Euler Systems
1
Heegner points (and BSD)
1.1 Introduction
Let
E
∼
Q, [F : Q] < ∞ ⇒
⊗QZr(E,F ) .
QE(F ) = E(F−s)tor
−1
−s
L-function, L(E, s) = p|N (1 − ap p )
+ p1−2s )−1
p-N (1 − ap p
be an elliptic curve over
E , we have an

p + 1 − #E(Fp ) p - N



0
E additive at p
, N = cond(E).
and ap =

1
split
multi
at
p



−1
non split multi at p
Q
Q p
Q
p
Note. L(E, 1) ” = ”
p|N ( ) ·
p Np where Np = #E(Fp )ns .
p-N p−ap +1 =
Given
Conjecture (Birch-Swinnerton-Dyer).
1.
ords=1 L(E, s) = rkZ E(Q) = r
2.
lims→1
L(E,s)
(s−1)r
= #X(E, Q)
Let
det(hPi ,Pj i)
#E(Q)2tor
E/Q
be an elliptic curve
Q
cv ,
v|N
where
{Pi }
are generators of
with
Re(s) > 3/2
E(Q).
that E is modular (now we know this is always true): there exists f ∈ S2 (Γ0 (N )) a newform f (q) =
P∞Suppose
P∞
n
−s
. Or equivalently ap (E) = ap (f ). This implies that L(E, s)
n=1 an q such that L(E, s) = L(f, s) =
n=1 an n
−s
has analytic continuation to C. It has a functorial equation ∧(E, s) = (2π)
Γ(s)N s/2 L(E, s). We have ∧(E, s) =
− ∧ (E, 2 − s) where ∈ {±1} and is determined by wN (f ) = · f where w is the Atkin-Lehner function. We shall
call − the sign(E, Q).
/ E/Q
non-constant morphism dened over Q.
f modular representation. φ : X0 (N )/Q
OO
AJ
(
Hecke−op
J0 (N ) = JacX0 (N )
1.2 Class eld theory
On ⊆ K be an order, On = Z + nOK , n ≥ 1 an integer.
rec : Pic(On ) ∼
= Gal(Kn /K) where Kn /K is an abelian extension
unramied away from n. Recall that Pic(On ) = I(n)/P (n) where I(n) = {fractional ideals coprime to n}, P (n) =
h(α) : α ∈ OK , α ≡ a mod nOK , a ∈ Zi. The map is dened by [p] 7→ Frob−1
p
Let
K⊆C
be an imaginary quadratic eld. Let
By class eld theory:
Lemma.
an
There is a map
∗
∗
Gn ∼
= Pic(On )/Pic(OK )(∼
= (OK /nOK ) / (Z/nZ) .
` + 1 ` inert in K
odd prime unramied in K , then G` is cyclic and [K` : K1 ] =
` − 1 ` split in K
Q
If n is square free then Gn ∼
G
.
= `|n `
Let
Gn = Gal(Kn /K1 ).
Then
1
In particular, if
`
is
1.3 Complex Multiplication
X0 (N ) classies (up to isomorphism) cyclic N -isogonies, A → A0 where ker(A → A0 ) is cyclic of order N . Let
K/Q imaginary eld satisfying Heegner Hypothesis: `|N ⇒ ` splits in K . Can pick an ideal N ⊆ OK such that
OK /N ∼
= Z/N Z (cyclic). Consider On = Z + nOK , χn = [C/On → C/Nn−1 ] ∈ X0 (N )(C) where Nn = N ∩ On
(⇒ On /Nn ∼
= Z/N Z) .
Theorem
σ|Kn
.
(Main Theorem of Complex Multiplication)
[aσ ].
Let
σ ∈ Aut(C/K)
σ|Kn ∈ Gal(Kn /K) ∼
= Pic(On )
−1 −1
χσn = C/a−1
σ → C/Nn aσ
σ|Kn = 1.
Remark. Suppose
Then
χσn = χn .
Hence
χn ∈ X0 (N )(Kn ).
χ0 (N )): For each P
` - N , T` is a correspondence on X0 (N )
[φ : A 7→ A0 ] 7→ C⊂A[`],cyclic subgp of order ` [A/C → A0 /φ(C)].
Trace map: Tr` : DivX0 (N )(Kn` ) → DivX0 (N )(Kn ).
Hecke action (On
DivX0 (N )(F )
T` : (DivX0 (N ))(F ) →
dened by
Also have the
Proposition.
so
1. As elements in
{χn }n , (n, N D) = 1
D = disc(K). Let ` - N D, then


if ` - n is inert in K
T` χn
DivX0 (N )(Kn ), Tr` (χn` ) = (T` − Frobλ − Frob−1
)χ
if ` = λλ - n split in K
n
λ


T` χn − χn/`
`|n
Consider
` - N is inert
X0 (N )(Fλn ).
2. If
in
K , λ = `OK , λn
where
is a prime in
Kn
above
λ.
Then
n
redλn` (χn` ) = redλn (χFrobλ
) ∈
n
1.4 Heegner points on E
Let
φ : X0 (N ) → E . yn := φ(χn ) is the Heegner point of conductor n
yK = TrK1 /K (y1 ) (in E(K)) is the basis Heegner point
Proposition.
E(Kn )).
` - ND :


a` yn
Tr` (yn` ) = (a` − Frobλ − Frob−1
λ )yn


a` yn − yn/`
Proposition.
(in
If
τ
is a complex conjugation, then there exists
if ` is inert in K, ` - n
redλn ` (yn` ) = redλn (ynFrobλn )
E(Fλn )
σ ∈ Gal(Kn /K) such that ynτ = ynσ
on
E(K)/E(K)tors
M 00 → M
is a continuous
.
2
Local Cohomology
2.1 Introduction to cohomology
Let
G
M a G-module. Both G and M have topology.
{f : G → M |f continuous and ∀g, h ∈ G, f (gh) = f (g) + gf (h)}
Are {f : G → M |f continuous and f (g) = g · m − m for some m ∈ M }
be a group and
1-cocycles: Are
1-coboundary:
Denition 2.1.
H 1 (G, M ) ={1-cocylce}/{1-coboundary}, H 0 (G, M ) = M G
Consider the short exact sequence of
G-modules, 0 → M 0 → M → M 00 → 0
and
section of sets. Then we get a long exact sequence
0 → M 0G → M G → M 00G → H 1 (G, M 0 ) → H 1 (G, M ) → H 1 (G, M 00 ) → H 2 . . .
We have some useful maps between cohomology groups:
2
•
Let
H ≤ G,
•
Let
H CG
we get a map
Res : H 1 (G, M ) → H 1 (H, M ).
be a normal closed subgroup of
G,
then we have
inf : H 1 (G/H, M H ) → H 1 (G, M )
We have the following relationship between inf and Res called the Hochschild-Serre spectral sequence
inf
Res
0 → H 1 (G/H, M H ) → H 1 (G, M ) → H 1 (H, M )G/H → H 2 (G/H, M H ) → . . .
2.2 Galois Cohomology
Fox prime
`,p
and
|K : Q` | < ∞
and let
be the Inertia group
Denote by
T
Let
GK = Gal(K/K)
to
If
un
∼
∼b
Gun
K = Gal(K /K) = GK /IK (= Z)
Fp -vector space with discrete GK action.
K
and let
IK = Gal(K/K un )
We say that
T
is unramied if
IK
acts
T.
There is a perfect pairing of
Fact.
be the maximal unramied extension of
and
be a nite dimensional
trivially on
K un
` 6= p and T
Fp -vector
is unramied then
spaces
H 1 (GK , T ) ⊗Fp H 1 (GK , T ∗ ) → Fp
where
T ∗ = Hom(T, Mp∞ ).
∗
1
un
H 1 (Gun
K , T ) and H (GK , T ) are exact orthogonal complements with respect
h , i.
Denition 2.2.
F for T is a choice of Fp -subspace of H 1 (GK , T )
1
:= H (GK , T )/Hf,F
(GK , T ) the singular quotient.
A local Selmer structure
1
We call the quotient HS,F (GK , T )
denoted
1
Hf,F
(GK , T ).
