Exercise 2 Solutions

Problem 2.1. (Local Galois cohomology computation) Let K be a finite extension of Qp or Fp((t)) with residue field Fq. Let V be a representation of GK on an F`-vector space. 1 GK ∗ GK (1) Show that when ` 6= p, H (GK ,V ) = 0 unless V 6= 0 or V (1) 6= 0. When ` = p 1 and K = Qp, what is dim H (GQp ,V ) “usually”? (2) When ` 6= p, compute without using Tate local duality and Euler charactersitic for- i mula, in an explicit way, dim H (GK , F`(n)). Your answer will depend on congruences of qn modulo `. Observe that the dimenions coincide with the prediction by local Tate duality and Euler characteristic formula. i (3) When ` = p and K a finite extension of Qp, compute the dimension of dim H (GK , Fp(n)). (Use Tate local duality and Euler characteristic formula.)

0 Solution. (1) Using Euler characteristic formula, we see that χ(GK ,V ) = dim H (GK ,V ) − 1 2 1 0 dim H (GK ,V ) + dim H (GK ,V ) = 0. So H (GK ,V ) = 0 unless either H (GK ,V ) = 0 or 2 GK H (GK ,V ) = 0; the first condition is just V = 0, and the second condition (by local Tate 0 ∗ ∗ ∗ GK duality) is equivalent to H (GK ,V (1)) = 0 or equivalently V (1) = 0. (2) (Let k denote the residue field of K.) We use Hoshchild–Serre spectral sequence, i,j i j i+j E2 = H (Gk,H (IK , F`(n)) ⇒ H (GK , F`(n))

We know that F`(n) is an unramified extension. As PK is pro-p and p 6= `, we have  (n) j = 0 F` j ∼ j ∼ H (IK , F`(n)) = H (IK /PK , F`(n)) = Hom(Zb(1), F`(n)) = F`(n − 1) j = 1 0 j > 1. From this, we deduce ( if ` | pn − 1 0 0 Gk F` H (GK , F`(n)) = H (Gk, F`(n)) = F`(n) = 0 otherwise.

( n−1 2 1 F`(n − 1) F` if ` | p − 1 H (GK , F`(n)) = H (Gk, F`(n − 1)) = = (φk − 1)F`(n − 1) 0 otherwise. The first cohomology sits in an exact sequence

1 1 1 Gk 0 / H (Gk, F`(n)) / H (GK , F`(n)) / H (IK , F`(n)) / 0 ( ( if ` | pn − 1 if ` | pn−1 − 1 F` Gk F` F`(n − 1) = 0 otherwise. 0 otherwise.

Combining these two, we obtain  2 if p ≡ 1 mod `  1 n n−1 dim H (GK , F`(n)) = 1 if exactly one of p − 1 and p − 1 is divisible by ` 0 otherwise. We can check that the numerics agree with the prediction by local Tate duality and the Euler characteristic formula. 1 2

0 (3) H (GK , Fp(n)) = 0 unless Fp(n) is the trivial representation, which could either because (p − 1)|n or because K contains Qp(µp). In general, if we denote d := [K ∩ Qp(µp): Qp] (which is a divisor of p−1), then Fp(n) is the trivial representation if and only if p−1 divides dn. 2 ∼ 0 ∗ By Tate local duality, H (GK , Fp(n)) = H (GK , Fp(1 − n)) which is zero unless p − 1 divides d(n − 1). Finally, by Euler characteristic formula, we deduce that  2 if p − 1|d or equivalently Qp(µp) ⊆ K i  dim H (GK , Fp(n)) = 1 if exactly one of dn and d(n − 1) is divisible by p 0 otherwise  Problem 2.2. (Dimension of local Galois cohomology groups) Let K be a finite extension of Qp, and let V be a representation of GK over a finite dimensional F`-vector space. Suppose that V is irreducible as a representation of GK and dim V ≥ 2. 1 2 (1) When ` 6= p, show that H (GK ,V ) = 0. (Hint: compute H using local Tate duality and then use Euler characteristic.) 1 (2) When ` = p, what is dim H (GK ,V )?

