A K,S =: K p. with the direct limit topology where. A K,S := K p O K p

Transcription

1 42 M. FLACH 3. The idele class grou and idelic class field theory We have seen that for a global field K the modulus m aears as a arameter for the tower of ray class fields K(m). The content of class field theory is that the K(m) have an exlicit Galois grou and that their union is the maximal abelian extension of K (the union of all abelian extensions in a given algebraic closure). This same information can be reackaged in a more canonical construction which at the same time makes the relation to local class field theory much more transarent and allows to comute the Brauer grou of the global field K. This is the role of the idele class grou. It is a more elegant but slightly more abstract aroach to class field theory since one has to oerate with rofinite or more general toological grous. Definition Let K be a global field. Define the idele grou of K as A K = S with the direct limit toology where A K,S =: K A K,S := K O K S / S has the roduct toology and S runs through all finite sets of laces containing all archimedean ones. This is a locally comact grou by Tychonoff s theorem. Remark As is suggested by the notation A K is the unit grou of the similarly defined adele ring A K for which we have no use in this course, excet for Lemma 3.1 below. Since any element α K is integral at all but finitely many laces we have a natural embedding K A K. We also have an absolute value ma (α ) := α and K lies in its kernel by the roduct formula. Definition The idele class grou C K of the global field K is the quotient grou C K := A K /K. It is a (slit) extension 1 CK 1 log C K R 0 if K is a number field and 1 CK 1 log / log(q) C K Z 0 if K has characteristic and F q is the algebraic closure of F in K. Here CK 1 = ker( ). Proosition 3.1. The toology on K induced from the locally comact toology of A K is the discrete toology. Therefore C K is a locally comact abelian grou. Moreover, the grou CK 1 is actually comact.

2 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 43 Proof. It suffices to find a neighborhood of 1 containing no other element of K. Take U = O K B(1,ɛ) where B(1,ɛ)={x K x 1 <ɛ} is the ball of radius ɛ<1. Now look at 1 α U K. Then0 α 1 O K and α 1 <ɛfor all archimedean laces. But this contradicts the roduct formula. Lemma 3.1. There is a constant C>0 with the following roerty: If a =(a ) A K with a >C then there exists α K such that α a. Proof. This is reminiscent of the Minkowski argument in the classical geometry of numbers. We use without roof that K is a discrete subgrou of A K (similar argument to the one just given for A K ) and that A K/K is comact. Let c 0 be the Haar measure of A K /K and and c 1 that of the set W = {ξ A K ξ ɛ } where ɛ = 1 for non-archimedean and ɛ =1/10 for archimedean. For C = c 0 /c 1 the set T = {ξ A K ξ ɛ a } has measure c 1 a >c 1 C = c 0 and therefore there must be a air of distinct oints t 1,t 2 of T with the same image in A K /K. Then α := t 1 t 2 K and α = t 1 t 2 a by the triangle inequality and its ultrametric sharening. For an idele a with a >C consider the set W = {ξ A K ξ a } which is comact since it is a roduct of comact sets. Now for b A K with b =1 we also have b 1 a >C, hence the Lemma gives α K with α b 1 a, i.e. αb W Class field theory using the idele grou. In section we reformulate the main theorems of section 1 in terms of the idele grou. We start by describing ray class grous H m in terms of the idele class grou. Given a modulus m = m f m = n we define an oen subgrou O K finite, n =0 1+ n O K finite, n > 0 U(n )= K 2 real and n =1 K infinite, n =0 of K for each lace, and then we define the oen subgrou U m = U(n )

3 44 M. FLACH of A K. To a finite Galois extension L/K we can also associate an oen subgrou N L/K (A L )= N LP /K L P where we choose a P and note that the local norm subgrou is indeendent of this choice. We remark that this is just the image of the norm ma for the finite flat ring extension A L /A K and this ma sends L to K. Hence there is also an induced ma N L/K : C L C K. Proosition 3.2. Denoting by Ū m the image of U m in C K we have C K /Ū m = A K /K U m J m /P m = H m and C K /N L/K (C L )=A K /K N L/K (A L ) = J m /P m N L/K J m L. Proof. We have A K /U m = J m S K /U(n ) where S is the union of the set of rimes dividing m and the infinite laces. By weak aroximation there is an exact sequence 1 K,m K S K /U(n ) 1 where is as in Lemma 1.6. Then K,m = {α K α 1 mod m} C K /Ū m = A K /U m K = (J m S K /U(n ))/ im(k ) = (J m 1)/ im(k,m ) = J m /P m = H m. For a finite extension L/K we can choose a modulus m so that U(n ) N LP /K L P for all laces and hence that U m N L/K (A L ). Weak aroximation gives us the isomorhism (16) K /N L/K L K,m K /N LP/K (L P ) S

4 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 45 as in the roof of Lemma 1.6. Then we have C K /N L/K (C L ) =A K /K N L/K (A L ) = J m K /U(n ) / N L/K JL m N LP/K (L P )/U(n ) im(k ) S S =(J m 1)/(N L/K JL m 1) im(k,m N L/K L ) =J m /(N L/K JL m P m ). Hence we can view the Artin ma for a finite abelian extension L/K as a ma ρ = ρ L/K : C K Gal(L/K) inducing an isomorhism C K /N L/K C L = Gal(L/K). In articular, viewing K as a subgrou of C K, via the embedding α (1,...,1,α, 1,...) A K we get homomorhisms for all rimes ρ : K Gal(L/K) which clearly coincide with the local Artin ma ρ defined in Ma160b if is unramified in L/K. It is not so clear whether ρ = ρ for all laces. One can now roceed in two ways. Either one shows that ρ takes values in Gal(L P /K ) Gal(L/K) and in fact induces an isomorhism K /N LP/K L P = Gal(L P /K ) for any lace. This is done, for examle, in Lang s Algebraic Number Theory (Thm. 3 in Ch. XI, 4) and this is how local class field theory was discovered in the first lace (by Hasse). However, the roblem still remains to identify this ma with ρ (and thereby show indeendence of the choice of K, for examle). Perhas, after one checks comatibilities with change of fields one can do this but I haven t seen it anywhere in the literature. Alternatively, one can show that the global Artin ma ρ = ρ : A K Gal(L/K), defined as the roduct of local Artin mas, is trivial on K. Since it is clearly trivial on N L/K (A L ) and coincides with ρ on the unramified rimes we must have ρ = ρ and therefore ρ = ρ for all rimes. Theorem 3.1. Let K be a global field. Then one has: a) (Recirocity) For every abelian extension L/K the ma ρ is trivial on K. b) (Isomorhism) For every abelian extension L/K the ma ρ is surjective with kernel K N L/K (A L ). c) (Existence) For every oen subgrou U C K of finite index there is a unique abelian extension L/K with norm subgrou U. d) (Classification) There is an inclusion reversing bijection between abelian extensions L/K and oen subgrous U of C K of finite index.