1
1
1
0 → Hf,F
(GK , T ) → H 1 (GK , T ) → HS,F
(GK , T ) → 0 (†)
We say
F
is an unramied structure if
Using the Tate pairing we dene the
IK
1
) (via inf ).
(GK , T ) = H 1 (Gun
Hf,F
K ,T
∗
∗
Dual Selmer structure F on T to be
In this case
(†) identies with inf-res.
the exact orthogonal complement of
1
Hf,F
(GK , T ).
In particular
1
HS,F
(GK , T ) ⊗Fp Hf,F ∗ (GK , T ∗ ) → Fp
we get an induced perfect pairing.
2.3 Local cohomology for elliptic curves
Let
(⇒
E be an elliptic curve over K . We let T = E[p] = E(K)[P ]. We
E[P ]∗ ∼
= E[P ]). We have a short exact sequence of GK -module
have the Weil pairing
E[P ] ⊗ Fp E[P ] → µp
p
0 → E[P ] → E(K) → E(K) → 0
If we take the associated long exact sequence
0 → E[P ]GK → E(K)GK → E(K)GK → H 1 (GK , E[P ]) → H 1 (GK , E(K)) → . . . (††)
From
(††)
we get:
δ
0 → E(K)/pE(K) → H 1 (GK , E[p]) → H 1 (GK , E(K))[p] → 0
where
sends
δ is dened as follows: for Q ∈ E(K)
σ ∈ GK to σ(L) − L ∈ E[P ].
Denition 2.3.
Fact.
x
L ∈ E(K)
The geometric local Selmer structure
F
on
such that
E[P ]
pL = Q.
Then
is the image of
δ
in
δ(Q)
is the
1-cocycle
which
H 1 (GK , E[p]).
The dual geometric local Selmer structure is the geometric local Selmer structure (it make sense since
E[P ]∗ ∼
= E[P ])
If E has good
reduction over
K
and
` 6= p.
Then
E[p]
is an unramied
agrees with the unramied structure.
3
GK -module
and the geometric structures
3
Global Section
In this talk
K
will be a number eld,
Gv .
GK × T → T is
v
K , Kv the completion of K at v , Gv = Gal(K v /Kv ) and Iv
T to be a nite dimensional Fp -vector space with a discrete
given the discrete topology. We choose an embedding K ,→ K v
a place of
the inertia subgroup of
As in the last talk we let
GK -action.
continuous when
Gv ,→ GK .
dimensional and GK acts
T
is
which gives us an embedding
Since
T
is nite
v
For each place
T is unramied
Resv : H 1 (K, T ) → H 1 (Kv , T ).
discretely, this implies that
we have a restriction map
almost everywhere.
3.1 Selmer Groups
Denition 3.1.
place
v
F
A global Selmer structure
on
T
is a choice of a local Selmer structure
1
Hf,F
(Kv , T )
for each
1
Hf,F
(Kv , T ) = H 1 (Kvunr , T Iv ) almost everywhere.
1
1
group SelF (K, T ) of F is dened to be ker H (K, T ) → ⊕v Hs,F (Kv , T ) .
such that
The Selmer
If we take
E
to be an elliptic curve over
Denition 3.2.
K
and
T = E[p].
The geometric global Selmer structure
obtained by setting
1
Hf,F
(Kv , T )
F
on
E[p]
is dened to be the global Selmer structure
to be the local geometric structure at each
v.
SelF (K, E[p]) = Sel(p) (E).
In this setting
3.2 Global Duality
Denition 3.3.
∗
∗
Let F be a global Selmer structure on T . The Cartier dual Selmer structure F
on T
=
1
∗
HomFp (T, µp ) is dened to be the global Selmer structure obtained by setting Hf,F
(K
,
T
)
to
be
the
local
Cartier
∗
v
dual Selmer structure.
F and F ∗ and start omitting it from notation.
1
1
Given an ideal a C OK , we dene Sela (K, T ) = {c ∈ H (K, T ) : cv ∈ Hf,F (Kv , T ) for all v - a}.
a
Sel (K, T ) = {c ∈ SelF (K, T ) : cv = 0, ∀v|a}. This gives us the following exact sequences
We now x an
1
0 → Sel(K, T ) → Sela (K, T ) → ⊕v|a HS,F
(Kv , T ) → 0
1
0 → Sela (K, T ∗ ) → Sel(K, T ∗ ) → ⊕v|a Hf,F
(Kv , T ∗ ) → 0
We have
1
1
∗ ∨
⊕v|a Hs,F
(Kv , T ) ∼
= ⊕v|a Hf,F
∗ (Kv , T ) . We can put the two sequences above together as
M
1
0 → Sel(K, T ) → Sela (K, T ) →
Hs,F
(K, T ) → Sel(K, T ∗ )∨ → Sela (K, T ∗ )∨ → 0
follows
v|a
Proposition 3.4.
The above sequence is exact.
The exactness of this sequence yields
1
0 → ⊕v|a Hs,F
(Kv , T ) im(Sela (K, T )) → SelF ∗ (K, T ∗ )∨ → Sela (K, T ∗ )∨ → 0
3.3 Bounding Sela
p 6= 2. If τ ∈ GK is an involution, we have a decomposition T = T + ⊕ T − , where T = {t ∈ T : τ (t) =
t}. We say that τ is non-scalor if T + 6= 0, T − 6= 0.
From now on, T will be assumed to be irreducible. L/K = nite Galois extension such that the action of GK
on T factors through Gal(L/K).
1
a
1
If S ⊆ H (K, T ) nite dimensional Fp -subspace S = {s ∈ S : sv ∈ Hf,F (Kv , T ) and (sV = 0∀v|a)}. We can
1
nd a nite Galois extension M/L such that S ⊆ inf(H (M/K, T )). Assume that τ ∈ Gal(L/K) is a non-scalar
involution and that it extends to an involution in Gal(M/K). Let {γ1 , . . . , γr } be a set of group generators for
Gal(M/L).
From now on
4
Proposition 3.5.
K
ω1 , . . . , ωr
Let
a = v1 · · · vr .
and set
M such that FrobM/K (wi ) = τ γi .
S ⊆ inf(H 1 (L/K, T )).
be places of
a
Then we have
Let
vi
be the restricted places to
E be an elliptic curve over Q without CM, T = E[p]. ρ : GK GL2 (Fp ), H 1 (Q(E[p])/Q, E[p]) = 0.
Let L0 /K be an extension such that GK factors through Gal(L0 /K), and we assume that K is a quadratic
extension of K0 . We also assume that the action Gal(L0 /K) is a restriction of a Gal(L0 /K0 )-action on T . We have
Let
M
L
L0
K
2
K0
S
H 1 (M/K, T ) and τ will be
Gal(K/K0 ). By the action of τ
will be a nite dimensional subspace of
projects to the non-trivial element of
a non-scalor involution in
on
H 1 (M/K, T ),
Gal(M/K0 )
which
we have the decomposition
H 1 (M/K, T ) = H 1 (M/K, T )+ ⊕ H 1 (M/K, T )−
We further assume that
again let
{γ1 , . . . , γr }
Proposition 3.6.
to
K.
4
Set
S ⊆ H 1 (M/K, T )
be a set of generators of
Let
a = v1 · · · vr ,
∈ {±}.
Gal(M/L).
for some
w1 , . . . , wr be places of M such
a
1
then S ⊆ inf(H (L/K, T ) ).
that
Let
σ ∈ Gal(M/L0 )
FrobM/K (wi ) = τ σγi
be such that
and let
vi
τ στ −1 = σ −1 .
We
be the restrictions of
wi
Calculations in Galois Cohomology
Let
E be an elliptic curve over Q.
K/Q an imaginary quadratic extension
Fix yk ∈ E(K) a Heegner point
Theorem 4.1.
Let
p≥3
E
2.
Gal(Q(E[p])/Q) ∼
= GL2 (Fp )
3.
yk ∈
/ pE(K)
has good reduction at
Sel(K, E[p])
is of
Fp
Condition 2.) implies
Cond(E)
split.
be such that
1.
