Central Simple Algebras and the Brauer Group

Home , Brauer group, Central simple algebra, Division algebra, Separable extension, Simple ring

Central Simple Algebras and the Brauer group

XVIII Latin American Algebra Colloquium

Eduardo Tengan (ICMC-USP) Copyright c 2009 E. Tengan

Permission is granted to make and distribute verbatim copies of this document provided the copyright notice is preserved on all copies.

The author was supported by FAPESP grant 2008/57214-4. Chapter 1

CentralSimpleAlgebrasandthe Brauergroup

1 Some conventions

Let A be a ring. We denote by Mn(A) the ring of n n matrices with entries in A, and by GLn(A) its group of units. We also write Z(A) for the centre of A×, and Aop for the opposite ring, which is the ring df with the same underlying set and addition as A, but with the opposite multiplication: a Aop b = b A a op ≈ op × × for a,b A . For instance, we have an isomorphism Mn(R) Mn(R) for any commutative ring R, given by∈M M T where M T denotes the transpose of M. → 7→ A ring A is simple if it has no non-trivial two-sided ideal (that is, an ideal different from (0) or A). A left A-module M is simple or irreducible if has no non-trivial left submodules. For instance, for any n field K, Mn(K) is a simple ring and K (with the usual action) is an irreducible left Mn(K)-module. If K is a field, a K-algebra D is called a division algebra over K if D is a skew field (that is, a “field” with a possibly non-commutative multiplication) which is finite dimensional over K and such that Z(D)= K. Let K L be an extension of fields and let A be an arbitrary K-algebra. We shall write A for the ⊂ L L-algebra A K L obtained by base change K L. We shall often refer to AL as the restriction of A to L (in geometric⊗ language, base change with−⊗ respect to Spec L Spec K corresponds to restriction in the flat topology). →

2 Cyclic Algebras Let L K be a cyclic extension, that is, a Galois extension of ﬁelds with cyclic Galois group G = Gal(L/K⊃ ). Let n = G = [L : K]. An element χ Hom(G, Z/n) will be called a character. Notice | | ∈ that to give a surjective character is the same as to give an isomorphism χ: G ≈ Z/n, i.e., to choose a speciﬁc generator σ of G, characterised by χ(σ)= 1.¯ → ≈ Now let a K × and let χ: G Z/n be a surjective character. Denote by σ the generator of G chosen by χ. We∈ now manufacture→ a K-algebra (χ,a) as follows. As an additive group, (χ,a) is an n-dimensional vector space over L with basis 1,e,e2,...,en−1:

(χ,a) =df Lei 1≤Mi

Multiplication is given by the relations

e λ = σ(λ) e for λ L · · ∈ en = a

Some boring computations show that (χ,a) is, in fact, an associative K-algebra, which is called the cyclic algebra associated to the character χ and the element a K. Observe that since dimL(χ,a)= 2 ∈ [L : K]= n we have that dimK (χ,a)= n is always a square. Example 2.1 (Real cyclic algebras) Consider the cyclic extension C R of degree n = 2 with G = Gal(C/R) = id, σ (here σ(z) = z is the conjugation automorphism)⊃ and let χ: G Z/2 be the character deﬁned by{ χ(σ})= 1.¯ Then, for a R×, → ∈ (χ,a)= z + w e z, w C { · | ∈ } 2 Central Simple Algebras and the Brauer group

where e2 = a and e z = z e for z C. For instance, we have that · · ∈ (z + we) (z we)= zz zwe + wez wewe · − − − = zz zwe + wze wwe2 ( ) − − ∗ = z 2 w 2a R | | − | | ∈

If a> 0 is positive then (χ,a) ∼= M2(R). In fact, we have that the standard embedding (φ: C ֒ M (R → 2 α β α + βi α, β R 7→ β α ∈ − √a 0 can be extended to a map of R-algebras φ: (χ,a) M (R) by setting φ(e)= , that is, → 2 0 √a − α + γ√a β δ√a φ(α + βi + γe + δie)= − α,β,γ,δ R β δ√a α γ√a ∈ − − − and since this map is injective and both (χ,a) and M2(R) have dimension 4 over R, it must be an isomorphism. Now suppose that a < 0. Then (χ,a) is a division algebra since for every z + we = 0 the real number z 2 w 2a> 0 is nonzero and hence 6 | | − | | z w (z + we)−1 = e z 2 w 2a − z 2 w 2a · | | − | | | | − | | in (χ,a) by the computation ( ). We have that (χ,a) is actually isomorphic to the usual real quaternion algebra H, which is the 4-dimensional∗ real algebra

H =df R + Ri + Rj + Rk

with basis 1, i, j, k satisfying the relations

i2 = j2 = k2 = 1 − ij = k, jk = i, ki = j ij = ji, jk = kj, ik = ki − − − =(In fact, an isomorphism φ: (χ,a) ≈ H is given by extending the inclusion φ: C ֒ H given by φ(α + βi → → α + βi, α, β R, by setting φ(e)= a j, as can be easily checked. ∈ | | · p The above example shows that the cyclic algebra construction encompasses both regular matrix algebras as well as quaternion algebras. Cyclic algebras can be thought of “twisted forms” of matrix algebras: by a suitable base extension, they can be easily “trivialised.” Theorem 2.2 (Splitting cyclic algebras) Let L K be a cyclic extension of order n. Let a K × ⊃ ∈ and let χ: G ≈ Z/n be a surjective character. Then → (χ,a) L = M (L) ⊗K ∼ n

Proof Let σ be the generator of G given by χ(σ)= 1.¯ Deﬁne a morphism of L-algebras φ: (χ,a) L ⊗K → Mn(L) by setting

λ 0 0 0 0 0 σ(λ) 0 · · · 0 0  2 · · ·  φ(λ 1) = 0 0 σ (λ) 0 0 for λ L ⊗  ·. · ·  ∈  .   .   00 0 0 σn−1(λ)   · · ·  Central Simple Algebras 3

and φ(e 1) to be the transpose of the “companion matrix” of xn a: ⊗ − 0 1 0 0 0 · · ·  0 0 1 0 0  ·. · · φ(e 1) =  .  ⊗    0 0 0 1 0   · · ·   0 0 0 0 1   a 0 0 · · · 0 0   · · ·  It is easy to check that φ is a well-deﬁne morphism of L-algebras. Since dimL(χ,a) K L = dimK (χ,a)= 2 ⊗ n = dimL Mn(L), in order to show that φ is an isomorphism it is enough to show that it is surjective. But by the lemma below one has L 0 0 · · ·  0 L 0  φ(L L)= . · · · ⊗K .  .   0 0 L   · · ·  and therefore 0 0 L 0 0 0 L 0 0 0 0 0 L · · · 0 0 0 L · · · 0     ·. · · ·. · · 2 . φ(Le K L)= . , φ(Le K L)=   , and so on ⊗   ⊗  0 0 0 0 L   0 0 0 L   · · ·   · · ·   L 0 0 0 0   L 0 0 0   · · ·   · · ·   0 L 0 0 0   · · ·  Hence im φ = Mn(L), as required.

Lemma 2.3 (Lemma K Lemma) Let L K be a Galois ﬁeld extension of degree n with G = Gal(L/K). Then we have⊗ an isomorphism of L⊃-algebras ≈ L L Maps(G, L) = Ln ⊗K → ∼ a b f(σ)= σ(a)b for σ G ⊗ 7→ ∈ Proof Let G = σ1,...,σn . Using the primitive element theorem, write L = K(θ) for some element θ L. Let p(t)={ (t } σ (θ)) K[t] be the minimal polynomial of θ. We have isomorphisms of ∈ 1≤i≤n − i ∈ L-algebras (where QL K L is viewed as an L-algebra via its second entry) ⊗ CRT K[t] L[t] L[t] n L K L = K L = = = Maps(G, L) = L ⊗ ∼ p(t) ⊗ ∼ p(t) ∼ t σ (θ) ∼ ∼ 1≤Yi≤n − i where CRT stands for Chinese remainder theorem. Following the above isomorphisms, if λ = h(θ) L with h(t) K[t], we have that ∈ ∈ λ 1= h(θ) 1 h(σ (θ)),h(σ (θ)),...,h(σ (θ)) = (σ (λ), σ (λ),...,σ (λ)) ⊗ ⊗ 7→ 1 2 n 1 2 n

Cyclic algebras are the simplest examples of central simple algebras, which are nothing else other than “twisted forms” of matrices. That is our next topic. 3 Central Simple Algebras We now introduce a very important class of algebras which, in some sense, are not too distant relatives of matrix algebras: Definition 3.1 Let K be a field and let Ω be an algebraically closed field extension of K. We say that a finite dimensional K-algebra A is a central simple algebra over K (CSA for short) if it satisfies one (hence all) of the following equivalent conditions: 1. A is a simple ring with centre K; 2. there is an isomorphism A ∼= Mn(D) where D is a division algebra with centre K; 3. AΩ is isomorphic to the matrix ring Md(Ω) for some d; op 4. the canonical map φ: A K A EndK-mod(A) is an isomorphism (φ sends a b to the K-vector space endomorphism x⊗ axb of→ A). ⊗ 7→ Before showing that the above are indeed equivalent, let us recall the well-known 4 Central Simple Algebras and the Brauer group

Theorem 3.2 (Wedderburn) Let K be a field and let A be a finite dimensional simple K-algebra (i.e. A has no non-trivial two-sided ideals). Then A ∼= Mn(D) for some division algebra D and some integer n. Moreover D is uniquely determined up to isomorphism as D = EndA(I) for any (non-zero) minimal ideal I of A (they are all isomorphic to Dn). Proof First, some pep talk. Secretly, we know that A ∼= Mn(D) for some division algebra D. How can we recover D intrinsically in terms of A? The idea is to notice that the minimal non-zero left ideals of Mn(D) are given by “column matrices” (as can be easily seen by a straightforward computation): D 0 0 0 D 0 0 0 D · · · · · · · · ·  D 0 0   0 D 0   0 0 D  I = . · · · I = . · · · I = . · · · 1  .  2  .  · · · n  .   .   .   .   D 0 0   0 D 0   0 0 D   · · ·   · · ·   · · ·        Being minimal, they are irreducible as left Mn(D)-modules. Besides they are all isomorphic to the n- n dimensional D-vector space D and therefore we have Mn(D) ∼= EndD(I1). Finally, we can identify D with the set of all A-endomorphisms of I via d R , where R : I I is the right multiplication by 1 7→ d d 1 → 1 d D. In fact, Rd is clearly an A-endomorphism of the left A-module I1; conversely, given φ EndA(I1), consider∈ ∈ 1 0 0 · · · df  1 0 0  v = . · · · I . ∈ 1  .   1 0 0   · · ·  Since I1 is irreducible, Mn(D)v = I1 and therefore since φ(av)= aφ(v) for all a M (D) ∈ n we conclude that φ is completely determined by the value of φ(v), which is of the form φ(v)= vd = Rd(v) for some d D, as one can see by taking a M (D) to be permutation matrices in the above formula. ∈ ∈ n Now we turn the above reasoning upside-down. Starting with a simple finite dimensional K-algebra A, let I =df any minimal non-zero ideal of A df D = EndA(I) Notice that I exists since A is artinian (it is finite dimensional over a field) and that D is a division ring by the so-called Schur’s lemma: given any non-zero d D, we must have ker d = 0 and im d = I since I is irreducible as a left A-module, hence d is an automorphism∈ of I. We can now view I as a D-vector space via d v =df d(v) for d D and v I. Everything is finite · ∈ n ∈ dimensional over K, a fortiori I is finite dimensional over D. Hence I ∼= D (non-canonically since it depends on the choice of a basis) for some n and EndD(I) ∼= Mn(D). We can now define a ring morphism ρ: A End (I) = M (D) → D ∼ n a L 7→ a where L : I I denotes the left multiplication by a, which is a D-endomorphism since L (d v) = a → a · a d(v) = d(av) = d La(v) for all a A, d D and v I. Now it is easy to check that ρ is a ring morphism,· which must· be injective since∈ its kernel∈ is a prope∈ r two-sided ideal of the simple ring A. We are left to show that ρ is also surjective. For that it is enough to show that ρ(A) is a left ideal of EndD(I) since ρ(A) contains the unity element id = L1 of EndD(I). First we show that ρ(I) is a left ideal. For w I, right multiplication R by w belongs to D = End (I), hence ∈ w A f(vw)= f(R v)= R f(v)= f(v)w for all f End (I) and v, w I, w w ∈ D ∈ that is, f L = L f ρ(v)= ρ(f(v)) for all f End (I) and v I, ◦ v f(v) ⇐⇒ · ∈ D ∈ showing that ρ(I) is a left ideal of EndD(I). Since A is simple, the two-sided ideal IA coincides with A, hence ρ(A)= ρ(I)ρ(A) is also a left ideal. This completes the proof that A ∼= Mn(D). This also proves that all minimal ideals I of A are isomorphic, and therefore D is uniquely determined by the equation D = EndA(I) (up to isomorphism). Central Simple Algebras 5

We also need a Lemma 3.3 Let K be a ﬁeld and A and B be two ﬁnite dimensional K-algebras. 1. Z(A B)= Z(A) Z(B); ⊗K ⊗K 2. if A and B are simple rings and Z(A)= K, then A B is also simple with centre Z(B). ⊗K

Proof 1. Clearly Z(A K B) Z(A) K Z(B). Conversely, let ω1,...,ωn be a basis of B over K. Since ⊗ ⊃ ⊗ A B = A Kω = Aω ⊗K ⊗K i i 1≤Mi≤n 1≤Mi≤n as K-vector spaces, any element z A K B can be uniquely written as z = a1 ω1 + + an ωn with a A. In particular, if z Z(∈A ⊗ B), then ⊗ · · · ⊗ i ∈ ∈ ⊗K (a 1) z = z (a 1) aa ω + + aa ω = a a ω + + a a ω ⊗ · · ⊗ ⇐⇒ 1 ⊗ 1 · · · n ⊗ n 1 ⊗ 1 · · · n ⊗ n for any a A, hence aa = a a for all i by the uniqueness of the above representation. Hence all ∈ i i ai Z(A), that is, z Z(A) K B A K B. Now switching the roles of the ﬁrst and second entries, we∈ conclude that z ∈Z(A) ⊗ Z(B)⊂ Z⊗(A) B, as was to be shown. ∈ ⊗K ⊂ ⊗K 2. Let I be a non-zero two-sided ideal of A B. Suppose ﬁrst that there is a “simple non-zero tensor” ⊗K a b I. Since A is simple, the two-sided ideal generated by a = 0 equals A, hence there exist ai and ′⊗ ∈ 6 ai in A such that ′ aiaai =1 1≤Xi≤n Hence 1 b = (a 1) (a b) (a′ 1) I. Reversing the roles of A and B, we conclude that ⊗ 1≤i≤n i ⊗ · ⊗ · i ⊗ ∈ 1 1 I as well,P i.e., I = A K B. ⊗ ∈ ⊗ Now let x = a b + + a b I 1 ⊗ 1 · · · n ⊗ n ∈ be an element with smallest n. We may assume that the bi are linearly independent over K, otherwise writing (say) b = λ b + + λ b as a linear combination of the other b (with λ K) and plugging 1 2 2 · · · n n i i ∈ in in the above expression, we can rewrite it to make it shorter. Similarly, we can assume that the ai are also linearly independent over K. Moreover, applying the reasoning of the special case above, we may assume that a1 = 1 as well. Now suppose that n > 1. We have that a2 / K (otherwise a1 and a2 would be linearly dependent over K and again we could shorten the above expression).∈ Since Z(A)= K there exists a A such that aa = a a. Consider the element ∈ 2 6 2 (a 1) x x (a 1)=(aa a a) b + + (aa a a) b I ⊗ · − · ⊗ 2 − 2 ⊗ 2 · · · n − n ⊗ n ∈

Since the bi are linearly independent over K and aa2 a2a = 0, this element is not 0, which contradicts the minimality of n. Therefore n = 1 and by the special− case6 proven, we are done.

