On p-torsion of p-adic elliptic curves with additive reduction

0.15 MB·English

by  René Pannekoek

On p-torsion of p-adic elliptic curves with additive
3
1 reduction
0
2

n
a Ren´e Pannekoek
J

0
January 31, 2013
3

]
G
A
1 Introduction
.
h
t
a In this article, we fix a prime p. If E/Q is an elliptic curve with additive reduction, and
p
m
one chooses for it a minimal Weierstrass equation over Z :
p
[

2 y2 +a xy +a y = x3 +a x2 +a x+a , a ∈ Z for each i,
1 3 2 4 6 i p
v
3
then we denote by E (Q ) ⊂ E(Q ) the subgroup of points that reduce to a non-singular
3 0 p p
8 point of the reduced curve. As is well-known, this subgroup does not depend on the choice
5
of minimal Weierstrass equation.
.
1
The purpose of this note is to investigate the structure of E (Q ) as a topological
1 0 p
2 group.
1
v: Theorem 1. Let E/Qp be an elliptic curve with additive reduction, such that it can be
i given by a minimal Weierstrass equation over Z :
X p
r
a y2 +a xy +a y = x3 +a x2 +a x+a ,
1 3 2 4 6
where the a are contained in pZ for each i. Then the group E (Q ) is topologically
i p 0 p
isomorphic to Z , except in the following four cases:
p
(i) p = 2 and a +a ≡ 2 (mod 4);
1 3
(ii) p = 3 and a ≡ 6 (mod 9);
2
(iii) p = 5 and a ≡ 10 (mod 25);
4
(iv) p = 7 and a ≡ 14 (mod 49).
6
In each of the cases (i)-(iv), E (Q ) is topologically isomorphic to pZ × F , where F
0 p p p p
has the discrete topology.
1
The proof of Theorem 1 will be given in Section 4.5. The case p > 7 of Theorem 1
was also mentioned in [3].
We will say a few words about the idea of the proof. It is a standard fact from
the theory of elliptic curves over local fields [2, VII.6.3] that E (Q ) admits a canonical
0 p
filtration
E (Q ) ⊃ E (Q ) ⊃ E (Q ) ⊃ E (Q ) ⊃ ...,
0 p 1 p 2 p 3 p
where for each i ≥ 1 the quotient E (Q )/E (Q ) is isomorphic to F . The quotient
i p i+1 p p
E (Q )/E (Q ) is also isomorphic to F by the fact that E has additive reduction. One
0 p 1 p p
has a natural isomorphism of topological groups j : E (Q ) →∼ p2Z given by the theory of
2 p p
formal groups. If p > 2, the same theory even gives a natural isomorphism j′ : E (Q ) →∼
1 p
pZ . These isomorphisms identify E (Q ) with pnZ for all n ≥ 2. The idea of the
p n p p
proof of theorem 1 is to start from j or j′ and, by extending its domain, to build up an
isomorphism between E (Q ) and either Z or pZ ×F .
0 p p p p
Rather than elliptic curves over Q with additive reduction, we consider the more
p
general case of Weierstrass curves over Z whose generic fiber is smooth and whose special
p
fiber is a cuspidal cubic curve. This allows more general results. Theorem 1 is derived as
a special case.
At the end of the note, we give examples for each prime 2 ≤ p ≤ 7 of an elliptic curve
E/Q with additive reduction at p such that E (Q ) contains a p-torsion point defined
0 p
over Q.
2 Preliminaries
2.1 Preliminaries on Weierstrass curves
All proofs of facts recalled in this section can be found in [2, Ch. IV, VII].