The
where all primes dividing
p
dimension
1
and generated by
κ(yk ).
Gal(K(E[p])/K) ∼
= GL2 (Fp )
Fact. Under complex conjugation yk 7→ yk (up to torsion) where is the global root number.
Recall: Sela (K, E[p]) restricted.
Sela (K, E[p])
(⇒
yk ∈ (E(K)/pE(K))
relaxed.
this gives the exact sequence
Sela (K, E[p]) →
M
Hs1 (Kv , E[p]) → Sel(K, E[p])∨ → Sela (K, E[p])∨ → 0
v|a
Complex conjugation commutes with the maps so splits into a
Let
L0 = K(E[p]).
For
z
such that
Lemma 4.2.
Gal(L/L0 )
There exists an ideal
−1
such that τ στ
= σ −1 .
We can choose
a
such that
a
a
pz = yk
of
K
let
+
part and the
−
part.
L = L0 (z).
such that
Sela (K, E[p])± ⊆ H 1 (L/K, E[p])± .
is divisible by primes lying above primes of
5
Q
with
Moreover if we x
FrobL/Q ∼ τ σ .
σ ∈
)
4.1 Preliminaries
Lemma 4.3.
Let
`
be a prime of
1.
E
2.
` 6= p
3.
FrobK(E[p])/Q ` ∼ τ
has good reduction at
HS1 (Kλ , E[p])±
Then
such that
`
is a 1-dimensional
Fp -vector-space.
mod p representation. Then Neron-Ogg-Shar implies that
K are unramied in K(E[p]). FrobK/Q ` ∼ τ 6= id, the residue
class of ` in K/Q is 2, and the residue class degree of ` in K(E[p])/K is 1. Hence ` splits completely in K(E[p])/K .
So K(E[p])λ = K` , i.e., E[p] ⊆ E(K` ).
Proof.
E[p]
K(E[p])
Q
is the xed eld of the kernel of the
is unramied at
`
and so all primes above
`
in
HS1 (K` , E[p])
unr
G
∼
= H 1 (I` , E[p]) K`
= HomGunr
(I` , E[p])
K
`
=
HomGunr
(I` /pI` , E[p])
K
`
I` /pI` ∼
= Gal(K`unr (`1/p )/K`unr ) ∼
= µp . HS1 (K` , E[p]) ∼
= Hom(µp , E[p]).
±
∓
complex conjugations. We also get Hom(µp , E[p]) → E[p] .
Everything done so far commutes with
4.2 − eigenspace
Lemma 4.4.
H i (L0 /K, E[p]) = 0
Lemma 4.5.
H 1 (L/K, E[p])
Lemma 4.6.
Gal(L/L0 )
Consider
H 1 (K, E[p])
is
for all
i
1-dimensional Fp
is isomorphic to
Gal(L/L0 ) ,→ E[p]
E[p]
as
-vector-space generated by
κ(yK ).
Gal(L0 /K)-modules.
Gal(L/L0 )±
σ : {z 7→ z + q} 7→ q .
H 1 (L0 , E[p])Gal(L0 /K) .
dened by
under the restriction to
Proposition 4.7.
exists
Let ` be a prime of Q with FrobL0 /Q ∼ τ which
c(`) ∈ Selλ (Kλ , E[p])± such that c(`)sλ 6= 0 in Hλ1 (Kλ , E[p]).
Theorem 4.8.
And
is
1-dimensional.
This map is actually the image of
does not split completely in
L/L0 .
κ(yK ) ∈
Then there
Sel(K, E[p])− = 0.
1 6= σ ∈ Gal(L/L0 ) such that τ στ = σ −1 . There exists a such that Sela (K, E[p])− ⊆ H 1 (L/K, E[p])− . But
the second thing is 1-dimensional and is generated by κ(yK ). If we pick a such that it is divisible only by primes
1
−
lying over `1 , . . . , `r ∈ Z with FrobL/Q Li ∼ τ σ . ⊕li |a Hi (K`i , E[p])
has dimension r , but we use the fact there
s
exists c(`1 ) such that c(`i )λ 6= 0 with linear independent images. Hence the cokernel is 0.
Proof.
5
Finishing the proof
5.1 Notation and Recap
Notation.
• E
an elliptic curve of conductor
• φ : X0 (N ) → E
• K
N
a modular parametrization
an imaginary quadratic eld with discriminant
D,
6
in which every prime dividing
N
splits.
• p
•
a prime at which
Assume
• τ
E
has good reduction.
Gal(Q(E[p])/Q) ∼
= GL2 (Fp ).
complex conjugation
• Kn
•
Let
K
the ring class eld of
unramied away from
` ∈ Z
of conductor
be a prime such that
completely in
Kn
n,
which (among other properties) is an abelian extension
Kn /K
n.
`
K/Q and splits in K(E[p])/K . Setting λ = `OK , λ
Kn` . In the case n and ` are coprime we get:
is inert in
and is totally ramied in
splits
λn` Kn`
tot ramified
K`
λn Kn
K1
tot splits
λ K
2
inert
` Q
•
Note that
• redλn
Kn,λn = Kλ .
the reduction map
E(Kn,λn ) → E(Fλ ).
We have the short exact sequence
0 → Hf1 (K, E[p]) → H 1 (K, E[p]) → Hs1 (K, E[p]) → 0,
imκ.
E
We also recall that if
has good reduction at
then
G
Hs1 (Kλ , E[p]) ∼
= Hom(Iλ , E[p]) Kλ ,
Theorem 5.1.
There
For any integer n not dividing N D there exists a point yn ∈ E(Kn )
Tr` yn` = a` yn ∈ E(Kn )
and is inert in K , redλn` (yn` ) = redλn (FrobKn /K λn · yn )
exists σ ∈ Gal(Kn /K) such that τ yn = σyn in E(K)/E(K)tors .
5
with
Gun
Kλ =
such that
then
• L0 = K(E[p]),
This
Hf1 (K, E[p]) =
un
`,
Gal(Kλun /Kλ ).
` - ND
` - ND
with
For
z
such that
pz = yk
let
L = L0 (z).
weeks have been building towards proving
Theorem.
• E
Let
p
be an odd prime such that:
has good reduction at
p
• Gal(Q(E[p])/Q) ∼
= GL2 (Fp )
• yK ∈
/ pE(K)
Then
Sel(K, E[p])
has order
p;
it is generated by the image of
yK
under the Kummer map.
This was done by using the cohomology tools that Chris and Pedro set up, to bound
restricted and relaxed
a
Sel (K, E[p])
and
Sela (K, E[p]).
Sel(K, E[p])±
using the
Alex proved this last week under the assumption of the
following two proposition, which we will prove this week.
Proposition.
in
L/L0 .
Proposition.
L0 /L.
Assume that
There there exists
yk ∈
/ pE(K). Let ` ∈ Q be a prime with FrobL0 /Q ` ∼ τ
c(`) ∈ Selλ (K, E[p])− with c(`)sλ 6= 0 in H 1 (Kλ , E[p])
yk ∈
/ pE(K). Let ` ∈ Q be a prime with FrobL/Q ` ∼ τ
c(q`) ∈ Selλ (K, E[p]) with c(q`)sλ 6= 0 in H 1 (Kλ , E[p]).
Assume that
There there exists
and
and
`
`
not splitting completely
not splitting completely in
We will prove this using the Heegner points that Marc introduced in the rst week to construct a class in
H 1 (K, E[p])
that satisfy those conditions.
7
5.2 The derivative operator
Let
R
K/Q
Lemma 5.2.
and splits in
For all primes
E[p],
Proof. As on
mod
gcd(n, pN D) = 1 and
K(E[p])/K , and that E has good
be the set of square free integers,
is inert in
` ∈ R, p|` + 1
`|n
then
FrobK(E[p])/Q ` ∼ τ .
`.