Now we are ready to show: Theorem 3.4 The above definition makes sense. Proof We have 1 2: the implication is just Wedderburn’s theorem and it is easy to check that ⇐⇒ ⇒ Mn(D) is simple with centre Z(D)= K. Next we show that there is no non-trivial division algebra D over an algebraically closed field Ω. Since dimΩ D < , for any a D, the subfield Ω(a) of D generated by a is finite dimensional over Ω, and hence algebraic∞ over Ω, that∈ is, Ω(a)=Ω a Ω. This proves that D = Ω. ⇒ ∈ Now let A be a simple ring with centre K. By the lemma, AΩ is also simple with centre Ω. By what we have just proven together with Wedderburn’s theorem we conclude that A = M (Ω), i.e., 1 3. Ω ∼ d ⇒ Next we prove that 3 1. First observe that Ω is free as a K-algebra, hence Ω is faithfully flat ⇒ over K. Recall that this means that K Ω is an exact functor and that M K Ω=0 M = 0 for any K-module M. In particular, −⊗ Ω preserves injective maps. Hence if⊗ there were⇐⇒ a non-trivial −⊗K two-sided ideal a of A, then a Ω would be a non-trivial two-sided ideal of A Ω = M (Ω), but that ⊗K ⊗K ∼ n 6 Central Simple Algebras and the Brauer group

would contradict the fact that Mn(Ω) is simple. Therefore A must be simple. On the other hand, by the lemma Z(A) Ω= Z(A Ω) = Z(M (Ω)) = Ω, hence Z(A)= K, as required. ⊗K ⊗K n We now show that 1 4. Observe that φ is a ring morphism since φ(1 1) = id and ⇒ ⊗ φ (a b)(a′ b′) = φ(aa′ b′b)= φ(a b) φ(a′ b′) ⊗ ⊗ ⊗ ⊗ ◦ ⊗ op Since ker φ is a two-sided ideal of A K A , a simple ring by the lemma, we conclude that ker φ = 0, that is, φ is injective. But since dim⊗ A Aop = dim End (A) = (dim A)2 (observe that K ⊗K K K-mod K EndK-mod(A) is isomorphic to a matrix algebra), the map of K-vector spaces φ must also be surjective. To prove 4 1, observe that by the lemma Z(A) Z(Aop)= Z(A Aop)= Z(End (A)) = ⇒ ⊗K ⊗K K-mod K, hence Z(A)= K. Besides EndK-mod(A) is simple, hence so is A, for any non-trivial two-sided ideal op op op a of A would give rise to a non-trivial two-sided ideal a K A of A K A (observe that A is free hence faithfully ﬂat over K). ⊗ ⊗

Corollary 3.5 If A is a CSA over a ﬁeld K, then dimK A is a square.

2 Proof With the notation of the theorem, we have dimK A = dimΩ AΩ = dimΩ Mn(Ω) = n .

Definition 3.6 The degree of a CSA A over a field K is defined to be the square root of dimK A. Just for the record, we specialise the previous lemma restating it as

Lemma 3.7 Let K be a ﬁeld and let A and B be a CSA over K. 1. (Stability under tensor products) A B is a CSA over K. ⊗K 2. (Stability under base change) AL is a CSA over L for any ﬁeld extension L of K.

Example 3.8 Matrix algebras and cyclic algebras are CSA’s, and so are tensor products thereof. Hence we know how to construct a plethora of CSA’s. A very deep theorem by Merkurjev and Suslin (K- cohomology of Severi-Brauer varieties and the norm residue homomorphism (Russian), Izv. Akad. Nauk SSSR Ser. Mat. 46 (1982), no. 5, 1011–1046, 1135–1136) says that, in the presence of enough roots of unity, all central simple algebras are “stably isomorphic” to tensor products of cyclic algebras. More on that when we talk about the Brauer group.

4 Splitting Fields

Of great importance will be the study of the so-called splitting ﬁelds of CSA.

Theorem 4.1 Let A be a CSA over a field K. A field L K is called a splitting field for A if it ⊃ trivialises (or splits) A, i.e., AL ∼= Mn(L) for some n. Example 4.2 The algebraic closure Kalg of K is always a splitting field for any CSA A over K. Actually, alg alg we can be more economical: since AKalg ∼= Mn(K ) and both rings are finite dimensional over K , this isomorphism (and its inverse) is defined by a finite amount of data, namely the images of the elements of a basis. Hence we can define this isomorphism over a finite extension L over K. To sum up, for any CSA A there exists a finite extension L K that splits A. ⊃ Example 4.3 If L K is a cyclic extension, then any cyclic algebra (χ,a) (for a K × and χ a surjective character of⊃ Gal(L/K)) is split by L. ∈

The last example is a particular instance of

Theorem 4.4 Let K be a ﬁeld and A be a CSA over K of degree n. If L is a ﬁeld contained in A such that n = [L : K] then L splits A.

op ≈ Proof The restriction of the isomorphism φ: A K A EndK-mod(A) to A K L induces an injective map φ: A L ֒ End (A): in fact, the⊗ image under→ φ of d λ for d⊗ A and λ L is the ⊗K → L-mod ⊗ ∈ ∈ (right) L-linear map x dxλ. Since dim A L = dim A = n2 = dim End (A), φ: A L ≈ 7→ L ⊗K K L L-mod ⊗K → EndL-mod(A) must be an isomorphism. Splitting Fields 7

Remark 4.5 It can be shown, using the double centraliser theorem, that for a division algebra D over K, the subfields L D with [L : K] = deg D are maximal subfields of D (with respect to the inclusion order). Therefore we⊂ shall refer to such splitting fields of degree as maximal ones, even though we are not going to use the fact that they are actually maximal. Do such maximal subfields exist? In fact, any division algebra is packed with them! We can even find maximal subfields which are separable over the base field! Recall that a finite extension of fields L K is separable if it satisfies any of the following three equivalent properties: ⊃ 1. every element λ L has a separable minimal polynomial over K, i.e., a polynomial with distinct roots; ∈ ;there are exactly n = [L : K] K-immersions φ: L ֒ Kalg .2 → 3. L Kalg = (Kalg)n where n = [L : K]. ⊗K ∼ Any finite extension of a perfect field (for instance, a finite field or a field of characteristic 0) is automatically separable. Hence separability may only pose an issue for infinite fields of positive characteristic. Theorem 4.6 (Separable maximal subfields) Let D be a division algebra over a field K. Then there exists an element θ D such that K[θ] K is a separable field extension of degree deg D. In particular, D can be split by∈ a separable extension⊃ of degree deg D.

Proof Let n = deg D and choose a basis ω1,...,ωn2 of D over K. Let L be the algebraic closure of K ≈ and so that there exists an isomorphism φ: D K L Mn(L). View the elements a1ω1 + an2 ωn2 D, ⊗ →n2 · · · ∈ ai K, as points (a1,...,an2 ) of the affine space K , and similarly regard the elements of Mn(L) as ∈ 2 points of the affine space Ln . As we shall see, there are no non-trivial division algebras over a finite field, hence we may assume K is infinite. Now consider the discriminant d(x ,...,x 2 ) L[x ,...,x 2 ] of the characteristic polynomial 1 n ∈ 1 n of a matrix in M (L) with entries x ,...,x 2 . By the general position lemma below, we can find a K n 1 n i ∈ such that the element θ = a ω + a 2 ω 2 D satisfies d φ(θ 1) = 0. In other words, the 1 1 · · · n n ∈ ◦ ⊗ 6 characteristic polynomial p(x) of matrix φ(θ 1) Mn(L) will have distinct roots, hence φ(θ 1) will have n distinct eigenvalues and will therefore⊗ be diagonalisable∈ . Now let E = K[θ] be the subfield⊗ of D generated by θ. By construction, this field has the property that E K L is isomorphic to the subalgebra ⊗ n of Mn(L) generated by φ(θ 1). But φ(θ 1) is diagonalisable, hence E K L ∼= L . This proves that [E : K]= n = deg D and that⊗ E is separable⊗ over K. ⊗

Lemma 4.7 (General Position) Let K L be an extension of fields and suppose that K is an infinite ⊂ field. Let p(x1,...,xn) L[x1,...,xn] be a non-zero polynomial. Then there exist infinitely many points (a ,...,a ) in Kn such∈ that p(a ,...,a ) =0. 1 n 1 n 6 Proof We proceed by induction on n. If n = 1, the result is obvious since p(x1) has only finitely many roots. Assume n> 1 and think of p(x1,...,xn) as a polynomial in xn with coefficients in L[x1,...,xn−1]. By induction hypothesis, there exists (a ,...,a ) Kn−1 such that at least one of the coefficients of 1 n−1 ∈ p(x1,...,xn) does not vanish, i.e., such that p(a1,...,an−1, xn) L[xn] is not identically 0. But now there exist infinitely many a K such that p(a ,...,a , x ) =∈ 0, and we are done. n ∈ 1 n−1 n 6 Corollary 4.8 (Galois Splitting) Any CSA A over a field K admits a splitting field L which is a finite Galois extension of K. Proof We may suppose that K is infinite, since any splitting field of a CSA over a finite field K is automatically a Galois (even cyclic) extension of K. Then, since A ∼= Mn(D) for some division algebra D over K, and D has a separable splitting field by the theorem, we conclude that A can also be split by a finite separable extension of K. By taking the normal closure, we we may assume this extension to be Galois as well.

Remark 4.9 Even though a division algebra admits a Galois splitting field, it does not necessarily contain a maximal Galois subfield. Division algebras without Galois maximal subfields are called noncrossed products, and are quite hard to come about. The first examples of such bizarre algebras were discovered by Amitsur (On central division algebras, Israel J. Math. 12 (1972), 408–420). Non-crossed products do not exist for division algebras over Q or Qp, but already appear for Q(t) and Q((t)) (see Brussel’s construction in Noncrossed products and nonabelian crossed products over Q(t) and Q((t)), Amer. J. Math. 117 (1995), no. 2, 377–393). 8 Central Simple Algebras and the Brauer group

5 Faithfully Flat Descent and Galois Cohomology As we just saw, any degree n CSA A over K can be split or “trivialised” by restricting it to some finite Galois extension L of K, hence A should be thought of a “twisted form” of Mn(L). The natural question is: how to get all such forms? In topology, we have a very similar situation. Recall that a (real) vector bundle of rank n over a topological space X is a family of “continuously varying” n-dimensional real vector spaces over X. More precisely, we have a continuous map π: E ։ X from a topological space E to X such that the fibre φ−1x E of each point x X has the structure of an n-dimensional R-vector space, and π is locally ⊂ ∈ n isomorphic to the first projection map p1: U R ։ U: each x X has an open neighbourhood U X ≈× ∈ ⊂ together with a homeomorphism ξ : π−1U U Rn making the diagram U → ×

ξU U Rn - π−1U ⊂ - E × ≈

p1 π π ? ? ? U ======U ⊂ - X

n ≈ −1 commute and such that ξU induces an R-linear isomorphism ξU,x: x R π x between the fibres 1 { }× 1 → 1 for each x U. For example, over the circle S , the projection map p1: S R S is a rank 1 vector bundle (the∈trivial rank 1 vector bundle). But also the projection π: M ×S1 of→ the “infinite Möbius” band M to the middle circle is a rank 1 vector bundle, which is not tri→vial: M is locally of the form U R, but not globally. ×

(Just to make sure we are talking about the same thing, M is the quotient of [0, 1] R obtained by identifying (0,y) (1, y) for all y R; the middle circle is the image of [0, 1] 0 in× this quotient and π: M S1 is given∼ by−π(x, y)= x) ∈ ×{ } → Two rank n vector bundles π1: E1 X and π2: E2 X are isomorphic if there is a homeomorphism ≈ → → φ: E E such that the diagram 1 → 2 φ - E1 E2 ≈

π1 π2 ? ? X ======X

commutes and such that the induced morphism of ﬁbres φ : π−1x ≈ π−1x is R-linear for each x X. x 1 → 2 ∈ By deﬁnition, any vector bundle π: E X over X becomes trivial (i.e., isomorphic to U Rn) → i × when restricted to some open cover = Ui of X. This observation allows for a very simple recipe to obtained all rank n vector bundles (upU to{ isomorphism)} that are “trivialised” by . First suppose that we are given a vector bundle π: E X and a trivialisation U → ξ : U Rn ≈ π−1U i i × → i Faithfully Flat Descent and Galois Cohomology 9

n ≈ n Then, for each pair (i, j), we have a vector bundle automorphism φij : (Ui Uj ) R (Ui Uj ) R over U U given by the composition ∩ × → ∩ × i ∩ j −1 ξ ξ φ : (U U ) Rn -i π−1(U U ) j- (U U ) Rn ij i ∩ j × i ∩ j i ∩ j × where we still write ξ and ξ−1 for their restrictions to (U U ) Rn and π−1(U U ) respectively. i j i ∩ j × i ∩ j These automorphisms clearly satisfy

1. φii = id for all i; −1 2. φij = φji for all pairs i, j; 3. (1-cocycle condition) for every triple i, j, k,

φ = φ φ ik jk ◦ ij over U U U . i ∩ j ∩ k The above data φ is called the Cechˇ 1-cocycle of the vector bundle E associated to the trivialisation { ij } ξi. Now how does a 1-cocycle change when we change the trivialisation? Or, more generally, given isomorphic vector bundles πE : E X and πF : F X with 1-cocycles φij and ψij , what is their relation? → → { } { } ≈ n ≈ −1 To answer that question, let λ: E F be an isomorphism and let ξi: Ui R πE Ui and n ≈ −1 → −1 −1 × → χi: Ui R πF Ui be trivialisations so that φij = ξj ξi and ψij = χj χi over Ui Uj . Then, over U , we× have→ a vector bundle automorphism µ : U Rn ◦ U Rn given by◦ the composition∩ i i i × → i × ξ λ χ−1 µ : U Rn -i π−1U - π−1U i- U Rn i i × E i F i i × Now a quick computation shows that, for all i, j,

ψ = µ φ µ−1 over U U ij j ◦ ij ◦ i i ∩ j In case there are automorphisms µ of U Rn such that the above relation holds, we say that the 1- i i × cocycles ψij and φij are cohomologous. As it is easily seen, this establishes an equivalence relation among the{ 1-cocycles,} { } and therefore we have showed that to each isomorphism class of rank n vector bundles that is trivialised by there is a well-defined 1-cocycle class, which is independent of the chosen trivialisation. U n Conversely, given a Cechˇ 1-cocycle φij (that is, automorphisms φij of (Ui Uj ) R satisfying the relations 1–3 above), we can construct{ a vector} bundle π: E X by “glueing”∩ trivial× vector bundles U Rn! Just take E to be the quotient → i × (U Rn) E = i i × F ∼ of the disjoint union of the U Rn. Here i × v w for v U Rn and w U Rn φ (v)= w over U U ∼ ∈ i × ∈ j × ⇐⇒ ij i ∩ j Now conditions 1–3 above ensure that is indeed an equivalence relation, so everything is well-defined. Now the projection maps U Rn U∼induce a global projection map π: E X, which is a bona fide i × → i → rank n vector bundle with associated 1-cocycle φij by construction! As easily checked, cohomologous 1-cocycles give rise to isomorphic vector bundles{ so} that we have shown that there is a bijection

isomorphism classes of rank n cohomology classes of vector bundles trivialised by ↔ -Cechˇ 1-cocycles U U Let us rephrase the above construction in such a way to make it more “algebraisable.” First recall that given two continuous maps f: A C and g: B C, we can form their ﬁbre product A C B, deﬁned as the subspace of A B given→ by → × × A B = (a,b) A B f(a)= g(b) ×C { ∈ × | } 10 Central Simple Algebras and the Brauer group

We have thus a commutative diagram

A B - B ×C

g ? ? A - C f

where A B A and A B B are induced by the projection maps from A B to A and ×C → ×C → × B, respectively. For instance, if A is a subspace of C and f is the inclusion map, then A C B is homeomorphic to g−1A and the left vertical arrow is just the restriction of g. × Now let X and be as above and let Y = U be the disjoint union of the U . Consider the map U i i i h: Y X induced by the inclusion maps Ui ֒ FX. Then we have homeomorphisms → → Y Y = U U and Y Y Y = U U U ×X i ∩ j ×X ×X i ∩ j ∩ k Gi,j i,j,kG and maps p 12 p1 h Y X Y X Y Y X Y Y X −→ −→p2 × × −→p23 × −→ −→ Here p1 and p2 are the projection maps, and p12(y1,y2,y3) = (y1,y2), p13(y1,y2,y3) = (y1,y3) (the “middle one” among the triple arrows, not labelled for lack of space), p23(y1,y2,y3) = (y2,y3). Now to n give a collection φij of vector bundle automorphisms of (Ui Uj ) R is the same as to give a single a vector bundle automorphism{ } over Y Y : ∩ × ×X φ: Y Y Rn ≈ Y Y Rn ×X × → ×X × The 1-cocycle condition is now summarised by a single relation

p∗ φ = p∗ φ p∗ φ 13 23 ◦ 12 where p∗ φ denote the pull-backs of φ with respect to p , that is, the morphism on Y Y Y Rn ij ij ×X ×X × induced by φ on the (i, j)-components and the identity on the remaining component: for instance, ∗ p13φ(y1,y2,y3, v) = (y1,y2,y3, w) where φ(y1,y3, v) = (y1,y3, w). Now two automorphisms φ and ψ are cohomologous if and only if there is an automorphism µ: Y Rn ≈ Y Rn such that ψ = p∗µ φ p∗µ−1. × → × 2 ◦ ◦ 1 Now we mimic the above algebraically. We work in the more general situation of faithfully flat extensions of commutative rings, whose proof is no more difficult than the case of fields. Recall that a commutative ring extension A B is faithfully flat if it is flat, i.e., the functor A B is exact, and M B =0 M = 0 for any⊂ A-module M. In particular, a complex of A-modules−⊗ ⊗A ⇐⇒ - M - M - M - M - · · · i i−1 i−2 i−3 · · · is exact if, and only if, the complex of B-modules