5 46 M. FLACH Proof. The first statement is unfortunately somewhat lengthy to rove, even though we already have the recirocity theorem in its classical formulation. However, Theorem 1.6 only imlies that ρ is trivial on K,m. We shall give a roof in conjunction with the comutation of the cohomology of the idele class grou in the next section. For b) note that ρ is trivial on N L/K (A L ) by local class field theory and hence trivial on K N L/K (A L ) by a). By Pro. 3.2 this subgrou has index [L : K] andρ is surjective since it restricts to the classical Artin ma on the unramified rimes. For c) and d) it suffices to note that the grous U m, res. Ū m, form a basis of oen subgrous of A K, res. a basis of oen subgrous of finite index of C K and aly the classical existence theorem, Theorem 1.7. Remark If K is a global field of characteristic 0 we could relace oen subgrou of finite index by oen subgrou in statements c) and d) since any oen subgrou of C K has finite index. Remark Another advantage of the idelic aroach is that it easily generalizes to infinite abelian extensions. The inverse limit of the recirocity mas ψ K : lim U is injective with image C K /U lim Gal(L/K) = Gal(K ab /K) L {σ Gal(K ab /K n Z,σ F =Frob n } if char(k) =>0 and it is surjective with kernel the connected comonent C 0 K of the identity in C K if char(k) =0. One has an isomorhism where C 0 K = R (S 1 ) r2 V r1+r2 1 V = lim S 1 n = (R Ẑ)/Z = Q D is the solenoid, the Pontryagin dual of the discrete grou Q. Note that the connected comonent A,0 K of the identity in A K easily comutes to A,0 K = C R,>0 = (S 1 ) r2 R r1+r2. comlex real 3.2. Galois cohomology of the idele class grou. In the next section on class formations we shall give the roofs of Theorem 3.1 a) and b) without using results from section 1, in articular without any use of ray class L-functions. We shall not rerove c) and d) since the roofs in the idelic setting are really identical to the classical ones. In order to verify that the idele class grou satisfies the axioms of a class formation axioms we shall comute the Galois cohomology of the idele class grou in this subsection, more recisely just the H 0,H 1 and H 2. This comutation goes hand in hand with the determination of the Brauer grou of the global field K about which the classical aroach had nothing to say. It turns out that the Galois cohomology of the idele class grou contains a lot of information which one does not see in the classical formulation. For a start, H m or J m /P m N L/K JL m are not naturally Galois cohomology grous whereas C K /N L/K C L is:

6 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 47 Lemma 3.2. For any Galois extension L/K with grou G we have and hence H 0 (G, C L )=C K Ĥ 0 (G, C L )=C K /N L/K (C L ). Proof. The short exact sequence of G-modules (29) 1 L A L C L 1 induces an exact cohomology sequence 1 K A K H0 (G, C L ) 1 since H 1 (G, L ) = 0 by Hilbert 90. The following result is in some sense the analogue of the norm index theorem in the idelic aroach. On the one hand it imlies general results on Galois cohomology of the idele class grou which are needed in the roof of Theorem 3.1 a) and b). On the other hand it has a direct roof using urely algebraic techniques, i.e. avoiding the analytic theory of ray class L-functions. At this oint we can deduce it rather quickly from Pro. 3.2 and the results of section 1. Theorem 3.2. If L/K is cyclic with grou G = Gal(L/K) then { [L : K] = G i =0 Ĥi (G, C L ) = 1 i =1. Proof. By Lemma 3.2 and Pro. 3.2 we get Ĥ 0 (G, C L )=C K /N L/K (C L ) = J m /P m N L/K JL m = G and so it suffices to show that q(c L )= G. Consider the oen subgrou A L,S = O L P L P P / S P S of A L, where S is a finite G-stable set of laces of L containing the archimedean laces and those ramified in L/K and large enough so that Cl(O L,S ) = 0. Corollary 1.6 imlies (30) H i (G, P O L P )=H i (G, O L P )=0 for i =0, 1and / S and since cohomology commutes with roducts we further obtain form Corollary 1.6 q(a L,S )= q(g,l P )= G. S S Then with m = P S P one has and A L /L A L,S = J m L / im(l ) = Cl(O L,S )=0 C L = A L /L = A L,S /L A L,S = A L,S /O L,S.

7 48 M. FLACH and therefore q(c L )=q(a L,S )/q(o L,S )= S G q(o L,S ) = G by Lemma 1.7. For the rest of this section we will develo consequences of Theorem 3.2 without using any other results from class field theory. In other words, Part 1 will only enter in this section via Theorem 3.2. Later we will then give an indeendent roof of Theorem 3.2 making all results indeendent of Part 1. First, using Theorem 3.2 we can comute H 1 of the idele class grou. It satisfies an analogue of Hilbert s theorem 90. Proosition 3.3. For any Galois extension L/K with grou G we have H 1 (G, C L )=0. Proof. Assume first that G is a -grou. Then G has a cyclic quotient of order. If L /K denotes the corresonding Galois extension we have an inflation restriction sequence 0 H 1 (Gal(L /K),C L ) H 1 (G, C L ) H 1 (Gal(L/L ),C L ) obtained form the Hochschild-Serre sectral sequence for the grou extension 1 Gal(L/L ) Gal(L/K) Gal(L /K) 1. By induction on G we deduce H 1 (G, C L ) = 0 from Thm In general let G G be a -Sylow subgrou. Then the comosite ma H 1 (G, C L ) res H 1 (G,C L )=0 cor H 1 (G, C L ) is multilication with [G : G ], hence induces a bijection on the -rimary subgrou H 1 (G, C L )[ ]ofh 1 (G, C L ) (recall that this is a G -torsion grou). So H 1 (G, C L )[ ] = 0 and since this holds for all G we get H 1 (G, C L )=0. Our next aim is to comute H 2 (G, C L ) but it is not easy to do this without assing to the limit and considering the full grou H 2 (K, C K) = lim H 2 (Gal(L/K),C L ). L Since the system of all finite extensions L/K is filtered, the direct limit 1 K A K C K 1 of the exact sequence (29) is an exact sequence of discrete G K -modules, and its cohomology in degree 2 is given by the following theorem. Theorem 3.3. One has a commutative diagram of short exact sequences 0 H 2 (K, K ) H 2 (K, A K) H 2 (K, C K) 0 = = inv K 0 Br(K) Br(K ) inv Q/Z 0 where Q/Z nonarchimedean inv :Br(K ) 1 = 2Z/Z real 0 comlex.

8 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 49 We first need some local as well as semilocal rearations. Lemma 3.3. For any finite Galois extension L/K with grou G and i Z one has Ĥ i (G, A L ) = Ĥ i (G,L P ) and for i>0 H i (K, A K) = H i (K, K ). Proof. Since cohomology commutes with roducts and using Shairo s Lemma we get Ĥ i (G, A L,S ) = Ĥ i (G, O L P ) Ĥ i (G,L P ). / S S But if S contains all ramified rimes the decomosition grous G for / S are cyclic. Hence from (30) and the fact that cohomology of finite grous commutes with colimits we get (31) Ĥ i (G, A L ) = lim Ĥ i (G, A L,S ) = lim Ĥ i (G,L P )= Ĥ i (G,L P ). S S S Now by definition A K = A L = lim A L L L and Ĥ i (G, A K) = lim L = Ĥ i (G, A L ) = lim L lim Ĥ i (G,L P )= L Ĥ i (G,L P ) H i (K, K ) since L L P = K by Krasner s Lemma. The following Lemma summarizes the statements from local class field theory we will need. Lemma 3.4. For any finite Galois extension L P /K of local fields with grou G there is an isomorhism inv LP/K : H 2 (G,L P ) 1 = [L P : K ] Z/Z which is related to the local recirocity ma ρ = ρ LP/K : K G ab by the formula χ(ρ (α))=inv LP/K (α δχ) where δ : H 1 (G, Q/Z) H 2 (G, Z) is the connecting homomorhism in the long exact cohomology sequence induced by 0 Z Q Q/Z 0.