This implies that
`
reduction at
p|a` = ` + 1 − #(F` )
and
complex conjugation and
if
Frob
are conjugate, their characteristic polynomial must be the same
p
Recall that, for
Fix a generator
` ∈ R, G` = Gal(K` /K1 ) is
σ` and for a prime ` dene
• D` =
P`
iσ`i ,
• T` =
Pl
σ`i .
i=1
i=0
•
For
n∈R
•
Let
γi
we dene
Dn =
Q
`|n
D`
For any
n∈R
we have
Q
Gn ∼
= `|n G` )
P
Gal(Kn /K). T = i γi ,
Gn
in
Q
Gn = `|n G` , we just need to show that (σ` − 1)Dn yn ∈ pE(Kn ).
(σ` − 1)D` = ` + 1 − T` , Theorem 5.1 and Lemma 5.2.
Pn = TDn yn ∈ E(Kn ).
`
is inert in
K ).
T
−1
=
P
i
γi−1 .
Dn γn ∈ (E(Kn )/pE(Kn ))Gn .
Proof. As
We dene
(as
(using the fact that
be a set of coset representative for
Lemma 5.3.
`+1
cyclic (Marc's talk) of order
If we consider
This is calculations using the identity
Pn in E(Kn )/pE(Kn ), then by the above it is Gal(Kn /K)-invariant
and independent of the choice made for T.
Consider the following, we want to get an element in
H 1 (K, E[p])
from
Pn ∈ E(Kn )/pE(Kn )
E(Kn )/pE(Kn )
H (Kn , E[p])
κ
1
H 1 (K, E[p])
As the Kummer map is equivariant,
Lemma 5.4.
/ H 1 (Kn , E[p])Gal(Kn /K)
res
κ(Pn ) ∈ H 1 (Kn , E[p])Gal(Kn /K) .
The restriction map is an isomorphism
H i (Kn /K, E[p]Gal(Kn /K) ) for i = 1, 2
for n ∈ R, these two groups are trivial.
Proof. Using long exact sequence have that the kernel and cokernel of res is
respectively. Using the non-trivial fact that
E
has no
Kn -rational p-torsion
Hence the restriction map is an isomorphism
So we let
Recap:
c(n) ∈ H 1 (K, E[p])
c(n) = κ(Pn ).
c(n) ∈ H 1 (K, E[p])
the + or − space
be such that
At this point we have constructed a class
yn ∈ E(Kn ).
Lemma 5.5.
We now show it lies in one of
Let
n∈R
have
k
prime factors. Then
which comes from an Heegner point
k
c(n) ∈ H 1 (K, E[p])(−1)
8
.
τ
Proof. As
res, we show τ Pn = (−1)k Pn
Y
= τT
D` yn Definition
commutes with
τ Pn
κ
and
in
E(Kn )/pE(Kn ).
`|n
=
T
−1
Y
T
−1
Y
τ D` yn by τ T = T−1 τ
`|n
=
(`T` − σ` D` ) τ yn
`|n
=
−1
T
Y
(`T` − σ` D` )σyn Lemma 5.1
`|n
Y
= T−1 (−σ` D` )σyn since T` yn = a` yn/` = 0 in E(Kn )/pE(Kn )
`|n
k
=
(−1) σ
Y
σ` T−1 Dn yn
`|n
Then using the fact that
Dn yn
is
Gn -invariant,
and
Pn
is
Gal(Kn /K)-invariant,
this collapse down to what we
want.
Lemma 5.6.
Proof. Let
ν
Fix
n = `1 . . . `r R,
be a place of
K
then
c(n) ∈ Selλ1 ...λr (K, E[p])
distinct from
λi
such that
E
has good reduction at
ν
(The proof of ν has bad
c(n)ν ∈ Hf1 (Kν , E[p]), we
w be a place of Kn above ν
reduction is more involved and uses tools we have not developed). Instead of showing
c(n)sν = 0 in Hs1 (Kν , E[p]). We
note that Kn,w /Kν is unramied.
show
and
unr
Hs1 (Kν , E[p]) = Hom(Iν , E[p])GKν . Let
Hence its inertia group is also Iν , and using the
have
exact sequence we get the
following commuting diagram:
E(Kν )/pE(Kν )
E(Kn,w )/pE(Kn,w )
/ Hom(Iν , E[p])
res
/ H 1 (Kn,w , E[p])
κ
/ Hom(Iν , E[p])
Gunr
unr
Hom(Iν , E[p])GKν ⊆ Hom(Iν , E[p]) Kn,w ⊆ Hom(Iν , E[p]).
s
s
denition resc(n)ν = κ(Pn ), so (resc(n)ν ) = 0. Hence c(n)ν = 0.
where we use
By
/ H 1 (Kν , E[p])
κ
Then it is jut a matter of diagram chasing.
As Pn` ∈ E(Kn` ) is Gal(Kn` /K)-invariant in E(Kn` )/pE(Kn` ), and G` = Gal(K` /K1 ) ≤ Gal(Kn` /K), we have
(σ` − 1)Pn` ∈ pE(Kn` ). Setting
`+1
a`
Qn,` =
TDn γn` − Pn (†)
p
p
(which makes sense by Lemma 5.2) we see
has no
Kn` -rational p-torsion
Lemma 5.7.
redλn` (Qn,` )
Proof. First we show
for
σ = 1,
but as
point,
Qn,l
is trivial in
pQn,` = (σ` − 1)Pn`
is the unique point in
E(Fλ )
if and only if
(using Lemma 5.3). Note that by the fact that
Kn`
Pn ∈ pE(Kλ )
redλn` (σyn` ) = redλn (FrobKn /K λn · σyn ) for all σ ∈ Gal(Kn /K).
5.1 with the ideal σ −1 λn .
This is true by Theorem 5.1
and we use Theorem
redσ−1 λn` (yn` )
=
redσ−1 λn (FrobKn /K (σ −1 λn ) · yn )
redλn` (σyn` )
=
redλn (σFrobKn /K (σ −1 λn ) · yn )
FrobKn /K (σ −1 λn ) = σ −1 FrobKn /K λn σ
we are done. By the denition of
Dn and
redλn` (TDn λn` ) = redλn (FrobKn /K λn · TDn yn )
9
E
with that property.
T, we have
Hence combining this with
(†)
we get
redλn` (Qn,` ) = redλn
`+1
a`
FrobKn /K λn −
p
p
Pn
(` + 1)FrobKn /K λn − a` annihilates E(Fλ ). Note that E(Fλ ) = E(Fλ )+ ⊕ E(Fλ )− as FrobKn /K λn is an
+
−
involution. But E(Fλ ) = E(F` ) which by denition has order ` + 1 − a` . Then by the Weil conjectures E(Fλ )
has order l + 1 + a` .
Now FrobKn /K λn is the reduction of a complex conjugation, so FrobKn /K λn Pn = τ Pn , which by the proof of
k
Lemma 2.4 τ Pn = (−1) Pn in E(Kn )/pE(Kn ), so FrobKn /K λn Pn = νPn + pQ for some ν ∈ {±1} and Q ∈ E(Kn ).
Claim:
redλn` (Qn,` ) =
(` + 1)ν − a`
redλn (Pn ) ∈ E(Fλ )ν /pE(Fλ )ν
p
p-primary part of E(Fλ )ν is cyclic. First we note that p|(l + 1) ± a` , so we have at least
p p-torsion point in E(Fλ )± . Suppose we had more than p p-torsion point in one of E(Fλ )± , then we would
2
2
+
−
have p of them, but p > |E(Fλ ) [p]| + |E(Fλ ) [p]| = |E(Fλ )[p]|which is a contradiction (as ` is a good prime,
E(Fλ )[p] = p2 ). We can conclude that E(Fλ )ν /pE(Fλ )ν is cyclic of order p.
Hence redλn` (Qn,` ) = 0 if and only if redλn (Pn ) ∈ pE(Kn,λn ) = pE(Kλ ). But the kernel of redλn is pro-`, which
is coprime to p, so Pn ∈ pE(Kλ )
We now claim that the
Lemma 5.8.
Let
n` ∈ R
with
`
c(n`)sλ = 0
prime. Then
if and only if
Pn ∈ pE(Kλ ).
un
Proof. Since
` ∈ R we have Hs1 (Kλ , E[p]) = Hom(Iλ , E[p])GKλ
We shall show this factors as follows:
so we can view
c(n`)sλ as a homomorphism Iλ → E[p].