GK 0 Solution. (1) When dim V ≥ 2 and V is irreducible, V must be trivial; so H (GK ,V ) = 0. ∗ 2 ∼ 0 ∗ ∗ Similarly, V (1) is also irreducible; so we deduce similarly that H (GK ,V ) = H (GK ,V (1)) = 1 0. Finally, by Euler characteristic formula, we deduce that H (GK ,V ) = 0. 0 2 (2) By the same argument as in (1), we deduce that H (GK ,V ) = H (GK ,V ) = 0. So Euler characteristic formula says that 1 H (GK ,V ) = −χ(GK ,V ) = [K : Qp] · dim V.  Problem 2.3. (An example of Poitou–Tate long exact sequence) Consider F = Q, and let S = {p, ∞} for an odd prime p. Determine each term in the Poitou–Tate exact sequence for 2 the trivial representation M = Fp. (Hint: usually H (GF,S, Fp) is difficult to determine; but one can use Euler characteristic to help.)

i Solution. As p is odd, the Tate cohomology HTate(GR, Fp) = 0 and the same for Fp(1). Write out the Poitou–Tate exact sequence for Fp as follows:

0 0 2 ∗ 0 −→ H (GQ,S, Fp) H (GQp , Fp) H (GQ,S, Fp(1))

1 α 1 1 ∗ H (GQ,S, Fp) H (GQp , Fp) H (GQ,S, Fp(1))

2 2 0 ∗ H (GQ,S, Fp) H (GQp , Fp) H (GQ,S, Fp(1)) −→ 0

0 0 It is easy to see that H (GQ,S, Fp) = Fp, H (GQp , Fp) = Fp, and 1 ∼ ∼
∼ H (GQ,S, Fp) = Hom(GQ,S, Fp) = Hom Gal(Q(µp∞ )/Qp), Fp = Fp; 1 ur
∼ ∼ ∞ ∼ H (GQp , Fp) = Hom(GQp , Fp) = Hom Gal(Qp (µp )/Qp), Fp = Fp ⊕ Fp. 3

2 ∼ 0 ∗ By Tate local duality, H (GQp , Fp) = H (GQp , Fp(1)) = 0, and by global Euler characteristic formula,

χ(GQ,S, Fp) = χ(GQp , Fp) + χ(GR, Fp) = −1 + 1 = 0. 2 So we deduce that H (GQ,S, Fp) = 0. Now the Poitou–Tate exact sequence has become

2 ∗ 0 −→ Fp Fp H (GQ,S, Fp(1))

α 1 ∗ Fp Fp ⊕ Fp H (GQ,S, Fp(1))

0 ∗ 0 0 H (GQ,S, Fp(1)) −→ 0

0 It is easy to see directly that H (GQ,S, Fp(1)) = 0. For the other two terms, we claim that 1 ∼ the restriction map α is injective. This is because the non-trivial element in H (G ,S, Fp) = ∼ Q Hom(GQ,S, Fp) = Fp corresponds to the subextension Q(ζp2 )/Q of degree p, which is totally ramified at p; so it induces a nontrivial element in Hom(GQp , Fp). Thus α is injective. 2 ∗ 1 ∗ From this, we deduce that H (GQ,S, Fp(1)) = 0 and H (GQ,S, Fp(1)) = Fp. 

1 × Problem 2.4. (Cohomology of OF S [ S ] ) Let F be a number field and S a finite set of places of F including all archimedean places and places above `. (1) Show that there is a natural exact sequence   (2.4.1) 1 → O [ 1 ]× Q F × × Q O× → F × × → Cl(O [ 1 ]) → 1, F S v∈S v v∈ /S Fv AF F S

1 1 where Cl(OF [ S ]) is the ideal class group of OF [ S ], namely the quotient of the ideal class group Cl(OF ) by the subgroup generated by ideals in S. (2) By studying the exact sequence