9 50 M. FLACH Proof. The existence of the invariant ma follow from the determination of the Brauer grou of the local field K, the key fact being that any central simle algebra has an unramified slitting field. So there is an isomorhism inv K : H 2 (K, K ) = H 2 (K ur /K, K ) v H 2 (K ur /K, Z) = Q/Z. For this isomorhism one roves directly that there is a commutative diagram with exact rows 0 H 2 (G,L P ) H2 (K, K ) H 2 (L P, K ) inv L P /K invk 1 0 [L P :K Z/Z Q/Z [L P:K ] ] Setting ( ) γ := inv 1 1 K H 2 (G,L P [L P : K ] ) one defines the recirocity isomorhism Ĥ 2 (G, Z) =H 1 (G, Z) =G ab invl P Q/Z. γ K /N LP/K L P = Ĥ0 (G,L P ). In other words, setting s α = ρ (α) one has by definition γ s α =ᾱ and hence ᾱ δχ = γ (s α δχ). Since δ commutes with cu roducts we get s α δχ = δ(s α χ) with s α χ Ĥ 1 (G, Q/Z) 1 = [L P : K ] Z/Z. For any finite grou G, s G ab and χ Hom(G, Q/Z) one checks that s χ = χ(s) (Serre, Local fields, Aendix to Ch. XI, Lemma 3) and also that δ(r/n) = r under the boundary ma Ĥ 1 (G, Q/Z) = ker ( n ) 1 Q/Z = n Z/Z δ Z/nZ = coker ( n Z ) = Ĥ0 (G, Z) where n = G =[L P : K ]. Finally then inv LP/K (ᾱ δχ) =inv LP/K (u r) =r/n. Remark In the next section we shall define a recirocity ma for any class formation of which the formula in Lemma 3.4 is a secial case. Proof. (of Theorem 3.3) We shall write H i (L/K, M) forh i (Gal(L/K),M). It is clear that the to sequence is exact at the left because Proosition 3.3 imlies H 1 (K, C K) = lim H 1 (L/K, C L )=0 L and by Lemma 3.4 the bottom sequence is then also exact at the left. It is also clear that the to sequence is exact in the middle and the bottom sequence is exact at the right but unfortunately it will not be easy to construct the ma inv K so that the right square commutes. In order to exloit Theorem 3.2 one first looks at the analogous situation for a finite and then cyclic extension L/K. For finite L/K the

10 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 51 vanishing of H 1 (L/K, C L ) gives the analogous commutative diagram (without the dotted arrow) (32) 0 H 2 (L/K, L ) H 2 (L/K, A L ) H 2 (L/K, C L ) 0 0 H 2 (L/K, L ) φ H 2 (L P /K,L P ) inv L/K 1 [L:K] Z/Z 0 with rows which are exact at the left and which we use to define the global invariant ma inv L/K (c) := inv LP/K (c ) as the sum of local invariant mas. Note that inv L/K is certainly not always surjective since there are global extensions for which the least common multile of all local degrees is a roer divisor of [L : K]. We first record the functoriality of the invariant ma. Lemma 3.5. Let L/K be Galois with grou G. a) For K K L there are commutative diagrams H 2 (L/K,L ) H 2 (L/K, A L ) res cor H 2 (L/K, L ) inf res cor H 2 (L/K, A L ) inf H 2 (K /K, K ) H 2 (K /K, A K ) inv L/K 1 [L:K ] Z/Z [K :K] incl [L:K] Z/Z inv L/K 1 incl inv K /K 1 [K :K] Z/Z 0 where the bottom ortion only alies if K /K is Galois. b) The columns are exact. c) Moreover, for each α A K and χ H1 (G, Q/Z) we have (33) χ(ρ L/K (α))) = inv L/K (α δχ) where ρ L/K : A K Gab is the global Artin ma defined as the roduct of local ones 0 0 ρ L/K ((α )) = ρ LP/K (α ). Proof. The commutativity involving the middle and right hand columns follows from the corresonding roerties of the local invariant ma. The commutativity involving the middle and left hand columns is just functoriality of res, cor and

11 52 M. FLACH inf. The exactness of the left hand column follows from the inflation-restriction sequence which extends to an exact six term sequence in view of the vanishing H 1 (K /K, H 1 (L/K,L )) = 0 by Hilbert 90. Exactness of the right hand column is clear and exactness of the middle column follows from exactness of the right hand column (for the various local field extensions) and the fact that the invariant ma is an isomorhism in the local case. Alternatively, it is a consequence of Hilbert 90 for the local fields, just like the exactness of the left hand column. To see (33) first note that the global Artin ma is well defined since for any rime such that α O K and is unramified in L/K we have ρ LP/K (α )=1. If χ is the restriction of χ to G then α δχ is the local comonent of α δχ and hence inv L/K (α δχ) = = inv LP/K (α δχ ) χ (ρ LP/K (α )) = χ(ρ L/K (α))). The next theorem verifies that the bottom row in (32) is a comlex. It is a recirocity theorem and just like Theorem 1.6 we will eventually reduce its roof to cyclotomic extensions of Q. For such extensions the roof will conclude by a direct verification of Theorem 3.1 a). Theorem 3.4. For finite L/K and any c H 2 (L/K, L ) one has inv L/K (c) =0. Proof. The key is to construct a diagram of fields E = L E L L K E Q where L /Q is Galois and E/Q is cyclic cyclotomic with certain local roerties. By Lemma 3.5 a) the three mas H 2 (L/K, L ) inf H 2 (L /K, L ) cor H 2 (L /Q,L ) inf H 2 (E /Q,E )

