1.4. Objetivos
2.1.4. Definiciones del debido proceso
duction. If 6e(K/Qp) < p−1, then E0(K) is topologically isomorphic to OK.
Proof. The statement that E0(K) is topologically isomorphic to OK only depends on the OK-isomorphism class ofE. By Lemma 1.18, there exists a Weierstrass curve E0 with a
i ∈mK that is OK-isomorphic to E. Now apply Proposition 1.20 to E0.
1.5
Proof of the main theorem
In this section, we gather some general properties of nice Weierstrass curves over Zp with additive reduction and finish the proof of Theorem 1.1.
Lemma 1.22. LetE/Zp be a nice Weierstrass curve with additive reduction.
Then there exists a topological isomorphism χ: Eb(pZp)
∼
→ pZp that, for all
n ∈Z≥1, identifies Eb(pnZp) with pnZp.
Proof. Forp > 2, this is standard; the proof may be found in [32, IV.6.4(b)]. We now treat the case p = 2. By Lemma 1.18, we may assume that the Weierstrass coefficients ai of E all lie in 2Z2. The multiplication by 2 on
b
E(2Z2) is given by the power series
[2](T) = FEb(T, T) = 2T −a1T2−2a2T3+ (a1a2−7a3)T4 −. . . , (1.9) whereFEbis the formal group law of E. By [32, IV.3.2(a)],Eb(2Z2)/Eb(4Z2) is
cyclic of order 2. By [32, IV.6.4(b)], there exists a topological isomorphism
b
E(4Z2)
∼
→4Z2. Hence there exists an extension 0→4Z2
i
→Eb(2Z2)→Z/2Z→0.
From Proposition 1.14 we see thatEb(2Z2) is topologically isomorphic either
to 2Z2 or to 4Z2 ×Z/2Z. Assume that the latter is the case, then there is an element z of order 2 in Eb(2Z2) that is not contained in Eb(4Z2). For such a z we have v2(z) = 1, where v2: Eb(2Z2) → Z≥1 ∪ {∞} is the 2- adic valuation on the underlying set 2Z2 of Eb(2Z2). Using that in the
duplication power series (1.9) we have ai ∈ 2Z2 for each i, it follows that
v2([2](z)) = 2, so [2](z) 6= 0. This is a contradiction, so there exists an isomorphism χ: Eb(2Z2)
∼
→2Z2 as topological groups. From this, and from the fact that Eb(2nZ2)/Eb(2n+1Z2)∼=Z/2Z for alln ∈Z≥1 [32, IV.3.2(a)], we see that χnecessarily respects the filtrations on either side.
16 Chapter 1. Elliptic curves overp-adic fields
Corollary 1.23. LetE/Zp be a nice Weierstrass curve with additive reduc-
tion. Then there exists an isomorphism E1(Qp)
∼
→ pZp which for n ∈ Z≥1
identifies En(Qp) with pnZp.
Proof. Such an isomorphism can be obtained by composing the isomorphism
χ from Lemma 1.22 with the isomorphism ψQp from Proposition 1.4.
1.5.1
The case
p= 2
Proposition 1.24. Let E/Z2 be a nice Weierstrass curve with its coeffi-
cientsai in2Z2. ThenE0(Q2)is topologically isomorphic toZ2 ifa1+a3 ≡0 (mod 4), and to 2Z2×Z/2Z otherwise.
Proof. Proposition 1.2 shows that there is a short exact sequence
0→ E1(Q2)→ E0(Q2)→Z/2Z→0.
By Lemma 1.22, we have E1(Q2) ∼= 2Z2, so Proposition 1.14 implies that E0(Q2) is topologically isomorphic either toZ2 or to 2Z2×Z/2Z.