Let K be a finite field extension of Q for some prime p, and let v : K → Z∪{∞}
p K
be its normalized valuation. Let O be the ring of integers, m its maximal ideal and
K K
k its residue field. By a Weierstrass curve over O we mean a projective curve E ⊂ P2
K O
K
defined by a Weierstrass equation
2 3 2
y +a xy +a y = x +a x +a x+a , (1)
1 3 2 4 6
such that the generic fiber E is an elliptic curve with (0 : 1 : 0) as the origin. The
K
coefficients a are uniquely determined by E. The discriminant of E, denoted ∆ , is
i E
defined as in [2, III.1]. The curve E is said to be minimal if v (∆ ) is minimal among
K E
vK(∆E′), where E′ ranges over the Weierstrass curves such that EK′ ∼= EK.
We will say that a Weierstrass curve E/O has good reduction when the special fiber
K
E is smooth, multiplicative reduction when E is nodal (i.e. there are two distinct tangent
k k
directions to the singular point), and additive reduction when E is cuspidal (i.e. one
k
tangent direction to the singular point). A non-minimal Weierstrass curve has additive
reduction. The reduction type of an elliptic curve E is defined to be the reduction type
2
of a minimal Weierstrass model of E over O , which is a minimal Weierstrass curve E/O
K K
∼
such that E = E. By the fact that the minimal Weierstrass model of E is unique up to
K
O -isomorphism, this is well-defined.
K
We have E(K) = E(K) = E(O ) since E is projective. Therefore, we have a reduction
K
map E(K) → E(k) given by restricting an element of E(O ) to the special fiber. By
K
E (K) we denote the subgroup E (K) ⊂ E(K) of points reducing to a non-singular point
0 0
of the special fiber E . By E (K) ⊂ E (K) we denote the kernel of reduction, i.e. the points
k 1 0
that map to the identity 0 of E(k). A more explicit definition of E (K) is
k 1
E (K) = {(x,y) ∈ E(K) : v (x) ≤ −2,v (y) ≤ −3}∪{0 }. (2)
1 K K E
More generally, one defines subgroups E (K) ⊂ E (K) as follows:
n 0
E (K) = {(x,y) ∈ E(K) : v (x) ≤ −2n,v (y) ≤ −3n}∪{0 }.
n K K E
We thus have an infinite filtration on the subgroup E (K):
1
E (K) ⊃ E (K) ⊃ E (K) ⊃ ··· (3)
1 2 3
For an elliptic curve E/K and an integer n ≥ 0, we define E (K) to be E (K), where E
n n
is a minimal Weierstrass model of E over O . The E (K) are well-defined, again by the
K n
fact that the minimal Weierstrass model of E is unique up to O -isomorphism.
K
Proposition 2. For E a Weierstrass curve over Z , there is an exact sequence
p
0 → E (K) → E (K) → E (k) → 0,
1 0 sm
e
where E is the complement of the singular points in the special fiber E.
sm
e e
Proof. This comes down to Hensel’s lemma. See [2, VII.2.1].
For any Weierstrass curve E, we can consider its formal group E [2, IV.1–2]. This is a
b
one-dimensional formal group over O . Giving the data of this formal group is the same
K
as giving a power series F = FEb in OK[[X,Y]], called the formal group law. It satisfies
F(X,Y) = X +Y +(terms of degree ≥ 2)
and
F(F(X,Y),Z)) = F(X,F(Y,Z)).
For E as in (1), the first few terms of F are given by:
F(X,Y) =
2 2 3 3 2 2
X +Y − a XY − a (X Y +XY ) − 2a (X Y +XY )+(a a −3a )X Y −
1 2 3 1 2 3
4 4 2 3 2 2 3
(2a a +2a )(X Y +XY )−(a a −a +4a )(X Y +X Y ) +...