- M B - M B - M B - M B - · · · i ⊗A i−1 ⊗A i−2 ⊗A i−3 ⊗A · · · is exact. In fact, if hi(M•) is the homology of the first complex, then hi(M•) A B is the homology of the second one since B is flat over A, and therefore tensor products commute with⊗ quotients. Therefore h (M )=0 h (M ) B = 0, as required. i • ⇐⇒ i • ⊗A It can be shown that B is faithfully flat over A if and only if it is flat and Spec B Spec A is surjective. In our algebraic setup, Y = Spec B will play the role of the disjoint union of→ a cover of X = Spec A. Since we work with rings rather than schemes, arrows will be reversed and fibre products will be replaced by tensor products. In what follows, we write

p : B B B p : B B B 1 → ⊗A 2 → ⊗A b b 1 b 1 b 7→ ⊗ 7→ ⊗ Faithfully Flat Descent and Galois Cohomology 11

Also, if N is a B-module, let p∗N = N (B B)= N B and p∗N = (B B) N = B N 1 ⊗B ⊗A ⊗A 2 ⊗A ⊗B ⊗A pull-backs of N, namely the B A B-modules obtained by tensoring N with respect to the first and second entries respectively. Then⊗ the p∗N will play the role of trivial bundles Y Rn. Let i × φ: p∗N ≈ p∗N 1 → 2 n be an isomorphism of B A B-modules (φ will play the role of the automorphism of Y X Y R ). Consider the “dual projection⊗ maps” × × p : B B B B B p : B B B B B 12 ⊗A → ⊗A ⊗A 13 ⊗A → ⊗A ⊗A b b b b 1 b b b 1 b 1 ⊗ 2 7→ 1 ⊗ 2 ⊗ 1 ⊗ 2 7→ 1 ⊗ ⊗ 2 p : B B B B B 23 ⊗A → ⊗A ⊗A b b 1 b b 1 ⊗ 2 7→ ⊗ 1 ⊗ 2 and the pull-backs of N to B A B A B, obtained by tensoring the latter ring with N over B with respect to the first, second and⊗ third⊗ entries of B B B, respectively: ⊗A ⊗A N B B and B N B and B B N ⊗A ⊗A ⊗A ⊗A ⊗A ⊗A Tensoring φ with id over B with respect to the third, second and first entries of B A B A B, we obtain maps of B B B-modules ⊗ ⊗ ⊗A ⊗A p∗ φ: N B B B N B p∗ φ (n 1 1) = b n 1 12 ⊗A ⊗A → ⊗A ⊗A 12 ⊗ ⊗ i i ⊗ i ⊗ P p∗ φ: N B B B B N p∗ φ (n 1 1) = b 1 n 13 ⊗A ⊗A → ⊗A ⊗A 13 ⊗ ⊗ i i ⊗ ⊗ i P p∗ φ: B N B B B N p∗ φ (1 n 1) = 1 b n 23 ⊗A ⊗A → ⊗A ⊗A 23 ⊗ ⊗ i ⊗ i ⊗ i P for φ(n 1) = i bi ni. These maps will play the roles of the pull-backs of the automorphism of ⊗ n ⊗ n Y X Y X R Pto Y X Y X Y R . We are now ready to state × × × × × Theorem 5.1 (Grothendieck’s Faithfully Flat Descent) Let B A be a faithfully flat extension of commutative rings. ⊃ 1. For any A-module M, the sequence

ǫ d0 d1 d2 0 - M - M B - M B B - M B B B - ⊗A ⊗A ⊗A ⊗A ⊗A ⊗A · · · is exact. Here ǫ(m)= m 1 and ⊗ dr(m b b )= ( 1)i m b b 1 b b ⊗ 0 ⊗···⊗ r − · ⊗ 0 ⊗···⊗ i−1 ⊗ ⊗ i ⊗···⊗ r 0≤Xi≤r 2. Let N be a B-module (or a B-algebra). Then there is a bijection between the set of isomorphism classes of A-modules (or A-algebras) M such that M B = N and the set of equivalence classes ⊗A ∼ of B A B-isomorphisms ⊗ ≈ φ: N B B N ⊗A → ⊗A satisfying the 1-cocycle condition p∗ φ = p∗ φ p∗ φ, so that we have a commutative diagram 13 23 ◦ 12 p∗ φ N B B 12- B N B ⊗A ⊗A ⊗A ⊗A p ∗ 13 φ ∗ p23φ - ? B B N ⊗A ⊗A Two B A B-isomorphisms φ and ψ are equivalent if and only if there is an automorphism ≈ ⊗ µ: N N such that → ψ = p∗µ φ p∗µ−1 2 ◦ ◦ 1 where p∗µ = µ id: N B ≈ N B and p∗µ = id µ: B N ≈ B N are the pull-backs 1 ⊗ ⊗A → ⊗A 2 ⊗ ⊗A → ⊗A of φ with respect to p1 and p2. 12 Central Simple Algebras and the Brauer group

Proof 1. First we show that the given complex is homotopic to 0 under the assumption that the map ǫ: M M B has a section s: M B M (that is, s ǫ = id). Define → ⊗A ⊗A → ◦ kr: M B⊗(r+1) M B⊗r ⊗A → ⊗A m b b b s(m b ) b b ⊗ 0 ⊗ 1 ⊗···⊗ r 7→ ⊗ 0 ⊗ 1 ⊗···⊗ r Now check that id = dr−1 kr + kr+1 dr for all r. This shows that the complex is exact. To tackle the general case, it is enough to◦ show that◦ the sequence obtained by base change B is exact since B −⊗A is faithfully flat over A, hence it is enough to show that the map ǫB: M A B M A B A B has a section. Just define s: M B B M B by s(m b b )= m ⊗b b and→ check⊗ that⊗ s ǫ = id. ⊗A ⊗A → ⊗A ⊗ 1 ⊗ 2 ⊗ 1 2 ◦ 2. Suppose that M is an A-module such that M A B = N. Then we may define an isomorphism ≈ ⊗ φ: N B B N of B B-modules by ⊗A → ⊗A ⊗A φ: (M B) B ≈ B (M B) ⊗A ⊗A → ⊗A ⊗A m b b b m b ⊗ 1 ⊗ 2 7→ 1 ⊗ ⊗ 2 n which, in our geometric discussion, corresponds to the vector bundle automorphism of Y X Y R n n × × that identifies the restrictions of Ui R and Uj R over Ui Uj . Now a straightforward computation shows that the above φ satisfies the× 1-cocycle condition.× Observe∩ that by part (1), we may identify M with the A-submodule of N given by m 1 m M . { ⊗ | ∈ } Now let M and M be two A-modules such that M B = M B = N, and let φ and 1 2 1 ⊗A 2 ⊗A 1 φ2 are the corresponding isomorphisms N A B B A N. If there is an isomorphism of A-modules ≈ ⊗ → ⊗ ∗ ∗ −1 ν: M1 M2, then it induces a B-automorphism µ = ν id of N and we have that φ2 = p2µ φ1 p1µ since → ⊗ ◦ ◦ p∗µ φ p∗µ−1(m b b ) = (id ν id) φ (ν−1 id id)(m b b ) 2 ◦ 1 ◦ 1 2 ⊗ 1 ⊗ 2 ⊗ ⊗ ◦ 1 ◦ ⊗ ⊗ 2 ⊗ 1 ⊗ 2 = (id ν id) φ (ν−1m b b ) ⊗ ⊗ ◦ 1 2 ⊗ 1 ⊗ 2 = (id ν id)(b ν−1m b ) ⊗ ⊗ 1 ⊗ 2 ⊗ 2 = b m b = φ (m b b ) 1 ⊗ 2 ⊗ 2 2 2 ⊗ 1 ⊗ 2 ∗ ∗ −1 ∗ ∗ for all m2 M2 and b1,b2 B. Conversely, if φ2 = p2µ φ1 p1µ φ2 p1µ = p2µ φ1 then ≈ ∈ ∈ ≈ ◦ ◦ ⇐⇒ ◦ ◦ µ: N N restricts to an isomorphism ν: M M since → 1 → 2 φ (µ(m ) 1) = φ p∗µ(m 1) = p∗µ φ (m 1) = p∗µ(1 m )=1 µ(m ), 2 1 ⊗ 2 ◦ 1 1 ⊗ 2 ◦ 1 1 ⊗ 2 ⊗ 1 ⊗ 1 i.e., µ(m ) M for all m M . 1 ∈ 2 1 ∈ 1 Now we have to show that given an isomorphism φ: N B ≈ B N satisfying the 1-cocycle ⊗A → ⊗A condition, there exists an A-module M such that M A B = N. Secretly, we know that M exists and that φ is given by the above “swapping formula,” hence⊗ it is natural to define

M =df m N φ(m 1)=1 m { ∈ | ⊗ ⊗ } which, in geometric terms, corresponds to the subset of Y Rn = U Rn of those elements that × i i × “agree on the overlaps” Ui Uj . F ∩ The subset M N is clearly an A-module. We now show that the natural map ⊂ λ: M B N ⊗A → m b bm ⊗ 7→ is an isomorphism of B-modules. For that, consider the following diagram

α id - - M A B N A B ⊗ - B A N A B ⊗ ⊗ β id ⊗ ⊗ ⊗ λ φ p∗ φ ≈ ≈ ≈ 23 ? ? e0 ? -ǫ - N B A N - B A B N ⊗ e1 ⊗ ⊗ Faithfully Flat Descent and Galois Cohomology 13

where the horizontal maps of the bottom row are the ones of the part (1), namely, e1(b n)=1 b n, and e0(b n)= b 1 n for b B, n N, and those of the top are obtained by the faithfully⊗ flat⊗ base⊗ change ⊗ B of⊗ the⊗ exact sequence∈ ∈ −⊗A - α- M N - B A N β ⊗ that defines the A-module M, where α(n) = φ(n 1) and β(n)=1 n. In particular, we have that ⊗ ⊗0 1 M A B is the equaliser of α id and β id, while N is the equaliser of e and e . Hence the isomorphism φ will⊗ induce the required isomorphism⊗ ⊗ λ once we show that the above diagram commutes. But it is easy to check that the right square formed by the bottom arrows β id and e1 commutes, and the same holds for the left square by the definition of M. The right square formed⊗ by the top arrows α id and e0 also commutes since p∗ φ α id(n b)= p∗ φ p∗ φ(n 1 b)= p∗ φ(n 1 b)= e0 φ⊗(n b). 23 ◦ ⊗ ⊗ 23 ◦ 12 ⊗ ⊗ 13 ⊗ ⊗ ◦ ⊗ If N is a B-algebra, we have also to show that the multiplication map m: N B N N also “descends,” which can be done by similar computations as the previous ones. Details⊗ are→ left to the reader.

After this little detour in faithfully flat descent, we go back to CSA. Let L K be a finite extension of fields (which is automatically faithfully flat since L is free over K). In order to⊃ apply the above to our study, we need to know the automorphism group of M (L) L. Fortunately, it is easy to describe: n ⊗K Lemma 5.2 (Weak Skolem-Noether) Let K be a field. All automorphisms of Mn(K) are inner, df × hence the automorphism group of Mn(K) is the projective linear group P GLn(K) = GLn(K)/K .

Proof Any automorphism φ of Mn(K) permutes its minimal non-zero left ideals, i.e., the column matrices I1,...,In. Conjugating φ by a suitable permutation matrix, we may assume that φ(I1) = I1. n In other words, φ restricts to an automorphism of I1 ∼= K , and as such it can be represented by an invertible matrix C GLn(K), i.e., φ(P )= CP for all P I1. Now if M Mn(K) and P I1, we have that ∈ ∈ ∈ ∈ φ(MP )= φ(M)φ(P ) CMP = φ(M)CP ⇐⇒ n −1 Since this holds for all P I1 ∼= K , we have that CM = φ(M)C φ(M)= CMC . This proves ∈ × ⇐⇒ that φ is inner. Since the centre of GLn(K) is K , the last statement also follows.

We saw in the last section that any CSA over K is split by some ﬁnite Galois extension L K, ⊃ and from now on we work in this setup. Let G = Gal(L/K) and n = [L : K]. By the Lemma K Lemma we have that L L = Maps(G, L), M (L) L = M (L L) = M (Maps(G, L))⊗ = ⊗K n ⊗K n ⊗K n Maps(G, Mn(L)) and similarly L K Mn(L) = Maps(G, L). Therefore an L K L-isomorphism Mn(L) K L L M (L) corresponds to⊗ a Maps(G, L)-automorphism of Maps(G, M⊗ (L)), that is, to an element⊗ → ⊗K n n of Maps(G, P GLn(L)) by the previous lemma. Putting everything together, we obtain a bijection

isomorphism classes of CSA elements of Maps(G, P GLn(L)) A such that AL ∼= Mn(L) ↔ satisfying the 1-cocycle condition We can be more explicit about the 1-cocycle condition. Applying the Lemma Lemma twice, we obtain ⊗K (L L) L = Maps(G, L) L = Maps(G, L L) = Maps(G, Maps(G, L)) = Maps(G G, L) ⊗K ⊗K ⊗K ⊗K × given by L L L ≈ Maps(G G, L) ⊗K ⊗K → × a b c f(τ, σ)= σ(τ(a) b) c for all σ, τ G ⊗ ⊗ 7→ · · ∈ and, similarly, M (L) L L = Maps(G G, M (L)) and so on. Now let f: G P GL (L) be n ⊗K ⊗K × n → n an element of Maps(G, P GLn(L)). In terms of the above isomorphisms, we have that the pull-backs of f correspond to the Maps(G G, L)-automorphisms of Maps(G G, M (L)) given by the elements of × × n Maps(G G, P GLn(L)) × ∗ p12f(τ, σ)= σf(τ) ∗ p23f(τ, σ)= f(σ) ∗ p13f(τ, σ)= f(στ) Hence the 1-cocycle condition now reads f(στ)= f(σ) σf(τ) for all σ, τ G. · ∈ 14 Central Simple Algebras and the Brauer group

Definition 5.3 Let G be a finite group and M be a (not necessarily abelian) G-module, written multi- plicatively. A 1-cocycle is a function f: G M such that → f(στ)= f(σ) σ f(τ) for all σ, τ G · ∈ Two 1-cocycles f and g are cohomologous (in symbols f g) if there exists an element m M such that ∼ ∈ f(σ)= m−1 g(σ) σ(m) for all σ G · · ∈ It is easy to check that defines an equivalence relation on the pointed set Z1(G, M) of all 1-cocycles (the distinguished point being∼ the constant function f(σ) = 1). We define the zeroth and first cohomology groups of G with coefficients in M as the pointed sets

H0(G, M)= M G =df m M σ(m)= m for all σ G { ∈ | ∈ } H1(G, M)= Z1(G, M)/ ∼ Notice that when M is abelian, the above definitions coincide with the usual cohomology of G with coefficients in M. Rewriting the faithful flat descent in terms of the above language we obtain: Theorem 5.4 (Twisted forms of matrices) Let L K be a finite Galois extension of fields with G = Gal(L/K). There is a bijection ⊃

isomorphism classes of CSA 1 H (G, P GLn(L)) A such that AL ∼= Mn(L) ↔

This isomorphism takes (the isomorphism class of) the CSA A to the class of the 1-cocycle

f(σ)= φ−1 σ φ σ−1 Aut (M (L)) = P GL (L) for σ G ◦ ◦ ◦ ∈ L n n ∈ ≈ where φ: Mn(L) AL is a ﬁxed L-isomorphism. The inverse map associates to each 1-cocycle f 1 × → ∈ Z (G, P GLn(L )) the K-subalgebra of Mn(L) given by

M M (L) f(σ) σ(M)= M for all σ G { ∈ n | ◦ ∈ }

Example 5.5 (Cyclic Algebras Revisited) Let L K be a cyclic extension of degree n with G = ⊃ 1 Gal(L/K) and let σ be a generator of G. Then an element of H (G, P GLn(L)) represented by a 1-cocycle f: G P GLn(L) is completely determined by the value of f(σ) since f(id σ)= f(id)f(σ) f(id) = 1 and → · ⇒ f(σ2)= f(σ) σf(σ), · f(σ3)= f(σ σ2)= f(σ) σf(σ2)= f(σ) σf(σ) σ2f(σ), · · · · . . Since f(σn)= f(id) = 1, f(σ) is also subject to

f(σ) σf(σ) σ2f(σ) . . . σn−1f(σ)=1 ( ) · · · · ∗ For instance, let a K × and consider “companion matrix” of xn a: ∈ − 0 0 0 a 1 0 · · · 0 0  · · ·  C =df 0 1 0 0 a  . · · ·   .   .   0 0 1 0   · · ·  We may form the 1-cocycle given by × f(σ)= Ca mod L The Brauer group 15

Since a K, the condition ( ) is indeed verified: ∈ ∗ f(σ) σf(σ) . . . σn−1f(σ)= Cn mod L× = aI mod L× = I mod L× · · · a Here I is the identity matrix. According to the above, the K-algebra corresponding to this 1-cocycle is i i i i −i the set of matrices M Mn(L) such that f(σ ) σ (M)= M Caσ (M)Ca = M for all i, which ∈−1 ◦2 n−1 ⇐⇒ amounts to σ(M) = Ca MCa. Clearly I, Ca, Ca ,...,Ca satisfy the latter identity. Moreover, since conjugation by Ca is “almost a cyclic permutation”, it is not difficult to guess and it is immediate to verify that the matrices b 0 0 0 σ(b) · · · 0 df   Sb = . · · · for b L . ∈  .   0 0 σn−1(b)   · · ·  also satisfy that identity as well, that is, S C = C σ(S ) = C S for all b L, or equivalently, b a a b a σb ∈ S −1 C = C S for all b L. Therefore σ b a a b ∈ A =df S S C S C2 S Cn−1 b ⊕ b a ⊕ b a ⊕···⊕ b a 2 is a K-subalgebra of Mn(L) of the correct dimension n , hence it must be the algebra defined by the 1-cocycle above. But clearly A is isomorphic to our old friend (χ,a), the cyclic algebra given by a and the character χ(σ)= 1 mod n. − We end this section with a partial generalisation of the long exact sequence of abelian cohomology to the non-abelian case: Lemma 5.6 (Not So Long Exact Sequence) Let 1 A B C 1 → → → → an exact sequence of Z[G]-modules with A Z(B). Then the sequence ⊂ 1 - H0(G, A) - H0(G, B) - H0(G, C) δ0 - H1(G, A) - H1(G, B) - H1(G, C) δ1 - H2(G, A) is an exact sequence of pointed sets. Here, all the maps but the connecting morphisms δi are induced by the corresponding maps of the short exact sequence. The connecting maps are defined as follows. Given c H0(G, C), δ0(c) is the class of the 1-cocycle f(σ)= b−1σ(b) A, σ G, where b B is a pre-image of∈c. On the other hand, for any φ H1(G, C), δ1(φ) is the class∈ of the∈ (abelian) 2-cocycle∈ g: G G A given by g(σ, τ) = f(σ) σ(f(τ)) ∈f(στ)−1 for σ, τ G, where f: G B is such that the composition× → with B C is a 1-cocycle· representing· φ. ∈ → → Proof Straightforward computation shows that the maps δi are well-defined. By an exact sequence of pointed sets we just mean that the image of the preceding map coincides with the pre-image of the distinguished point of the next term of the sequence. With this definition, exactness of the above long sequence is a series of boring checks that are better done in private, when no one is looking, so I leave it to the (patient) reader!

