GALOIS COHOMOLOGY GEUNHO GIM

GALOIS COHOMOLOGY GEUNHO GIM Abstract. This note is based on the 3-hour presentation given in the student seminar on Fall 203. We will basically follow [HidMFG, Chapter IV] and [MilADT, Chapter I 0,, 2]. Contents. Group Cohomology 2. The main duality theorem 4 3. Local/global case 8 3.. Local case 8 3.2. Global case 0 4. Euler characteristic formula 0 References. Group Cohomology Let G be a group and M be a G-module (i.e., an abelian group which G acts on). For G- modules M and N, a G-module homomorphism α : M N is a group homomorphism satisfying g α(m) = α(g m) for g G, m M. If we define the action of G on Hom Z (M, N) by (g φ)(m) = g(φ(g m)), then we have Hom G (M, N) = Hom Z (M, N) G. We define G-Mod to be the category of G-modules with G-homomorphisms. Note that G-Mod is equivalent to the category of modules over the ring Z[G], G-Mod = Z[G]-Mod. There are several equivalent ways to construct the cohomology groups. Construction : Extension groups Note that G-Mod has enough injectives. For a G-module M, we have an injective resolution: 0 M ɛ M 0 0 M (exact sequence with injective M i s for all i) and we write this 0 M (M, ) in short. Given M, M G-Mod, and injective resolutions 0 M (M, ) and 0 N (N, δ), we define Ext r G (M, N) = {homotopy classes of deg r chain maps (M, ) (N, δ)} = H r (Hom G (M, N ), δ ) Here deg r maps f, g : (M, ) (N, δ) are homotopic if there is a deg(r ) map satisfying f g = δ +. We define the r-th cohomology group by H r (G, M) = Ext r G (Z, M) Date: April 8, 204.

where G acts on Z trivially. Construction 2 : Right derived functor For a G-module M, we define M G = {m M gm = m for all g G}. Since ( ) G = Hom Z (Z, ) G = Hom G (Z, ), ( ) G : G-Mod Ab is a left exact functor. We apply this to an injective resolution and define as the right derived functor of ( ) G. 0 M G (M 0 ) G d 0 (M ) G d H r (G, M) = ker dr im d r Construction 3 : Cochain complex This is the most explicit construction because we can actually see what the elements of the cohomology groups are. We first define C r (G, M) = {φ : G n M} as the set of set-theoretic maps. Also define the boundary maps r : C r C r+ by ( r φ)(g,..., g r+ ) = g φ(g 2,..., g r+ ) + r j= ( ) j φ(g,..., g j g j+,..., g r+ ) + ( ) r+ φ(g,..., g r ) Now we can define H r ker r (G, M) = im r as the cohomology group. Note that all three constructions are equivalent. (See for example, [HidMFG, 4.3.].) Example.. () H 0 (G, M) = Hom G (Z, M) = M G (2) H {u : G M u(gh) = gu(h) + u(g)} (G, M) = {α m : G M m M} g (g )m (3) If G acts on M trivially, then H (G, M) = Hom group (G, M). Definition.2. Let H G be a subgroup and M H-Mod. We define the induced module Ind G H(M) G-Mod by Ind G H(M) = {φ : G M φ(hg) = hφ(g) for all h H} and the G-action is given by (gφ)(m) = φ(mg). Definition.3. Let M G-Mod and M G -Mod. Consider G G α M M β. The maps α and β are called compatible if β(α(g )m) = g (β(m)) for all g G and m M. In this case, there is a natural map between cohomology groups: H r (G, M) H r (G, M ) [φ : G n M] [β φ (α α) : G n M ] 2