12 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 53 do not change inv L/K (c). Denoting by c H 2 (E /Q,E ) the image of c, Lemma 3.6 alied to S = { inv (c ) 0} and m a common denominator of the inv (c ), gives an extension E = L E so that res(c )=0 H 2 (E /E, E ). Lemma 3.5 b) then imlies that c = inf(c ) for some c H 2 (E/Q,E ). By Lemma 3.7 we have inv E/Q (c ) = 0 and another alication of Lemma 3.5 a) gives inv E /Q(c )=inv L/K (c) =0. Lemma 3.6. Given a number field L, finite set of laces S and an integer m then there exists a cyclic cyclotomic extension E /L so that m divides [E P : L ] for all S. Proof. Let E(q r ) Q(ζ q r) be the cyclic subextension of degree q r 1 for q odd (for q = 2 we leave it as an exercise that there is a totally comlex cyclic subextension E(2 r )ofq(ζ 2 r) of degree 2 r 2 ). For any rime (equal to q or not) the local degree [E(q r ) : Q ] tends to as r since this is true for the local degrees [Q(ζ q r) : Q ]and[q(ζ q r) : E(q r ) ] is bounded by q. If m = q n1 1 qnk k the extension E = E(q r1 1 ) E(qrk k ) will have local degree divisible by m for all S for r i >n i large enough. By ossibly enlarging the r i further we get the same conclusion for E /L where E = EL and L is any number field. Lemma 3.7. For any cyclotomic extension E/Q Theorem 3.1 a) holds, and if E/Q is cyclic we have inv E/Q (c) =0for each c H 2 (E/Q,E ). Proof. Let ρ = ρ : A Q Gal(Q(ζ m)/q) = (Z/mZ) = (Z/q nq Z) be the recirocity ma. Recall that the ma ρ has image in the decomosition grou for all laces. For we have ρ : Q Gal(Q (ζ m )/Q ) = (Z/ n Z) <> (Z/ n Z) q q (Z/q nq Z) and in terms of this roduct decomosition and the decomosition Q = Z Z the ma ρ is given by ρ (u k )=(u 1, k ). This is the content of Dwork s theorem (see Serre, Local fields). For = (and m>2) the ma ρ : R Gal(C/R) =< ±1 > (Z/mZ). is given by ρ (u) = sgn(u). It suffices to show that ρ( 1) = 1 and ρ(l) =1 for every rime number l. For a (Z/mZ) = q Z) denote by (a) (Z/qnq q its q-comonent. Then we have 1 m, l (l 1 ) m, l ρ (l) = q m,q l (l) q = l 1 =

13 54 M. FLACH and so ρ(l) = q m,q l (l) q m, l (l 1 ) = 1. Moreover 1 m ρ ( 1) = ( 1) m 1 = and so ρ( 1) = ( 1) m ( 1) = 1. If G = Gal(E/Q) is cyclic ick χ H 1 (G, Q/Z) so that δχ H 2 (G, Z) = Z/ G Z is a generator. Then δχ : Ĥ0 (G, M) Ĥ2 (G, M) is an isomorhism for both M = E and M = A E. Writing c = α δχ with α Q Lemma 3.5 c) and the first art of this Lemma give inv E/Q (c) =inv L/K (α δχ) =χ(ρ(α)) = 0. We now come back to diagram (32) and assume that L/K is cyclic. Then the following holds: Ste 1. The to row is exact since H 3 (G, L ) = H 1 (G, L )=0. Ste 2. There is a ma φ so that the diagram commutes since inv L/K (im(h 2 (G, L ))) = 0 by Theorem 3.4. Ste 3. The ma inv L/K is surjective. Indeed, if L/K is cyclic of rime degree then we know C L /N L/K C K =[L : K], hence there must be a rime of K with G = G since otherwise N L/K : A L A K would be surjective. The same conclusion holds if L/K is cyclic of rime ower degree by looking at the unique subextension of rime degree. So in this case inv L/K is surjective and surjectivity for general cyclic L/K follows from Lemma 3.5. Ste. 4 The ma φ is an isomorhism since it is surjective by Ste 3 and H 2 (L/K, C L ) = Ĥ0 (L/K, C L ) =[L : K], again by Theorem 3.2. We conclude that for cyclic L/K both rows in (32) are exact and φ exists and is an isomorhism. As remarked above, for general finite L/K the rows in (32) need not be exact at the right. Instead we ass to the direct limit over all cyclic (or even just cyclotomic) extensions L/K and note that H 2 (K, A K) = lim H 2 (L/K, A L )= H 2 (L/K, A L ) L/K cyclic L/K cyclic in view of the fact that any element c H 2 (K, K ) in the local Brauer grou has an unramified slitting field, hence any (necessarily finite) family (c ) H 2 (K, A K)

14 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 55 has a suitable global cyclic cyclotomic slitting field L/K. Passing to the limit over all finite L/K in Lemma 3.5 a) we obtain the inclusions 0 H 2 (K, K ) H 2 (K, A K) 0 H 2 (K, K ) H 2 (K, A K) 0 H 2 (K /K, K ) H 2 (K /K, A K ) 0 0 we also obtain (diagram chase) H 2 (K, K ) = lim H 2 (K /K, K )= H 2 (K /K, K ). K /K cyclic K /K cyclic This then imlies exactness of the bottom row in Theorem 3.3 by taking the direct limit over all cyclic L/K of the bottom row of (32) (the system of cyclic L/K is not filtered but at this oint we only need to show right exactness which is ok for any index category since colimits commute with colimits). This concludes the comutation of the Brauer grou of the global field K. One also obtains Q/Z as a subgrou (and hence direct summand) of H 2 (K, C K) but at this oint we still don t know equality of the two grous. For any finite extension L/K one has the commutative diagram with exact columns 0 Q/Z L H 2 (L, C K) [L:K]=res res ι 0 Q/Z K H 2 (K, C K) 0 1 [L:K] Z/Z ι L/K H 2 (L/K, C L ) where Q/Z K = 0 0 K /K cyclic H 2 (K /K, C K ) is maed to Q/Z L under the restriction ma in view of the commutative diagram G L Gal(LK /L) G K Gal(K /K) and the fact that LK /L is again cyclic. A simle diagram chase shows that 1 ι( [L:K] Z/Z) H2 (L/K, C L ) and hence the existence of ι L/K. To show that ι

15 56 M. FLACH is an isomorhism it suffices to show that ι L/K is an isomorhism for all L/K. This in turn will follow if we can rove that (34) H 2 (L/K, C L ) [L : K]. We know equality for cyclic L/K. For a tower K K L there is an exact inflation-restriction sequence 0 H 2 (K /K, C K ) H 2 (L/K, C L ) H 2 (L/K,C L ) in view of the vanishing of H 1 (L/K,C L ). Hence by an easy induction we get (34) for all solvable L/K. Finally recall that for any finite grou G with Sylow subgrous G,anyG-module M and any i 0 H i (G, M) res H i (G,M) is injective. Hence, if K denotes the fixed field of the Sylow -subgrou of Gal(L/K), we get H 2 (L/K, C L ) H i (L/K,C L ) [L : K ]=[L : K]. This finally comletes the roof of Theorem 3.3. We summarize our comutation of the cohomology of the idele class grou in the following theorem. Theorem 3.5. Assume the statement of Theorem 3.2 holds. Then for any Galois extension L/K of number fields one has C K i =0 H i (L/K, C L )= 0 i =1 and for C K = lim L C L one has G 1 [L:K] H i (K, C K) = lim L H i (L/K, C L )= Z/Z i =2 C K i =0 0 i =1 Q/Z i =2. 4. Class formations and duality Definition A class formation consists of a rofinite grou G and a discrete G-module C together with isomorhisms inv U/V : H 2 (U/V,C V ) 1 = [U : V ] Z/Z for each air V U G of oen subgrous such that the following hold a) H 1 (U/V,C V )=0 b) If W U and W V the diagram inf H 2 (U/V,C V ) H 2 (U/W,C W ) H 2 (V/W,C W ) inv inv res inv 1 [U:V ] Z/Z 1 [U:W ] Z/Z [U:V ] 1 [V :W ] Z/Z