6 The Brauer group In this section, for each field K we define a very important group, the Brauer group of K, whose elements classify all division algebras over K. We begin with a Definition 6.1 Let K be a field. Two CSA A and B over K are Brauer equivalent if they satisfy the two following equivalent conditions: 1. A is stably isomorphic to B, that is, there is an isomorphism

M (A)= A M (K) = B M (K)= M (B) n ⊗K n ∼ ⊗K m m for some m and n. 2. the division algebras associated to A and B in Wedderburn’s theorem are isomorphic. 16 Central Simple Algebras and the Brauer group

The equivalence of the two conditions above is a consequence of Wedderburn’s theorem. It is easy to see that Brauer equivalence is an equivalence relation. We write [A] for the equivalence class of A. When A and B have the same degree, Brauer equivalence [A] = [B] is the same as the plain old isomorphism A ∼= B of CSA (again by Wedderburn’s theorem). An important feature of the Brauer equivalence is that it is compatible with tensor products: [A] = [A′] [A B] = [A′ B′] [B] = [B′] ⇒ ⊗K ⊗K Hence the tensor product induces a binary operation

[A] + [B] =df [A B] ⊗K on the set Br(K) of all Brauer equivalence classes of CSA over K. It turns out that Br(K) is an abelian group: the operation is associative and commutative since the tensor product has the same properties; the class [K] of the trivial CSA over K is clearly an identity for this operation; and since op ≈ op A K A EndK-mod(A) [A] + [A ] = 0 for every CSA A over K, each element [A] Br(K) has an⊗ inverse [→Aop]. ⇒ ∈ Definition 6.2 Let K be a field. The set Br(K) of all Brauer equivalence classes of CSA over K, equipped with the product induced by the tensor product, is called the Brauer group of K. For any field extension L K, the restriction map is the group morphism given by ⊃ res: Br(K) Br(L) → [A] [A ] 7→ L In particular, if L is algebraic over K, we also write Br(L/K) for the kernel of the restriction map, that is, the subgroup of Br(K) consisting of Brauer classes split by L. Observe that the elements of Br(K) are exactly the isomorphism classes of division algebras over K. Knowledge of the Brauer group is thus of central importance in the study of CSA. As we shall see, Br(K) is also a very important arithmetic invariant of the field K. Example 6.3 If K is algebraically closed (separably closed suffices), then Br(K) = 0 since there are no non-trivial division algebras over K. Example 6.4 In Br(R), the class [H] of the quaternion algebra H satisfies 2[H]=0 [H]= [H]= op ⇐⇒ − [H ]. In fact, viewing H as the real subalgebra of M2(C) given by matrices of the form α β α, β C β α ∈ − we see that the transpose isomorphism M (C) ≈ M (C)op given by M M T restricts to an isomorphism 2 → 2 7→ H ≈ Hop. → The great thing about the Brauer classes of CSA is that they are much easier to deal with compared to the isomorphisms classes of CSA, thanks to df Theorem 6.5 (Cohomological Brauer group) Let K be a field and GK = Gal(Ksep/K) be its absolute Galois group, where Ksep denotes the separable closure of K. Then there is a natural isomorphism

≈ - 2 × Br(K) H (GK ,Ksep)

compatible with restriction maps: for any ﬁeld extension L K, one has ⊃ - 2 × Br(L) ≈ H (GL,Lsep) 6 6 res res

- 2 × Br(K) ≈ H (GK ,Ksep) (Note that L K = L so that the restriction map of cohomology groups is well-deﬁned). · sep sep The Brauer group 17

Proof Let L K be a ﬁnite Galois extension with G = Gal(L/K), and let B(n,L) Br(K) be the subset whose elements⊃ are the classes of CSA over K of degree n that are split by L. Then⊂ we have that

Br(L/K)= B(n,L) and Br(K)= Br(L/K) n[≥1 L finite[ Galois over K since every CSA is split by some ﬁnite Galois extension. On the other hand, by Wedderburn’s theorem two degree n CSA are Brauer equivalent if and only if they are isomorphic, hence by what we showed ≈ 1 in the last section there is a natural bijection B(n,L) H (G, P GLn(L)). From the exact sequence of G-modules → - × - - - 1 L GLn(L) P GLn(L) 1 we obtain an exact sequence

δ1 1= H1(G, GL (L)) - H1(G, P GL (L)) - H2(G, L×) ( ) n n ∗ ≈ 1 where the ﬁrst term is trivial by Satz 90 Hilberts. Hence, composing B(n,L) H (G, P GLn(L)) with δ1, we obtain a map → ι : B(n,L) H2(G, L×) n → We have that ιn([A])=0 A ∼= Mn(K) by the exactness of ( ). Now an explicit computation shows that for any A B(m,L)⇐⇒ and A B(n,L) we have that ∗ 1 ∈ 2 ∈ ι ([A A ]) = ι ([A ]) + ι ([A ]) ( ) mn 1 ⊗K 2 m 1 n 2 ∗∗ In particular, if A = M (K) then ι ([A M (K)]) = ι ([A ]), which shows the compatibility of 2 n mn 1 ⊗K n m 1 the maps ιm with Brauer equivalence, expressed by the following commutative diagram: ι B(mn,L) mn- H2(G, L×) 6 - ι m

∪ B(m,L) Hence taking the union we obtain a map

ι: Br(L/K) H2(G ,L×) → L which is group morphism by ( ). It is injective by the exactness of ( ). We now show that it is ∗∗ ∗ also surjective. For that it is enough to show that ιn is surjective when n = [L : K]. Consider a 2-cocycle g: G G L×, and let V be an n-dimensional L-vector space with basis e , σ G. For × → σ ∈ σ G, let f(σ) P GLn(L) be the image of the matrix corresponding to the V -automorphism given by e ∈ g(σ, τ)e .∈ Then a computation shows that f is a 1-cocycle and that τ 7→ στ g(σ, τ)= f(σ)σ(f(τ))f(στ)−1 for all σ, τ G ∈ 1 1 Hence f deﬁnes an element of H (G, P GLn(L)) with δ ([f]) = [g], proving that ιn is surjective. The CSA algebra deﬁned by f is the opposite of the usual cross product given by the 2-cocycle g. Finally, for L′ L it is easy to check that ⊃ ι′ Br(L′/K) - H2(G′,L′×) 6 6 inf

∪ ι ∪ Br(L/K) - H2(G, L×) commutes, where the primes denote the corresponding objects of L′ instead of L. Hence, passing to the ≈ 2 × limit, we obtain the desired isomorphism Br(K) H (GK ,Ksep), which is easily shown to be compatible with restriction maps. → 18 Central Simple Algebras and the Brauer group

Remark 6.6 A more detailed analysis shows that if K′ K is a separable ﬁeld extension of degree n, then the connecting maps induce an isomorphism ⊃

1 1 ′ ≈ 2 × 2 ′ × ker H (GK ,PGLn(Ksep)) H (GK′ ,PGLn(Ksep)) ker H (GK ,Ksep) H (GK′ ,Ksep ) → → → between the kernels of the restriction maps, and the right hand side is isomorphic to Br(K′/K) by the last theorem.

Corollary 6.7 Let L K be a Galois extension with G = Gal(L/K). Then we have an isomorphism ⊃ Br(L/K) ≈ H2(G, L×) →

Proof × GL × 1 × Since Lsep = Ksep, (Ksep) = L , G = GK /GL, and H (GL,Lsep) is trivial by Hilbert 90, the inﬂation-restriction exact sequence reads

inf res - 2 × - 2 × - 2 × 0 H (G, L ) H (GK ,Ksep) H (GL,Lsep)

Hence we can identify H2(G, L×) with the kernel of res: Br(K) Br(L), namely with Br(L/K). →

Corollary 6.8 The Brauer group is torsion. Proof With the above notation, we have that H2(G, L×) is killed by G by the restriction-corestriction argument. | |

Corollary 6.9 If n is prime to char K, the n-torsion of Br(K) is given by

Br(K) ≈ H2(G , µ ) n → K n

where µn denotes the GK -module of all n-th roots of unity in Ksep.

Proof We have an exact sequence of GK -modules (the so-called Kummer sequence)

n - - × - × - 1 µn Ksep Ksep 1

1 × where the superscript n denotes “exponentiation by n.” Since H (GK ,Ksep) is trivial by Hilbert 90, we obtain the following fragment of the associated long exact sequence:

n 1 × - 2 - 2 × - 2 × 1= H (GK ,Ksep) H (GK , µn) H (GK ,Ksep) H (GK ,Ksep)

2 2 × Hence we can identify H (GK , µn) with the n-torsion of Br(K)= H (GK ,Ksep).

Theorem 6.10 (Cyclic Algebras) Let L K be a cyclic extension of degree n with G = Gal(L/K). ⊃ 1. Let χ H1(G, Z/n) = Hom(G, Z/n) be a surjective character, viewed as an element of H1(G, Q∈/Z) = Hom(G, Q/Z). Consider the connecting map δ: H1(G, Q/Z) H2(G, Z) associated to the exact sequence of G-modules (with trivial action) →

0 - Z - Q - Q/Z - 0

Then for any a K × = H0(G, L×), we have that the cup product ∈ a δχ H2(G, L×) = Br(L/K) ∪ ∈ equals the class of the opposite of the cyclic algebra (χ,a). 2. A CSA A over K is Brauer equivalent to a cyclic algebra if and only if there exists a cyclic extension of K splitting A. The Brauer group 19

3. In Br(K), we have

[(χ,a)] + [(χ,b)] = [(χ,ab)] and [(χ,a)] + [(ψ,a)] = [(χ + ψ,a)]

for all a,b K × and χ, ψ H1(G, Z/n). ∈ ∈ Proof 1. Let σ be the generator of G such that χ(σ)= 1.¯ By the explicit formulas of the appendix, a δχ is represented by the 2-cocycle ∪ 1 if i + j

1 But that is just the coboundary of the element of H (G, P GLn(L)) given in example 5.5, and the result follows. 2. We have already seen that a cyclic algebra (χ,a) is split by the cyclic extension deﬁned by χ. Conversely, suppose that L splits A, that is, [A] Br(L/K) = H2(G, L×). Let χ Hom(G, Z/n) be a surjective character. Since G is cyclic, the cup∈ product with δχ induces an isomorphism∈ of Tate cohomology groups × K ∪δχ = H0 (G, L×) - H2 (G, L×) = Br(L/K) × T ≈ T NL/KL

In particular, [A]=¯a δχ for some a K ×, hence A is cyclic by the ﬁrst part. ∪ ∈ 3. Follows from the bilinearity of the cup product.

Now we come to two of the most important invariants associated to a CSA.

Definition 6.11 Let A be a CSA over a field K. 1. The period or exponent of A (or its class in Br(K)), denoted by per A or exp D, is the order of [A] in Br(K). 2. The (Schur) index of A (or its class in Br(K)), denoted by ind A, is the smallest degree [L : K] of a separable splitting field L of A. Equivalently, it is the degree of the unique division algebra D over K which is Brauer equivalent to A.

The last equivalence is a consequence of the next theorem and the fact that a degree n division algebra D is split by a maximal subﬁeld of degree n over K.

Theorem 6.12 If D is a division algebra over K and L K is a ﬁnite separable ﬁeld extension splitting D then deg D [L : K]. ⊃ | Proof Let n = [L : K]. By the remark, we have that

[D] Br(L/K) = ker H1(G ,PGL (K )) H1(G ,PGL (K )) ∈ K n sep → L n sep Hence [D] can be represented by a degree n CSA A over K, i.e., A ∼= Mr(D) with deg A = n = r deg D. This shows that deg D n, as required. · |

Let K be a field and let α Br(K). From their definitions, it is easy to check that the period and index have the following properties:∈ 1. For any CSA A over K, ind A deg A, with equality if and only if A is a division algebra. | 2. If L K is any field extension then ind α ind α. ⊃ L | 3. ind(α + β) ind α ind β for any α, β Br(K). | · ∈ 4. per(α + β) per α per β, with equality if per α and per β are relatively prime. | · That they are not unrelated quantities is shown by 20 Central Simple Algebras and the Brauer group

Theorem 6.13 (Period-Index) The period always divides the index, and both integers have the same prime factors. Proof Let K be a field and α Br(K). If n = ind α then α is split by a separable field extension L K of degree n, i.e., α Br(L/K∈ ). But then nα = 0 by the restriction-corestriction argument (see appendix).⊃ Since per α is the∈ order of α Br(K), we must have per α n. ∈ | Now let L K be a Galois splitting field of α, and suppose that p is a prime number that divides [L : K] but not⊃ per α. Let H be the p-Sylow subgroup of G = Gal(L/K) and let M = LH . Then, since [L : M]= H is a power of p, by the restriction-corestriction argument res: Br(M) Br(L) is injective | | → on the prime-to-p part, hence αM = 0 since αL = 0 and gcd(p, per α) = 1. In other words, M is a splitting field of α, and therefore ind α [M : K] = [G : H], i.e., ind α is not divisible by p. Together with the first part, this shows that per α| and ind α have the same prime factors.