1 3 4 1 3 2 4
3
Treating the Weierstrass coefficients a as unknowns, we may consider F as an element of
i
Z[a ,a ,a ,a ,a ][[X,Y]]calledthegenericformalgrouplaw. IfwemakeZ[a ,a ,a ,a ,a ]
1 2 3 4 6 1 2 3 4 6
into a weighted ring with weight function wt, such that wt(a ) = i for each i, then the
i
coefficients ofF indegree narehomogeneous ofweight n−1 [2, IV.1.1]. Foreach n ∈ Z ,
≥2
we define power series [n] in O [[T]] by [2](T) = F(T,T) and [n](T) = F([n−1](T),T)
K
for n ≥ 3. Here also, we may consider each [n] either as a power series in O [[T]] or as a
K
power series in Z[a ,a ,a ,a ,a ][[T]] called the generic multiplication by n law. We have:
1 2 3 4 6
Lemma 3. Let [p] = b Tn ∈ Z[a ,a ,a ,a ,a ][[T]] be the generic formal multiplica-
Pn n 1 2 3 4 6
tion by p law. Then:
1. p | b for all n not divisible by p;
n
2. wt(b ) = n−1, considering Z[a ,a ,a ,a ,a ] as a weighted ring as above.
n 1 2 3 4 6
Proof. (1) is proved in [2, IV.4.4]. (2) follows from [2, IV.1.1] or what was said above.
The series F(u,v)converges toanelement of m forall u,v ∈ m . To E oneassociates
K K
the group E(m ), the m -valued points of E, which as a set is just m , and whose group
K K K
operation +b is given by u + v = F(u,v) fobr all u,v ∈ E(m ). The identity element of
K
E(m ) is 0 ∈ m . If n ≥ 1 is an integer, then by E(mn )bwe denote the subset of E(m )
K K K K
cborresponding to the subset mn ⊂ m , where mn isbthe nth power of the ideal m obf O .
K K K K K
The groups E(mn ) are subgroups of E(m ), and we have an infinite filtration of E(m ):
K K K
b b b
E(m ) ⊃ E(m2 ) ⊃ E(m3 ) ⊃ ··· (4)
K K K
b b b
Proposition 4. The map
ψ : E (K) →∼ E(m )
K 1 K
b
(x,y) 7→ −x/y
0 7→ 0
is a isomorphism of topological groups. Moreover, ψ respects the filtrations (3) and (4),
K
i.e. it identifies the subgroups E (K) defined above with E(mn ).
n K
b
Proof. See [2, VII.2.2].
It follows from the proof given in [2, VII.2.2] that there exists a power series w ∈
O [[T]], with the first few terms given by
K
3 4 2 5 3 6
w(T) = T +a T +(a +a )T +(a +2a a +a )T +...,
1 1 2 1 1 2 3
such that the inverse to ψ is given by z 7→ (z/w(z),−1/w(z)). Given a finite field
K
extension K ⊂ L, we have an obvious commutative diagram
E (K) ψK // E(m )
1 K
b
incl incl
(cid:15)(cid:15) (cid:15)(cid:15)
E (L) ψL // E (m )
1 OL L
d
4
Here E (m ) is the set of m -valued points of the formal group of E , the base-change
OL L L OL
d
of E to Spec(O ).
L
2.2 Extensions of topological abelian groups
Proposition 5. Suppose X is a topological abelian group and we have a short exact
sequence
0 → Zd → X → F → 0.
p p
of topological groups where the second arrow is a topological embedding. Then X is iso-
morphic as a topological group to either Zd or Zd ×F .
p p p
It is indeed necessary to require Zd → X to be a topological embedding, i.e. a homeo-
p
morphism onto its image, since otherwise we could take X to be the product (Zd)ind×F ,
p p
where the first factor is the abelian group Zd endowed with the indiscrete topology.