Let [2](T)∈ OK[[T]] be the formal duplication formula (1.9) onE. Let Ψ be the map from Proposition 1.20. Since Ψ is an isomorphism of topological groups, we have for all P ∈ E0(Q2):
Ψ(2P) = [2](Ψ(P)). (1.10) By Corollary 1.16, we have E0(Q2)∼=Z2 if and only if for all P ∈ E0(Q2)− E1(Q2) we have 2P ∈ E1(Q2)− E2(Q2), which by (1.10) is true if and only if for all z ∈ Eb(Z2)−Eb(2Z2) we have v2([2](z)) = 1, where v2: Eb(Z2) →
Z≥0∪ {∞} is the 2-adic valuation on the underlying set Z2 of Eb(Z2). This
condition may be checked using the duplication power series [2](T) = 2T −a1T2−2a2T3+ (a1a2−7a3)T4−. . .=
∞ X
i=1
biTi.
In deciding whether v2([2](z)) = 1 for z ∈ Eb(Z2) −Eb(2Z2), we do not need to consider those parts of terms whose coefficients have valuation ≥2. The non-linear parts of each coefficient bi will contribute only terms with valuation≥2, so may ignore these and keep only the linear parts. The terms
bizi with i odd and greater than 1 we may discard altogether; by Lemma 1.3, all their coefficients have valuation ≥ 2. Finally, we may discard all terms bizi with i even and ≥6: a polynomial in Z[a1, . . . , a6] whose weight
1.5. Proof of the main theorem 17
is odd and at least 5 does not contain a linear term (there being no a5), so the terms involving z6, z8, z10, . . . will have valuation ≥2.
We thus get that, if z ∈Eb(Z2)−Eb(2Z2),
v2([2](z)) = 1 ⇔ v2(2z−a1z2−7a3z4) = 1. The last statement is true for all z ∈Eb(Z2)−Eb(2Z2) if and only if
v2 z− a1 2 z 2−7a3 2 z 4 = 0⇔a1+ 7a3 ≡0 mod 4⇔a1+a3 ≡0 mod 4 since z ≡z2 ≡z4 (mod 2). This proves the proposition.
1.5.2
The case
p= 3
Proposition 1.25. Let E/Z3 be a nice Weierstrass curve with its coeffi-
cients ai in 3Z3. Then E0(Q3) is topologically isomorphic to Z3 if a2 6≡ 6 (mod 9), and to 3Z3×Z/3Z otherwise.
Proof. We proceed as in the proof of Proposition 1.24, using the formal
triplication formula: [3](T) = 3T −3a1T2+ (a21−8a2)T3+ (12a1a2−39a3)T4+. . .= ∞ X i=1 biTi. (1.11) We consider the usual exact sequence for E0(Q3):
0→ E1(Q3)→ E0(Q3)→Z/3Z→0.
We see from E1(Q3) ∼= 3Z3 and Corollary 1.16 that E0(Q3) is topologically
isomorphic to 3Z3×Z/3Zif and only if for all elementsz ∈Eb(Z3)−Eb(3Z3),
[3](z) has valuation greater than 1. On the other hand, E0(Q3) is topo- logically isomorphic to Z3 if for all such z, the valuation of [3](z) is 1. Reasoning as in the proof of Proposition 1.24, we see that we may ignore all terms whose degree is not 1 and not a multiple of 3, since these have coefficients divisible by 3 and of positive weight. Also we may ignore the terms of degree both equal to a multiple of 3 and greater than 3, since their coefficients do not contain parts that are linear in a1, . . . , a6. Finally, we may ignore the non-linear part of the term of degree 3. We see that for
z ∈Eb(Z3)−Eb(3Z3), we have
18 Chapter 1. Elliptic curves overp-adic fields The last statement is true for all such z if and only if
v3 z−8a2 3 z 3 = 0 ⇔1− 8a2 3 6≡0 mod 3⇔a2 6≡6 mod 9 since z ≡z3 (mod 3). This proves the proposition.
1.5.3
The case
p= 5
Proposition 1.26. Let E/Z5 be a nice Weierstrass curve with its coeffi-
cients ai in 5Z5. Then E0(Q5) is topologically isomorphic to Z5 if a4 6≡ 10 (mod 25), and to 5Z5×Z/5Z otherwise.
Proof. For simplicity, we give the formal multiplication by 5 power series in the case where a1, a2, a3 are zero:
[5](T) = 5T −1248a4T5+. . .=
∞ X
i=1
biTi (1.12) This formula suffices for our purposes, since the same arguments as in the proofs of Propositions 1.24 and 1.25 show that the terms that are canceled by setting a1 =a2 =a3 = 0 could have been ignored anyway.