16 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 57 is commutative. If (G, C) is a class formation then C G i =0 H i (G, C) = 0 i =1 1 1 G Z/Z := lim U [G:U] Z/Z i =2. So if for any integer n there is an oen subgrou U G of index divisible by n we have H 2 (G, C) = Q/Z. This is the case if there is a surjection G Ẑ like in examles a)- e) below. Here are some examles of class formations. a) (G, C) =(G K,C K) where K is a global field. b) (G, C) =(G K, K ) where K is a local field. c) (G, C) =(Ẑ, Z) d) (G, C) =(G K,C K ) where K/k is an extension of transcendence degree one over an algebraically closed field k of characteristic zero, i.e. a function field of a smooth roer curve X over k, and C K = lim Hom Z (Pic(X L, Q/Z) L/Kfinite where X L is the smooth roer curve over k with function field L. e) There is class field theory for higher local fields. A 2-local field is a comlete discretely valued field K with residue field a local field, for examle K = Q ((T )), F q ((T ))((S)) or } { Q {{T }} = a i T i a i Q, i= inf v (a i ) >, i Z lim v (a i )= i which has residue field F ((T )). Then one has to extend the notion of class formation to allow a comlex of G K -modules, and it turns out (G, C )= (G K, K L K Z ) is a class formation. f) (G S,C S ) where S is a set of laces of the global field K containing all archimedean ones, G S = Gal(K S /K) the Galois grou of the maximal extension of K unramified outside S and C S (L) =A L /L U S where U S is the comact subgrou U S = O L P P / S P S{1}. Note that by (30) we have Ĥi (G, U S )=0fori =0, 1 (and L/K unramified outside S) and in fact for any i Z since the decomosition grous at P / S are cyclic. By the long exact cohomology sequence induced by 0 ŪS C L C S (L) 0 we get isomorhisms Ĥi (G, C L ) = Ĥi (G, C S (L)) and also H 0 (G, C S (L)) = C S (K).

17 58 M. FLACH We shall now develo a sequence of duality theorems for finite grous which will culminate in the Tate-Nakayama duality for class formations. Throughout we set A = Hom(A, Q/Z) for any abelian grou A. This is an exact contravariant functor from abelian grous to abelian grous. If A is finite then A coincides with the Pontryagin dual of A. Theorem 4.1. Let Γ be a finite grou and A a Γ-module. Then for all i Z the airing Ĥ i (Γ,A ) Ĥ i 1 (Γ,A) Ĥ 1 (Γ, Q/Z) = 1 Z/Z Q/Z Γ induces an isomorhism Ĥ i (Γ,A ) = Ĥ i 1 (Γ,A). Proof. We first show the statement for i = 0. A homomorhism f : A Q/Z is a Γ-homomorhism if and only if f(i Γ A) = 0, hence we obtain an isomorhism H 0 (Γ,A ) = H 0 (Γ,A). If f N Γ A, i.e. f = σ Γ σh, then for a A with N Γa = 0 we have f(a) = (σh)(a) = h(σ 1 a)=h(n Γ a)=0 so we obtain a ma (A ) Γ /N Γ A ( NΓ A/I G A). If g : NΓ A Q/Z is a homomorhism that vanishes on I Γ A then g can be extended to a homomorhism g : A Q/Z which is a Γ-homomorhism since g(i Γ A)=0. So our ma is surjective. If f (A ) Γ vanishes on NΓ A there exists g (N Γ A) with f(a) =g(n Γ a) since N Γ : A/ NΓ A N Γ A is an isomorhism. Again g can be extended to a homomorhism g : A Q/Z and then f = N Γ g since (N Γ g)(a) = σ Γ g(σ 1 a)=g(n Γ a)=f(a). For arbitrary i one can use dimension shifting which gives a commutative diagram Ĥ i (Γ,A ) Ĥ i 1 (Γ,A) δ i Ĥ 0 (Γ, Hom(A, Q/Z) i ) δ i Ĥ 1 (Γ,A i ) H 1 (Γ, Q/Z) ( 1) i(i+1)/2 H 1 (Γ, Q/Z) Since Hom(A, Q/Z) i = Hom(A i, Q/Z) the desired result follows. We recall that for any Γ-module A one defines A 1 by the exact sequence 0 A A Z Z[Γ] A 1 0 and since the middle term is cohomologically trivial we get an isomorhism δ : Ĥi (Γ,A 1 ) = Ĥi+1 (Γ,A). For >0one defines A =(A 1 ) 1 and for <0one uses induction and the module A 1 defined by the exact sequence 0 A 1 Hom Z (Z[Γ],A) A 0..

18 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 59 Remark For an arbitrary (discrete) grou Γ and (discrete) Z[Γ]-module A one always has a duality between cohomology and homology H i (Γ,A ) = H i (Γ,A) for any i 0. This follows from the Hom- -adjunction Hom Z[Γ] (P, Hom Z (A, Q/Z)) = Hom Z (P Z[Γ] A, Q/Z) which is a secial case of the adjunction Hom S (P, Hom R (A, B)) = Hom R (P S A, B) for a right S-module P, S-R-bimodule A and right R-module B. Theorem 4.1 extends this to all i Z in case Γ is finite. Proosition 4.1. Let Γ be a finite grou and A a Z-free Γ-module. Then for all i Z the airing Ĥ i (Γ, Hom(A, Z)) Ĥ i (Γ,A) H 0 (Γ, Z) = Z/ Γ Z induces an isomorhism Ĥ i (Γ, Hom(A, Z)) = Ĥ i (Γ,A). Proof. Since A is Z-free we have an exact sequence 0 Hom(A, Z) Hom(A, Q) Hom(A, Q/Z) 0 and Hom(A, Q) is cohomologically trivial. So we get a commutative diagram Ĥ i 1 (Γ,A ) δ Ĥ i (Γ,A) Ĥ i (Γ, Hom(A, Z)) Ĥ i (Γ,A) where the vertical arrows are isomorhisms. id H 1 (Γ, Q/Z) δ H 0 (Γ, Z) Remark One can combine Theorem 4.1 and Pro. 4.1 into an isomorhism in the derived category of abelian grous RΓ(Γ,RHom(A, ˆ Z)) = R Hom( RΓ(Γ,A), ˆ Q/Z) = RΓ(Γ,A) ˆ for any abelian grou A. If A is free abelian then R Hom(A, Z) = Hom(A, Z) and we recover Pro If A is arbitrary the exact triangle R Hom(A, Z) R Hom(A, Q) R Hom(A, Q/Z) induces an isomorhism RΓ(Γ,RHom(A, ˆ Q/Z)) = RΓ(Γ,RHom(A, ˆ Z))[1] which imlies Ĥ i (Γ,A ) = Ĥi+1 (Γ,RHom(A, Z)) and we recover Theorem 4.1. Theorem 4.2. (Nakayama-Tate) Let Γ=U/V be a layer in a class formation and A a Z[Γ]-module, finite free over Z.