p
Proof. First, weclaimthat Ext1(F ,A) = A/pAforanyabeliangroupA. Taking thelong
Z p
p
exact sequence associated to Hom (−,A) for the exact sequence 0 → Z → Z → F → 0
Z p
results in the exact sequence
Hom(Z,A) →p Hom(Z,A) → Ext1(F ,A) → Ext1(Z,A) = 0
Z p Z
where the last equality follows from the fact that Hom(Z,−) is an exact functor. Using
that Hom(Z,A) = A, we get
Ext1(F ,A) = A/pA,
Z p
which proves the claim. Putting A = Zd, we find Ext1(F ,Zd) = Fd. We conclude that
p Z p p p
theequivalence classes ofextensions ofZ-modules 0 → Zd → X → F → 0 areinbijective
p p
correspondence with the elements of Fd. The element 0 ∈ Fd corresponds to the split
p p
extension. The non-split ones are obtained as follows. For v ∈ Zd−pZd, we construct an
p p
extension
0 → Zd → X →fv F → 0.
p v p
by defining the subgroup X ⊂ Qd as X = Zd +hv/pi and letting f : X → F be the
v p v p v v p
unique group homomorphism that is trivial on Zd ⊂ X and that sends v/p to 1. The
p v
equivalence class of the above extension only depends on the class of v modulo pZd. Note
p
that if we take any element x ∈ X mapping to 1 ∈ F , we have px = v +pv ∈ Zd for
v p 1 p
some v ∈ Zd. Note further that X is topologically isomorphic to Zd, if we give it the
1 p v p
subspace topology.
A diagram chase shows that this construction gives us pd − 1 different equivalence
classes of extensions. Suppose that v,w ∈ Zd −pZd and φ : X →∼ X are such that
p p v w
0 //Zd // X fv //F //0
p v p
id φ id
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
0 //Zd // X fw //F //0
p w p
5
is a commutative diagram. Consider an element x ∈ X such that f (x) = 1. Then
v v
f (φ(x)) = 1. Furthermore, px = v+pv for some v ∈ pZd, andφ(px) = pφ(x) = w+pw
w 1 1 p 1
for some w ∈ pZd. Hence v +pv = φ(v +pv ) = w+pw , so v ≡ w (mod pZd).
1 p 1 1 1 p
Let X be a topological group sitting inside an extension of topological groups 0 →
Zd →i X →f F → 0, with i a topological embedding and f continuous. This means that
p p
there exists an extension of topological groups 0 → Zd → Y → F → 0 that is either
p p
split or equal to one of the form 0 → Zd → X →fv F → 0, an isomorphism of groups
p v p
∼
φ : X → Y, and a commutative diagram:
0 //Zd //X f //F //0
p p
id φ id
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
0 //Zd //Y // F //0.
p p
We claim that φ must also be a homeomorphism. Since both X and Y are topological
disjoint unions of the translates of their subgroups Zd, and φ respects the disjoint union
p
decomposition, this is clear. So X is topologically isomorphic to Y, and hence to either
Zd or Zd ×F .
p p p
Remark 6. By repeatedly applying Proposition5, we see that if we have a finite filtration
Zd = M ⊂ M ⊂ ... ⊂ M
p n n−1 1
of topological groups, in which all quotients are isomorphic to F , then M is torsion-free
p 1
if and only if it is topologically isomorphic to Zd.
p
The following is a strengthening of Proposition 5 in the case d = 1, which will be
important for us.
Corollary 7. Let p be a prime and suppose we have a short exact sequence
i
0 → pZ → X → F → 0
p p
of topological abelian groups where the second arrow is a topological embedding. If X is
topologically isomorphic to Z , then v (i−1(px)) = 1 for all x ∈ X−i(pZ ), where v is the
p p p p
p-adic valuation. If X is not topologically isomorphic to Z , it is topologically isomorphic
p
to pZ ×F , and we have v (i−1(px)) > 1 for all x ∈ X −i(pZ ).
p p p p
Proof. If X is topologically isomorphic to Z , the map i is given by multiplication by
p
some unit α ∈ Z∗ followed by the inclusion pZ ⊂ Z . The conclusion follows.
p p p
If X is not topologically isomorphic to Z , then by Proposition 5 we must have X ∼=
p
pZ ×F . But then if x = (y,c), we have v (i−1(px)) = v (py) > 1.
p p p p
6
Lemma 8. Let K be a finite extension of Q with ring of integers O . Then O is
p K K
topologically isomorphic to Zd, where d = [K : Q ].
p p
Proof. O is a free Z -module of rank d, so there is a group isomorphism Zd →∼ O .
