of Assumption 6.1 (3) that ρ¯ is ramified at 2. In fact, since A[λ] = λ−1Λ/Λ, we know
that λ−1(a) generates a ramified extension of Q
2. On the other hand, for any b∈T(Q4),
λ−1(N(b)) = λ′(pN(b)), where λ′ is an integral ideal of O such that λλ′ = (2). Since
Q2(
p
N(b))/Q2 is unramified, we know that λ−1(N(b)) generates an unramified extension
ofQ2. Thereforea is not of the form N(b), as desired.
6.4
Parity of 2-Selmer ranks
Since we are working in characteristic 2, to prove the Parity Conjecture 6.15, we not only need the local Tate pairing
h, iv:H1(Kv, V)×H1(Kv, V)→k(1),
but also a quadratic formQv giving rise to it. To defineQv, first recall that the line bundle
L=OE(2∞) onE induces a degree 2 map
E→P1=P(H0(E,L)).
For P ∈ E, let τP be the translation by P on E. Since for P ∈ E[2], τP∗L ∼= L, the
translation by E[2] induces an action of E[2]on P1, i.e., a homomorphism E[2]→PGL 2.
The short exact sequence
0→Gm→GL2→PGL2→0
induces the connecting homomorphism in nonabelian Galois cohomology
6.4. Parity of 2-Selmer ranks
Definition 6.21. We defineQto be the composition
Q:H1(K, E[2])→H1(K,PGL2)→H2(K,Gm).
For a placevofK, we denote its restriction by
Qv :H1(Kv, E[2])→H1(Kv,PGL2)→H2(Kv,Gm).
By local class field theory, H2(K
v,Gm)∼=Q/Zand soQv takes value inH2(Kv,Gm)[2]∼=
Z/2Z. By [O’N02, §4], Qv is a quadratic form and extending scalars we obtain a quadratic
form
Qv:H1(Kv, V)→k,
whose associated bilinear form (x, y) 7→ Qv(x+y)−Qv(x)−Qv(y) is equal to h , iv by
[O’N02, 4.3].
Definition 6.22. We say a subspaceW ⊆H1(Qv, V)istotally isotropic forQvifQv|W = 0.
We say W ismaximal totally isotropic if it is totally isotropic and W =W⊥.
Remark 6.23. Aschar(k) = 2, the requirementQv|W = 0is stronger thanh, iv|W = 0. For
example, for the 2-dimensional quadratic space(k2, Q)withQ((x, y)) =xy, the associated bilinear form is given by
h(x1, y1),(x2, y2)i=x1y2+x2y1.
In particular h(x, y),(x, y)i = 2xy = 0and hence all three lines in k2 are maximal totally isotropic for the bilinear form h , i. But only the two lines x = 0and y = 0 are maximal totally isotropic for the quadratic formQ.
Remark 6.24. The local conditionLv(E)is maximal totally isotropic forQvby [PR12, Prop.
6.4. Parity of 2-Selmer ranks
Lemma 6.25. Suppose Frobq∈GQq has order 2 acting on V. Then for any placev ofK,
Lv(A) is maximal totally isotropic for Qv.
Proof. The claim for v 6= q follows immediately from Lemma 6.20 and Remark 6.24. It remains to check the case v = q. We provide an explicit way to compute the image of a cocycle c ∈H1(Kq, E[2]) under Qq. Recall that H1(Kq,PGL2) classifies forms of P1, i.e.,
algebraic varieties S/Kq which become isomorphic to P1 over Kq. For any cocycle c, the
corresponding formScan be described as follows. As a set,S =P1(Kq). The Galois action
of g ∈GKq on x ∈S is given by g.x = c(g).g(x). The cocycle c gives the trivial class in
H1(Kq,PGL2)if and only if S(Kq)6=∅.