19 60 M. FLACH a) If γ H 2 (Γ,C V ) is the canonical generator, i.e. the unique element with inv U/V (γ) = 1 [U:V ], then is an isomorhism for all i Z. b) For all i Z the cu roduct Ĥ i (Γ,A) γ Ĥi+2 (Γ,A Z C V ) Ĥ i (Γ, Hom(A, C V )) Ĥ2 i (Γ,A) H 2 (Γ,C V ) = induces an isomorhism of finite abelian grous Ĥ i (Γ, Hom(A, C V )) = Ĥ2 i (Γ,A). 1 [U : V ] Z/Z Proof. Let 0 C V C(γ) Z[G] Z 0 be a Yoneda 2-extension corresonding to The comosite ma γ H 2 (Γ,C V ) = Ext 2 Z[G](Z,C V ). Z/ Γ Z = Ĥ0 (Γ, Z) δ1 Ĥ1 (Γ,I Γ ) δ2 Ĥ2 (Γ,C V ) induced by the short exact sequences (35) 0 I Γ Z[Γ] Z 0 and (36) 0 C V C(γ) I Γ 0 also coincides with the cu roduct γ (at least u to sign) by general homological algebra. Since C was a class formation γ is an isomorhism. Since δ 1 is always an isomorhism this imlies that δ 2 is an isomorhism. The long exact sequence H 1 (Γ,C V ) H 1 (Γ,C(γ)) H 1 (Γ,I Γ ) δ2 H 2 (Γ,C V ) H 2 (Γ,C(γ)) H 2 (Γ,I Γ ) induced by (36) together with H 1 (Γ,C V )=0andH 2 (Γ,I Γ ) = H 1 (Γ, Z) = 0 then shows that H 1 (Γ,C(γ)) = H 2 (Γ,C(γ)) = 0. The same holds for all subgrous, so C(γ) is cohomologically trivial. Statement a) then follows from tensoring (35) and (36) with A, Lemma 4.1 below, and the descrition of the cu roduct as the comosite ma Ĥ i (Γ,A) δ1 Ĥi+1 (Γ,A Z I Γ ) δ2 Ĥi+2 (Γ,A Z C V ). Similarly, since A is Z-free alying Hom(A, ) to the exact sequences (35) and (36) yields exact sequences whose middle terms are again cohomologically trivial

20 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 61 by Lemma 4.1 below. This then imlies that the mas δ in the diagram Ĥ i 2 (Γ, Hom(A, Z)) Ĥ i 1 (Γ, Hom(A, I Γ )) δ Ĥ 2 i (Γ,A) id Ĥ 2 i (Γ,A) Ĥ 0 (Γ, Z) δ H 1 (Γ,I Γ ) δ Ĥ i (Γ, Hom(A, C V )) Ĥ 2 i (Γ,A) H 2 (Γ,C V ) are isomorhisms and the asserted duality follows from Pro Lemma 4.1. If X is a cohomologically trivial module over a finite grou Γ and A is any finitely generated Z-free Γ-module then the Z[Γ]-modules Hom Z (A, X) and A Z X are cohomologically trivial. Proof. A Z[Γ]-module is cohomologically trivial if and only if it has finite rojective dimension if and only if it has rojective dimension one. So take a resolution 0 P 1 P 2 X 0 by Z[Γ]-rojective modules. We get induced exact sequences 0 Hom Z (A, P 1 ) Hom Z (A, P 0 ) Hom Z (A, X) 0 and 0 A Z P 1 A Z P 2 A Z X 0. Since A is finitely generated Hom Z (A, P ) = Hom Z (A, Z) Z P with diagonal action which is well known to be isomorhic to Hom Z (A, Z) Z P with trivial action on the first factor if P is free, hence to P dimz A. Similarly, A Z P = P dimz A if P is free. So if P is free then Hom Z (A, P )anda Z P are again free, hence c.t. By assing to direct summands we deduce that Hom Z (A, P )anda Z P are c.t., hence so are Hom Z (A, X) anda Z X. Corollary 4.1. Given any layer Γ=U/V in a class formation there is an isomorhism ρ = ρ U/V : C U /N Γ C V Γ ab given by χ(ρ(a)) = inv U/V (a δχ) for χ H 1 (Γ, Q/Z) δ H 2 (Γ, Z) and a C V. The ma ρ is called the recirocity ma, or norm residue homomorhism. Proof. Taking A = Z and i = 0 in Theorem 4.2 gives an isomorhism ρ : C U /N Γ C V = Ĥ0 (Γ,C V ) = Ĥ2 (Γ, Z) characterized by the roerty inv U/V (a δχ) = ρ(a)(δχ) and the isomorhism Ĥ 2 (Γ, Z) = H 1 (Γ, Q/Z) = Γ ab sends the character ψ to the grou element σ ψ Γ ab with ψ(δχ) =χ(σ ψ ). We then define ρ(a) :=σ ρ(a) and obtain the above formula for χ(ρ(a)). id δ

21 62 M. FLACH Recall from Lemma 3.4 that the recirocity ma for local fields was defined via this corollary, and this is the only way I know to define this ma in the ramified case. For global fields one has the alternative definition using only the Frobenius automorhisms for the unramified laces, i.e. the classical Artin ma. The following corollary finally roves Theorem 3.1 a) and b). Corollary 4.2. Let K be a global field, L/K a finite abelian extension and ρ L/K = ρ : A K Gal(L/K) the recirocity ma. If α K then ρ L/K (α) =1. Moreover, ρ L/K isomorhism A K /N L/KA L K = C K /N L/K C L = Gal(L/K). induces an Proof. By Lemma 3.5 c) for each α A K and χ H1 (G, Q/Z) wehave χ(ρ L/K (α))) = inv L/K (α δχ) which means that ρ L/K coincides with the ma ρ of Corollary 4.1. Corollary 4.3. For a Galois extension L/K with subfield K K L there are commutative diagrams cor C K C K res ρ 1 ρ 1 Gal(L/K) ab cor res Gal(L/K ) ab. Proof. Since ρ 1 is given by γ and res(γ) =γ it suffices to show res(x y) = res(x) res(y) and the rojection formula cor(x res(y)) = cor(x) y. By dimension shifting this reduces to the degree 0 case where we have cor(x) y = s s(x) y = s s(x y) = cor(x y). Note that in negative degrees, in articular for i = 2, we have homology grous, so the natural ma is the corestriction ma whereas the restriction is the oosite or Umkehr ma. For i = 2 it is called the transfer Direct roof of Theorem 3.2. In this section we give a roof of Theorem 3.2 that does not use results from Part 1 excet for some Lemmas which had a self-contained roof. Other Lemmas from Part 1 we will have to rerove but we will not use any results of section 1.4, the analytic theory of ray class L-functions. In the above short derivation of Theorem 3.2 we already showed directly, only using Corollary 1.6 and Lemma 1.7, that (37) q(c L ):= Ĥ0 (G, C L ) = G =: n Ĥ1 (G, C L ) for a cyclic extension L/K with grou G. Hence it suffices to show that Ĥ0 (G, C L ) divides n, or that Ĥ1 (G, C L ) = 0 in order to rove Theorem 3.2. By induction on