K p p K
Since both groups are topologically finitely generated, any isomorphism between them is
bicontinuous [1, 1.1].
3 Weierstrass curves with additive reduction over O
K
As in section 2, let K be a finite extension of Q . Let O again be the ring of integers of
p K
K, with maximal ideal m and residue field k.
K
In this section, we gather some general properties of Weierstrass curves over O with
K
additive reduction.
Lemma 9. Let E/O be a Weierstrass curve with additive reduction. Then E is O -
K K
isomorphic to a Weierstrass curve of the form
2 3 2
y +a xy +a y = x +a x +a x+a ,
1 3 2 4 6
where all a lie in m .
i K
Proof. We construct an automorphism α ∈ PGL (O ) that maps E to a Weierstrass
3 K
curve of the desired form. Consider a translation α ∈ PGL (O ) moving the singular
1 3 K
point of the special fiber E to (0 : 0 : 1). The image E = α (E) is a Weierstrass
k 1 1
curve with coefficients satisfying a ,a ,a in m . There exists a second automorphism
3 4 6 K
α ∈ PGL (O ), of the form x′ = x,y′ = y +cx, such that in the special fiber of α (E )
2 3 K 2 1
the unique tangent at (0 : 0 : 1) is given by y′ = 0. The Weierstrass curve E = α (E )
2 2 1
now has all its coefficients a ,a ,a ,a ,a in m . One may thus take α = α ◦α .
1 2 3 4 6 K 2 1
Suppose that E/O is a Weierstrass curve given by (1), and suppose that the a
K i
are contained in m . In particular, E has additive reduction. If we let F denote the
K
formal group law of E, then the assumption on the a implies that F(u,v) converges to
i
an element of O for all u,v ∈ O . Hence F can be seen to induce a group structure on
K K
O , extending the group structure on E(m ). The same statement holds true when we
K K
b
replace K by a finite field extension L.
Definition 10. Let E/O be a Weierstrass curve given by (1), and assume that the a
K i
are contained in m . For any finite field extension K ⊂ L, we denote by E(O ) the
K L
b
topological group obtained by endowing the space O with the group structure induced
L
by F.
ThefollowingpropositionwillbefundamentalindeterminingofthestructureofE (Q )
0 p
as a topological group for Weierstrass curves with additive reduction.
7
Proposition 11. Let E/O be a Weierstrass curve given by (1), and assume that the a
K i
are contained in m .
K
1. The map Ψ : E (K) → E(O ) that sends (x,y) to −x/y is an isomorphism of
0 K
b
topological groups.
2. If 6e(K/Q ) < p − 1, where e denotes the ramification degree, then E (K) is also
p 0
topologically isomorphic to O equipped with the usual group structure.
K
Proof. Let π be a uniformizer for O . Consider the field extension L = K(ρ) with ρ6 = π.
K
Then define the Weierstrass curve D over O by
L
2 3 2 4
y +α xy +α y = x +α x +α x x+α ,
1 3 2 4 6
where α = a /ρi. There is a birational map φ : E × O 99K D, given by φ(x,y) =
i i OK L
(x/ρ2,y/ρ3). The birational map φ induces an isomorphism on generic fibers, and hence
a homeomorphism between E(L) and D(L). Using (2) and the fact that we have (x,y) ∈
E (L) if and only if v (x),v (y) are both not greater than zero, one sees that φ induces
0 L L
∼
a bijection E (L) → D (L), that all maps (a priori just of sets) in the following diagram
0 1
are well-defined, and that the diagram commutes:
E (K) incl //E (K) incl // E (L) φ //D (L)
1 0 0 1
(cid:15)(cid:15)ψK (cid:15)(cid:15)Ψ (cid:15)(cid:15)ΨL (cid:15)(cid:15)ψL
E(m ) incl //E(O ) incl //E(O ) ·ρ // D(m )
K K L L
b b b b
Here the map Ψ : E (L) → O is defined by (x,y) 7→ −x/y, the rightmost lower hori-
L 0 L
zontal arrow is multiplication by ρ, and the maps labeled incl are the obvious inclusions.
