Galois Cohomology (Study Group)
1 Cohomology of global elds and Poitou-Tate duality (Marc Masdeu)
Notation.
• K
is a global eld (think number eld).
• S a set of places of K containing ∞K
Hom(GQ , Z/2Z), which by class eld
• KS ⊂ K is the maximal
S = µK then KS = K .
theory, is the dual
subextension which is unramied outside
• RK,S = {a ∈ K : ordν (a) ≥ 0∀ν ∈ S}.
•
Fix embeddings of
1.1
K ) (Mostly S will be nite: if not H 1 (GQ , Z/pZ) =
∗
∗2
of Q /Q
everywhere nite)
(the innite place of
K ,→ Kν
for all
S = {∞K }
If
ν ∈ S.
then
S.
RK,S = OK ,
This gives embeddings
if
If
S = {∞K }
S = µK
then
then
KS = K unr ,
if
RK,S = K .
GKν ∼
= Dν ,→ GK → GS := Gal(KS /K)
Localisation
H r (GS , M ) → H r (Kν , M ) for all ν .
Q
Q
r
Denition 1.1. PSr (K, M ) = 0ν∈S H r (Kν , M ) = {(cν ) ∈ ν∈S H r (Kν , M ) : cν ∈ Hunr
(Kν , M ) almost all ν}.
0
0
0
(Convention for ν archimedean, (r = 0), take HT = H /NG H instead (Tate cohomology groups)).
Q
• PS0 = ν∈S H 0 (Kν , M )
Q
• PS1 = ν∈S H 1 (Kν , M )
Let
M
be any
GS -module
(nite). We have maps
• PS2 = ⊕ν∈S H 2 (Kν , M )
Since each class in
H r (GS , M )
H r (GS , M ) → PSr (K, M ).
PSr (K, M ). Dualising (and
arises from
We dene
if
M
X
is nite,
H r (Gal(L/K), M )
for some
L,
localisations induces a map:
by the following short exact sequence
∗
#M ∈ RK,S
)
PS2 (K, M D )∗
=
see Pedro0 s talk ∼
β r∗
βr
0 → XrS (K, M ) → H r (GS , M ) →
we get
/ H 2−r (GS , M D )∗
6
/ X2−r (K, M D )∗
S
γ n (K,M )
PSr (K, M )
The upshot is there exists maps:
4pc PSr (K, M )
Assume
M
γ r (K,M )
/ H 2−r (GS , M D )∗
is nite.
1
βr :
/ X2−r (K, M D )
S
/0
Proposition 1.2. The map β1 : H 1 (GS , M ) → PS1 (K, M ) is proper (i.e., preimages of compact is compact). (Note
the topology in H 1 is the discrete topology and the topology of PS1 is induced from the product topology)
Proof.
Q
Q
1
PS1 is contained in some PT = ν∈S\T Hunr
(Kν , M ) × ν∈T H 1 (Kν , M ) (where T ⊂ S is
1 −1
nite). Let XT = (β )
(PT ), we want XT is nite. First there exists a nite extension L/K such that GL acts
1
1
trivially on M , H (GK , M ) → H (GL , M ) has nite kernel. Without loss of generality assume M has trivial action.
ker(f )
1
Then H (GS , M ) = Hom(GS , M ) 3 f , this gives Lf = KS
, so f ∈ XT if and only if Lf /K is unramied outside
T
Any compact in
Hermite - Minkowski. Given n and nite T , there exists nitely many extension L/K such that [L : K] = n and
L is unramied outside T .
Theorem 1.3. Let M be nite, #M
∗
∈ RK,S
.
1. There exists a canonical nondegenerate pairing h , i : X1S (K, M ) × X2S (K, M D ) → Q/Z.
2. For all r ≥ 3, β r is a bijection, β r : H r (GS , M ) →
Q
ν arch
H r (Kν , M )
3. There exists a g -term exact sequence:
/ H 0 (GS , M )
0
β0
/ P 0 (K, M )
S
γ0
/ H 2 (GS , M D )∗
h,i
H 1 (GS , M D )∗ o
0
(
t
X1S (K, M D )∗
0
γ1
s
H 1 (GS , M )
β1
h,i
*
v
H 2 (GS , M )
Remark. X0S (K, M ) = 0
X1S (K, M )
PS1 (K, M ) o
+
t
X1s (K, M ) ∼
= X2S (K, M D )∗
is nite because it is equal to
β2
/ P 2 (K, M )
S
γ2
/ H 0 (GS , M D )∗
(β 1 )−1 ({1})
2
By duality, XS (K, M ) is also nite.
1.2
Euler - Poincaré pairing
Warning:
So we
H r (GS , M ) may be
∗
look at χ (GS , M ) =
r!
#H 0 (GS ,M )#H 2 (GS ,M )
. We now assume
#H 1 (GS ,M )
non zero for innitely many
S
is nite.
Theorem 1.4. We have
Y #H 0 (Gν , M )
χ∗ (GS , M ) =
=
|#M |ν
ν inf
(
1
Q
0
#HT
(Gν ,M D )
ν inf #H 0 (Gν ,M D )
2
if K is function field
if K is a number field
/0
1.3
Let
Set
Applications
E be an elliptic curve over Q and x a prime p ≥ 3. S be a nite set containing {∞, 2, p, prime of bad reductions} .
G = GS and V = Vp (E) = Qp ⊗Zp limn E[pn ]. Note that Mn is nite, #Mn = p2n and MnD = Mn . We get
←− | {z }
=:Mn
#H 0 (G, Mn )#H 2 (G, Mn )
#H 1 (G, Mn )
=
=
=
#HT0 (R, Mn )
#H 0 (R, Mn )
1
#(1 + c)Mn
1
pn
#H 0 (G, Mn ) will stabilise. Hence we get dim H 1 (G, Vp (E)) − dim H 2 (G, Vp (E)) = 1.
1
2
D
1
0
2
D
Let us look back at h , i : XS (K, M ) × XS (K, M ) → Q/Z. Take a ∈ XS (K, M ) and a ∈ XS (K, M ). So
1
0
0
2
D
0
0
a correspondPto α ∈ H (GS , M ), αν = dβν for all ν , and a correspond to α ∈ H (GS , M ), αν = 0dβν for all ν .
0
0
0
Check that
ν invν (βν ∪ αν − ν ) =: ha, a i ∈ Q/Z, where α ∪ α = d.
Note that in the limit
3