1 × Q S × 1 × Q S × (2.4.2) 1 → OF S [ S ] → v∈S(Fv ⊗ F ) → OF S [ S ] v∈S(Fv ⊗ F ) → 1, show that   1 1 × ∼ 1 H GF,S, OF S [ S ] = Cl(OF [ S ]), and there is an exact sequence  Q`/Z` v non-arch 2 1 × M  1 0 → H GF,S, OF S [ S ] ⊗ Z` → 2 Z/Z v = R and ` = 2 → Q`/Z` → 0. v∈S 0 otherwise For i ≥ 3, we have (   1 / ` = 2 and i even, i 1 × ∼ M i × ∼ M 2 Z Z H GF,S, OF S [ ] ⊗ Z` = H (R, C ) = S 0 otherwise. v real v real

1 × 1 × Remark: Using Kummer theory 1 → µ` → OF S [ S ] → OF S [ S ] → 1, we can then use this 1 to further compute H (GF,S, µ`). 4

Solution. (1) By definition, Cl(O ) ∼ F ×\ × / Q O× . So we have F = AF,f v fin Fv . Y Y  Cl(O [ 1 ]) ∼ F × × O× F × F S = AF,f Fv v v∈ /S v∈S fin . Y Y  ∼ F × × O× F × = AF Fv v v∈ /S v∈S This implies an exact sequence Y Y 1 → Ker → O× × F × → F ×\ × → Cl(O [ 1 ]) → 0. Fv v AF,f F S v∈ /S v∈S   Here Ker = F × ∩ Q O× × Q F × ∼ O [ 1 ]×. The exact sequence (2.4.1) follows. v∈ /S Fv v∈S v = F S (2) Taking Galois cohomology of (2.4.2), we deduce

1 × Q × 0 1 × Q S ×
1 −→ OF [ S ] v∈S Fv H GF,S, OF S [ S ] v∈S(Fv ⊗ F )

1 1 ×
1 1 × Q S ×
H GF,S, OF S [ S ] 0 H GF,S, OF S [ S ] v∈S(Fv ⊗ F )

2 1 ×
L 2 S,× 2 1 × Q S ×
H GF,S, OF S [ S ] v∈S H (GFv ,Fv ) H GF,S, OF S [ S ] v∈S(Fv ⊗ F ) −→ 0

S S The red zero follows from Hilbert 90, and Fv denote the completion of F along any place of F S above v. i 1 × Q S ×
Plugging in the input of cohomology of H GF,S, OF S [ S ] v∈S(Fv ⊗ F ) from class field theory, we deduce that

1 −→ O [ 1 ]× Q F × F ×\ ×/ Q O× F S v∈S v AF v∈ /S Fv

1 1 ×
H GF,S, OF S [ S ] 0 0

2 1 ×
L 2 S,× H GF,S, OF S [ S ] ⊗ Z` v∈S H (GFv ,Fv ) ⊗ Q`/Z` Q`/Z` −→ · · ·

Here, for the last row, we tensored with Z` to take only the `-part. 1 1 × ∼ Comparing this exact sequence with the one in (1), we deduce that H (GF,S, OF S [ S ] ) = 1 2 3 Cl(OF [ S ]). The description of H and H follow from this as well.  Problem 2.5. (A step in the proof of local Euler characteristic formula) Consider the fol- lowing situation. Let K be a finite extension of Qp such that K = K(µp). Let L/K be a finite cyclic extension with Galois group H of order relatively prime to p. Let N be a finite Fp[H]-module. Our goal is to compute H  × × p
 dim OL /(OL ) ⊗ N(−1) 5

(1) Consider the logarithmic map

× logp : OL /µ(L) / L 1 n a / n logp(a )

pn 2 where n is taken sufficiently divisible so that a ∈ 1 + p OL so that logp(1 + x) = x2 x3 x − 2 + 3 − · · · makes sense. Show that logp is well-defined homomorphism (and independent of the choice of n), and that it induces an isomorphism ×
∼ logp : OL /µ(L) ⊗Zp Qp = L.