Example.4. () Let H G and M G-Mod. Note that M is also an H-module by restricting the action of G to H. The maps H G and M id M are compatible. Thus, we get the restriction map The map M m (g gm) res : H r (G, M) H r (H, M) Ind G H M induces the following commutative diagram. res H r (G, M) H r (H, M) H r (G, Ind G H M) where the isomorphism on the right is from Shapiro s lemma. (2) Let H G and M G-Mod. Note that M H (G/H)-Mod. The maps G G/H and M H M are compatible, thus inducing the inflation map inf : H r (G/H, M H ) H r (G, M) (3) Suppose the index (G : H) is finite and M G-Mod. The map Ind G H M M φ sφ(s ) s G/H defines the corestriction map by the commutative diagram cor H r (H, M) H r (G, M) H r (G, Ind G H M) where the isomorphism on the left is from Shapiro s lemma. Proposition.5. Let H G be a subgroup of a finite group G and M G-Mod. () The composition cor G/H res G/H : H r (G, M) H r (G, M) is multiplication by (G : H). (2) G kills H r (G, M). (3) Suppose H G, r > 0, and we have H i (H, M) = 0 for all 0 < i < r. Then there is the following inflation-restriction exact sequence 0 H r (G/H, M H ) inf H r (G, M) res H r (G, M) Proof. () The composition of the maps defined earlier M Ind G H(M) M is multiplication by (G : H). (2) (G:) H r (G, M) H r (G, M) H r ({}, M) = 0 (3) If r =, it is clear. For r >, we use induction. See [MilCFT, p. 69]. 3

Definition.6. Let G be a finite group and M G-Mod. We define Tate cohomology groups as follows: H r (G, M) r > 0 HT(G, r M G /N M) = G M r = 0 ker N G /I G M r = H r (G, M) r < where N G : M m g G gm M and I GM = (g )M. This connects homology and cohomology. g G Example.7. H 2 T (G, Z) = Gab. Theorem.8 (Tate). Let G be a finite group and C G-Mod. Suppose () H (H, C) = 0 (2) H 2 (H, C) is cyclic of order H for all subgroups H G. Then we have H r T(G, Z) γ H r+2 T (G, C) where [γ] is a generator of H 2 (G, C). (Note that the isomorphism depends on the choice of γ.) In general, if M G-Mod is torsion-free, then we have H r T(G, M) Proof. Use dimension-shifting. See [MilCFT, p. 8]. γ H r+2 T (G, M Z C) 2. The main duality theorem Let G be a profinite group and M be a discrete G-module (i.e., M has discrete topology and the action G M M is continuous.) This is equivalent to say that M = M U. We define the continous cochain complex Cct(G, n M) = {φ : G n M, continuous} and by using this, define Hct(G, r M). Note that for φ Cct(G, n M), we have G n = φ (m i ) by compactness. Thus, φ is locally constant and factors through finite UG G n φ M (G/U) n for some U G. Therefore we have Hct(G, r M) = lim H r (G/U, M U ) UG where the limit is over inf U/V : H r (G/U, M U ) H r (G/V, M V ) for V U G subgroups. From now on, G-Mod means the category of discrete G-modules (which has enough injectives) and H = H ct. 4

Definition 2.. Suppose C G-Mod and we have inv U : H 2 (U, C) Q/Z for all subgroups U G (called the invariant map relative to U). (G, C) is called a class formation if () H (U, C) = 0 for all U G (2) for all V U G, we have H 2 (U, C) H 2 (V, C) inv U Q/Z res (U:V) inv U Q/Z Theorem 2.2. Let (G, C) be a class formation. Then, there is a map rec G : C G G ab called the reciprocity map such that im(rec G ) G ab is dense and ker(rec G ) = N G/U : C U x g G/U gx CG. UG N G/U C U where Proof. Let V U G. From the inflation-restriction exact sequence 0 H (U/V, C V ) inf H (U, C) res H (V, C) we have H (U, C) = 0 and H (V, C) = 0 by assumption (), thus H (U/V, C V ) = 0. From assumption (2), inf 0 H 2 (U/V, C V ) H 2 (U, C) H 2 (V, C) res 0 inv U/V inv U inv V (U:V) Z/Z Q/Z Q/Z 0 (U : V) we get inv U/V := H 2 (U/V, C V ) u U/V (U : V) Z/Z where H2 (U/V, C V ) = u U:V. (U:V) Now we apply Tate s theorem to G/U to get H 2 Tate T (G/U, Z) H0 T (G/U, CU ) (G/U) ab C G /N G/U C U Now we define rec G : C G G ab = lim (G/U) ab by using these maps. Then we get ker(rec G ) = UG UG N G/U C U and im(rec G ) is dense in G ab. 5