22 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 63 the number of rime factors of n it furthermore suffices to show this for rime degree n in view of the long exact inflation restriction sequence 0 H 1 (Gal(L /K),C L ) H 1 (G, C L ) H 1 (Gal(L/L ),C L ) and the fact that H 1 = Ĥ1. We can further assume that K contains a rimitive n-th root of unity by the following argument. The field K := K(ζ n ) has degree [K : K] dividing n 1, hence rime to n, the field L := LK is cyclic of degree n over K and G := Gal(L /K ) is isomorhic to G by restriction. We have mas of G -modules N L /L C L C L CL whose comosite is multilication by [L : L] =[K : K]. Hence the induced ma H 1 (G,C L ) H 1 (G,C L ) N K /K H 1 (G,C L ) is multilication by [K : K] but is also the zero ma if we know that H 1 (G,C L )= 0. Since H 1 (G,C L )isann-torsion grou this imlies H 1 (G,C L )=H 1 (G, C L )= 0. Since ζ n K, our field L = K( n a) is a Kummer extension. We lace ourselves in the situation of Proosition 1.6, making sure that L S,n contains L. SoletS be a finite set of laces of K such that S n S { 1,..., k } S where 1,..., k =Cl(O K ). v (a) 0 S and define ( ) n L S,n := K O K,S. Clearly, L L S,n since a O L,S. By the roof of Pro. 1.6 we also know that [L S,n : K] =n s where s = S. What we don t know at this oint is whether L or L S,n are class fields. In fact we don t even know surjectivity of the Artin ma since this was deduced from the analytic theory. So we rerove here Corollary 1.3. Lemma 4.2. For any abelian extension L/K and modulus m divisible by all ramified rimes the Artin ma is surjective. Proof. We again look at the fixed field L of the image of the Artin ma and conclude that all m slit comletely in L /K. Let E L be a subextension so that E/K is cyclic. Then again all m slit comletely in E/K and in articular = N EP/K (E P ).SoifS = { m } we have K K 1 N E/K A E / S S and by the weak aroximation theorem K surjects onto K /N EP/K (E P ), i.e. S / S 1 S K K N E/K A E.

23 64 M. FLACH So we have Ĥ 0 (E/K,C E )=A K /(K N E/K A E )= K /(K N E/K A E )=1 which together with (37) for the cyclic extension E/K imlies E = K. Hence L = K, i.e. the Artin ma is surjective. Using similar ideas we rerove Lemma 1.13, actually a slightly strengthened version of it. Lemma 4.3. Let S S be a subset also satisfying the above four conditions and ut O K,S,n := O K,S (K ) n. S Then one has O K,S,n =(O K,S )n and therefore by (23) O K,S /O K,S,n = O K,S /(O K,S )n = n s. Proof. The inclusion O K,S,n (O K,S )n is clear. Now take α O K,S,n and consider the cyclic extension E = K( n α). Then all S are slit comletely in E/K and all / S are unramified in E/K. So A K,S := O K K N E/K A E. / S S By the fact that S contains generators of the ideal class grou we have A K = K A K,S. Therefore Ĥ 0 (E/K,C E )=A K /(K N E/K A E )=K A K,S /(K N E/K A E )=1 which together with (37) for the cyclic extension E/K imlies E = K. This means α O K,S (K ) n =(O K,S )n. In the situation of Lemma 4.3 we now assume that the rimes { 1,..., t } = S \ S are chosen so that {Frob 1,...,Frob t } is a Z/n-basis of Gal(L S,n/L) Gal(L S,n/K). This we can do by Lemma 4.2. In articular, the rimes i slit comletely in L/K and we have t = s 1ands = s + t =2s 1. We therefore have U := (K ) n K O K N L/K (A L ) S S\S / S and Ĥ 0 (G, C L )=A K /(K N L/K (A L )) = K A K,S /(K N L/K (A L )) has order dividing [K A K,S : [A K K,S U)] = : U)] [K A K,S : K U] = S [K :(K ) n ] [O K,S : O K,S,n ] = n2s n s = n2s n 2s 1 = n where we have also used Lemma 1.14 which had a self-contained roof. This finishes our second roof of Theorem 3.2.

24 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/ More examles and comlements 5.1. The quadratic recirocity law for number fields. Recall the quadratic recirocity law first roven by Gauss. For an integer a and rime number 2a define the Legendre symbol ( ) { a 1 if a is a square modulo = 1 otherwise. More generally, for an integer b = 1 r relatively rime to 2a one defines the Jacobi symbol ( a ) ( ) ( ) a a =. b 1 r Clearly, ( ) a b only deends on a modulo b. Theorem 5.1. (Gauss) For ositive, odd, relatively rime integers a, b one has ( a ) ( ) b =( 1) a 1 b b a Moreover, for odd b there are the sulementary statements ( ) ( ) 1 =( 1) b ; =( 1) b b b We shall rerove the quadratic recirocity law from the Artin recirocity law and at the same time generalize it to number fields. Fix a number field K. Forα O K and a rime ideal 2α one can define an obvious analogue of the Legendre symbol ( ) { α 1 if α is a square modulo = 1 otherwise and an obvious analogue of the Jacobi symbol for β O K relatively rime to 2α ( ) ( ) ( ) α α α = β where (β) = 1 r is the rime factorization of β. One clearly has ( ) { α 1 if slits in K( α)/k = 1 if is inert in K( α)/k 1 with the understanding that any rime slits in K( α)/k if α is a square. This condition has a straightforward reformulation in terms of the Frobenius automorhism Frob Gal(K( α)/k) {±1} which in turn is the image of a local uniformizer π under the local recirocity ma K So for 2α we have ρ Gal(K( α)p /K ) Gal(K( α)/k) {±1}. ( ) α = ρ (π ). r

25 66 M. FLACH In order to rove a recirocity law one needs to look at ρ for all laces of K. Note that the extension K ( α)/k can be ramified at laces 2α (or can be trivial if α is locally a square). If α but 2 then K ( α)/k unramified v (α) 0 mod 2 and if v (α) 1 mod 2 this extension is tamely ramified. By local class field theory the ma ρ then induces an isomorhism ρ : O K /(O K ) 2 = (OK /) /((O K /) ) 2 = Gal(K( α)p /K ) = {±1} and so we have for α O K ( α ρ (α )= ) v(α) for the entirely different reason that the Legendre symbol is the only nontrivial quadratic character of (O K /). If is real then ρ is also easy to analyze. We have K ( sgn σ(α) 1 α) K σ (α) < 0 ( 1) 2 = 1 and in that case we have ρ (α )=( 1) sgn σ(α ) 1 2. So altogether ρ (α )=( 1) sgn σ(α ) 1 2 sgn σ(α) 1 2. The hardest rimes to analyze are those dividing 2 since there are several ramified quadratic extensions of K. Let s do this when K = Q2.Wehave Q 2 /(Q 2 )2 = [2] Z/2Z [ 1] Z/2Z [5] Z/2Z and this grou both arametrizes the quadratic extensions by Kummer theory and is the source of the Artin ma for all quadratic extension. So we can define a bilinear symbol on this Z/2Z-vector sace, the Hilbert symbol ( α ),α := ρ (α ) {±1}. ( ) Then α,α = 1 if and only if α is a norm of the quadratic extension Q 2 ( α)/q 2 if and only if x 2 αy 2 = α has a solution (x, y) Q 2 2 which (for nonsquare α) is true if and only if x 2 αy 2 α z 2 =0 has a nonzero solution (x, y, z) Z 3 2. But this condition is symmetric in α, α which means that our bilinear form is symmetric. It is given by the following table Q 2 ( 2) Q 2 ( 1) Q 2 ( 5) which is justified by the following comutations Q 2 ( 2) = = 1 1 Q 2 ( 1) 1 a 2 + b 2 1 mod =5 Q 2 ( 5) =5.

