2+λT.
7→
2nd case: b = 0: In this case, the vanishing of d and d determines a and a . In
3 11 10 1 0
6
consequence,
2
d = (b b a +b b b2 b2)b +.... (6)
9 −b 3 2 3 1 3− 3− 2 0
3
h
The coefficient h of b does |not vanish{bzecause oth}erwise d = 1b3 = 0. Hence we
0 9 2 3 6
obtain b . This leaves us with polynomials d ,d ,d in five variables where we have
0 6 7 8
still two normalisations left. In the next instance, we note that
1
d 2a d = b (3b b +3b2a2 2b2a 6b a b +3b2)
7− 3 8 2 3 1 3 3 3− 3 2− 3 3 2 2
This factorisationprovidesus with a . Atthis point, wewantto analysethe vanishing
2
of d independently from d . The numerator of d is a complicated polynomial. We
8 6 8
shall sketch two ways to solve it. The first normalisation is ad-hoc, while the second
will use some extra knowledge.
ThefirstnormalisationconsistsinalineartransformationinT suchthata =1,b =0.
3 1
(This is possible unless char(k)=3 and b =0. In this special case, we find five single
2
solutions to d which are not roots of d .) Then the numerator of d is a polynomial
8 6 8
ofdegree12inb ,b . Since everysummandhasdegreeatleast10,this polynomialhas
2 3
degree 2 in the homogenising variable. Hence it can be solved explicitly. We obtain
a one-dimensional rational parametrisationof elliptic K3 surfaces with I fibres. The
18
general surface has no non-trivial sections.
The second solution is much more efficient. It was motivated by a private correspon-
dance with N. Elkies who kindly informed us about an explicit 1-parameter family of
elliptic K3 surfaces with an I fibre and generically trivial group of section which he
18
had found independently. We therefore decided to choose the normalisation in such a
way that it would meet his example after a change of variables.
Claim: For any solution, there are two linear transformations T αT +β such that
7→
after rescaling:
a =1, b =a b , b =2b .
4 2 3 3 1 3
4 Fibre I in characteristic =2 8
19
6
Proof: We rescale by X,Y to meet our first normalisation a = 1. Then the linear
4
transformation results in the new coefficients
a +4β b b +3b β b +2b β+3b β2
′ 3 ′ 3 ′ 2 3 ′ 1 2 3
a = , b = , b = , b = .
3 α 3 α5 2 α6 1 α7
The first requirement
′ ′ ′
b =a b b +3b β =(a +4β)b
2 3 3 ⇔ 2 3 3 3
gives β = b2 a . Then the second condition b′ =2b′ implies
b3 − 3 1 3
h(a ,b ,b ,b )
α2 = 3 1 2 3
− b2
3
with the polynomial h as in (6). By assumption, h=0, so the claim follows. 2
6
Applying one of the above linear transformations to our elliptic surface, we obtain
d =b9(2a +b +4)(2a +b 4).
8 3 3 3 3 3−
It is easily checked that both solutions are identified under the scaling T T,a
3
7→− 7→
a (which exchanges the two choices of α). Setting b = 2λ,a = 2 λ, we obtain
3 3 3
− −
the second family of elliptic K3 surfaces with a fibre of type I :
18
Y2 = X3+(T4 (λ 2)T3+3T2 (2λ 2)T +1)X2
− − − − (7)
+2λ(T3 (λ 2)T2+2T λ+1)X +λ2((T +1)2 λT).
− − − −
This has discriminant
∆ = λ3[(4λ 4)T6+(24λ 8λ2 12)T5+( 20λ2+4λ3 24+45λ)T4
− − − − −
+( 28 30λ2+76λ)T3+( 24+13λ3+66λ 52λ2)T2
− − − −
+( 46λ2+72λ 12)T 96λ2+32λ3 4+37λ].
− − − −
Hence the elliptic K3 surface with a fibre of type I is uniquely obtained as the
19
specialisation at λ=1. This gives the claimed equation and discriminant. 2
Remark 14
To relate the models in Prop. 13 and in [11], we go through the twist in [9]. Here we
have to correct the following typo: One summand of B in [9, 2] ought to be 15t4.
§
Thenthe equationinProp.13is obtainedfromthe translationx x+t+t2 after the
7→
variable change t T 1.
7→− −
Corollary 15
LetS be anelliptic K3surface overk witha fibre oftype I . ThenS is supersingular
19
if and only if p=2 is not a quadratic residue mod 19 or p=19.
6
Proof: Asbefore,S arisesfromasingularK3surfaceS overQbyreductionbyThm.1.
0
Hence the claim follows as in the proof of Cor. 12. (See e.g. [9, Cor. 2.2 & 3].) 2
§
5 Wild ramification of singular fibres 9
5 Wild ramification of singular fibres
In the absence of wild ramification, the j-invariant gives us complete control of the
singular fibres of an elliptic surface (cf. [13, Table 4.1]). Although there are papers
investigating the case of wild ramification in wide generality (e.g. [4]), it seems that
thereisnoreferencewhichgivesexplicitlowerboundsfortheindexofwildramification.