Example 2.3 (trivial example). Suppose G = σ = Ẑ and G C = Z trivially. Let U G be an subgroup. By using the fact that H r (G, Q) = lim H r (G/U, Q) is a torsion group for UG r > 0 and Q n Q, we get H r (G, Q) n H r (G, Q) and H r (G, Q) = 0 for all r > 0. By applying H (U, ) to the exact sequence 0 Z Q Q/Z 0 we can define inv U : H 2 (U, Z) H (U, Q/Z) = Hom ct (U, Q/Z) Q/Z and H (U, Z) = Hom ct (U, Z) = 0 since the image of a continuous map from U is a compact subgroup of Z. Therefore, (G, Z) is a class formation. If (G : U) = m, then we have C G /N G/U C U (G/U) ab and rec G : C G (= Z G = Z) G ab = Ẑ. Z/mZ Z/mZ Definition 2.4. Let M be an abelian group. We define M = Hom Z (M, Q/Z). (If M is a torsion group, then this is the Pontryagin dual of M.) Definition 2.5. Let M G-Mod and (G, C) be a class formation. We have the following Ext pairing: Ext r G (M, C) Ext2 r G (Z, M) Ext2 G (Z, C) H 2 r (G, M) H 2 (G, C) Q/Z where the map is the composition of chain maps. We define by using the Ext pairing. α r (G, M) : Ext r G (M, C) H2 r (G, M) Example 2.6. Let M = Z and G M trivially. Then, we have Ext r G (Z, C) = Hr (G, C). () (r = 0) α 0 (G, Z) : H 0 (G, C)(= C G ) H 2 (G, Z)(= Hom ct (G, Q/Z) = G ab ) and we have α 0 (G, Z) = rec G. (2) (r = ) α (G, Z) : H (G, C)(= 0) H (G, Z) (= 0) since H (G, Z) = Hom ct (G, Z) = 0. (3) (r = 2) α 2 (G, Z) : H 2 (G, C) H 0 (G, Z) (= Z = Q/Z) and we have α 2 (G, Z) = inv G. Example 2.7. Let M = Z/mZ and G M trivially. By using the exact sequence 0 Z m Z Z/mZ 0, we get the following: 6 inv G

() (r = 0) α 0 (G, Z/mZ) : Ext 0 G (Z/mZ, C) H2 (G, Z/mZ) C G [m] G ab [m] and the composition is rec G : C G [m] G ab [m]. (2) (r = ) α (G, Z/mZ) : Ext G (Z/mZ, C) H (G, Z/mZ) C G /mc G G ab /mg ab and we have α (G, Z/mZ) = rec G : C G /mc G G ab /mg ab. (3) (r = 2) α 2 (G, Z/mZ) : Ext 2 G (Z/mZ, C) H0 (G, Z/mZ) H 2 (G, C)[m] m Z/Z and we have α 2 (G, Z/mZ) = inv G H2 (G,C)[m]. Theorem 2.8 (Main Theorem : Tate). Let (G, C) be a class formation and M G-Mod be a finitely generated G-module. Then, () α r (G, M) : Ext r G (M, C) H2 r (G, M) is bijective for r 2. α (G, M) is bijective for torsion-free M. Ext r G (M, C) = 0 and for torsion-free M and r 3. (2) If α (U, Z/mZ) is bijective for all U G and m, then α (G, M) is bijective. (3) If α 0 (U, Z/mZ) is bijective for all U G and m, then α 0 (G, M) is bijective for finite M. sketch of proof. Step : Ext r G (M, C) = 0 for r 4 and Ext3 G (M, C) = 0 for torsion-free M. (Thus, for large r, we have α r (G, M) : 0 0. Step 2 : Show this when G M trivially. It is enough to show for M = Z, Z/mZ. Step 3 : General case. Since M = U M U is finitely generated, M = M U for some U G. Define π : Z[G/U] a g g a g We get the following diagram: Z and M = Hom Z (Z, M) π Hom Z (Z[G/U], M) =: I S = coker π Ext r G (S, C) Extr G (I, C) Extr G (M, C) Extr+ G (S, C) α r (G,S) α r (U,M) α r (G,M) α r+ (G,S) H 2 r (G, S) H 2 r (G, I) H 2 r (G, M) H r (G, S) 7