We can also show that Lemma 6.14 Let D a division algebra over a ﬁeld K and let L be an extension of K with [L : K] prime to ind D. Then DL is a division algebra over L. Proof Let n = ind D. We know that ind D n and we have to show that equality holds. But L | if M L is a ﬁeld extension splitting DL with [M : L] = ind DL, then M also splits D, hence n = ind⊃D [M : K] = ind D [L : K]. But n is prime to [L : K], hence n ind D , as required. | L · | L

Theorem 6.15 (Sylow factors) Let D be a division algebra over a ﬁeld K and let

e1 er ind D = p1 ...pr be the canonical decomposition of ind D into powers of distinct primes. Then

D = D D ∼ 1 ⊗K ···⊗K r

ei where Di are division algebras over K with ind Di = pi . Proof We may write [D] Br(K) into its primary components, [D] = [D ]+[D ]+ +[D ] for division ∈ 1 2 · · · r algebras Di with per[Di] (and hence ind Di) a power of pi. Hence everything follows if we can show that D D is a division algebra. For that, we show that if D and D are two division algebras 1 ⊗K ···⊗K r 1 2 over K with m = ind D1 relatively prime with n = ind D2 then D1 K D2 is also a division algebra, that is, ind D D = mn. Let L and L be separable splitting ﬁelds⊗ of D and D with [L : K]= m and 1 ⊗K 2 1 2 1 2 1 [L2 : K]= n. Then the compositum L1L2 splits D1 K D2, hence ind D1 K D2 mn = [L1L2 : K]. On the other hand, we have that ind(D D ) = ind(⊗D ) = ind D = n⊗by the| previous lemma, hence 1 ⊗K 2 L1 2 L1 2 n = ind(D1 K D2)L1 ind D1 K D2. Similarly, m divides ind D1 K D2, and since gcd(m,n) = 1, this proves that ⊗mn divides| ind D ⊗ D . ⊗ 1 ⊗K 2 Chapter 2

BrauergroupofLocalFields

A local field K is a finite extension of either Qp or Fp((t)). The discrete valuations of Qp and Fp((t)) extend uniquely to K, turning it into a complete discretely valued field. A local field arises naturally as the completion of a global field (i.e. a finite extension of either Q or Fp(t)) with respect to its maximal ideals. The main result of this chapter is the following computation: Theorem 0.1 (Brauer group of a local field) Let K be a local field. We have an isomorphism

≈ 2 × - invK : H (GK ,Ksep) Q/Z

This isomorphism takes the class of the cyclic algebra (χ, π ) of degree n to the element 1 mod Z. K − n Here χ denotes the character given by χ(ΦL/K )= 1¯, where ΦL/K is the Frobenius automorphism of the degree n unramiﬁed extension L K, and π is a uniformiser of K. ⊃ K 1 Notation Throughout this chapter, we adopt the following notations and conventions. For any ﬁeld K we denote by df Ksep = separable closure of K df GK = Gal(Ksep/K) = absolute Galois group of K

ab df [GK :GK ] K = Ksep = maximal abelian extension of K

= compositum of all finite abelian extensions of K inside Ksep ab ab GK = Gal(K /K) Now let K be a local field with residue field k and normalised valuation v: K Z . We write → ∪ {∞} df πK = uniformiser of K (that is, an element of valuation 1) O =df ring of integers of v = x K v(x) 0 K { ∈ | ≥ } U =df group of units of O = x O v(x)=0 K K { ∈ K | } U (i) =1+(df πi ) = closed ball x K v(x 1) i centred at 1 K K { ∈ | − ≥ } df Knr = maximal unramified extension of K in Ksep

= compositum of all ﬁnite unramiﬁed extensions of K inside Ksep Gnr =df Gal(K /K) G = Zˆ K nr ≈ k df Φ = Frobenius automorphism of K K K nr ⊃ where df Zˆ = lim Z/(n)= Zp ←−n∈N Yp and p runs over all prime integers. The last isomorphism follows from the Chinese Remainder Theorem nr ˆ (check!). The Frobenius map ΦK is a topological generator of GK , corresponding to the element 1 Z under the above isomorphism. ∈ 22 Brauer group of Local Fields

2 Unramified Cohomology In this section, L K will be a finite unramified extension with G = Gal(L/K), and l k will be the corresponding extension⊃ of residue fields. Recall that we have a canonical isomorphism G ⊃= Gal(L/K) Gal(l/k), and so G is cyclic. ≈ We now compute H2(G, L×). The starting point is the exact sequence of G-modules

v 0 - U - L× - Z - 0 ( ) L †

where v denotes the normalised valuation of L. We need to compute the cohomology of UL and Z. To compute the cohomology of Z, we use the exact sequence of G-modules (with trivial G-action)

0 Z Q Q/Z 0 → → → → r Since HT (G, Q) is torsion and multiplication by any non-zero integer is an automorphism of Q, we r r r−1 conclude that HT (G, Q) = 0 for all r. Hence we have that HT (G, Z)= HT (G, Q/Z) for all r. Next we compute the cohomology of UL. Theorem 2.1 For all r we have that r HT (G, UL)=0

Proof We have exact sequences of G-modules (via the isomorphism G Gal(l/k)) ≈ 0 U (1) U l× 0 → L → L → →

0 U (r+1) U (r) l+ 0 → L → L → → r + + The group HT (G, l ) is trivial for all r since l is an induced module by the normal basis theorem r × (see appendix). We now show that the group HT (G, l ) is also trivial for all r. By the periodicity of cohomology of cyclic groups, it is enough to prove that for r = 0 and r = 1. The case r = 1 is just × × Hilbert 90, while the case r = 0 follows from the surjectivity of the norm map Nl/k : l k : if k = Fq n−1 → and b is a generator of the group l× then N (b)= b1+q+···+q has order q 1 and hence is a generator l/k − of k×. Hence, from the long exact sequences associated to the two short ones above, we conclude that r (i+1) r (i) r HT (G, UL ) = HT (G, UL ) for all r and all i 0. Now to show that HT (G, UL) is trivial, again r ≥ by periodicity we may assume r > 0. Let f: G UL be an r-cocycle and denote by d the (r 1)- r (1) r → − th coboundary map. Since H (G, UL ) = H (G, UL), f diﬀers by a coboundary from an r-cocycle r (1) r−1 −1 f1: G UL , i.e., there exists g0: G UL such that f1 = f d(g0) . Proceeding in this manner, we → r−1 → (i) · −1 inductively construct functions gi: G UL such that f d(g0g1 ...gi) is an r-cocycle with values (i+1) → · r−1 in UL . Then the product g0g1 ...gi converges to a function g: G UL and we have that dg = f, r → that is, the class [f] is trivial in H (G, UL).

The case r = 0 is of special interest, since it gives a proof of Corollary 2.2 (Norm groups of unramiﬁed extensions) Let L K be the unramiﬁed extension ⊃ of degree n and let π K be a common uniformiser. Then the norm map NL/K: UL UK is surjective and hence ∈ → N L× = πnZ U L/K · K

Back to the computation of H2(G, L×). From ( ) and the fact that U has trivial cohomology, we † L conclude that the valuation v induces an isomorphism H2(G, L×) = H2(G, Z). On the other hand, we have another isomorphism given by connecting map δ: H1(G, Q/Z) ≈ H2(G, Z). Putting everything together, we obtain a canonical isomorphism, called invariant map, →

1 ≈ Z inv : H2(G, L×) - |G| L/K Z Unramiﬁed Cohomology 23

given by the composition

1 v δ f7→f(Φ) Z H2(G, L×) - H2(G, Z) H1(G, Q/Z) = Hom(G, Q/Z) - |G| ≈ ≈ Z

where we write Φ G for the Frobenius automorphism. ∈ If M L is another unramiﬁed extension with H = Gal(M/K) then following the isomorphisms above we obtain⊃ a commutative diagram

1 invM/K Z H2(H,M ×) - |H| Z 6 ≈ 6 inf ∪ ∪ 1 invL/K Z H2(G, L×) - |G| Z ≈ where the left vertical arrow is given by inflation and the right vertical one is the inclusion map. Hence the invariant maps for the various unramified extensions of K fit together into a single invariant map

≈ 2 nr × - invK : H (GK ,Knr) Q/Z

nr Here GK denotes the Galois group of the maximal unramified extension of K. Theorem 2.3 (Functorial property of the invariant map) Let K′ K be an arbitrary (possibly ramified) finite extension of local fields. We have a commutative diagram ⊃

2 nr ′× invK′ - H (GK′ ,K nr) Q/Z 6 ≈ 6 res [K′ : K]

inv 2 nr × K - H (GK ,Knr) Q/Z ≈ where the left vertical map is restriction and the right vertical one is multiplication by [K′ : K]. Hence if K′ K is an arbitrary finite Galois extension with G = Gal(K′/K) then H2(G, K′×) contains a cyclic group⊃ of order G . | | Proof ′ ′ 2 First we note that Knr = Knr K so that the restriction map on H is well-defined. Let e and f be the ramification and inertia degrees· of K′ K and denote by v and v′ the normalised valuations ′ ′ ⊃ f ′ of K and K respectively so that v K = e v. Observe that ΦK Knr =ΦK . Hence, from the definition of the invariant map, we obtain a commutative| · diagram |

′ 2 nr ′× v- 2 nr - nr - H (GK′ ,K nr) H (GK′ , Z) Homct(GK′ , Q/Z) Q/Z 6 6 6 6 res e res e res ef · · v 2 nr × - 2 nr - nr - H (GK ,Knr) H (GK , Z) Homct(GK , Q/Z) Q/Z

Since ef = [K′ : K], the first result follows. To show the second result, assume that K′ K is finite Galois with G = Gal(K′/K). Using the inflation-restriction sequence and Hilbert 90, we conclude⊃ that the inflation maps

2 nr × 2 × 2 nr ′× 2 × ( inf: H (G ,K ) ֒ H (G ,K ) and inf: H (G ′ ,K ) ֒ H (G ′ ,K K nr → K sep K nr → K sep 24 Brauer group of Local Fields

2 nr × 2 nr ′× are injective. Identifying H (GK ,Knr) and H (GK′ ,K nr) with Q/Z via invK and invK′ and using the result just proven, we obtain a commutative diagram inf res - 2 ′× - 2 × - 2 × 0 H (G, K ) H (GK ,Ksep) H (GK′ ,Ksep) 6 6 inf inf

∪ res ∪ 2 nr × - 2 nr ′× H (GK ,Knr) H (GK′ ,K nr)

inv w winv ′ Kw w K w w w ′ w w [K : K] - w Qw/Z Qw/Z 1 Z 2 ′× [K′:K] We conclude that H (G, K ) contains a subgroup isomorphic to Z , that is, a cyclic group of order G = [K′ : K]. | | 1 Z 2 ′× [K′:K] 2 ′× In order to show that H (G, K ) actually equals Z , we shall bound the order of H (G, K ) from above using a counting argument. Since G is solvable, it will be enough to do that assuming G cyclic. This is done in the next section. 3 Brauer group of a Local Field We begin by introducing a very useful tool in the cohomology of cyclic groups that will help simplify our counting argument. Definition 3.1 Let G be a cyclic group and M be a G-module whose Tate cohomology groups are all finite. We define its Herbrand quotient as

H0 (G, M) h(G, M)= | T | H1 (G, M) | T | The Herbrand quotient plays the same role as the Euler characteristic in Topology. We have two main computational lemmas: Lemma 3.2 (Multiplicativity) Let G be a cyclic group and consider an exact sequence of G-modules 0 M ′ M M ′′ 0 → → → → If two of the Herbrand quotients h(G, M), h(G, M ′), h(G, M ′′) are deﬁned (i.e. have ﬁnite Tate cohomology groups) then so is the third and

h(G, M)= h(G, M ′) h(G, M ′′) ·

Proof From the periodicity of cohomology of cyclic groups, the long exact sequence associated to the above short one becomes an “exact hexagon” 0 0 0 ′ f- 0 g- 0 ′′ HT (G, M ) HT (G, M) HT (G, M ) 6 δ1 δ0

1 1 ? 1 ′′ g 1 f 1 ′ HT (G, M ) HT (G, M) HT (G, M ) The ﬁrst result follows directly from from this hexagon. The second result also follows from this hexagon by elementary counting, as one has H0 (G, M ′) = ker f 0 ker g0 H1 (G, M ′) = ker f 1 ker g1 | T | | | · | | | T | | | · | | H0 (G, M) = ker g0 ker δ0 H1 (G, M) = ker g1 ker δ1 | T | | | · | | | T | | | · | | H0 (G, M ′′) = ker δ0 ker f 1 H1 (G, M ′′) = ker δ1 ker f 0 | T | | | · | | | T | | | · | | Brauer group of a Local Field 25

Lemma 3.3 (Finite Index Invariance) Let G be a cyclic group, M be a G-module and M ′ be a G-submodule of finite index. Then h(G, M ′) is defined if and only if h(G, M) is defined, in which case h(G, M ′)= h(G, M). Proof By the last lemma, it suffices to show that if M is a finite group then h(G, M) = 1. Let σ be a generator of G. Since M is a finite group one has that H0 (G, M) = M G / N (M) and | T | | | | G | H1 (G, M) = ker N / (σ 1) M . But M = ker N N (M) and similarly (since M G is the | T | | G| | − · | | | | G| · | G | kernel of multiplication by σ 1) one has that M = M G (σ 1) M , and the result follows. − | | | | · | − · | Let L K be a cyclic extension of local fields with Galois group G of order n. We apply the above to the exact⊃ sequence of G-modules

0 U L× Z 0 → L → → → induced by the valuation of L. It is easy to compute h(G, Z)= n and we have

Theorem 3.4 With the above notation and hypotheses, h(G, UL)=1.

Proof By the previous lemma, it is enough to show that UL contains an induced G-submodule of finite index. First assume that char K = 0 and let π be a uniformiser of K. But we have an isomorphism of (i) mi G-modules UL ∼= L for i sufficiently large, given by the log map. On the other hand, for j sufficiently mi j (i) mi j large L π OL ∼= OL. Since UL/UL and L/π OL are finite, it is enough to show that OL contains an induced⊃ module of finite index. Now let ω ,...,ω be a normal basis of L K; multiplying by a 1 n ⊃ convenient power of π we may assume that ω1,...,ωn OL. Since OL is a finite OK -module, we have that the induced module M = O ω + + O ω has∈ finite index in O . K 1 · · · K n L Now we sketch a proof that works even if char K = 0. Let M be as above. Multiplying the ωi by a sufficiently large power of π we may assume that M M6 πM. Then we may consider the submodule of · ⊂(i) i finite index V =1+M of UL and the filtration given by V =1+π M for i 0. It is easy to show that (i) (i+1) ≥ we have an isomorphism of G-modules V /V ∼= M/πM, which has trivial cohomology since the latter is induced. As in the proof of theorem 2.1, this implies that V itself has trivial cohomology.

Hence we get h(G, L×) = h(G, U ) h(G, Z) = n. Since H1(G, L×) is trivial by Hilbert 90, we L · conclude that H2(G, L×) = n when G is cyclic. In general, for an arbitrary Galois extension we have | | Theorem 3.5 Let L K be an arbitrary finite Galois extension of local fields with G = Gal(L/K). ⊃ Then the group H2(G, L×) is cyclic of order G . | | Proof Since H2(G, L×) contains a cyclic group of order G by theorem 2.3, it is enough to show that | | the order of H2(G, L×) divides G . For that, we use the special cyclic case above together with the fact that G is solvable. Alternatively,| | one may use the fact that res: H2(G, L×) H2(G ,L×) defines an → p injection on the p-primary components, where Gp is any p-Sylow subgroup of G (see appendix), and we may work with the solvable group Gp instead of G. The proof is by induction on G . We already know the result for G cyclic. In general, let H ⊳ G be a normal subgroup such that H| is| cyclic and non-trivial. By Hilbert 90, we have an exact inflation- restriction sequence 0 H2(G/H, (LH )×) H2(G, L×) H2(H,L×) → → → Hence the order of H2(G, L×) divides the product of the orders of H2(H,L×) and H2(G/H, (LH )×), which in turn divides H G/H = G by induction hypothesis, and we are done. | | · | | | | As a corollary, we obtain the following result that allows us to extend the invariant map of last 2 × section to the whole group H (GK ,Ksep). This map plays an important role in the study of division algebras over a local field. Theorem 3.6 (Brauer group of a local field) Let K be a local field. We have an isomorphism

≈ 2 × - invK : H (GK ,Ksep) Q/Z

2 nr × ≈ 2 × obtained by composing the inﬂation map inf: H (GK ,Knr) H (GK ,Ksep) and the invariant map ≈ → inv : H2(Gnr,K × ) Q/Z of last section. K K nr → 26 Brauer group of Local Fields

Proof 2 nr × ֒ (By Hilbert 90 and the inﬂation-restriction sequence, we have an inclusion inf: H (GK ,Knr 2 × → H (GK ,Ksep), which we now show to be also surjective. Since

2 × 2 ′ ′× H (GK ,Ksep) = lim H Gal(K /K),K , K−→′ finite Galois over K given any γ H2(G ,K × ) we can find a finite Galois extension K′ such that ∈ K sep res 2 ′ ′× 2 × - 2 × γ H Gal(K /K),K = ker H (G ,K ) H (G ′ ,K ) ∈ K sep K sep (by Hilbert 90 and the inflation-restriction sequence the inflation maps in the above limit are injective, 2 ′ ′× 2 × so we may view H Gal(K /K),K as a subgroup of H (GK ,Ksep) and we write γ also for the corresponding element in this subgroup). Now by theorem 2.3 and the above theorem, we have that inf: H2(Gnr,K × ) ֒ H2(G ,K × ) allows us to make the identification K nr → K sep res res 2 nr × - 2 nr ′× 2 × - 2 × ker H (GK ,Knr) H (GK′ ,K nr) = ker H (GK ,Ksep) H (GK′ ,Ksep) (see the second diagram of the proof of theorem 2.3). Hence γ belongs to this kernel and a fortiori to 2 nr × (the inflation of) H (GK ,Knr).

Corollary 3.7 All division algebras over a local ﬁeld K are cyclic. More precisely, if χ H1(G , Z/n) ∈ K is the character given by χ(ΦK )= 1¯, then 1 inv (χ, π)= mod Z K −n

Proof Since Br(K) = Br(Knr/K), every division algebra is split by some ﬁnite unramiﬁed extension L K. But every such extension is cyclic, and therefore the division algebra is cyclic. For the formula ⊃ 1 above, observe that the class of (χ, πK ) in Br(K) is just πK δχ, and invL/K (πK δχ)= n mod Z as can be checked by an explicit calculation. − ∪ ∪

4 Local Reciprocity The computation of the Brauer group of a local ﬁeld K gives also a very simple and nice description of the abelian extensions of K as a byproduct, which will be needed in the next chapter.

