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Exercise 2 Solutions Problem 2.1. (Local Galois Cohomology

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Exercise 2 Solutions Problem 2.1. (Local Galois Cohomology 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 8 (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 ` j 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 ` j 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 ` j pn − 1 if ` j pn−1 − 1 F` Gk F` F`(n − 1) = 0 otherwise. 0 otherwise. Combining these two, we obtain 8 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)jn 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 8 >2 if p − 1jd 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 = fp; 1g 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(µp1 )=Qp); Fp = Fp; 1 ur ∼ ∼ 1 ∼ 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 v2S v v2 =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 ] ! v2S(Fv ⊗ F ) !OF S [ S ] v2S(Fv ⊗ F ) ! 1; show that 1 1 × ∼ 1 H GF;S; OF S [ S ] = Cl(OF [ S ]); and there is an exact sequence 8 >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: v2S :>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 ×n × = 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 v2 =S v2S fin . Y Y ∼ F × × O× F × = AF Fv v v2 =S v2S This implies an exact sequence Y Y 1 ! Ker ! O× × F × ! F ×n × ! Cl(O [ 1 ]) ! 0: Fv v AF;f F S v2 =S v2S Here Ker = F × \ Q O× × Q F × ∼ O [ 1 ]×. The exact sequence (2.4.1) follows. v2 =S Fv v2S v = F S (2) Taking Galois cohomology of (2.4.2), we deduce 1 × Q × 0 1 × Q S × 1 −! OF [ S ] v2S Fv H GF;S; OF S [ S ] v2S(Fv ⊗ F ) 1 1 × 1 1 × Q S × H GF;S; OF S [ S ] 0 H GF;S; OF S [ S ] v2S(Fv ⊗ F ) 2 1 × L 2 S;× 2 1 × Q S × H GF;S; OF S [ S ] v2S H (GFv ;Fv ) H GF;S; OF S [ S ] v2S(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 ] v2S(Fv ⊗ F ) from class field theory, we deduce that 1 −! O [ 1 ]× Q F × F ×n ×= Q O× F S v2S v AF v2 =S Fv 1 1 × H GF;S; OF S [ S ] 0 0 2 1 × L 2 S;× H GF;S; OF S [ S ] ⊗ Z` v2S H (GFv ;Fv ) ⊗ Q`=Z` Q`=Z` −! · · · Here, for the last row, we tensored with Z` to take only the `-part.
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