26 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 67 The remaining 1 simly results from the fact that Q 2 ( 5) is the unramified quadratic extension of Q 2, so 2 is sent to the Frobenius, i.e. to 1. Of course it also follows from a comutation modulo 8. One verifies directly that ρ Q2( 2)/Q 2 (α )=( 1) α for α Z 2 and ρ Q2( α)/q 2 (α )=( 1) α 1 α for α, α Z 2 by checking that both sides are multilicative and agree on the generators listed in the above table. These formulas are just a neat way to condense the table, for otherwise one would have to make many case distinctions. For an extension K /Q 2 of degree d the sace K /(K ) 2 has dimension d +2 over Z/2Z and the Hilbert symbol also gives a symmetric bilinear form which is fairly straightforward to comute in any given examle. The quadratic recirocity law for K would deend on this table as is clear from the following theorem. Theorem 5.2. Let K be a number field. For α, β O K relatively rime to each other and relatively rime to 2 we have ( )( ) α β = ρ (β) sgn σ(α) 1 sgn σ(β) 1 ( 1) 2 2 β α 2 real where ρ is the recirocity ma for the quadratic extension K ( α)/k.if2slits comletely in K/Q this law simlifies to ( )( ) α β (38) = ( 1) σ(α) 1 σ(β) 1 sgn σ(α) 1 sgn σ(β) ( 1) 2 2 β α 2 real and we have the following two sulementary laws. If α O K and β is rime to 2 then ( ) α (39) = ( 1) σ(α) 1 σ(β) 1 sgn σ(α) 1 sgn σ(β) ( 1) 2 2. β 2 real If α is 2-rimary and β is rime to 2 then ( ) α = σ(β) 2 1 v(α) ( 1) 8 ( 1) 2 v(α) σ(α) 1 σ(β) 1 (40) 2 2 β 2 sgn σ(α) 1 sgn σ(β) 1 ( 1) 2 2. real Proof. We have and and ( ) α = ρ (π ) v(β) = β 2α real ( ) β = α α 2α ρ (β) ( ) v(α) β = ρ (β) α sgn σ(α) 1 sgn σ(β) 1 ( 1) 2 2 = ρ (β). real

27 68 M. FLACH Theorem 5.2 now follows from Theorem 3.1 a) (41) ρ(β) = ρ (β) =1. ( ) The sulementary law (39) follows from (38) since β α =1if(α) =O K is the unit ideal. The sulementary law (40) is again a direct consequence of (41) but there are no odd rimes dividing α to consider. At each 2 we write σ (α) =2 v(α) u with u Z 2 and note that ρ (β) is multilicative in σ (α). Remark Note that (38), (39) and (40) include a generalization of the classical recirocity and sulementary laws to negative b. For examle if K = Q and β = b <0 we get ( ) 1 b =( 1) b 1 2 ( 1) 1 1 =( 1) b+1 2 =( 1) b 1 2 = ( ) 1 b which is in accordance with the fact that ( ) a b only deends on the ideal generated by b. Examle In K = Q( 7) decide if 5+2ω is a square modulo 73 where ω = Note we have N K/Q (a + bω) =a 2 + ab +2b 2 and in articular N(5 + 2ω) =43. The rime 73 3 mod 7 is inert in K/Q since 3 is a nonsquare modulo 7. SinceK has no real laces and 73 1 mod 4 we have ( ) ( ) ( ) ( )( )( ) 5+2ω = = = ω 5+2ω 5+2ω 5+2ω 5+2ω Since 5 1 mod 4 we have with =(ω) ( ) ( ) ( ) 5 5+2ω ω2 ω = = =( 1) σ( ω) ( 1) 2 1 σ (ω) ( 1) 2 2 = ω 5 5 Since 3 1 mod 4 and (5 + 2ω 1)/2 =2+ω is even at =(ω) and odd at we get ( ) ( ) 3 2+2ω = 5+2ω 3 = [( 1) 32 1 = [1 1] = 1 8 ( 1) (1+σ(ω)) 1 2 ] [ ( 1) ( 1) (σ (ω)) 3 8 since (1+σ(ω)) 1 2 =1/σ ( ω) is odd and so is (σ (ω)+1)/4 1 = (σ (ω)) since =8 0 mod 8 but not mod 16. Finally (5 + 2ω) 2 1 = ( ω +4ω 2 )/8 =(24+20ω +4(ω 2))/8 =2+3ω 8 is even at =(ω) and odd at =( ω) and hence ( ) 2 =( 1) (5+2σ(ω)) ( 1) (5+2σ (ω))2 1 8 =1 ( 1) = ω ] 3 1 2

28 COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 69 So we find ( ) 5+2ω =( 1) ( 1) ( 1) = Actually ( ) one could have saved quite a bit of work in this examle by noting that a 5+2ω = ( a 43) for a Z rime to 43. We just use this remark to double check our results. ( ) ( ) 5 3 = = ( ) ( ) 3 1 = = ( ) 2 43 =( 1) = 1 Remark Note that also for K = Q( ( 7) we can verify that ) 1 β only deends on the ideal generated by β but this time the comensating sign changes occur at the two rimes dividing 2, not at 2 and as was the case for K = Q. One has ( 1 β ) =( 1) σ(β) 1 σ (β) 1 2 ( 1) 2 =( 1) σ(β) 1 σ (β) 1 2 ( 1) 2 = ( ) 1. β We continue with some remarks on more general recirocity laws. If a local field K contains a rimitive n-th root of unity, the combination of Kummer theory and local class field theory leads to the definition of the local Hilbert symbol. For α, α K let ρ : K Gal(K ( n α)/k ) μ n be the recirocity ma for the extension K ( n α)/k followed by the canonical homomorhism (42) σ ( n α) σ 1 of Kummer theory, and define the Hilbert symbol ( α ),α := ρ (α ) μ n. This symbol is a bilinear ma K /(K ) n K /(K ) n μ n and one can show that it actually coincides with the cu roduct H 1 (K,μ n ) H 1 (K,μ n ) H 2 (K,μ 2 n ) = H 2 inv μ n (K,μ n ) μ n μn. Aart from bilinearity it has the following roerties ( ) a) α,α = 1 if and only if α is a norm from K ( n α)/k. ) b) c) d) ( α,α ( α,1 α ( α,α ( ) α,α 1. = ) ( ) = 1 and α, α =1. ) = 1 for all α K if and only if α (K ) n.

29 70 M. FLACH Theorem 5.3. Let K be a number field containing a rimitive n-th root of unity. Then for α, β K ( ) α, β =1. Proof. This is just a rewriting of the Artin recirocity law ρ(β) = ρ (β) =1 where ρ : A K Gal(K( n α)/k) is the global recirocity ma. ForanumberfieldK containing a rimitive n-th root of unity one can define the n-th ower residue symbol. Given α O K and nα let ( ) α μ n n ( ) be the unique root of unity congruent to α N 1 α n modulo. Then =1if n and only if there is β O K with β n α mod. From (42) and the fact that ρ (π )=Frob where π is a uniformizer of K on deduces that ( ) ( ) α π,α =. n This is the starting oint for deducing a recirocity law for the n-th ower residue symbol from Theorem 5.3.