Since Frobq ∈ GQq has order 2 acting on V, we know that E[2](Qq) ∼= Z/2Z. Let
P ∈E[2](Qq) be its generator. Let σ ∈ GKq be a Frobenius and let τ be a generator of
the tame quotient Gal(Kqt/Kqur). Then by Lemma 6.20 (3), Lq(A)is generated by the two
cocycles
c(σ) = 0, c(τ) =P
and
c′(σ) =P, c′(τ) = 0.
For the cocycle c, the corresponding form S has a Kq-rational point if and only if there
exists x∈P1(Kt
q) such that
σ(x) =x, P.τ(x) =x.
Suppose E has a Weierstrass equation y2 = F(x), where F(x) ∈ Q(x) is an irreducible
cubic polynomial. Letα1, α2, α3 be the three roots ofF(x). We fix a embeddingK ֒→Kq
and view αi as elements in Kq. Without loss of generality, we may assume that α1 ∈Kq
and thus P = (α1,0). Then the action of P on P1 is an involution that swaps α1 ↔ ∞,
6.4. Parity of 2-Selmer ranks
transformation
x7→ α1x+ (α2α3−α1α2−α1α3) x−α1
.
Therefore Qq(c) = 0 if and only if there existsx∈P1(Kqt)such that
σ(x) =x, (τ(x)−α1)(x−α1) = (α1−α2)(α1−α3). (6.2)
The right hand side is the image ofα1−α2 under the norm mapKq×→Q×q, hence has even
valuation. Taking x =α1+
p
(α1−α2)(α1−α3) ∈Kq we see that Qq(c) = 0. Similarly,
Qq(c′) = 0 if and only if there exists x∈P1(Kqur)such that
(σ(x)−α1)(x−α1) = (α1−α2)(α2−α3).
Taking x=α2 ∈Kq we see thatQq(c′) = 0. Hence Lq(A)is maximal totally isotropic for
Qv.
Remark 6.26. The local condition Lq(A) may fail to be maximal totally isotropic for Qq
when workingoverQinstead ofoverK. This is expected since there is no parity prediction overQ(Remark 6.16).
Now we are ready to prove the Parity Conjecture 6.15 on the 2-Selmer ranks.
Theorem 6.27. Assume Assumption 6.1 and 6.13. Suppose Frobq ∈ GQq has order 2
acting on V. Then
s2(E/K) =s2(A/K)±1.
Moreover,
s2(E/K) =s2(A/K)−1
if and only ifresq(Sel2(E/K)⊗k)⊆Hur1(Kq, W), whereW is the unique GQq-stable line in
6.4. Parity of 2-Selmer ranks
Proof. Define the strict local conditions S bySv=Lv(E) =Lv(A)forv6=q and
Sq=Lv(E)∩ Lv(A) =Hur1(Kq, W)
(by Lemma 6.20). Similarly, define the relaxed local conditionsR byRv=Lv(E) =Lv(A)
forv6=w andRq=Sq⊥. Then we have
HS1(V)⊆HL(1 E)(V)⊆HR1(V), HS1(V)⊆HL(1 A)(V)⊆HR1(V).
SinceS⊥=R, we use [DDT97, Theorem 2.18] to compare the dual Selmer groups:
#H1 S(V) #H1 R(V) =Y v #Sv #H0(K v, V) , #H 1 R(V) #H1 S(V) =Y v #Rv #H0(K v, V) . It follows that dimHR1(V)−dimHS1(V) = 1 2(dimRq−dimSq) = 1,
sinceSqis 1-dimensional andRqis 3-dimensional. By global class field theory, for any class
c∈H1
R(V), we have
X
v
Qv(resv(c)) = 0.