4.1 Statements of the main theorem

Theorem 4.1.1 (Local Artin Reciprocity) Let K be a local ﬁeld. There exists a unique group morphism, called local Artin map, θ : K × Gab K → K × such that the following holds: for any ﬁnite abelian extension L K, the map θL/K : K Gal(L/K) (also referred to as local Artin map) given by the composition ⊃ →

θ canonical × K- ab -- K GK Gal(L/K) satisﬁes:

× 1. θL/K is surjective with kernel given by the norm group NL/KL . Thus we have an induced isomorphism (which we still denote by θL/K )

K × θ : Gal(L/K) L/K × NL/K(L ) ≈

2. if L K is unramiﬁed, ΦL/K Gal(L/K) denotes the corresponding Frobenius map, and v: K ×⊃ Z denotes the normalised∈ valuation of K, then for all a K × → ∈

v(a) θL/K(a)=ΦL/K Local Reciprocity 27

√ √ Example 4.1.2 Consider the Galois extension M = Q3( 2, 3) of Q3 with Galois group Gal(M/Q3) ∼= Z/2 Z/2, generated by automorphisms σ and τ given by × σ(√2) = √2 τ(√2) = √2 − σ(√3) = √3 τ(√3) = √3 − The lattice of subﬁelds is

M = Q3(√2, √3) τ στ σ

L0 = Q3(√2) L1 = Q3(√3) L2 = Q3(√6) L2 σ L = τ | 0 τ σ L2 | |L1 | K = Q3

Now we now identify the corresponding norm subgroups in Q× = 3Z 1 U (1). Observe that 3 Q3 (1) × {± }× (1) since the indices of these subgroups divide 4 and U = Z3 is 2-divisible, all of them contain U . Q3 ∼ Q3 × 2Z × Since L Q is unramiﬁed we know that N Q (L )=3 UQ . Moreover 3 N Q (L ) and 0 ⊃ 3 L0/ 3 0 × 3 − ∈ L1/ 3 1 6 N (L×), thus 3 N (L×) since 2 U (1). Also since M is the compositum of L and L2/Q3 2 L2/Q3 2 Q3 0 − ∈ ∈× × − ∈ × L2 we have that NM/Q3 (M )= NL0/Q3 (L0 ) NL2/Q3 (L2 ). Putting everything together, we obtain the × ∩ following lattice of subgroups of Q3 , drawn upside down:

× 2Z (1) NM/Q (M )=3 U 3 × Q3

× 2Z (1) × Z (1) × Z (1) NL /Q (L )=3 1 U NL /Q (L ) = ( 3) U NL /Q (L )=3 U 0 3 0 × {± }× Q3 1 3 1 − × Q3 2 3 2 × Q3

Q× =3Z 1 U (1) 3 × {± }× Q3 Finally, we have that

× × θ Q 1 N Q (L ) =1 θ Q 1 N Q (M ) = τ L0/ 3 − · L0/ 3 0 M/ 3 − · M/ 3 × ×  θ Q 3 N Q (L ) =1  θ Q 3 N Q (M ) = σ  L1/ 3 − · L1/ 3 1 ⇒  M/ 3 − · M/ 3  ×  × θ Q 3 N Q (L ) =1 θ Q 3 N Q (M ) = στ L2/ 3 · L2/ 3 0 M/ 3 · M/ 3  

which completely determines θM/Q3 .

4.2 Tate-Nakayama theorem Let G be a ﬁnite group. In this section, we prove a purely group theoretic result giving an isomorphism

CG Gab ≈ NG(C)

G where C is a G-module satisfying some conditions and NG: C C denotes the norm map of C (see appendix). The methods of this section are inspired in compu→tations of Algebraic Topology, with the ab starting point being the well-known relation H1(X, Z) = π1(X) between the ﬁrst singular homology ab group of a topological space X and the maximal abelian quotient π1(X) of the fundamental group of X. Since Galois extensions are algebraic analogs of covering spaces in Topology, this turns out to be a quite natural point of view (if, of course, you’ve studied Algebraic Topology before, but don’t worry if you haven’t, the proofs below are purely algebraic, but it’s a good idea to eventually look at the source of inspiration for them). 28 Brauer group of Local Fields

Theorem 4.2.1 (Twin number vanishing criterion) Let G be a ﬁnite group and M be a G-module. If there are two consecutive numbers i and i +1 for which

Hi (H,M)= Hi+1(H,M)=0 for all subgroups H G T T ≤ then Hr (G, M)=0 for all r Z. T ∈ Proof First observe that by dimension shifting (see appendix) it is enough to show the “weaker” conclusion Hr (G, M) = 0 for all r 1. In fact, writing an exact sequence of G-modules T ≥ 0 N P M 0 → → → → for some induced G-module P (which is thus also induced as an H-module for all H G) we obtain an j+1 j ≤ r isomorphism HT (H,N)= HT (H,M) for all H G and j Z. Hence if we know that HT (H,M)=0 r ≤ ∈ for all r 1 then we know that HT (H,N) = 0 for all r 2, and applying the “weak twin number ≥ 1 ≥ 0 criterion” to N in place of M we conclude that HT (H,N) = 0 as well, that is HT (H,M) = 0, so that the conclusion of the weak criterion holds also for r 0. Proceeding inductively in this manner, we extend the result to all r Z. ≥ ∈ A similar proof using dimension shifting (check!) also allows us to assume that i = 1 (or any other ﬁxed number we deem convenient). Hence from now on we assume that H1(H,M)= H2(H,M)=0 for all H G and prove that these conditions imply that Hr(G, M) = 0 for all r 1. ≤ ≥ Since Hr(G, M) is a torsion abelian group, it is enough to show that its p-primary component vanishes for all prime numbers p. Let Gp be a p-Sylow subgroup of G. A restriction-corestriction r r argument shows that res: H (G, M) H (Gp,M) is injective on p-primary components (see appendix), r → thus it is enough to show that H (Gp,M)=0. Hence we may assume that G is solvable and proceed by induction on the order of G. If G is cyclic, the theorem follows from the periodicity of cohomology. Now let H ⊳ G be a proper normal subgroup such that G/H is cyclic. By induction Hr(H,M) = 0 for all r 1 and hence we have an exact inﬂation-restriction sequence ≥

0 Hr(G/H,M H ) Hr(G, M) Hr(H,M)=0 → → → for all r 1. Therefore Hr(G/H,M H ) = Hr(G, M) for all r 1 and in particular H1(G, M) = H2(G, M)≥ = 0 implies that H1(G/H,M H ) = H2(G/H,M H ) = 0.≥ But G/H is cyclic, so periodicity yields Hr(G/H,M H ) = 0 for all r 1, and hence Hr(G, M) = 0 for all r 1 too. ≥ ≥ Now we can prove the main result of this section. Theorem 4.2.2 (Tate-Nakayama) Let G be a ﬁnite group and let C be a G-module such that for all subgroups H G ≤ 1. H1(H, C)=0 2. H2(H, C) is cyclic of order H | | Let γ be a generator of H2(G, C). Then for all r Z the cup product with γ gives an isomorphism ∈

∪γ Hr (G, Z) - Hr+2(G, C) T ≈ T

Observe that if γ H2(G, C) is a generator then res(γ) H2(H, C) is also a generator since cor res(γ) = [G : H] γ∈has order H , hence res(γ) must have order∈ H as well in view of 2. ◦ · | | | | Proof The key idea of the proof is to apply a “double dimension shifting” (wow!) and for that we construct a cohomologically trivial G-module C(γ) ﬁtting into an exact sequence

0 C C(γ) I 0 ( ) → → → G → ∗

Here IG = σ 1 σ G is the kernel of the augmentation map Z[G] Z (see appendix), so that we have an exacth − sequence| ∈ ofiG-modules →

0 I Z[G] Z 0 ( ) → G → → → ∗∗ Local Reciprocity 29

Once C(γ) is constructed, the proof of the theorem follows easily: from ( ), using the fact that Z[G] r ∗∗ r+1 has trivial cohomology, we conclude that the connecting map δa: HT (G, Z) HT (G, IG) is an isomor- r+1 ≈ r+2 phism; similarly, from ( ), we have that the connecting map δc: HT (G, IG) HT (G, C) is also an ∗ r ≈ r+2 isomorphism. The composition δc δa gives the desired isomorphism HT (G, Z) HT (G, C), which equals to the cup product with γ, as◦ we later show. ≈ Let c be a 2-cocycle representing γ. Since we wish H2(G, C(γ)) to vanish, the idea is to construct C(γ) so that c becomes a coboundary in C(γ). Take C(γ) to be the direct sum of C with the free abelian group with basis x , σ G, σ = 1: σ ∈ 6 C(γ) =df C Zx ⊕ σ Mσ∈G σ6=1 We extend the G-action from C to C(γ) in such a way that c becomes the coboundary of σ x : 7→ σ σ x = x x + c(σ, τ) · τ στ − σ where we interpret “x1” to be c(1, 1). The 2-cocycle relation then guarantees that 1 xτ = xτ and (ρσ) x = ρ (σ x ) hold for all ρ, σ, τ G (check!), turning C(γ) into a G-module containing· C as a · τ · · τ ∈ G-submodule. Finally the map C(γ) IG in ( ) is given by xσ σ 1 for σ = 1, and is identically zero on C. Another easy check shows→ that the latter∗ map preserves7→ the−G-action.6 In order to show that C(γ) is cohomologically trivial we apply the twin number vanishing criterion: we need to verify that H1(H, C(γ)) = H2(H, C(γ)) = 0 for all subgroups H G. Here the hypotheses 1 and 2 of the theorem come into play. First observe that from ( ) and the≤ explicit description of the connecting map in terms of the standard resolution (see appendix)∗∗ we obtain 1 0 (i) H (H, IG)= HT (H, Z) is cyclic of order H with generator given by the class [f] of the 1-cocycle f(σ)= σ 1, σ H. | | 2 − 1∈ (ii) H (H, IG)= H (H, Z) = Hom(H, Z)=0 From ( ) we have an exact sequence ∗ 1 - 1 - 1 0= H (H, C) H (H, C(γ)) H (H, IG) δ- 2 - 2 - 2 H (H, C) H (H, C(γ)) H (H, IG)=0

Hence everything falls through if we can show that the connecting map δ is an isomorphism. At least we 1 2 know that both H (H, IG) and H (H, C) are cyclic groups of order H , so it is enough to show that δ is surjective. We show by explicit computation that δ([f]) = res(γ) where| | [f] H1(H, I ) is as in (i). First ∈ G we lift f to the function f˜: H C(γ) given by f˜(σ)= x . Now (df˜)(σ, τ)= σ x x + x = c(σ, τ) → σ · τ − στ σ for all σ, τ G, where d denotes the coboundary map, hence δ([f]) = [df˜] = [c] = res(γ), as required. ∈ r ≈ r+1 Finally, we verify that the composition δc δa of the two connecting maps δa: HT (G, Z) HT (G, IG) r+1 ≈ r+2 ◦ 0 → and δc: HT (G, IG) HT (G, C) is indeed the cup product with γ. Denote by µ HT (G, Z)= Z/ G → 1 ∈ | | the generator 1 mod G and by φ = [f]= δa(µ) H (G, IG) where f is as in (i). By the compatibility of cup products with| the| connecting maps (see appendix)∈ we obtain, for all α Hr (G, Z), ∈ T α γ = α δ (φ) = ( 1)r δ (α φ) = ( 1)r δ α δ (µ) ∪ ∪ c − · c ∪ − · c ∪ a = ( 1)r δ ( 1)r δ (α µ) = ( 1)r δ ( 1)r δ (α) = δ δ (α) − · c − · a ∪ − · c − · a c ◦ a since µ is the identity on Hr (G, Z). − ∪ T −2 −1 2 ab 2 Recall that (see appendix) HT (G, Z) = HT (G, IG) = IG/IG and that G = IG/IG via the isomorphism σ [G : G] (σ 1) I2 . As a consequence we obtain · 7→ − · G Corollary 4.2.3 (“Abstract” Reciprocity) With the notation and hypotheses of the Tate-Nakayama theorem, we have an isomorphism

G C 0 −2 ab = HT (G, C) HT (G, Z)= G NG(C) ≈ 30 Brauer group of Local Fields

4.3 Reciprocity law We wish to apply the Tate-Nakayama’s theorem to G = Gal(L/K) and C = L× where L K is a finite Galois extension of local fields. Hence we need to verify that, for all subgroups H G, ⊃ ≤ 1. H1(H,L×)=0; 2. H2(H,L×) is cyclic of order H . | | The first condition is just Hilbert Satz 90 (see appendix), and the second one follows from the computation of Br(K). We can now make the following

Definition 4.3.1 For any finite abelian extension L K of local fields, we define the local reciprocity × ⊃ × ։ × × map θL/K : K Gal(L/K) as the composition of the natural projection map K K /NL/KL with the inverse of the→ Tate-Nakayama isomorphism

K × Gal(L/K) ≈ × → NL/K L

2 × given by the cup product with the unique element γ H (Gal(L/K),L ) such that invL/K(γ) = 1 ∈ [L:K] mod Z. Such element γ is called a fundamental class.

To prove that the above isomorphism satisﬁes the two properties of theorem 4.1.1, we describe θL/K in terms of characters. Let L K be as above with G = Gal(L/K). Recall that we have two exact sequences of G-modules ⊃ 0 Z Q Q/Z 0 ( ) → → → → ∗ 0 I Z[G] Z 0 ( ) → G → → → ∗∗ where Q and Z[G] have trivial cohomology, hence the connecting maps

δ: H1(G, Q/Z) ≈ H2(G, Z) → δ : H−2(G, Z) ≈ H−1(G, I ) a T → T G

ab −1 are isomorphisms. Besides we have a natural isomorphism G = HT (G, IG) given by

ab ≈ −1 IG G HT (G, IG)= 2 → IG σ [G : G] (σ 1) I2 · 7→ − · G

Theorem 4.3.2 (Local Reciprocity Revisited) With the above notation, for any a K × and any character χ H1(G, Q/Z) = Hom(G, Q/Z) one has ∈ ∈

χ θ (a) = inv (¯a δχ) L/K K ∪

0 × × × where a¯ is the image of a in HT (G, L )= K /NL/KL . In other words, the cup product gives a perfect pairing

inf H0 (G, L×) H2 (G, Z) ∪- H2 (G, L×) ⊂ - H2 (G ,K × ) T ⊗ T T T K sep 6

id δ invK ⊗ ≈ ≈ ? ≈

× 1 ? K |G| Z Hom(G, Q/Z) - ⊂ - Q/Z × NL/KL ⊗ Z Local Reciprocity 31

Proof Let γ H2(G, L×) be the fundamental class and n = G . Given a K × and χ H1(G, Q/Z), df ∈ i | | ∈ 1 ∈ write σ = θL/K (a) G and χ(σ) = n mod Z with 0 i

id ⊗δ ∪ H0 (G, L×) H1 (G, Q/Z) - H0 (G, L×) H2 (G, Z) - H2 (G, L×) T ⊗ T ≈ T ⊗ T T

1 × is perfect. In fact, if χ HT (G, Q/Z) is such thata ¯ δχ =0 χ(θL/K (a)) = 0 for all a K then × ։∈ ∪ ⇐⇒ × ∈ χ = 0 since θL/K: K Gal(L/K) is surjective. On the other hand, if a K is such thata ¯ δχ = 1 ∈ ∪ × 0 χ(θL/K (a)) = 0 for all χ HT (G, Q/Z) then clearly θL/K(a)=0 a ker θL/K = NL/KL and⇐⇒ thusa ¯ = 0, showing that the∈ left kernel of this pairing is also trivial. ⇐⇒ ∈

Lemma 4.3.3 Let G be a ﬁnite group and A and B be G-modules. Let f: G B be a 1-cocycle. → N 1. for a ker(A -G A), the cup product of [a] H−1(G, A) and [f] H1 (G, B) is given by ∈ ∈ T ∈ T [a] [f]= σ(a) f(σ) H0 (G, A B) ∪ − ⊗ ∈ T ⊗ hσX∈G i Here brackets denote the corresponding cohomology classes. −2 ab −2 2. for σ G denote by σ¯ HT (G, Z) the image of σ under the isomorphism G HT (G, Z). Then ∈ ∈ ≈ σ¯ [f] = [f(σ)] H−1(G, B) ∪ ∈ T Proof We use dimension shifting. Write an exact sequence of G-modules