where we have α r (U, M) : Ext r G (I, C) = Ext r U(M, C) H 2 r (U, M) = H 2 r (G, I) We know that α r (U, M) is an isomorphism because U M trivially. We can use the five lemma to show the general case. Detailed proof can be found in [MilADT, p. 2] or [HidMFG, p. 208]. 3. Local/global case 3.. Local case. Let K/Q p be a finite extension and G = Gal(K/K) (fix K), G C = K. For L/K finite, we write H 2 (L/K) = H 2 (Gal(L/K), L ). We have the following exact sequence: U K un (K un ) Z 0 where K un is the maximal unramified extension and U K un = O K un. Note that H r T(Gal(L/K), U L ) = 0 for a finite unramified extension L/K and all r, thus H r T(Gal(K un /K), U K un) = 0. We define where inv G = inv K : H 2 (G, C) = H 2 (K/K) inf H 2 (K un /K) H 2 (Gal(K un /K), Z) H (Gal(K un /K), Q/Z) Q/Z inf : H 2 (K un /K) = H 2 (Gal(K/K)/ Gal(K/K un ), (K ) Gal(K/Kun) ) H 2 (Gal(K/K), K ) For U G, we have U = Gal(K/L) = Gal(L/L) for some L/K finite. Thus we can define inv U = inv L : H 2 (U, C) = H 2 (L/L) Q/Z similarly. Also note that H (U, K ) by Hilbert theorem 90. Consider the following: 0 H 2 (L/K) H 2 (K/K) H 2 (L/L) res 0 inv K [L:K]=(G:U) Z/Z Q/Z Q/Z [L : K] inv L This shows that (G, K ) is a class formation. By Tate s theorem, there is a map and in fact, we have K = Gal(K ab /K). rec G : (K ) G = K G ab = Gal(K ab /K) Theorem 3. (nonarchimedean case). Let (G, C) = (Gal(K/K), K ) be a class formation and M be a finitely generated G-module. Then, () α r (G, M) : H r (G, Hom(M, K )) Ext r G (M, K ) H 2 r (G, M) is bijective for r. (2) α 0 (G, M) : Hom G (M, K ) H 2 (G, M) is bijective for finite M. (3) For finite M, H r (G, M) is finite for r = 0,, 2 and zero otherwise. (This is also true for function field case if we use K sep /K and if ( M, char K) =.) 8

Proof. (), (2) Let I = Gal(K/K un ). We have the following commutative diagram: O K K Z 0 rec G Tensoring Z Z/mZ gives I ab G ab Ẑ 0 α (G, Z/mZ) : K /(K ) m G ab /(G ab ) m Ext G (Z/mZ, K ) H (G, Z/mZ) and applying Hom Z (Z/mZ, ) gives α 0 (G, Z/mZ) : µ m (K ) G ab [m] Hom G (Z/mZ, K ) Therefore α (G, M) is bijective and so is α 0 (G, M) for finite M by Theorem 2.8. (3) Note that H r (G, M) = 0 for r > 2 by duality. Since α 0 (G, M) : Hom G (M, K ) H 2 (G, M) and Hom G (M, K ) is finite, H 2 (G, M) is finite. Also, H 0 (G, M) = M G is finite. Apply ( ) G to to get µ m K m K µ m (K) K m K H (G, µ m ) H (G, K ) = 0 From this, we have H (G, µ m ) = K /(K ) m, which is finite. Choose m such that mm = 0. Take a finite extension L/K satisfying µ m L and M U = M for U = Gal(K/L). Then, M = µ n as a U-module. We have finite n m 0 H (G/U, M) inf H (G, M) res H (U, M) Here H (G/U, M) is finite because G/U and M are finite, and H (U, M) = H (U, µ n ) is also finite by previous argument. Therefore, H (G, M) is finite. Theorem 3.2 (archimedean case). G-module. Then, () Let G = Gal(C/R), C = C and M be a finitely generated HT(G, r Hom Z (M, C )) HT 2 r (G, M) H 2 (G, C ) 2 Z/Z is a perfect pairing for all r Z. (2) Let K = R or C, and H = Gal(C/K). Let M be a finite H-module. Then, Proof. See [MilADT, p. 35]. M [K:R] = H0 (H, M) H 0 (H, Hom Z (M, C )) H (H, M) 9