We apply Corollary 1.16 to:
0→5Z5 → E0(Q5)→Z/5Z→0.
In (1.12) we may ignore terms of degree not equal to 1 or 5, by the same reasoning as in the proofs of Propositions 1.24 and 1.25. We see that for
z ∈Eb(Z5)−Eb(5Z5) we have
v5([5](z)) = 1 ⇔ v5(5z−1248a4z5) = 1. The last statement is true for all such z if and only if
v5 z− 1248a4 5 z 5 = 0⇔1−1248a4 5 6≡0 mod 5⇔a4 6≡10 mod 25 since z ≡z5 (mod 5). This proves the proposition.
1.5. Proof of the main theorem 19
1.5.4
The case
p= 7
Proposition 1.27. Let E/Z7 be a nice Weierstrass curve with its coeffi-
cients ai in 7Z7. Then E0(Q7) is topologically isomorphic to Z7 if a6 6≡ 14 (mod 49), and to 7Z7×Z/7Z otherwise.
Proof. For simplicity, we give the formal multiplication by 7 power series
with a1, a2, a3 set to zero:
[7](T) = 7T −6720a4T5−352944a6T7+. . . (1.13) As before, the terms that have disappeared as a result could have been ignored anyway.
We apply Corollary 1.16 to:
0→7Z7 → E0(Q7)→Z/7Z→0,
In (1.13) we may ignore terms of degree not equal to 1 or 7, by the same reasoning as in the proofs of Propositions 1.24 and 1.25. We see that for
z ∈Eb(Z7)−Eb(7Z7) we have
v7([7](z)) = 1 ⇔ v7(7z−352944a6z7) = 1. The last statement is true for all such z if and only if
v7 z− 352944a6 7 z 7 = 0⇔1−352944a6 7 6≡0 mod 7⇔a6 6≡14 mod 49 since z ≡z7 (mod 7). This proves the proposition.
1.5.5
The proof
We are now ready to derive Theorem 1.1 from our previous results. In fact, we state a more general version of that theorem, since it is also valid for non-minimal Weierstrass equations.
Theorem 1.28. Let E/Zp be a nice Weierstrass curve given by
y2+a1xy+a3y=x3+a2x2+a4x+a6,
where the ai are contained in pZp for each i. Then there is a topological
20 Chapter 1. Elliptic curves overp-adic fields
(i) p= 2 and a1+a3 ≡2 (mod 4);
(ii) p= 3 and a2 ≡6 (mod 9);
(iii) p= 5 and a4 ≡10 (mod 25);
(iv) p= 7 and a6 ≡14 (mod 49).
Moreover, every isomorphism betweenE0(Qp)and Zp identifiesEn(Qp)with
pnZp for all n ∈ Z≥0. In each of the cases (i)-(iv), E0(Qp) is topologically
isomorphic to pZp×Z/pZ, where Z/pZ has the discrete topology.
Proof. The isomorphism type of E0(Qp) follows from applying part (ii) of Proposition 1.20 if p >7, or one of Propositions 1.24–1.27 if p≤7.
We claim that, if E0(Qp) ∼=Zp, then the isomorphism can be chosen in such a way that En(Qp) is identified with pnZp for all n ∈ Z≥0. For this, we choose the topological isomorphism χ: E1(Qp)
∼
→pZp from Lemma 1.22. By Corollary 1.17, the map χ extends to a topological isomorphism
e
χ: E0(Qp)
∼
→Zp
and we have pE0(Qp) =E1(Qp). It follows from Lemma 1.22 that pnE0(Qp) equals En(Qp); hence every group isomorphism E0(Qp)
∼
→ Zp will identify En(Qp) with pnZp. This concludes the proof.
Proof of Theorem 1.1. Theorem 1.1 follows by applying Theorem 1.28 to a
minimal Weierstrass equation of E
y2+a1xy+a3y=x3+a2x2+a4x+a6,
where the ai are contained in pZp for each i. Such an equation exists by
Lemma 1.18.