(2) Show that for any two OL-lattices Λ1, Λ2 ∈ L, we have H H dim(Λ1/pΛ1 ⊗ N) = dim(Λ2/pΛ2 ⊗ N) . (3) Recall that L ∼= K[H] as H-modules by Hilbert 90. From this and (2), deduce that H  × × p
 H dim OL /(OL ) ⊗ N(−1) = dim N + dim N · [K : Qp].

× × n Solution. (1) Take n to be divisible by #kL , so that for every a ∈ OL , a ≡ 1 mod $L. n n 1 n 2 1 n 3 Then logp(a ) := (a − 1) − 2 (a − 1) + 3 (a − 1) − · · · converges to an element in L, and the sequence convergences uniformly. By algebraic relations, it is easy to see that n n n × logp(a )+logp(b ) = logp((ab) ) for a, b ∈ OL . From this, it is immediate to see that defining 1 n logp(a) as n logp((a )) is a well-defined homomorphism. (2) The proof is similar to that of Exercise 1.1. By multiplication by a power of p, one −n may assume that Λ1 ⊆ Λ2 ⊆ p Λ1 for some n ∈ N. By further consider a sequence of n−1 n−2 modules Λ1 +p Λ2,Λ1 +p Λ2, ... ,Λ1 +Λ2 = Λ2, in turn, we may assume (at each step), −1 Λ1 ⊆ Λ2 ⊆ p Λ1. Now consider the tautological exact sequences

(2.5.1) 0 → pΛ2/pΛ1 → Λ1/pΛ1 → Λ1/pΛ2 → 0

(2.5.2) 0 → Λ1/pΛ2 → Λ2/pΛ2 → Λ2/Λ1 → 0 As the order of H is relative prime to p, taking H invariants is exact. So we have

H (2.5.1) H H dim(Λ1/pΛ1 ⊗ N) = dim(pΛ2/pΛ1 ⊗ N) + dim(Λ1/pΛ2 ⊗ N) H H = dim(Λ2/pΛ1 ⊗ N) + dim(Λ1/pΛ2 ⊗ N) (2.5.2) H = dim(Λ2/pΛ2 ⊗ N) . (3) From the exact sequence × × 1 → µ(L) → OL → OL /µ(L) → 1 we deduce that p × × p × p p 1 → µ(L)/µ(L) → OL /(OL ) → (OL /µ(L))/(OL/µ(L)) → 1 Thus, H  × × p
 dim OL /(OL ) ⊗ N(−1) H H  × × p
  p  = dim OL /(µ(L))/(OL /(µ(L))) ⊗ N(−1) + dim µ(L)/µ(L) ⊗ N(−1) . 6

p ∼ #µ(L)/p It is not hard to see that µ(L)/µ(L) = µp by sending ζ to ζ . So the last term is equal
H H to µp ⊗ N(−1) = dim N . H  × × p
 To compute dim OL /(µ(L))/(OL /(µ(L))) ⊗ N(−1) , we may use (1) and (2) to choose a different lattice in L ' K[H] for the convenience of the computation. From this, we deduce H H  × × p
   dim OL /(µ(L))/(OL /(µ(L))) ⊗N(−1) = dim OK /p[H]⊗N(−1) = [K : Qp]·dim N.

∼ dim N Here the in the last step, we used the fact that Fp[H]⊗N(−1) = Fp[H] as Fp[H]-module. Combining this with the above computation, we deduce (3).  Problem 2.6. (Comparing first Galois cohomology classes and extensions of Galois repre- sentations) If you are unfamiliar with the background of this problem, one can consult the short note on this topic, available on the webpage. Let k be a field with discrete topology, and let G be a finite group acting k-linearly on a finite dimensional k-vector space M. Let ρ : G → GLk(M) be the representation. 1 (1) Given a cohomology class [c] ∈ H (G, M), represented by cocycle g 7→ cg ∈ M, show that the following map defines a representation of G on Ec := M ⊕ k: ρ(g) c  g 7→ g . 0 1