0 B B′ B′′ 0 → → → → with B′ induced, and such that it is split as a sequence of abelian groups, so that

0 A B A B′ A B′′ 0 → ⊗ → ⊗ → ⊗ → is still exact (check the appendix for more details). Then the connecting map δ: H0 (G, B′′) ≈ H1 (G, B) T → T is an isomorphism and thus we may write [f]= δ[b′′] for some b′′ B′′G and ∈ [a] [f] = [a] δ[b′′]= δ[a b′′] ∪ ∪ − ⊗ −1 ′′ 0 On the other hand, the connecting map δ: HT (G, A B ) HT (G, A B) is induced by the norm. There exists a pre-image b′ B′ of b′′ such that f(σ)⊗ = σ(b′→) b′ B for⊗ all σ G. Since a b′ is a pre-image of a b′′ under A∈ B′ A B′′, we conclude that−δ[a ∈ b′′] is represented∈ by ⊗ ⊗ ⊗ → ⊗ ⊗ N (a b′)= σa σ(b′)= σa f(σ)+ σa b′ G ⊗ ⊗ ⊗ ⊗ σX∈G σX∈G σX∈G = σa f(σ)+ N (a) b′ = σa f(σ) ⊗ G ⊗ ⊗ σX∈G σX∈G

To show 2, observe that tensoring the augmentation sequence ( ) with B we obtain an exact sequence ∗∗ 0 I B Z[G] B Z B = B 0 → G ⊗ → ⊗ → ⊗ → 32 Brauer group of Local Fields

Z since Z is a free Z-module and hence Tor1 (B, Z) = 0. Since Z[G] B is induced (see appendix), we have −1 ≈ 0 ⊗ that the connecting map δa: HT (G, B) HT (G, IG B) is an isomorphism, hence it suﬃces to show → ⊗ −1 that δa(¯σ [f]) = δa[f(σ)]. Since δa(¯σ [f]) = δa(¯σ) [f] and δa(¯σ) = [σ 1] HT (G, IG), applying 1 we obtain∪ ∪ ∪ − ∈ δ (¯σ [f]) = [σ 1] [f]= τ(σ 1) f(τ) a ∪ − ∪ − − ⊗ hτX∈G i = τ f(τ) τσ f(τ) = τσ f(τσ) τσ f(τ) ⊗ − ⊗ ⊗ − ⊗ hτX∈G τX∈G i hτX∈G τX∈G i = τσ τf(σ) H0 (G, I B) ⊗ ∈ T G ⊗ hτX∈G i

−1 ≈ 0 On the other hand, since δa: HT (G, B) HT (G, IG B) is induced by the norm and 1 f(σ) Z[G] B is a pre-image of f(σ) B = Z B we→ have that ⊗ ⊗ ∈ ⊗ ∈ ⊗ δ [f(σ)] = N 1 f(σ) = τ τf(σ) H0 (G, I B) a G ⊗ ⊗ ∈ T G ⊗ hτX∈G i Hence the two classes δ (¯σ [f]) and δ [f(σ)] are equal in H0 (G, I B) since a ∪ a T G ⊗ τσ τf(σ) τ τf(σ)= τ(σ 1) τf(σ)= N (σ 1) f(σ) N (I B) ⊗ − ⊗ − ⊗ G − ⊗ ∈ G G ⊗ τX∈G τX∈G τX∈G

Using the description of the reciprocity map given in theorem 4.3.2, it is easy to show that for an unramiﬁed extension of local ﬁelds L K one has θ (a)=Φv(a) , where v denotes the normalised ⊃ L/K L/K valuation of K and ΦL/K is the Frobenius automorphism of L K. In fact, let n = [L : K] and ⊃ 1 G = Gal(L/K) and consider the character χ: G Q/Z given by χ(ΦL/K ) = n mod Z. Thena ¯ δχ is represented by the 2-cocycle c: G G L× given→ by ∪ × → i j a if i + j 0 c(ΦL/K, ΦL/K )= ≥ 0 i,j

× ab This will prove that the maps θL/K ﬁt together into a single map θK : K GK , which will then satisfy all the properties required in theorem 4.1.1. → Let a K ×, let χ Hom(Gal(L/K), Q/Z) and denote by χ′ = inf(χ) Hom(Gal(M/K), Q/Z). Then inf(¯a∈ δχ)=¯a δχ∈ ′ and thus ∈ ∪ ∪ χ(θ (a)) = inv (¯a δχ) = inv (¯a δχ′)= χ′(θ (a)) = χ(θ (a) ) L/K K ∪ K ∪ M/K M/K |L Since this holds for all χ Hom(Gal(L/K), Q/Z), we conclude that θL/K (a)= θM/K (a) L, as was to be shown. ∈ | With the above, we ﬁnish the proof of the local reciprocity theorem. Hurray! Chapter 3

BrauergroupofGlobalFields

By a global field we mean either a function field or a number field, that is, a finite extension of Fp(t) or a finite extension of Q, respectively. The main result of this chapter is the following computation: Theorem 0.1 (Brauer group of global fields) Let K be a global field. One has an exact sequence

resv invKv 0 - Br(K) v - Br(K ) v - Q/Z - 0 P v P Mv

Here K denotes the completion of K with respect to the place v of K, res : Br(K) Br(K ) and v v → v invKv :Br(Kv) Q/Z are the usual restriction and invariant maps, and the direct sum runs over all places of K, including→ the archimedean ones. The proof of the above theorem is the main step of the proof of the global reciprocity law. In terms of central simple algebras, it describes the local-global or Haße principle for division algebras: a division algebra D over a global field K is completely determined by its restrictions D K Kv to the local fields K . In particular, D is trivial if and only if D K is trivial for all v. ⊗ v ⊗K v 1 The geometric case In the next section, we will present a proof that works over a general global field K. However the geometric case, that is, the case when K is a finite extension of Fp(t), and therefore the function field of a non-singular projective curve C, is much easier to deal with, and serves as inspiration for the general method. We thus give a sketch of the proof of this special case first, assuming a few facts from the theory of algebraic curves. Here is a quick review of the facts we shall need. Given a non-singular projective curve C over a field k, we can view it as a collage of two Dedekind domains, namely the integral closures of k[t] and 1 k[ t ] in its function field K = k(C). For technical reasons, we assume that k is algebraically closed in K from now on. Each closed point P C0 (not necessarily defined over k) corresponds to a maximal ideal m of one of these two Dedekind domains,∈ and hence P defines a discrete valuation v : K Z : P P → ∪ {∞} vP (h) equals the exponent of mP in the factorisation of the principal fractional ideal (h). In geometric terms, vP (h) measures the order of the zero or pole of the function h at the point P . A divisor in C is just an element of the free abelian group generated by the closed points of C:

Div(C)= ZP

PM∈C0

Roughly, Div(C) corresponds to the multiplicative group of non-zero fractional ideals of a Dedekind domain, written additively. The degree deg D of a divisor D = n P Div(C), n Z, is defined P P ∈ P ∈ as a weighted sum of the finitely many non-zero coefficients nP Pof D: deg D = nP fP Z, where P ∈ fP = [k(P ) : k] is the residue degree of P , that is, the residue degree of the discreteP valuation vP over 1 its restriction to k[t] or k[ t ]. Each element h K × defines a principal divisor ∈ df div(h) = vP (h)P

PX∈C0 since vP (h) = 0 for all but ﬁnitely many P . A principal divisor corresponds to a principal fractional ideal in a Dedekind domain. An explicit computation shows that deg div(h) = 0 for any principal divisor 34 Brauer group of Global Fields

1 div(h) (do it ﬁrst for Pk and then use the formula i eifi = n for extension of primes in Dedekind domains). It is also easy to check that div(h)=0 P h k× (only constant functions have no zeros or poles). ⇐⇒ ∈ The Picard group Pic(C) of C is deﬁned as the quotient of Div(C) by the subgroup of principal divisors. It corresponds to a globalised form of the usual class group of a Dedekind domain. We have thus an exact sequence

div 0 - k× - K × - Div(C) - Pic(C) - 0

Since the degree of a principal divisor is zero, the degree of an element of Pic(C) is well-deﬁned. The kernel of the degree map is denoted by Pic0(C). If C has k-rational points (i.e. points with residue degree 1, or points “deﬁned over k”), the degree map is surjective and we have an exact sequence

deg 0 - Pic0(C) - Pic(C) - Z - 0

End of the review. From now on, let C be a non-singular projective curve defined over a finite field k, and let K = k(C) be its function field, which is a global field. Let us assume for simplicity that C has a k-rational point (even though the proof can be harnessed to eliminate this hypothesis). The computation of Br(K) is based on the following theorem: Theorem 1.1 Let C be a non-singular projective curve defined over a finite field k. Write C = C ×Spec k Spec k for the curve over k obtained by base change from C, and K = K k for the function field sep sep · sep of C. Then 1. Pic0(C) is a torsion abelian group; i 0 2. (Lang) H (Gk, Pic (C))=0 for i> 0; 3. (Tsen) Br(K)=0.

Proof Omitted. The proof of (1) follows from the fact that there is a natural bijection Pic0(C) = JC (ksep) where JC is the Jacobian variety of C (see Milne’s notes “Abelian varieties”, for instance), and JC (ksep) is the union of the finite groups JC (l) where l runs over all finite extensions of k. On the other hand, (2) holds more generally for any connected algebraic group instead of Pic0 by Lang’s article Algebraic groups over finite fields, Amer. J. Math. 78 (1956), 555–563. For a proof of Tsen’s theorem, see Shatz’s book for instance.

From Lang’s theorem and the long exact sequence associated to the short one

deg 0 - Pic0(C) - Pic(C) - Z - 0

i ≈ i we obtain an isomorphism deg: H (Gk, Pic(C)) H (Gk, Z) for i > 0 (actually, these groups are 0 for ˆ → i> 2 since Gk ∼= Z has cohomological dimension 1). Finally, from the exact sequence of Gk-modules

- × × - - - 0 K /ksep Div(C) Pic(C) 0 and the above computations, we obtain an exact sequence

1 deg- 1 H (Gk, Div(C)) H (Gk, Z) × - H2(G , K /k× ) - H2(G , Div(C)) - H2(G , Z) ( ) k sep k k ∗ - 3 × × H (Gk,K /ksep)=0

Here the last term vanishes since Gk has cohomological dimension 1. Since we are assuming that C has a k-rational point, we have that the map deg is surjective. On the other hand, we already know from 2 the computations of last chapter that H (Gk, Z)= Q/Z. We also have that

2 2 2 H (Gk, Div(C)) = H (Gk, Ind ZP )= H (Gk, Z)

PM∈C0 PM∈C0 The general case 35

2 and each factor H (Gk, Z) = Q/Z can be identiﬁed with Br(KP ), the Brauer group of the completion of K with respect to the valuation vP . Finally, using once more the fact that Gk has cohomological dimension 1, from the exact sequence

- × - × - × × - 0 ksep K K /ksep 0 we obtain an isomorphism 2 × 2 × × H (Gk, K )= H (Gk, K /ksep) But the ﬁrst group is Br(K/K), which equals Br(K) by Tsen’s theorem. Putting everything together, we obtain from ( ) the exact sequence ∗ - - - - 0 Br(K) Br(KP ) Q/Z 0

PM∈C0 Following the isomorphisms, it is easy to check that the second map is just the restriction map while the third is the sum of the invariant maps.

2 The general case Now we turn to the general case. Let K be an arbitrary global field. One of the key steps in the 1 0 computation of Br(K) of the geometric case was Lang’s theorem, which shows that H (Gk, Pic (C)) = 1 H (Gk, JC (ksep)) = 0. For number fields, we do not have an “arithmetic Jacobian” at our disposal. Instead, we shall define another object, the idéle class group, which will play an analogous role to the Jacobian (actually J Z) in the geometric case. C × In what follows, K will denote an arbitrary global field, and for any place v of K (archimedean or not), we write Kv for the completion of K with respect to v. For K a number field, an archimedean place v corresponds to an embedding σ: K ֒ Kalg, which defines an absolute value → df σ(a) if σ is a real embedding a v = | | a K k k σ(a) 2 if σ is a complex embedding ∈ | | and Kv is isomorphic to R and C, respectively. On the other hand, when v is non-archimedean, that is, v: K ։ Z is a (surjective) valuation, we define its normalised absolute value by ∪ {∞} a =df N(v)−v(a) a K k kv ∈ where N(v) is the cardinality of the residue field of v (which is a finite field). This definition is made so that the product formula holds: a =1 a K × k kv ∈ Yv Here v runs over all places of v, both archimedean or not. For non-archimedean places we also write

= x K v(x) 0 Ov { ∈ v | ≥ } U = × = x K v(x)=0 v Ov { ∈ v | } for its ring of integers and corresponding unit group, respectively. Now let L K be a finite Galois extension of global fields with G = Gal(L/K). Recall that G acts transitively on the⊃ set of places w ,...,w of L that divide v, via σ(w )= w σ−1. For w v, denote by 1 g i i ◦ | G = σ G σw = w w { ∈ | } the decomposition group of w. Since L is dense in Lw, each σ G extends uniquely to a continuous ≈ ∈ isomorphism σ: Lw Lσw. In particular, if σ Gw, σ extends to an automorphism of Lw, and this → ≈ ∈ gives a natural map G Gal(L /K ), which is an isomorphism. w → w v It will be convenient to write Lv for the completion of L with respect to any place w dividing v. They will all be isomorphic to each other by the above, and the corresponding decomposition groups will be conjugate in G. In particular, if G is abelian, the decomposition groups will be all equal and we v v will be able to identify Gal(L /Kv) with a subgroup of G = Gal(L/K). In particular, if L Kv is v ⊃ unramified, we will be able to view the Frobenius automorphism Φv of L as an element of G. 36 Brauer group of Global Fields

2.1 Id´eles We begin with a

× Definition 2.1.1 The idéle group IK of K is the multiplicative subgroup of v Kv given by Q I = (a ) K × a U for all but finitely many v K { v ∈ v | v ∈ v } Yv

where v runs over all places (both archimedean and non-archimedean) of K. We turn IK into a topological group as follows: as S runs over all finite sets of places of K containing all archimedean ones, the subsets of the form W U I , where W is an open subset of K ×, v × v ⊂ K v v vY∈S vY∈ /S define a basis of neighbourhoods of 1. The idéle group plays the role of the divisor group Div(C) of the geometric case. Note however that it is actually larger when K = k(C) is the function field of a curve C, since in that case one has a surjective map IK ։ Div(C) (a ) v (a )P P P ∈C0 7→ P P PX∈C0 with a nontrivial kernel. Yet IK is not too large: it can be shown that IK is a locally compact topological group, and we will manage to compute its cohomology without too much effort. × The embeddings K ֒ Kv allow us to view K as a subgroup of IK via the diagonal map a → × 7→ a,a,a,...). An element of the image of K ֒ IK is called a principal idéle; principal idéles will,...) play the role of principal divisors of the geometric→ case.

Definition 2.1.2 The idéle class group CK of K is defined to be the quotient

IK CK = K ×

The id´ele class group will play the role of the Jacobian (actually JC Z) of the geometric case. It is not a compact group nor a countable union of compact ones (as one would× expect from a generalisation of a projective group variety), but it is “almost compact,” which will do for our purposes.

× × × × Example 2.1.3 The group IQ consists of the tuples (a∞,a2,a3,a5,...) R Q2 Q3 Q5 with a U for all but ﬁnitely many primes p. ∈ × × × ×··· p ∈ p Given an id´ele α = (a ,a ,a ,...) IQ, let ∞ 2 3 ∈ r = (sign a ) pvp(ap) Q× ∞ ∈ Yp

where sign a∞ = 1 depending on whether a∞ is positive or negative. This is is well-deﬁned since ± −1 vp(ap) = 0 for all but ﬁnitely many primes p. We have αr R>0 U2 U3 . On the other hand, if r Q× (R U U ), then r = 1. Hence ∈ × × ×··· ∈ ∩ >0 × 2 × 3 ×··· × CQ = IQ/Q = R U U >0 × 2 × 3 ×···

Each Up is a compact group, but R>0 is not (the archimedean place is the “culprit” for the non- compactness of CQ). Now let L K be a finite Galois extension of number fields with G = Gal(L/K). There is an ⊃ × × injection IK ֒ IL which maps the v-th component av to (av,...,av) L L where the wi → ∈ w1 ×···× wg denote the places of L dividing v. Hence we can view IK as a subgroup of IL. Moreover, the G-action on the wi induces a G-action on IL, defined by σ(aw) = (bw), where bσw = σaw for σ G and (aw) IL. Pictorially, ∈ ∈ (a ,...,a ) L× L× 1 g ∈ w1 ×···× wg σ ↓ × × (σa1,...,σag) L L ∈ σw1 ×···× σwg The general case 37

Hence IL is a G-module, with “induced action” on each component over v. This fact will be quite useful later since it will allow us to apply Shapiro’s lemma during the cohomology computations. × If (aw) IL is ﬁxed by all σ G, taking σ Gw = Gal(Lw/Kv) we have that aw Kv for each w v, and hence∈ IG = I . Therefore∈ we may deﬁne∈ a norm map ∈ | L K N : I I L/K L → K α σ(α) 7→ σY∈G

We can also deﬁne a norm map between id´ele class groups thanks to Lemma 2.1.4 (Norm compatibility) The diagram

× ⊂ - L IL

NL/K NL/K ? ? × ⊂ - K IK

commutes. Hence N : C C is well-deﬁned. L/K L → K df df Proof Let w = w1, w2,...,wg be the places dividing a ﬁxed place v of K, and let σi G = Gal(L/K) be such that σ w = w. Since τ G τw = w = G σ , we have that the w-th component∈ of the norm i i { ∈ | i } w i of b L×, viewed as an element of I , is just ∈ L

τ(b)= τ(b)= NL/K(b)

1≤Yi≤g τ∈YGwσi τY∈G

Definition 2.1.5 For any global field K and any finite set S of places of K containing all archimedean ones, define the following subgroup of IK :

IS =df (a ) I a U for v / S = K × U K { v ∈ K | v ∈ v ∈ } v × v vY∈S vY∈ /S

The next lemma is just an idelic version of the finiteness of the class number. Lemma 2.1.6 (Idelic Class Group Finiteness) Let K be a global field. There exists a finite set S of places of K such that × S IK = K IK Hence C = IS /(IS K ×). K K K ∩ Proof Take S to be the set of all archimedean places together with enough non-archimedean ones to generate the whole class group of K. Then, given any (av) IK , consider the corresponding fractional v(av) ∈ ideal a = v pv , where now v runs over all non-archimedean places of K, and pv denotes the prime × ideal of QK corresponding to v. By the choice of S, there exists f K and nv Z such that O ∈ ∈ (f) a = pnv · v vY∈S v non-arch.