/ E[p]
O
Iλ
(
Iλ /Iλn` ∼
= G`
To see this we consider the diagram
H 1 (Kλ , E[p])
E(Kn` )/pE(Kn` )
Then
c(n`)
κ
res
/ H 1 (Kn`,λ , E[p])
n`
c(n`)(Iλn` ) = 0.
` - n), we see that G`
lies in the image of the Kummer map, hence
Looking at the diagram in the introduction (since
λ.
/ Hom(Iλ , E[p])
nl
is the inertia group of
Gal(Kn` /K)
at
Hence considering the diagram
Kλ
Iλn
un
Kn`,λ
n`
Iλ
Kn`,λn`
G`
Kλun
G`
Kλ = Kn,λn
Iλ /Iλn` ∼
= G` . Hence c(n`)sλ = 0 if and only if c(n`)(σ` ) = 0.
1
Fix Q ∈ E(K) such that pQ = Pn` , then we claim that the cocycle GK → E[p] given by σ 7→ (σ − 1)Q − (σ −
p
1
1)Pn` represents c(n`) (where p (σ − 1)Pn` is the unique point in E(Kn` ) which is a pth root of (σ − 1)Pn` ). This
we have
10
E[p]. To see it represents c(n`) one checks that resc(n`) = κ(Pn` ), but κ(Pn` ) is just
σ 7→ (σ − 1)Q and for all σ ∈ GKn` (σ − 1)P = 0.
Now c(n`)(σ` ) = (σ` − 1)Q − Qn,` (as we had pQn,` = (σ` − 1)P ).
Fix a prime λ of K over λ and consider red : E(K) → E(k λ ). As E has good reduction at p, this map is injective
on p-torsions, so c(n`)(σ` ) = 0 if and only if red((σ` − 1)Q − Qn,` ) = 1. Note that red((σ` − 1)Q) and red(Qn,` ) are
p-torsion (we had red(Qn,` ) = p1 ((l + 1)ν − a` )red(Pn )), and σ` lies in inertia and hence acts trivially, we have that
red((σ` − 1)Q) = 0. So red((σ` − 1)Q − Qn,` ) = −red(Qn,` ).
Hence c(n`)(σ` ) = 0 if and only if red(Qn,` ) = 0 if and only if Pn ∈ pE(Kλ )
expression is easy to see is in
the cocycle
Proposition.
in
L/L0 .
Assume that
There there exists
yk ∈
/ pE(K). Let ` ∈ Q be a prime with FrobL0 /Q ` ∼ τ
c(`) ∈ Selλ (K, E[p])− with c(`)sλ 6= 0 in H 1 (Kλ , E[p])
and
`
not splitting completely
Proof. We let c(`) be the class dened by the Heegner point yk . Then Lemma 5.5 and 5.6 shows that c(`) ∈
Selλ (K, E[p])− . Lemma 5.8 tells us that c(`)sλ 6= 0 if and only if P1 ∈
/ pE(Kλ ).
But P1 ∈ E(K) is the point yk by denition. Since by denition L is the minimal extension of L0 which yk is
divisible by p, we have that yk is divisible by p in E(Kλ ) if and only if λ splits completely in L/L0 . By assumption
it does not, hence yk ∈
/ pE(Kλ ).
For the
Sel(K, E[p])
space we need to reintroduce some work done by Alex. Let
c(q) ∈ Selq (K, E[p])− and c(q)sq 6= 0.
1
0
dened, i.e., c(q) ∈ H (L , E[p]).
previous proposition, so
c(q) ∈ H 1 (K, E[p])
Lemma 5.9.
Proposition.
L0 /L.
Let
is
`
be a prime with
FrobL/Q ` ∼ τ .
Then
We let
L0
Pq ∈ pE(Kλ )
if and only if
yk ∈
/ pE(K). Let ` ∈ Q be a prime with FrobL/Q ` ∼ τ
c(q`) ∈ Selλ (K, E[p]) with c(q`)sλ 6= 0 in H 1 (Kλ , E[p]).
Assume that
There there exists
q
be a prime satisfying the
be the smallest extension of
`
and
L
such that
splits completely in
`
L0 /L.
not splitting completely in
Proof. By the same argument as above we have c(q`) ∈ Selqλ (K, E[p]) . Then using the previous lemma and Lemma
s
1
5.8 we have that c(q`)λ 6= 0 in Hs (Kλ , E[p]).
s
It remains to show that c(q`) ∈ Selλ (K, E[p]) or equivalently that c(q`)q = 0. As FrobL/Q ` ∼ τ we have that `
−
splits in L/L0 , hence γK ∈ pE(Kλ ) using the proof of the above proposition. So considering c(`) ∈ Selλ (K, E[p])
,
s
−
= 0 as Alex proved last week. Hence c(`) = 0
by Lemma 2.7, we have c(`)λ = 0 and hence c(`) ∈ Sel(K, E[p])
itself, and since by construction resc(`) = κ(P` ) = 0, we must have P` to be 0 in E[K` ]/pE[K` ], i.e., P` ∈ pE(K` ).
s
So P` ∈ pE(K`,q ), hence by Lemma 2.7 c(q`)q = 0.
6
__ (Angelos)
6.1 Notation
• p, `
• K
are primes such that
is a eld and
• GK = Gal(K
sep
Kv
p
is xed
p 6= `
denotes the completion of
K
at
v
/K)
• A is a GK -module and A∗ = Hom(A, µ∞ ).
∗
∗ m
(A = A1 ), Am = A [p ]
In general,
A is a free Zp -module of nite rank then Am = A/pm A
• H i (K, A) = H i (Gm , A)
Comment: Since,
i > 1.For A∗
A
may not be a discrete
GK -module,
then it doesn't hold that
everything is ne since it has discrete topology
11
H i (GK , A) = lim H i (GK , Am )
←−
for
6.2 Dualing
Theorem 6.1.
A
Suppose
is nite
GK -module,
then the pair
invK
∪
H 1 (GKv , A) × H 1 (GKv , A∗ ) → H 2 (GKv , µ∞ ) → v Q/Z
is perfect
Remark.
Hf1 (Q` , E[pm ])⊥ = Hf1 (Q` , E[pm ])
F is Selmer structure of A then we get a Selmer structure for Am just
H (Kv , Am ) . This denes a Selmer structure for A∗m and from that we
HF1 ∗ (Kv , A∗ ) = lim HF1 ∗ (Kv , A∗m ).
←−
If
1
Proposition 6.2.
It holds that
Denition 6.3.
F, G
Note that if
If
G⊂F
HF1 (K, A) = lim HF1 (K, Am )
←−
are Selmer structure of
A,
and
we say that
using the canonical map
dene Selmer structure
H 1 (Kv , A) →
F ∗ for A∗ by
HF1 ∗ (K, A∗ ) = lim HF1 (K, A∗m ).
−→
G⊂F
if
HG1 (Kv , A) ⊂ HF1 (Kv , A)
for all
v.
then
• HG1 (K, A) ⊂ HF1 (K, A)
• F ∗ ⊂ G∗
Theorem 6.4.
F1 , F2
Suppose
are Selmer structure for a nite
⊕res
L
⊕res
L
• 0 → HF1 1 (K, A) → HF1 2 (K, A) →
• 0 → HF1 2∗ (K, A) → HF1 1∗ (K, A) →
Summing over
v
such that
GK -module A
and
F1 ⊂ F2 .
Then
HF1 2 (Kv , A)/HF1 1 (Kv , A)
HF1 1∗ (Kv , A)/HF1 2∗ (Kv , A)
HF1 1 (Kv , A) 6= HF1 2 (Kv , A).
The images of
⊕res
are orthogonal complement of each other
with respect to the Tate pairing.
From now on
K
Denition 6.5.
will be
Q.
The canonical Selmer structure
Fcon
for
A
is dened to be
•
If
v ∈ {∞, p}
then
HF1 con (Qv , A) = H 1 (Qv , A)
•
If
v∈
/ {∞, p}
then
HF1 con (Qv , A) = ker[H 1 (Qv , A) → H 1 (Qun
v , A) ⊗ Qp ].