Note that the horizontal and vertical outer maps are all continuous. Since ψ , φ and
L
multiplication by ρ are homeomorphisms (for ψ one uses Proposition 4), so is Ψ . Hence
L L
Ψ must be a homeomorphism onto its image. By Galois theory, Ψ is surjective, so it is
itself a homeomorphism.
Let Fb be the formal group law of D. One calculates that
D
ρF(X,Y) = Fb(ρX,ρY).
D
Hence all maps in the diagram are group homomorphisms. This proves the first part of
the proposition.
Now assume 6e(K/Q ) < p−1, so that v (p) = 6v (p) = 6e(K/Q ) < p−1. Now [2,
p L K p
IV.6.4(b)] implies that E (K) is topologically isomorphic to m , and D (L) to m . Since
1 K 1 L
E has additive reduction, we have E (k) ∼= k+ ∼= Ff, where f = f(K/Q ) is the inertia
sm p p
e
degree of K/Q and E is the smooth locus of the special fiber of E. Proposition 2 shows
p sm
e
we have a short exact sequence
0 → m → E (K) → Ff → 0.
K 0 p
8
In the diagram above, the topological group E (K) is mapped homomorphically into the
0
torsion-free group D (L), hence it is itself torsion-free. It follows from Remark 6 that
1
E (K) is topologically isomorphic to O . This proves the second part.
0 K
The following corollary is worth noting, but will not be used in what follows.
Corollary 12. Let E/O be a Weierstrass curve with additive reduction. If 6e(K/Q ) <
K p
p−1, then E (K) is topologically isomorphic to O .
0 K
Proof. The statement that E (K) is topologically isomorphic to O only depends on the
0 K
O -isomorphismclassofE. ByLemma 9, thereexists aWeierstrass curve E′ witha ∈ m
K i K
that is O -isomorphic to E. Now apply Proposition 11 to E′.
K
4 Weierstrass curves with additive reduction over Z
p
In this section, we gather some general properties of Weierstrass curves over Z with
p
additive reduction and finish the proof of theorem 1.
Lemma 13. Let E/Z be a Weierstrass curve with additive reduction. Then there exists
p
a topological isomorphism χ : E(pZ ) →∼ pZ such that for n ∈ Z , χ identifies E(pnZ )
p p ≥1 p
with pnZ . b b
p
Proof. For p > 2, this is standard; the proof may be found in [2, IV.6.4(b)]. We now treat
the case p = 2. By Lemma 9, we may assume that the Weierstrass coefficients a of E all
i
lie in 2Z . The multiplication by 2 on E(2Z ) is given by the power series
2 2
b
2 3 4
[2](T) = FEb(T,T) = 2T −a1T −2a2T +(a1a2 −7a3)T −..., (5)
where FEb is the formal group law of E. By [2, IV.3.2(a)], E(2Z2)/E(4Z2) is cyclic of order
b b ∼
2. By [2, IV.6.4(b)], there exists a topological isomorphism E(4Z ) → 4Z . Hence there
2 2
b
exists an extension
i
0 → 4Z → E(2Z ) → F → 0.