0 (2) Show that if (cg)g∈G and (cg)g∈G define the same cohomology class, the representations Ec and Ec0 defined in (1) are isomorphic. (3) By definition, there exists an exact sequence 0 → M → Ec → k → 0. Taking the G-cohomology gives a connecting homomorphism

kG = k −−−−→δ H1(G, M) Show that δ(1) = [c]. (4) (Optional) Given an exact sequence of k[G]-modules

0 → M → E1 → E2 → k → 0, we may write F as the image of E1 → E2 and thus get two short exact sequences

0 → M → E1 → F → 0, and 0 → F → E2 → k → 0 This way, the boundary maps of the group cohomology defines two maps δ : k = H0(G, k) −→ H1(G, F ) −→ H2(G, M) and thus the image δ(1) defines a cohomology class [c] in H2(G, M). Now, suppose that we have a commutative diagram

0 / M / E1 / E2 / k / 0

  0 0 0 / M / E1 / E2 / k / 0. Show that the second cohomology class defined by these two exact sequences are the same. 7

Solution. (1) It is enough to check that ρ(g) c  ρ(h) c  ρ(gh) c  g · h = gh 0 1 0 1 0 1

The only non-trivial equality spells cg + ρ(g)ch = cgh. This is precisely the cocycle condition. 0 (2) Let m ∈ M be taken so that cg = cg + gm − m for every g ∈ G. Then we can check immediately that ρ(g) c  1 −m ρ(g) c0  1 m g = g 0 1 0 1 0 1 0 1 0 This means that the representation on M ⊕ k given by cg and cg are conjugate and hence isomorphic. (3) We following the definition of the connecting homomorphism: given 1 ∈ k, we lift it to 0
column vector 1 ∈ Ec. Then apply the differential map in group cohomology, we deduce a cycle 0 0 c  0 c0 := g − = g − = c ∈ M. g 1 1 1 1 g This verifies that δ(1) = [c]. (4) The commutative diagram of long exact sequences splits into two commuting short exact sequences:

0 / M / E1 / F / 0 0 / F / E2 / k / 0

    0 0 0 0 0 / M / E1 / F / 0 0 / F / E2 / k / 0 Then we have a commutative diagram

k = H0(G, k) / H1(G, F ) / H2(G, k)

 k = H0(G, k) / H1(G, F 0) / H2(G, k) So the image of 1 along the top row is the same as the image of 1 along the bottom row, i.e. the cohomology class defined by “equivalent” extensions are the same.  Problem 2.7. (An explicit computation of local Galois cohomology when ` 6= p) Let K be a finite extension of either Qp or Fp((t)), with ring of integers OK and residue field kK . Let ` ∞ be a prime different from p. Let M be a finite GK -module that is ` -torsion. Following the instruction below to give another proof of the Euler characteristic for local Galois cohomology when ` 6= p: 2 X (2.7.1) χ(G ,M) := (−1)i · length Hi(G ,M) = 0. K Z` K i=0

(1) Let IK and PK denote the inertia subgroup and the wild inertia subgroup of GK . >0 ∞ Show that H (PK ,M) = 0 for any GK -module M that is ` -torsion. Using the Hoshchild–Serre spectral sequence to deduce that, for every i ≥ 0,

i ∼ i PK H (IK ,M) = H (IK /PK ,M ). 8

tξ,` i ∼ (2) Let PK,` denote the kernel of IK → IK /PK −−→ Z`(1). Show that we have H (IK ,M) = i PK,` H (Z`(1),M ). PK,` (3) Put N := M , and write τ for a generator of IK /PK , then we have 0 ∼ τ=1 1 ∼ H (IK ,M) = N ,H (IK ,M) = N/(τ − 1)N. Note also that the second isomorphism is given by evaluating the cochain at τ; so the Frobenius action on N/(τ − 1)N is twisted by the inverse of cyclotomic character. Thus, we should have wrote N(−1)/(τ − 1)N(−1) instead. (4) Let φK denote a Frobenius element. Show that we have isomorphisms: N(−1) . H0(G ,M) ∼= (N τ=1)φK =1,H2(G ,M) ∼= (φ − 1). K K (τ − 1)N(−1) K 1 For H (GK ,M), we may describe the unramified part and singular part of it as follows:

G 1 IK 1 1 kK 0 / H (GkK ,M ) / H (GK ,M) / H (IK ,M) / 0.