Proposition 6.6.
If
Proposition 6.7.
If
F = Fcon
F = Fcon
for
A = E[pm ] and ` 6= p, then HF1 (Q` , E[pm ]) = HF1 ∗ (Q` , E[pm ]∗ ) = Hf1 (Q` , E[pm ])
. The following are the same
0 → Selpm (E/Q) → HF1 (Q, E[pm ]) → HF1 (Qp , E[pm ])/Hf1 (Qp , E[pm ])
0 → HF1 ∗ (Q, E[pm ]) → Selpm (E/Q) → Hf1 (Qp , E[pm ])/HF1 ∗ (Qp , E[pm ])
6.3 The Hypotheses
H.1
A
H.2
There is
H.3
H 1 (Q(A)/Q, A) = H 1 (Q(A∗ )/Q, A ) = 0
H.4
Either
is an absolutely irreducible
τ ∈ GQ
such that
Fp [GQ ]-module,
τ =1
on
∗
A 6∼
= A∗
or
µp∞
and
not isomorphic to
A/(τ − 1)A
where
p≥5
12
Q(A)
Fp
or
µp
is free of rank one over
Zp
is the xed eld of the kernel of
GQ → Aut(A).
Let Σ be a nite set of primes which contains ∞, p and ` such that A is ramied at `
1
HF1 (Q` , A) 6= Hun
(Q` , A). For every ` ∈ Σ, H 1 (Qv , A)/HF1 (Qv , A) is torsion-free.
H.5
Fcon
Remark.
`
such that
satises H.5
A = Tp (E)
Let
and all
with
Fcon
Selmer structure. Assuming
• p≥5
•
p-adic representations GQ → Aut(E[p∞ ]) = GL2 (Zp )
Aut(E[p]) is surjective). Then Tp (E) satises H.1 to H.5.
the
is surjective (for
p≥5
this is the same as
GQ →
6.4 Euler - Kolyvagin Systems
Σ be a nite set of primes containing ∞, p and all primes where A ramies.
Zp [x]. Let N = {npk : n is square free product of primes ` ∈
/ Σ, k ≥ 0}.
Let
Denition
6.8.
(
P` (Frob−1
` )Fn
Fn` =
7
Fn
` 6= p
.
otherwise
A
`∈
/ Σ, P` (x) = det(1−Frob` x|A ) ∈
F = {Fn ∈ H 1 (Q(µn ), A), n ∈ N }
is a collection
Let ES(A) denote the
Zp [GQ ]-module
such that for
of Euler system of
n` ∈ N
A.
Kolyvagin Systems (Céline)
Notation.
A
An Euler system for
If
K
a non-archimedian local eld of characteristic
0. IK ⊂ GK , φ ∈ Gal(K un /K) , k
residue eld,
|k| = q .
GK -Module
7.1 Transverse and Unramied cohomology groups
Denition 7.1.
Suppose L/K is totally tamely ramied extension of degree q − 1. Then there is a canonical
Gal(L/K) = k ∗ .
In this talk, K = Q` and L = Q` (µ` ).
1
1
1
1
We dene the L-transverse cohomology subgroup Ht (K, A) ⊂ H (K, A) to be Ht (K, A) = ker[H (K, A) →
1
1
GL
H (L, A)] = H (L/K, A ).
isomorphism
Proposition 7.2.
Suppose that
A
is a nite unramied
1.
Ht1 (K, A) ∼
= Hom(Gal(L/K), Aφ=1 )
2.
Ht1 (K, A) ∼
= Aφ=1
3. Direct sum decomposition:
GK -module
such that
(q − 1)A = 0.
Then
H 1 (K, A) = Hu1 (K, A) ⊕ Ht1 (K, A)
Denition 7.3.
Suppose that n|q − 1, let R = Z/mnZ. Suppose that A is an unramied GK -module, free of
R, det(1 − φ|A ) = 0. Consider P (x) = det(1 − φ|A x) ∈ R[x]. By assumption P (1) = 0 so
P (x) = (x − 1)Q(x) with Q(x) ∈ R(x). By Cayley-Hamilton, P (φ−1 ) = 0, then Q(φ−1 )A ⊂ Aφ=1 . Dene the
nite rank over
unramied-transverse comparison map by
∼
Q(φ−1 )
Hu1 (K, A) → A/(φ − 1)A →
{z
|
φut
Example.
case
φ
ut
Under the same hypotheses:
Aφ=1
Aφ=1 → Ht1 (K, A)
}
is free of rank one if and only if
is an isomorphism.
13
A/(φ − 1)A
is also as well. In this
7.2 Kolyvagin Systems: Denition
GQ . Assume that A is ramied at only nitely
m, we have nite GQ -module An := A/pn A. For a given Selmer structure
F for A, let Σ be a nite set of places of Q containing p, ∞, ` where A ramies, all ` where HF1 (Q` , A) 6= Hu1 (Q` , A).
Let
A
be a
Zp -module
of nite rank with a continuous action of
many primes. For every positive integer
Denition 7.4.
follows,
c be
( a positive integer such
HF1 (Qv , A) v - c
.
HF1 (c) (Qv , A) =
Ht1 (Qv , A) v|c
Let
that
c
is not divisible by any prime in
Σ.
F(c)
We dene
for
A
as
Fix a Zp [GQ ]-module A and a Selmer structure F for A satisfying H.1 to H.5. If ` ∈
/ Σ, let ν(`) = max{m|l ≡ 1
mod pm and Am /(φ` − 1)Am is free of rank 1 over Z/pm Z}. Dene NA =be the set of square free product of primes
`∈
/ Σ. If n ∈ NA then we dene ν(n) = min{ν(`)|`|n} and set ν(1) = ∞.
n
o
Denition 7.5. A Kolyvagin System is a collection κn ∈ HF1 (n) (Q, Aν(n) )|n ∈ NA such that if n` ∈ NA the
following commutes
κn ∈ HF1 (n) (Q, Aν(n) )
Hu1 (Q` , Aν(n) )
φut
κn` ∈ HF1 (n`) (Q, Aν(n`) )
Let
KS(A)
be the
Zp -module
/ Ht1 (Q` , Aν(nl) )
res`
of the Kolyvagin system for
A.
7.3 Kolyvagin system construction
There exists a canonical map
Construction:
For every
ES(A) → KS(A, Fcan )
n ∈ NA , let Γn = Gal(Q(µn )/Q).
For ` ∈
/ Σ, Q
x a generator σ`
n ∈ NA , Dn = `|n D` ∈ Z[Γn ].
Proposition 7.6.
in
H
1
If ξ
(Q(µn ), Aν(n) )Γn .
Proposition 7.7.
If
under the restriction
Denote
Recall:
For
ξ0
ξ0
for
Γ` ,
If
such that if
ξ→κ
then
n = n1 n2 , Γn = Γn1 × Γn2 .
κ1 = ξ1
then the image of
Dn ξn
ξ ∈ ES(A), n ∈ NA , then the image of Dn ξn
1
1
Γ
map H (Q, Aν(n) ) → H (Q(µn ), Aν(n) ) n .
under
HF1 can (Q, A)
With this identication
dene the Kolyvagin's Derivative operator
∈ ES(A), n ∈ NA ,
in
D` :=
Γn =
P`−2
i
i=1 iσ`
Q
`|n
Γ` .
∈ Z[Γ` ].
If
H 1 (Q(µn ), A) → H 1 (Q(µn ), Aν(n) )
has a canonical inverse image in
lies
H 1 (Q, Aν(n) )
ξn0 to be the canonical inverse image of Dn ξn as above. Let ξ 0 = {ξn0 |n ∈ NA }.
H 1 (Q` , Aν(n) ) = Hu1 (Q` , Aν(n) ) ⊕ Ht1 (Q` , Aν(n) ), Res` (ξn0 ), (ξn0 )`,u , (ξn0 )`,t .
to be a Kolyvagin system we need:
1.
Res` (ξn0 ) ∈ HF1 can (Q` , Aν(n) )
2.
Res(ξn0 ) ∈ Ht1 (Q` , Aν(n) )
3.