2 2 2
b
From Theorem 5 we see that E(2Z ) is topologically isomorphic either to 2Z or to 4Z ×
2 2 2
b
F . Assume that the latter is the case, then there is an element z of order 2 in E(2Z ) that
2 2
b
isnotcontainedinE(4Z ). Forsuchaz wehavev (z) = 1,wherev : E(2Z ) → Z ∪{∞}
2 2 2 2 ≥1
b b
is the 2-adic valuation on the underlying set 2Z of E(2Z ). Using that in the duplication
2 2
b
power series (5) we have a ∈ 2Z for each i, it follows that v ([2](z)) = 2, so [2](z) 6= 0.
i 2 2
∼
This is a contradiction, so there exists an isomorphism χ : E(2Z ) → 2Z as topological
2 2
groups. From this, and from the fact that E(2nZ )/E(2n+1Zb ) ∼= F for all n ∈ Z [2,
2 2 2 ≥1
b b
IV.3.2(a)], we see that χ necessarily respects the filtrations on either side.
Corollary 14. Let E/Z be a Weierstrass curve with additive reduction. Then there exists
p
an isomorphism E (Q ) →∼ pZ which for n ∈ Z identifies E (Q ) with pnZ .
1 p p ≥1 n p p
Proof. Such an isomorphism can be obtained by composing the isomorphism χ from
Lemma 13 with the isomorphism ψ from Proposition 4.
Qp
9
4.1 p = 2
Proposition 15. Let E/Z be a Weierstrass curve with its coefficients a in 2Z . Then
2 i 2
E (Q ) is topologically isomorphic to Z if a +a ≡ 0 (mod 4), and to 2Z ×F otherwise.
0 2 2 1 3 2 2
Proof. Proposition 2 shows that there is a short exact sequence
0 → E (Q ) → E (Q ) → F → 0.
1 2 0 2 2
ByLemma13, wehaveE (Q ) ∼= 2Z , soProposition5impliesthatE (Q )istopologically
1 2 2 0 2
isomorphic either to Z or to 2Z ×F .
2 2 2
Let [2](T) ∈ O [[T]] be the formal duplication formula (5) on E. Let Ψ be the map
K
from Proposition 11. Since Ψ is an isomorphism of topological groups, we have for all
P ∈ E (Q ):
0 2
Ψ(2P) = [2](Ψ(P)). (6)
By Corollary 7, we have E (Q ) ∼= Z if and only if for all P ∈ E (Q )−E (Q ) we have
0 2 2 0 2 1 2
2P = E (Q ) − E (Q ), which by (6) is true if and only if for all z ∈ E(Z ) − E(2Z )
1 2 2 2 2 2
b b
we have v ([2](z)) = 1, where v : E(Z ) → Z ∪ {∞} is the 2-adic valuation on the
2 2 2 ≥0
b
underlying set Z of E(Z ). This condition may be checked using the duplication power
2 2
b
series
∞
[2](T) = 2T −a1T2 −2a2T3 +(a1a2 −7a3)T4 −... = XbiTi.
i=1
In deciding whether v ([2](z)) = 1 for z ∈ E(Z ) − E(2Z ), we do not need to consider
2 2 2
b b
those parts of terms whose coefficients have valuation ≥ 2. The non-linear parts of each
coefficient b will contribute only terms with valuation ≥ 2, so may ignore these and keep
i
only the linear parts. The terms b zi with i odd we may discard altogether; by Lemma
i
3, all their coefficients have valuation ≥ 2. Finally, we may discard all terms b zi with i
i
even and ≥ 6: a polynomial in Z[a ,...,a ] whose weight is odd and at least 5 does not
1 6
contain a linear term (there being no a ), so the terms involving z6,z8,z10,... will have
5
valuation ≥ 2.
We thus get that, if z ∈ E(Z )−E(2Z ),
2 2
b b
2 4
v ([2](z)) = 1 ⇔ v (2z −a z −7a z ) = 1.
2 2 1 3
This is true for all z ∈ E(Z )−E(2Z ) if and only if:
2 2
b b
a 7a
1 2 3 4
v (z − z − z ) = 0 ⇔ a +7a ≡ 0 (mod 4) ⇔ a +a ≡ 0 (mod 4)
2 1 3 1 3
2 2
since z ≡ z2 ≡ z4 (mod 2). This proves the proposition.
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