Chapter 3 Application to unit groups
3.2 Calculation of the Yakovlev diagram
3.2.2 Proof of Proposition 3.2.4
Now assume thatK/F is as at the start of Section 3.2.
Lemma 3.2.12.There is an NG(P)/P-equivariant isomorphism
H1(P,O×K/µK)∼=T0(K)P/T0(F0).
0 O×F/µF0 V0 F0 T0(F0) 0 0 (O×K/µK)P (VK0)P T0(K)P H1(P,O × K/µK) 0 a (3.4)
Applying snake lemma at cokeragives the desired isomorphism. That this isNG(P)-
equivariant is standard, see e.g. [NSW08, Prop. 1.5.2].
It remains to describe T0(K)P/T0(F0) (which we now know to be a finite group of exponent p). Note that T0(K)P need not be divisible, but since T0(F0) is of finite index and a divisible subgroup, T0(F0) must be the maximal divisible subgroup. The basis for calculatingT0(K)P/T0(F0) is thatT0(K)P/T0(F0) appears as a cokernel in the kernel/cokernel sequence of the diagram:
0 T0(F0) Pic0(F0) Cl(F0) 0 0 T0(K)P Pic0(K)P Cl(K)P H1(P, T0(K))
a b c
Here, both a:T0(F0)→ T0(K)P and b: Pic0(F0) → Pic0(K)P are induced by the map Div(F)→Div(K) described in Remark 3.2.10.
Lemma 3.2.13.The following is an exact sequence of finiteFp[NG(P)/P]-modules:
0→kera→kerb→kerc→T0(K)P/T0(F0)→ker(cokerb→γ cokerc)→0. (3.5)
Proof. We must check that each term is a finite Fp[NG(P)/P]-module. For kera
and kerγ, this will follow from checking the other terms, whilst forT0(K)P/T0(F0), this follows from Lemma 3.2.12. For kerb, consider following diagram obtained from the middle column of Figure 3.1:
0 F0×/µF0 Div0(F0) Pic0(F0) 0
0 (K×/µK)P Div0(K)P Pic0(K)P H1(P, K×/µK) b
(3.6) As Div0(F0),→Div0(K), we find that
is injective, but the latter is, by Hilbert Theorem 90, isomorphic toH1(P, µK) and
thus finite and of exponent at mostp. For kerc, finiteness is clear. By considering the right hand column of Figure 3.1, we find ker(Cl(F0)→Cl(K)),→P(K)G/P(F0),
which is a quotient of ˆH0(P, P(K)) and so has exponent dividingp. In particular, in the notation of Notation 3.2.2,
T0(K)P/T0(F0)∼= kera⊕kerc⊕kerγ kerb. (3.7) As such, to determine the isomorphism class of T0(K)P/T0(F0) it suffices to de- scribe the isomorphism class of each summand individually. For example, kerc = ker(Cl(F0)→Cl(K)) is already a classically studied invariant.
Lemma 3.2.14.kera∼= coker(OF×0/µF0 →(O×
K/µK)P) as NG(P)/P-modules.
Proof. This is immediate from the kernel/cokernel sequence of (3.4). The calculation of kerγ requires more work.
Lemma 3.2.15.There is a commutative diagram of finite Fp[NG(P)/P]-modules: 0 kerb H1(P, µK) Div0(K)P/Div0(F0) cokerb H1(P, K×/µK) 0 0 kerc P(K)P/P(F0) I(K)P/I(F0) cokerc H1(P, P(K))
d
γ z
(3.8)
with the horizontal rows being exact. Proof. By Hilbert Theorem 90, H1(P, µ
K) ∼= coker(F0×/µF0 → (K×/µK)P). As a
result, the kernel/cokernel sequence of (3.2.2) is given by
0→kerb→H1(P, µK)→Div0(K)P/Div0(F0)→cokerb (3.9) →H1(P, K×/µK)→H1(P,Div0(K)). Claim. H1(P,Div0(K)) = 0.
Proof of Claim. By the degree map it suffices to show that H1(P,Div(K)) = 0. As aP-module, Div(K) decomposes as a direct sum of terms indexed by places of
F0. Each of these is a permutation module and so has trivial cohomology in degree 1. As group cohomology commutes with arbitrary direct sums, the same is true of Div(K).
Claim. Every term of (3.9) is a finite Fp[NG(P)/P]-module.
Proof of Claim. That each term has exponent dividingp is clear as all terms other than kerbare eitherP-cohomology groups or quotients ofP-Tate cohomology groups. As H1(P, µK) is finite, it suffices to check finiteness for Div0(K)P/Div0(F0) and
cokerb. By considering the degree map, Div0(K)P/Div0(F0)∼= Div(K)P/Div(F0). The Galois stable elements of Div(K) are those divisors whose terms at all places
wlying over each fixed place ofF0 are diagonal. Using this description we find that Div(K)P/Div(F0)∼= M
p∈r(K/F0)
Fp =:W,
(in the notation of Notation 3.2.1) so that Div0(K)/Div0(F0) is finite. Finally, Hilbert Theorem 90 shows thatH1(P, K×/µK),→H2(P, µK) so is finite.