0
φut ◦ Res` (ξn/`
) = (ξn0 )`,t ∈ Ht1 (Q` , Aν(n) )
satises
1 and 3,
but
if
if
`-n
`|n
(ξn0 )`,u 6= 0 in general.
for
`|n
But we can dene a Weak Kolyvagin System by ignoring 2. and then
rening it to create a Strong Kolyvagin System.
7.4 Application
A, if
HF1 (Q, A)}.
Give a Kolyvagin system for
j
δ(K) = max{j|κ1 ∈ p
κ1 6= 0,
then
HF1 ∗ (Q, A∗ )
14
is nite and has length or equal to
δ(K)
where
8
Kato's Euler system
Let
p ≥ 5 be
T = Tp E
prime.
E/Q
modular Elliptic Curve over
Q
conductor
N. f
newform weight
2,
level
N.
(ζn ∈ H 1 (Q(ζn ), T ))n∈N , Σ ⊃ {∞, p} ∪ primes(N )
k
N = {squarefree product
( of ` ∈ Σ × p }
−1
P` (Frob` )ζn ` 6= p
CoQ(ζn` )/Q(ζn ) ζn` =
ζn
`=p
Euler system:
8.1 Overview of construction
(O(Y (n, L))∗ )
2
/ H 2 (Y1 (N )Q(ζ ) , Zp (1))2
n
ét
/ H 1 (Y (n, L), Zp (1))2
ét
2
Hét
(Y1 (N )Q(ζn ) , Zp (2))
1
H 1 (Q(ζn ), Hét
(Y1 (N )Q , Zp (2))
H 1 (Q(ζn ), T (1))
Proposition 8.1.
Let
E/S
be an elliptic curve,
c
prime to
6.
Then there exists a unique c θE
∈ O(E\E[c])∗
such
that
1. divisors
2. For all
c2 (0) − E[c]
a
coprime to
c, [a]∗ : O(E\E[ac])∗ → O(E\E[c])∗ , [a]∗ c θE = c θE .
Proof.
Uniqueness: Let
Existence:
2
2
f = ug , u ∈ O(S)∗ , f = [a]∗ f = [a]∗ ug = ua [a]∗ g = ua g ,
[a]∗ f = ua f
with
u ∈ O(S)∗ .
If we let
2
uba
−1
2
= uab
−1
,
Hence
2
ua
−1
=1
so
a = 2, 3
and
u = 1.
g = u−3
2 u3 f .
(
(E, e1 , e2 ) E/S
Denition 8.2. N ≥ 3, Y (N ) modular curve over Q represent S →
.
e1 , e2
Z/N Z − basis of N torsion
Y (N )(C) ∼
= (Z/N Z)∗ × Γ(N )\H. If N |N 0 , there is a map Y (N 0 ) → Y (N ) so O(Y (N )) ⊂ O(Y (N 0 )).
Denition 8.3.
Lα,β
N ≥ 3, c coprime
= ae1 + be2 : Y (N ) → E[N ].
Let
to
6N . (α, β) = ( Na , Nb ) ⊂ (Q/Z)2 \ {(0, 0)}.
Then c gα,β
:= L∗α,β ( c θE ),
GL2 (Z/N Z) Y (N )
Proposition 8.4.
1.
2.
σ ∈ GL2 (Z/N Z), σ ∗ c gα,β = c g(α,β)σ
Q
0
0
if a prime to c: then c gα,β =
aα0 =α,aβ 0 =β c gα ,β .
a b
1 0
Denition 8.5. Let M, N ≥ 3, M |L, N |L and dene Y (M, N ) := G\Y (L) where G = c d ≡ 0 1 mod M
N
GL2 (Z/LZ), S → {(E, e1 , e2 ) : E/S, e1 is M torsion,e2 is N torsion, (a, b) ∈ Z/N Z × Z/N Z 7→ ae1 + be2 is injective}.
Denition 8.6.
pn
∗
∗
1
1 → µpn → OX
→ OX
→ 1. O(X)∗ → Hét
(X, µpn ), (Z/pn Z)(k) := (µpn )⊗k .
i
i
∗
1
Hét (X, Zp (k)) = lim Hét (X, µpn (k)). O(X) → Hét (X, Zp (1)).
←−
étale sheaves:
15
Then
M
N
⊂
∪
j
i+j
i
2
Hét
(X, Zp (k)) × Hét
(X, Zp (k 0 )) → Hét
(X, Zp (k + k 0 )), f, g ∈ O(X)∗ , {f, g} := κ(f ) ∪ κ(g) ∈ Hét
(X, Zp (2)).
Lemma 8.7.
f :U →V
Denition 8.8.
nite étale,
u ∈ O(U )∗ , v ∈ O(V )∗ , f∗ {u, f ∗ v} = {f∗ u, v}, f∗ {f ∗ v, u} = {v, f∗ u}.
M, N ≥ 3, (c, 6M ) = 1 and (d, 6N ) = 1.
Dene c,d ZM,N
2
:= { c g M1 ,0 , d g0, N1 } ∈ Hét
(Y (M, N ), Zp (2)).
Proposition 8.9.
primes(N ) =
0
0
0
0
0
If we take M |M , N |N , (c, 6M ) = 1 and (d, 6N ) = 1 with primes(M ) = primes(M ) and
0
0
0
primes(N ). Then Y (M , N ) → Y (M, N ). The push-forwards of c,d ZM 0 ,N 0 is c,d ZM,N .
`-N
If
/ Y (M (`), N ) T 0 (`)
∼
Y (M, N (`)
)
u
Y (M, N )
Proposition 8.10.
M |M 0 , N |N 0 , (c, 6M 0 ) = 1
If we take
dened by
c,d ZM `,N ` 7→
1 − T 0 (`)
1/` 0
0 1
and
(d, 6N 0 ) = 1,
∗
+
1/` 0
0 1/`
there is a map
∗ ` ·
c,d ZM,N
) = Y (N, 1), n, N ≥ 3, n|L
Y1 (N
, N |L and primes(L) = primes(nN ).
∗ ∗
G=
mod N, det ≡ 1 mod n , Y (n, L) → Y1 (N )Q(ζn )
0 1
Let
Denition 8.11.
Proposition.
c,d Z1,n,N
2
∈ Hét
(Y1 (N )Q(ζn ), Zp (2))
σ` ∈ Gal(Q(ζm )/Q)
Dene c,d ζn
:=map
Theorem 8.12.
9
Then
Y1 (N )Q(ζn ) ∼
= G\Y (L).
push-forward of c,d Zn,N
` - nN
`
1 − T 0 (`)σ`−1 +
0
c,d Z1,n`,N ` 7→
where
Y (M `, N `) → Y (M, N )
and
`
0
∗
σ`−2 `
c,d Z1,n,N
∗
0
= σ`
1
of c,d Z1,np,N p .
( c,d ζn )
0
1/`
Σ = primes(6cdN p).
is an Euler system
Euler Systems and BSD
Notation. Let
E
be an elliptic curve over
all primes at which
T
Q, T = Tp E Tate module. Σ =nite set of prime containing p, ∞,
{cn }, cn ∈ H 1 (Q(µn ), T )} with corestriction conditions
and
ramies. Euler system
9.1 Recap: What can we do with Euler systems?
Mantra: Existence of Euler systems leads to bounds on Selmer groups.
Sel(Q, E[p])
ζ ∈ KS(A), ζ1 6= 0 ⇒ Sel(Q, A∗ ) is nite.
Pedro, Alex and Florian: Used Heegner points to bound
Céline: Kolyvagin systems,
Today we'll use a variant,
Theorem 9.1.
then
Let
∞
Sel(Q, E[p ])
Proposition 9.2.
Lemma 9.3.
If
E/Q be an elliptic curve, T
a Tate module,
c ∈ ES(T ).
If
locsp (c1 ) 6= 0 ∈ H 1 (Qp , T )/Hf1 (Qp , T ),
is nite.
There exists an exact sequence
locsp (c1 ) 6= 0,
then
E(Q)
and
0 → E(Q) ⊗ Qp /Zp → Sel(Q, E[p∞ ]) → X(E/Q)p∞ → 0
X(E/Q)p∞
16
are nite.