=∼ =∼  φK =1   N(−1)  τ=1 τ=1 (τ−1)N(−1) N /(φK − 1)N (5) From this, deduce Euler characteristic formula (2.7.1) directly. (6) Using the discussion above to prove the following isomorphism of exact sequences:

G 1 IK 1 1 kK 0 / H (GkK ,M ) / H (GK ,M) / H (IK ,M) / 0

=∼ =∼ =∼    G ∗ ∗ 1 ∗ kK
1 ∗ 1 ∗ IK
0 / H (IK ,M (1)) / H (GK ,M) / H (GkK , (M (1)) ) / 0.

1 IK 1 ∗ IK (Hint: first show that the subgroup H (GkK ,M ) and H (GkK , (M (1)) ) annihi- late each other. This is because such pairing factors through the cup product.

1 IK 1 ∗ IK 2 ∞ H (GkK ,M ) × H (GkK , (M (1)) ) → H (GkK , µ` ) = 0. G 1 IK 1 ∗ kK After this, it is enough to show that #H (GkK ,M ) = #H (IK ,M (1)) , which makes use of the discussion above.)

j j H Solution. (1) Since PK is pro-p, by definition H (PK ,M) = lim H (PK /H, M ) = 0 −→HCPK when j > 0 as M is `∞-torsion. Thus, the Hoshchild–Serre spectral sequence degenerates: ( i j 0 if j > 0 H (IK /PK ,H (PK ,M)) = i PK H (IK /PK ,M ) if j = 0. So we deduce that for every i ≥ 0,

i ∼ i PK H (IK ,M) = H (IK /PK ,M ).

(2) By exactly the same argument as in (1) with PK replaced by PK,`, we deduce that for every i ≥ 0, i ∼ i PK,` ∼ i PK,` H (IK ,M) = H (IK /PK,`,M ) = H (Z`(1),M ). (3) By cohomology of procyclic group, we deduce that 0 ∼ 0 τ=1 H (IK ,M) = H (Z`(1),N) = N , 9

1 ∼ 1 H (IK ,M) = H (Z`(1),N) = N(−1)/(τ − 1)N(−1). i (4) This follows immediately from the description of H (IK ,M) above and Hoshchild spec- tral sequence. (5) We note the following two exact sequences

φ −1 τ=1 φK =1 τ=1 K τ=1 τ=1 τ=1 0 → (N ) → N −−−→ N → N /(φK − 1)N → 0,

 N(−1) φK =1 N(−1) φ −1 N(−1) N(−1) . 0 → → −−−→K → (φ −1) → 0. (τ − 1)N(−1) (τ − 1)N(−1) (τ − 1)N(−1) (τ − 1)N(−1) K From these, we deduce immediately that 0 τ=1 φ =1 τ=1 τ=1
1 G length H (G ,M) = length (N ) K = length N /(φ −1)N = length H (I ,M) kK . Z` K Z` Z` K Z` K

2 N(−1) . length H (GK ,M) = length (φK − 1) Z` Z` (τ − 1)N(−1)  N(−1) φK =1 1 IK = length = length H (Gk ,M ). Z` (τ − 1)N(−1) Z` K Euler characteristic follows from this immediately. (6) First note that we have the following commutative diagram

1 I 1 ∗ I 2 K K ∞ H (GkK ,M ) × H (GkK , (M (1)) ) / H (GkK , µ` ) = 0

   1 1 ∗ 2 H (GK ,M) × H (GK ,M (1)) / H (GK , µ`∞ )

1 IK 1 ∗ IK So under the local Tate duality, H (GkK ,M ) annihilates H (GkK , (M (1)) ). As the local Tate duality is a perfect pairing, it is enough to verify that 1 I 1 ∗ G K kK #H (GkK ,M ) = #H (IK ,M (1)) . Using the above description of the terms, this is equivalent to prove that