Since Rv is totally isotropic for Qv for any v 6= q by Lemma 6.25, we know that that
Qq(resq(c)) = 0. In other words, the image resq(HR1(V)) is a totally isotropic subspace
for Qq. On the other hand, since Rq/Sq is a 2-dimensional quadratic space obtained by
extending scalars from a 2-dimensional quadratic space over F2 and Rq/Sq contains an
isotropic line with respect to Qv, we know that it has Arf invariant 0 and is isomorphic
to (k2, xy) as a quadratic space. By Remark 6.23 we know that there are exactly two
maximal totally isotropic subspaces ofRq containingSq, which must beLq(E)and Lq(A).
It follows that either H1
6.5. Parity of 2-Selmer ranks
hold simultaneously since
HL(1 A)(V)∩HL(1 E)(V) =HS1(V)(HR1(V).
So either
HL(1 A)(V) =HR1(V), HL(1 E)(V) =HS1(V),
or
HL(1 E)(V) =HR1(V), HL(1 A)(V) =HS1(V).
Moreover, the first case happens if and only ifresq(HL(1 E)(V))⊆ Sq. The desired result then
follows.
Remark 6.28. The conclusion of Theorem 6.27 may fail when Assumption 6.1 (4) is not satisfied due to the uncertainty of the local condition at 2. For example, the elliptic curve
E= 2351a1 :y2+xy+y=x3−5x−5
has trivialρ¯|GQ2. The elliptic curve
A= 25861i1 :y2+xy+y=x3+x2−17x+ 30
is obtained fromE via level raising at q= 11. For K=Q(√−111),
ralg(E/K) =s2(E/K) = 1 and ralg(A/K) =s2(A/K) = 4
6.5. Obstruction for rank lowering
6.5
Obstruction for rank lowering
It follows from Theorem 6.27 that if s2(E/K) = 1, then s2(A/K) = 0or 2. However, the
Chebotarev density argument forp≥5in [Zha14] fails in this case and does not show that one can always gets2(A/K) = 0. In fact, we prove the following very surprising obstruction
for rank lowering: s2(A/K) can never be lowered to zero!
Theorem 6.29. Assume Assumption 6.1 and 6.13. Suppose Frobq ∈ GQq has order 2
acting on V. Then
s2(E/K) = 1 =⇒s2(A/K) = 2.
Proof. By Theorem 6.27, we need to show that resq(Sel2(E/K)⊗k)⊆Hur1(Kq, W), where
W is the uniqueGQq-stable line inV. Notice that the action ofGal(K/Q)onE[2]induces
an action on Sel2(E/K). By assumption, we have Sel2(E/K) ∼= Z/2Z. So the action of
Gal(K/Q) onSel2(E/K) must betrivial. It follows that
resq(Sel2(E/K)⊗k)⊆Hur1(Kq, V)Gal(Kq/Qq)= Hom(Gal(Kqur/Kq), V)Gal(Kq/Qq)
= Hom(Gal(Kqur/Kq), W) =Hur1(Kq, W),
as desired.
We end with an example illustrating Theorem 6.29.
Example 6.30. Consider E = X0(11). In Table 6.2 we list the first few level raising
primes q and corresponding level raising abelian varieties A (dimA = 1). For each choice ofK=Q(√dK), we find thats2(A/K) = 2 always.
In most cases, this is explained by the fact thatralg(A/K) = 2. In the remaining cases,
there is a jump coming from the 2-part ofШfor the base changeA/K, dimШ(A/K)[2] = 2,
q A dK ralg(A/K) dimШ(A/K)[2] s2(A/K) 7 77a −8 2 0 2 7 77b −8 0 2 2 13 143a −7 2 0 2 13 143a −8 2 0 2 17 187a −7 2 0 2 17 187a −24 0 2 2 19 209a −7 2 0 2 19 209a −19 2 0 2 29 319a −8 2 0 2 29 319a −19 0 2 2
Table 6.2: Obstruction for rank lowering
though in all such cases the 2-part ofШforAand its quadratic twist AK are both trivial,
Ш(A/Q)[2] = 0, Ш(AK/Q)[2] = 0.
This is a phenomenon unique top= 2because for oddpit is always true that
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