9.2 Twisting
Recall: Aurel constructed an Euler System for the wrong module,
T (1), i.e., T
twisted by the cyclotomic character
χp .
Motivation: Bloeh-Kato conjecture: existence of 'nice' cohomology classes
Euler Systems means lots of nice cohomology.
systematic vanishing of
Fact.
vanishing of
L-values
L-functions.
For all elliptic curves over
modular form
↔
So we might expect Euler systems to exists where there is a
Q, L(E, S)
has a simple zero at
s = 0.
(More generally,
L(f, 0) = 0
for all
f)
Twisting by characters of nite order.
Let
χ : GQ → Z∗p
of nite order. Let
ker χ
L := Q
L/Q
,
nite. For all
n, GLQ(µn ) ⊂ ker(χ),
hence the natural
map on cocycles induces an isomorphism
∼
=
cm H 1 (LQ(µn ), T ) ⊗ χ
O
/ H 1 (LQ(µn ), T ⊗ χ)
⊗Zχ
H (Q(µn ), T ) 3 cχn
cor
cn c ∈ H 1 (LQ(µ ), T )
m
n
1
c1
Theorem 9.4.
cχ = {cχn }
is an Euler Systems for
P
∪ {` : `|cond(χ)}
χp
Twisting by
kerχ
L := Q
= Q∞ = ∪k Q(µpk ) so corestriction doesn't exist.
n
k
Idea: Z/p Z(1) is a trivial GQ(µ k ) -module, hence T /p ∼
= T /pk ⊗ Z/pk (1) ∼
= T /pk (1)
p
1
k
1
k
H (Q(µpk ), T /p T ) ∼
= H (Q(µpk ), T /p T (1))
Problem:
Denition.
The Iwasawa cohomology group for
T
over
to corestriction.
as
GQ(µpk ) -modules.
So
1
(Q(µn ), T ) = limk H 1 (Q(µnpk ), T ) with respect
Q(µn ) is H∞
←−
Proposition 9.5.
1
H 1 (Q(µnpk ), T /pk T )
(Q(µn ), T ) ∼
H∞
= lim
←−k
1
1
(Q(µn ), T (1))
(Q(µn ), T ) ∼
H∞
= H∞
Corollary 9.6.
Note: If
c ∈ ES(T ),
Theorem 9.7.
then for any
χ
cχp = {cnp }
n,
we can dene
1
cn,∞ = {cnpk }k≥0 ∈ H∞
(Q(µn ), T ).
is an Euler System.
9.3 BSD
General Euler systems machinery doesn't require conditions at
p.
But: For the arithmetic applications, we'll need to be precise about conditions at
p.
Hence we will need
p-adic
Hodge Theory.
Fact.
There exists a specic
Hf1 (Qp , V ) ⊂ H 1 (Qp , V ),
which is a
1-dimensional Qp -vector
such that its Selmer group is equal to the usual Selmer group.
∗
1
There exists an isomorphism, exp : Hs (Qp , V ) → Qp · ωE , where
Theorem 9.8
−1
c0 = cχp
((Kato))
.
Let
. Then there exists
corresponding to
ωE .
Corollary 9.9.
If
is the regular dierential for
V = T ⊗ Qp ,
E.
E/Q be an elliptic curve, T a Tate module, c the Euler system described by Aurel,
r L (E,1)ωE
rE ∈ Z \ {0} such that exp∗ (locsp (c01 )) = E N pΩE
, where ΩE = real period of E
LN p (E, 1) 6= 0,
then
Proof. Euler factors are non-zero at
Theorem 9.10.
ωE
space where
E(Q)
`|N p
and
X(E/Q)
are nite.
.
L/Q be an abelian number eld, χ : Gal(L/Q) → C∗ .
E(L) ⊗ C : σ(P ) = χ(σ)P ∀σ ∈ Gal(L/Q)} and X(E/L)χ are nite.
Let
17
If
L(E, χ, 1) 6= 0
then
E(L)χ = {P ∈
10
Euler Systems: Some Further Topics
10.1 Recap
Let
E/Q
be an elliptic curve.
Theorem 10.1.
If
L(E, 1) 6= 0,
then
E(Q)
is nite, and
Xp∞ (E/Q)
is nite for many primes
p.
E(Km ) where K is imaginary quadratic eld, Km ring class eld
Gal(Km /K) as −1)
Kato: Proof using Siegel units - cohomology classes for Tp E over cyclotomic elds.
Kato can tell you about L(E, χ, 1) where χ is a Dirichlet character
σ
−1
Kolyvagin: can tell you about L(E/K, ψ, 1) where ψ is a character of a ray class group of K such that ψ = ψ
.
Kolyvagin: Proof using Heegner points in
(abelian extension of
K
such that
Gal(K/Q)
acts on
1. Can we say something about arbitrary abelian extension of
character
K (↔L-functions L(E/K), ψ, 1)
for any ray class
ψ )?
2. Are there Euler systems for other Galois representations (not just
Tp E )?
10.2 An Euler System for two modular forms
Theorem 10.2.
Let
f, g
exists an Euler system for
be two modular forms of weight
2.
Galois representations
Tp (f ), Tp (g).
Then there
T = Tp (f ) ⊗ Tp (g).
Starting point: Siegel units,
= c2 gα,β − gcα,cβ
Y
Y
= qw
1 − q n q a/N e2πib/N
1 − q n q −a/N e−2πib/N
c gα,β
ga/N,b/N
n≥0
n≥1
1
a
a2
12 − 2N + 2N 2 .
∗
Not a modular form, it has poles at cusps (like the j -function). This lives in O(Y (N )) . In particular can take
∗
1
a = 0, b = 1, c g0,1/N ∈ O(Y1 (N )) → Hét (Y1 (N ), Zp (1)) using a Kummer map.
1
Consider ∆ : Y1 (N ) ,→ Y1 (N ) × Y1 (N ) diagonally.
We get the pushfoward ∆∗ : Hét (Y1 (N ), Zp (1)) →
3
2
1
Het (Y1 (N ) , Zp (2)) → H (Q, Tp (f ) ⊗ Tp (g)) for all f, g weight 2 level N .
where
w=
This idea is due to Beilinson in 1984 (Beilinson - Flach elements)
Theorem 10.3
.
(Lei-Loer-Zerbes)
Y1 (m2 N ) → Y1 (N )2
not Q.
dened by
Can extend this to an Euler System as follows: for
1
z 7→ (z, z + m
).
Then take
cm = (∆m )∗ ( c g0,1/m2 N ).
m ≥ 1,
∆m :
Q(µn )
consider
This is only dened over
(40 pages later) Some version of norm-compatibility relation holds.
10.3 Twisting
Tp (E)(1)
of Q(µpr ).
Recall: In Kato's setting, had to twist from
of
p, Tp (E)(1) ∼
= Tp (E)
as representations
to
Tp (E).
This was possible because, modulo any power
pr
Tp (f ) ⊗ Tp (g) are not the ones we've immediately get at. Their elements are related to
(like L(E, 0) = 0 in Kato's case).
Want to consider weight f is 2 and weight g is 1.
r
r
Idea: The Galois representation of g mod p shows up in cohomology of Y1 (N p ).
Special case: K imaginary quadratic eld, ψ ray class character of K . ψ gives a weight 1 modular form θψ . This
Interesting cases for
L(f, g, 1)
always being
0
answers Q1.
18
10.4 More Euler systems?
Suppose we have some variety
where
Y
X and
we want Euler Systems in cohomology of
X.We
look for subvariety
Y ,→ X
is a modular curve (or product of modular curves). If you are lucky and pushforward of Siegel units lands
in the right degree, then maybe that will give an Euler System.
This is reasonable if
SL2
over
X
and
Y
are Shimura varieties (comes from Matrix groups, like modular curves come from
Q)
Eg,
• GL2 ,→ GL2 × GL2
• SO2 ,→ GL2
(BF elts)
(Heegner points)
• GL2 × GL2 ,→ GSp4
• SL2 ∼
= SU (1, 1) ,→ SU (2, 1)
Problem: Very hard to show that these Euler Systems are not
19
0.