Now consider:
0 P(F0) I(F0) Cl(F0) 0 0 (K×/O×K)P I(K)P Cl(K)P H1(P, P(K))
b
Figure 3.2 This has kernel/cokernel sequence
0→kerc→P(K)G/P(F0)→I(K)P/I(F0)→cokerc→H1(P, P(K)). (3.10)
Claim. Every term of (3.10) is a finite Fp[NG(P)/P]-module.
Proof of Claim. As before, that each term has exponentpis automatic, as is finite- ness of kerc and cokerc. Now finiteness of I(K)P/I(F0), and so P(K)G/P(F0) follows as it is isomorphic toW =L
r(K/F0)Fp, whilstH1(P, P(K)),→H2(P,O×K)
soH1(P, P(K)) must be finite.
The maps between the middle column and right hand column of Figure 3.1 induce maps between Figure 3.9 and Figure 3.2 and so, by naturality of the snake lemma, of their associated kernel/cokernel sequences. Putting this all together gives (3.8).
Lemma 3.2.16.We have that
kerγ ∼= ker(W →Cl(K)P/im Cl(F0))⊕kerb
⊕ker(H1(P, K×/µK)→H1(P, P(K)) H1(P, µK)
as NG(P)-modules.
Proof. We argue from (3.8) of Lemma 3.2.15. After truncating we obtain a diagram of short exact sequences
0 Div0(K)P/ Div0(F0)·d(H1(P, µK))
cokerb H1(P, K×/µK) 0
0 I(K)P/I(F0)P(K)P cokerc H1(P, P(K))
t γ z
where the left hand arrow is surjective as Div0(K)P/Div0(F0) ∼=I(K)P/I(F0) (∼=
W). As a result
kerγ ∼= kert⊕kerz.
Returning to (3.8),
d(H1(P, µK))⊆ker(Div0(K)P/Div0(F0)→I(K)P/I(F0)P(K)P)
so kert can be rewritten as
ker(Div0(K)P/Div0(F0)→I(K)/I(F0)P(K)) H1(P, µK)⊕kerb.
Now simply note thatI(K)/I(F0)P(K) = Cl(K)/c(Cl(F0)).
Proof of Proposition 3.2.4. By Lemmas 3.2.12, 3.2.13, we need only substitute our calculations into (3.7). Lemmas 3.2.14, 3.2.16 give
T0(K)P/T0(F0)∼= kera⊕kerc⊕kerγ kerb
∼
= coker(OF×0/µF0 →(O×
K/µK)P)⊕ker(Cl(F0)→Cl(K)) ⊕ker(W →Cl(K)P/c(Cl(F0)))⊕kerb
⊕ker(H1(P, K×/µK)→H1(P, P(K))) H1(P, µK) kerb
Cancelling the kerb terms and applying Lemma 3.2.17 i), we obtain the desired formula.
For later use, we give some alternative descriptions of the terms appearing in Proposition 3.2.4.
Lemma 3.2.17. i) As NG(P)/P-modules we have
ker(H1(P, K×/µK)→H1(P, P(K)))∼= ((µF0 ∩NK/F0O×
K)/NK/F0µK)(1),
(here (−)(1) is as defined in Notation 3.2.3). ii) As NG(P)/P-modules we have
ˆ H−1(P, µK) = µK[p](1) µK[p∞] =µF0[p∞] 0 µK[p∞]6=µF0[p∞]and K 6=F0(ζ4) Z/2Z(1) K =F0(ζ4) , (3.11)
where for an abelian group A, A[n] denotes the n-torsion subgroup and A[p∞]
denotes S
nA[pn].
Proof. For i), consider the following diagram, which is exact by Hilbert Theorem 90:
0 Hˆ−1(P, K×/µK) Hˆ0(P, µK) Hˆ0(P, K×)
0 Hˆ−1(P, K×/O×K) Hˆ0(P,O×K) Hˆ0(P, K×)
Applying the definition of ˆH0, this becomes:
0 (µF0∩NK/F0K×)/NK/F0µK µF0/NK/F0µK F0×/NK/F0K× 0 (O×F0 ∩NK/F0K×)/NK/F0O× F0 O×F0/NK/F0O× K F 0×/N K/F0K× So that ker( ˆH−1(P, K×/µ K)→Hˆ−1(P, P(K)))∼= (µF0∩NK/F0O× K)/NK/F0µK.
Twisting now gives the desired formula. Forii), consider
ˆ
whereσ is a generator of P. We claim that ker(NK/F0:µK[p∞]→µK[p∞]) = µK[4] ifp= 2 and K =F0(ζ4)6=F0 µK[p] otherwise .
IfµK[p∞] =µF0[p∞], this is clear asNK/F0 is raising to the powerp. Otherwise, we
may assume that, on a generatorζpn of µK[p∞], σ acts by ζpn 7→ζ1+p n−1 pn . So NK/F0(ζpn) = Y a∈Z/pZ ζp1+napn−1 =ζ (p+pn−1·P a∈Z/pZa) pn . Since X a∈Z/pZ a= 1 ifp= 2 0 ifp≥3 ,
we find the kernel of NK/F0 is µK[p] unlessp = 2 and n = 2, where ζ4 also lies in
the kernel.
It is also easy to see that (σ−1)µK[p∞] = µK[p] µK[p∞]6=µF0[p∞] 0 µK[p∞] =µF0[p∞] .
Putting these together gives the final formula of the lemma.