3.2. Global case. Let K be a number field, G = Gal(K/K) and C = lim L/K formation, but it is too big. finite C L. (G, C) is a class Definition 3.3. Let P be a set of prime numbers. For a G-module C, (G, C) is called a P-class formation if for all U G, there is a map inv U : H 2 (U, C) Q/Z such that () H (U, C) = 0 (2) for V U G, we have the following commutative diagram 0 H 2 (U/V, C V ) H 2 (U, C) H 2 (V, C) res 0 inv U/V inv U inv V (U:V) Z/Z Q/Z Q/Z (U : V) (3) for l P, we have between l-primary parts. inv U [l] : H 2 (U, C)[l ] (Q/Z)[l ] Let P be a finite set of primes of Q including and S be the set of primes of K above P. Let K S /K be the maximal unramified extension outside S and G S = Gal(K S /K). For a finite extension F/K inside K S, let Σ be the set of primes above S. Define OF S = F ( O F,p ), C F,S = F S /(OS F ) and C S = lim C F F,S. Then, (G S, C S ) is a P-class formation. Theorem 3.4. Let (G S, C S ) be the P-class formation defined above and M be a finitely generated G S -module. Suppose l P. Then, () α r (G S, M)[l] : Ext r G (M, C S)[l ] H 2 r (G S, M) [l ] is bijective for all r. (2) Let M be finite, and F/K be a totally imaginary extension. Suppose Gal(K S /F) M, µ M trivially, and M O S = O S = lim F p/ Σ OF S. Then, there is the following exact sequence: Hom(M, D S (F)) N F/K Hom GS (M, C S ) α0 (G S,M) H 2 (G S, M) 0 where D S (F) is the identity component of C F (S) = C F /U F,S where U F,S = (O F Z Z l ). l P Proof. See [MilADT, p. 52]. 4. Euler characteristic formula Definition 4. (Local case). Let K/Q p be a finite extension and G = Gal(K/K). Let M be a finite G-module. Then we define the Euler characteristic of M by χ(m) = χ(g, M) = H0 (G, M) H 2 (G, M) H (G, M) Theorem 4.2. χ(m) = O K : M O K 0

Remark 4.3. If M has l-power order for a prime l, then H j (G, M) is a Z l -module of finite length for all j. We have log l χ(m) = 2 j=0 ( ) j log l H j (G, M) = (H j (G, M) = 0 for j = 0,, 2) and ( ) log l = O K : M O K proof of Theorem 4.2. See [HidMFG, p. 228]. j=0 ( ) j length Zl H j (G, M) { 0 l = p [K : Q p ] length Zp M l = p Definition 4.4 (Global case). Let K/Q be a finite extension and G S = Gal(K S /K). Let M be a finite G S -module and assume that M O S = O S. Then we define χ(m) = χ(g S, M) = H0 (G S, M) H 2 (G S, M) H (G S, M) Remark 4.5. It is NOT true that we have H r (G S, M) = 0 for r 3 like in local case. H Theorem 4.6. χ(m) = 0 (G v, M) where G v = Gal(K v /K v ) and M = M [Kv:R]. v: arch M v v Proof. See [MilADT, p. 67]. References [HidMFG] H. Hida, Modular Forms and Galois Cohomology, Cambridge University Press, 2000 [MilADT] J. S. Milne, Arithmetic Duality Theorems, 2006, available at www.jmilne.org/math [MilCFT] J. S. Milne, Class Field Theory (v4.02), 203, available at www.jmilne.org/math