For convenience and future reference, we decided to include such a list.
Proposition 16
Let E be an elliptic curve over a complete valuation ring R of residue characteristic
p=2or3. LetπdenoteauniformizerofR. Letwdenotetheindexofwildramification
the special fibre of E at π. Depending on the reduction type, the following table lists
whether we always have w =0 or gives a sharp lower bound for w:
fibre type p=2 p=3
I (n 0) 0 0
n
≥
II 2 1
≥ ≥
III 1 0
≥
IV 0 1
I∗(n=1) 1 ≥0
n ≥
(n=1) 2 0
IV6∗ ≥0 1
III∗ 1 ≥0
II∗ ≥1 1
≥ ≥
The examples in [1] show that even over a field like F (T) there is in generalno upper
q
bound on the index of wild ramification of a wild fibre. This is in contrast with the
situation over p-adic fields or number fields. (Compare [13, Theorem 10.4].)
Corollary 17
In the notation of Prop. 16, let the special fibre at π be additive. Then the vanishing
order of the discriminant ∆ at π satisfies
3, if p=3,
v (∆)
π
≥(4, if p=2.
Prop.16iseasilyverifiedwithTate’salgorithm[16]. Hereweshallonlygivetheproofs
for the three cases which we will need in this paper: fibre types II,III and I∗ in
n
characteristic 2. We work with the general Weierstrass equation
y2+a xy+a y =x3+a x2+a x2+a (a R).
1 3 2 4 6 i
∈
In characteristic 2, the discriminant reads
∆=a4(a2a +a a a +a a2+a2)+a4+a3a3. (8)
1 1 6 1 3 4 2 3 4 3 1 3
The sharpness of the given bounds follows immediately from the proofs.
Proof of Prop. 16 for fibre types II,III and p=2: Recall that at a II fibre v (∆)=
π
2+w. Ata fibreoftype III,v (∆)=3+w. Hence wehaveto provethatπ4 ∆. Bya
π
|
change of variables, we can move the singular point to (0,0). Hence π a ,a ,a . The
3 4 6
|
singular fibre is additive if and only if π a . By (8), this implies π4 ∆. Fibre type III
1
moreover requires π2 a . This still only|gives v (∆) 4. | 2
6 π
| ≥
5 Wild ramification of singular fibres 10
∗ ∗
Proof of Prop. 16 for fibre type I and p = 2: Let the singular fibre have type I
n n
for some n 0. Then v (∆) = n + 6 + w. Following [13], we can assume that
π
≥
π a ,a , π2 a ,a , π4 a . In particular, we see that π8 ∆. This proves the claim for
1 2 3 4 6
fib|re type I∗|. | |
0
Fibre type I∗ with n > 0 requires that after a change of coordinates π3 a , but π2 ∤
n | 4
a . More precisely, the integer n is determined by the conditions that after further
2
coordinate changes
πl+2 a , π2l+2 a , πl+1 a if n=2l 1 is odd, (9)
4 6 3
| | || −
πl+2 a , π2l+3 a , πl+2 a if n=2l is even. (10)
3 6 4
| | ||
Here denotes exact divisibility. If n=1,we deduce from the a4 in (8) the possibility
|| 3
that π8 ∆. To prove the general claim, we shall use two-step induction: To start the
||
induction,weneedthatforn=2,theapriorilowestordertermin(8)isa4a2,soπ10 ∆.
1 4 |
If n = 3, then we have to consider the term a4a a2 in (8), so π11 ∆. To complete the
1 2 3 |
induction, we note that n n+2 increasesthe π-divisibility of everysummand in (8)
by at least two by inspecti→on of (9), (10). Hence w 2 if n 2. 2
≥ ≥
AnellipticcurveoverafieldK ofcharacteristic2withj(E)=0canbegiveninnormal
6
form
1
Y2+XY =X3+a X2+ . (11)
2
j(E)
Note, however, that this form in general is not integral or minimal. A twist replaces
the coefficient a by a +D while preserving j(E). Such a twist is trivial (i.e. the two
2 2
curves are isomorphic over K) if D is of the form β2+β with β K. If a = 0, the
2
∈
above normal form has multiplicative fibres at all poles of j(E).
In characteristic 2 a fibre I∗ does not imply that the j-invarianthas a pole at the cor-
ν
responding place. Actually, every twist with sufficiently wild ramificationwill produce
a fibre of type I∗. We describe this locally.
ν
Lemma 18
Let k be an algebraically closed field of characteristic 2. Consider the elliptic curve
πrQ
E : Y2+XY =X3+DX2+
π6e
∞
over k((π)) where e Z, 0 r 5 and Q= ω πi is a unit in k[[π]]. Let
i
∈ ≤ ≤
i=0
P
δ 1
D = +... k[[π]]
π2d−1 ∈ π2d−1
with δ k× and d>0.
∈
If d>e then E has a special fibre I∗ with
ν
ν =8d 4 6e+r =8d 4 v(j(E))
− − − −
and index of wild ramification ω =4d 2.
−
Proof: Essentially this is proved in [10, Lemma 2.1], although the type of the fibre is
not stated explicitly there. 2