 N(−1) φK =1 # = #(N ∗(1))τ=1/(φ − 1)(N ∗(1))τ=1 (τ − 1)N(−1) K This follows from the fact that the following two exact sequences are dual to each other

 N(−1) φK =1 N(−1) φ −1 N(−1) 0 → → −−−→K (τ − 1)N(−1) (τ − 1)N(−1) (τ − 1)N(−1)

∗ τ=1 φK −1 ∗ τ=1 ∗ τ=1 ∗ τ=1 (N (1)) −−−→ (N (1)) → (N (1)) /(φK − 1)(N (1)) → 0  Problem 2.8. Fix a prime number `. Let F be a number field and S a finite set of places that includes all archimedean places and places above `. For an extension L of F , we write S SL for the set of places of L that lies over places in S. Let F denote the maximal Galois extension of F that is unramified outside S. We compare the cohomology groups (for i = 1, 2)

i  S,× ×  i  1 × Q S × H GF,S, F AF S ⊗ Z` with H GF,S, OF S [ S ] v∈S(Fv ⊗ F ) ⊗ Z`. 10

(1) For a finite extension L ⊂ F S, show that we have an exact sequence   (2.8.1) 1 → O [ 1 ]× Q L× × Q O× → L× × → Cl(O [ 1 ]) → 1, L S w∈SL w w∈ /SL Lw AL L S 1 1 where Cl(OL[ S ]) is the ideal class group of OL[ S ]. (This is Problem 2.4(1) earlier.) (2) Show that the limit lim Cl(O [ 1 ]) is trivial. (Hint: A property of Hibert class ←−L⊂F S L S field theory is that, if LHilb/L is the Hilbert class field of L, namely the maximal unramified abelian extension of L, every ideal of L becomes principal in LHilb. Using the commutative diagram for compatibility of Artin maps with ideal class groups

Art 0× × L0 ab L \AL0 / GL0 O O natural Ver

× × ArtL ab L \AL / GL , to show that this boils down to the following group theoretic statement: Let G be a pro-finite group and H its commutator group, then the transfer map Gab → Hab is the zero map. This is known as the Artin’s principal ideal theorem. The class field theory by Artin–Tate has a proof of this, on their page ) (3) Deduce that

0 1 × Q S × × × Q × H G , O S [ ] (F ⊗ F ) ∼ F \ / O , F,S F S v∈S v = AF v∈ /S Fv

1 1 × Q S × H GF,S, OF S [ S ] v∈S(Fv ⊗ F ) = 0 and    2 1 × Q S × ∼ H GF,S, OF S [ S ] v∈S(Fv ⊗ F ) ⊗ Z` = Q`/Z`. Solution. (1) This is problem 2.4 earlier. (2) As LHilb is a subfield of F S (as it is everywhere unramified), by the cited Artin’s 1 1 principal ideal theorem, the natural map Cl(OL[ S ]) → Cl(OLHilb [ S ]) is the zero map. So the limit is zero. (3) Taking limit of (2.8.1), we deduce that 1 × Q S × Q × ∼ S × S × O S [ ] (F ⊗ F ) × O S = (F ) ( ⊗ F ) . F S v∈S v v∈ /S F ⊗Fv AF >0 × Note that H (G , O S ), so we immediately deduce that for i ≥ 1, F,S F ⊗Fv ( i 1 × Q S ×

∼ i S × S ×
∼ 0 if i = 1 H GF,S, OF S [ S ] v∈S(Fv⊗F ) ⊗Z` = H GF,S, (F ) (AF ⊗F ) ⊗Z` = Q`/Z` if i = 2. For H0, we have instead an isomorphism ∼ 0 1 × Q S ×

Q × = × × H GF,S, OF S [ S ] v∈S(Fv ⊗ F ) × v∈ /S Fv −→ F \AF . 0 We deduce the description of H immediately from this.