But that amounts to saying that f a U for v / S, that is, f (a ) IS . · v ∈ v ∈ · v ∈ K 2.2 Global reciprocity In this section, we want to give a description of the abelian extensions of a global ﬁeld in terms of the id´ele class group. 38 Brauer group of Global Fields

Definition 2.2.1 Let L K be a finite Galois extension of global fields with abelian G = Gal(L/K). ⊃ The global Artin reciprocity map θL/K is defined as the product of the local reciprocity maps:

θ : I G L/K K → (a ) θ v (a ) v 7→ L /Kv v Yv

v Here, if v is archimedean and L K is isomorphic to C R, we deﬁne θ v (a ) to be ⊃ v ⊃ L /Kv v the conjugation map if av < 0 and the identity map otherwise, so that we have a trivial reciprocity isomorphism (the “sign map”)

× × R R ≈ θ : = Gal(C/R) C/R × NC/RC R>0 →

Also, observe that for all but finitely many v, the extension Lv K is unramified and a U , hence ⊃ v v ∈ v v θL /Kv (av) = 1. Hence the above product does make sense. We know from the local reciprocity theorem that NL/KIL is contained in the kernel of θL/K. We actually want to show that the principal idéles are also killed by θL/K and that Theorem 2.2.2 (Global Artin Reciprocity) Let L K be a finite Galois extension of global fields with abelian G = Gal(L/K). The global Artin map induces⊃ an isomorphism

CK ≈ θL/K: G NL/KCL →

The relation between the Brauer group and the Artin reciprocity map is given by the following Lemma 2.2.3 Let L K be a ﬁnite abelian Galois extension of global ﬁelds with G = Gal(L/K). Consider the following⊃ statements: 1. the sequence invv 0 - Br(L/K) - Br(Lv/K ) v - Q/Z - 0 v P Mv is a complex. 2. K × ker θ ⊂ L/K Then 1 2 unconditionally, and 2 1 if G is cyclic. ⇒ ⇒ Proof Let χ H1(G, Q/Z) = Hom(G, Q/Z) and denote by δχ H2(G, Z) the image of χ under the isomorphism given∈ by the connecting map associated to the short∈ exact sequence

0 - Z - Q - Q/Z - 0

Consider the following diagram:

θv K × - I v - G - G K Q v Q Mv δχ δχ χ χ ∪ ∪ v ? ? ?P ? invv Br(L/K) - Br(Lv/K ) v - Q/Z ======Q/Z v P Mv

Cup product with δχ gives a morphism K ×/N L× = H0 (G, L×) H2(G, L×) = Br(L/K), which L/K T → is an isomorphism if G is cyclic and χ generates H1(G, Q/Z) (the “periodicity isomorphism”, see the appendix). Similarly, still denoting by χ its restriction to Lv K , cup product with δχ gives morphisms ⊃ v K × Br(Lv/K ), which induce the second vertical map of the diagram: given (a ) I , for all but v → v v ∈ K The general case 39

v v v finitely many v, av is a unit and the extension L Kv is unramified, hence av UKv = NL /Kv UL and a δχ = 0 (see corollary II.2.2 and theorem II.4.1.1).⊃ ∈ v ∪ Now the whole diagram commutes: the left square, by the compatibility of the restriction maps with the cup product; the middle square, by the formula χ(θv(av)) = invv(av δχ) of the local reciprocity law (see theorem II.4.3.2); and finally, the right square, by the definition of∪ group morphism. Now assume (1) holds. To show that θ (a)=1 for a K ×, we have to show that χ(θ (a))=0 L/K ∈ L/K for all χ H1(G, Q/Z). But from the commutativity of the diagram, χ(θ (a)) = inv (a δχ), ∈ L/K v v ∪ which vanishes since the bottom row is a complex by assumption. Conversely, assumeP that G is cyclic and (2) holds. Let χ be a generator of H1(G, Q/Z). Then the leftmost vertical arrow is surjective, hence any element of Br(L/K) can be written as a δχ for some a K ×. But then again from the diagram we conclude that inv (a δχ)= χ(θ (a∪)) = 0, and hence∈ the bottom row is a complex. v v ∪ L/K P In other words, the computation of the Brauer group of a global field is essentially equivalent to proving Artin’s reciprocity law! As a first step toward this common goal, let us prove Theorem 2.2.4 (Reciprocity Law for Cyclotomic Fields) Let K be a number field and let L K ⊃ be a subextension of the cyclotomic extension K(ζn) K, where ζn denotes a primitive n-th root of unity. Then statement (2) of the last lemma holds. ⊃

Proof First we show that it is enough to prove the theorem for K = Q and L = Q(ζn). This follows from the functorial properties of the local reciprocity map, which imply that the following diagram commutes for any abelian extension L Q (use the fact that the local reciprocity map is given by the cup product with the fundamental class⊃ and that the cup product commutes with corestriction):

θKL/K- IK Gal(KL/K) ∩

NK/Q ? ? θL/Q - IQ Gal(L/Q)

× Since the right vertical arrow is injective, if we can show that Q ker θL/Q then θKL/K (a) L = × ⊂ | θ Q(N Q(a))=1 θ (a) = 1 for all a K . K/ L/ ⇒ KL/K ∈ mc df Let Q = n≥1 Q(ζn) be the maximal cyclotomic extension of Q. We have that S mc × × Gal(Q /Q) = lim Gal(Q(ζn)/Q) = lim(Z/n) = Zˆ ←−n∈N ←−n∈N

mc We now construct a continuous map ψ: IQ Gal(Q /Q), which we secretly know to coincide with mc → the inverse limit θQ: IQ Gal(Q /Q) of the reciprocity maps θQ Q: IQ Gal(Q(ζ )/Q) (they fit → (ζn)/ → n together by the functorial properties of the local reciprocity maps). By construction, we will have that × Q ker ψ, and then we will have to show that ψ = θQ. ⊂ mc The map ψ: IQ Gal(Q /Q) will be the inverse limit of ψn: IQ Gal(Q(ζn)/Q), defined as follows. Let S be a finite→ set of places containing and the ramified ones→ (i.e., those dividing n). For ∞ × (a ) IQ, we use the approximation theorem to find f Q such that v ∈ ∈ f a < min na , a 1 f −1a < n for each v S k − vkv {k vkv k vkv}⇒k − vkv k kv ∈ Set −1 df v(f av) −1 ψn(av) = Φv = θQv (ζn)/Qv (f av) vY∈ /S vY∈ /S where Φv denotes the Frobenius automorphism of Qv(ζn), namely the element of Gal(Qv(ζn)/Qv) p ⊂ Gal(Q(ζn)/Q) given by Φv(ζn)= ζn, where p is the prime corresponding to the place v. In other words, ψ (a ) is the automorphism given by ζ ζr , where r = f −1a −1 mod n. n v n 7→ n v∈ /S k vkv The product formula Q a =1 a Q× k kv ∈ Yv 40 Brauer group of Global Fields

ensures that the deﬁnition of ψ does not depend on the choice of S or f. For instance, if g Q× is n ∈ another element such that g av v < min nav v, av v for all v S, then since f v = g v = av v for all v S we have that k − k {k k k k } ∈ k k k k k k ∈ f −1a −1 g−1a −1 (mod n) f −1a g−1a (mod n) k vkv ≡ k vkv ⇐⇒ k vkv ≡ k vkv vY∈ /S vY∈ /S vY∈S vY∈S

holds. It is also easy to check that ψm(av) = ψn(av) Q(ζm) for m n, so that we may define ψ(av) = | | mc lim ψ (a ), and another routine check shows that the resulting map ψ: IQ Gal(Q /Q) is continuous. n v → ←− In particular, we get that Q× ker ψ for free since for a principal idéle a Q× we may choose f = a. In other words, ψ is defined⊂ so as to coincide with the product of the local∈ reciprocity maps at the unramified places, and so that Q× ker ψ: this allow us to replace an arbitrary idéle by an equivalent one which is “very close to 1” at the⊂ ramified places, so that the corresponding local reciprocity maps also vanish there. Besides, for any finite subextension L Q of Qmc Q we also have that ⊃ ⊃ N Q I ker ψ (which is no surprise if we believe ψ equals the reciprocity map θQ). In fact, for L/ L ⊂ |L β I , by the approximation theorem we may find b L× such that bβ is arbitrarily close to 1 at the ∈ L ∈ archimedean and ramified places of L. Since the norm map is continuous, NL/Q(bβ)= NL/Q(b) NL/Q(β) will also be arbitrarily close to 1 at the archimedean and ramified places of Q, and since we already· know × that N Q(b) Q belongs to ker ψ, in order to prove that N Q(β) ker ψ it suffices to show that L/ ∈ L/ ∈ |L N Q(bβ) ker ψ . But now ψ N Q(bβ) coincides with the product of the local maps at the L/ ∈ |L L/ |L unramified local extensions, and the result follows from the corresponding property of the local maps. × For each v, let ιv: Qv IK be v-th inclusion map given by ιv(a) = (..., 1, 1, a, 1,...) which equals → mc df the identity map at the v-th component. Let v be a fixed place of Q and Qv = n≥1 Qv(ζn) be the maximal cyclotomic extension of Q . We now show that the composition ψ =df ψ ι inducesS a continuous v v ◦ v map ψ : Q× Gal(Qmc/Q ) which coincides with the local reciprocity map θ . This will prove that v v → v v v ψ = θQ, finishing the proof of the theorem. mc mc We now show that the image of ψv indeed lies in Gal(Qv /Qv) Gal(Q /Q). We have to show that, for any finite subextension L Q of Qmc Q with G = Gal(⊂ L/Q), ψ (a) belongs to the ⊃ ⊃ v |L decomposition group G = Gal(Lv/Q ) of v for all a Q . Let M = LGv ; then M v = Q , and hence v v ∈ v v ιv(a) NM/QIM . Thus ψv(a) M is trivial, that is, ψv(a) L Gv, as required. ∈ | | ∈ nr mc Next we show that ψv = θv. First assume that v = . Let Qv Qv be the maximal unramified × 6 ∞ ⊂π extension of Qv, and for a uniformiser π Qv of v, denote by Qv the fixed field of θv(π). Then mc nr π ∈ × Qv = Qv Qv . Since the uniformisers π generate Qv , to prove that ψv = θv it suffices to show that · nr π nr ψv(π) and θv(π) agree on Qv and Qv . But both equal the Frobenius automorphism Φv on Qv , so π we are left to show that ψv(π) is trivial on Qv , or equivalently, on any finite subextension M Qv of π mc mc mc v ⊃ Qv Qv. Let L Q be the fixed field of Gal(Qv /Qv) Gal(Q /Q) so that M = L . We have that ⊃ v ×⊂ v ⊂ v π NL /Qv (L ) since θv(π) is trivial on L , therefore ιv(π) NL/QIL, showing that ψv(π) is trivial on M∈= Lv as well. ∈ Now if v = , θv(a) is the conjugation map if a < 0 and equals the identity otherwise. Since × ∞ 2 × R = NC RC is contained in ker ψ as well, ψ (a) = id for all a R , and so ψ (a) must be the >0 / v v ∈ v conjugation map if it is not trivial. Hence we just have to check that ψv is not trivial, and for that we look at its restriction to Q(i), which ramifies only at 2 and over Q. Therefore ∞ 1= ψ( 7) = ψ ( 7) ψ ( 7) ψ ( 7) − ∞ − · 2 − · 7 −

We have that 7 is a 2-adic norm so ψ2( 7) = 1. We have thus to show that ψ7( 7) is the conjugation map. But ψ (−7) is the Frobenius map i− i7 = i, so we are done. − 7 − 7→ − 2.3 The Brauer group of a global field The computation of the Brauer group Br(K) of a global field and the proof of the reciprocity law is a consequence of the following computation, which we shall dub “class field theory axioms.” Theorem 2.3.1 (Class Field Theory Axioms) Let L K be a Galois extension of global fields with abelian G = Gal(L/K). Then ⊃ 1. (Cohomology of Idéles) For all i 0, ≥

i i v × HT (G, IL)= HT (Gv, (L ) ) Mv The general case 41

2. (Cohomology of C ) Suppose further that G is cyclic of prime order p, and that µ K ×. Then L p ⊂ H0 (G, C ) = G and H1 (G, C ) =1 | T L | | | | T L |

We postpone the proof of this theorem to the next section. We now derive some important conse- 1 quences from it. Note that the vanishing of H (G, CL) is the analogue of Lang’s theorem for a general global field. Corollary 2.3.2 Let L K be a Galois extension of global fields with G = Gal(L/K). Then ⊃ 1. H0 (G, C ) is finite, and its order divides G ; T L | | 1 2. H (G, CL)=0; 3. H2(G, C ) is finite, and its order divides G . L | | 1 Proof Let us first show that H (G, CL) vanishes. First observe that the class field theory axiom 1 H (G, CL) = 1 holds for all cyclic extensions L K of prime degree p without the extra hypothesis × ⊃ 1 µp K . In fact, since the degree [K(ζp) : K] divides p 1, it is prime to p. But H (G, CL) is p-torsion ⊂ 1 1 − -it is killed by G ), hence res: H (G, CL) ֒ H (Gal(L(ζp)/K(ζp)), CL(ζp)) is injective by the restriction) | | → 1 corestriction argument (see appendix). By the class field theory axiom, H (Gal(L(ζp)/K(ζp)), CL(ζp))= 1 1, and therefore H (G, CL) = 1 as well. 1 For the general case, since H (G, CL) is torsion, it is enough to show that its p-primary part vanishes for all primes p. If Sp G is a p-Sylow subgroup, then by the restriction-corestriction argument, we 1 ⊂ 1 have that res: H (G, CL) ֒ H (Sp, CL) is injective on p-primary parts, hence it is enough to consider the case where G = S is a→p-group, hence solvable. Now we proceed by induction on G , the base case p | | G = p already solved. Let H ⊳ G be a non-trivial normal subgroup, and denote by M = LH the fixed | | 1 field of L by H. By induction hypothesis, H (H, CL) = 1, and hence from the short exact sequence

- × - - - 1 L IL CL 1

we have that - × - - H - 1 1 M IM CL H (H, CL)=1 H is exact and thus CL = CM . By the inflation-restriction sequence (see appendix), we thus obtain an exact sequence - 1 inf- 1 res- 1 0 H (G/H, CM ) H (G, CL) H (H, CL) We conclude that the middle term vanishes since the left and right terms vanish by induction hypothesis. 0 The proof of (1) proceeds along similar lines. Since HT (G, CL) is torsion, it is enough to show that its p-primary part is finite and divides S where S G denotes once more a p-Sylow subgroup of | p| p ⊂ G. Restriction-corestriction allows us to assume that G = Sp, and shows that we may dispense with 0 the roots of unity condition of the class field theory axiom HT (G, CL) = G for G cyclic. The only difference is that induction on the order of the p-group G utilises| the following| | | observation instead of the inflation-restriction sequence: if H ⊳ G then since

- 2 inf- 2 res- 2 0 H (G/H, CM ) H (G, CL) H (H, CL)

1 since we have already shown that H (H, CL)=1. 42 Brauer group of Global Fields

Theorem 2.3.3 Let L ) K be a solvable extension of global ﬁelds with G = Gal(L/K). Then

1. there exist infinitely many prime ideals p in K that do not split completely in L; O v O 2. if G is abelian, the Frobenius automorphisms Φv Gal(L /Kv) G generate G, as v runs over all valuations that do not belong to a fixed finite set∈ S of places of⊂K containing all archimedean and ramified ones. Proof 0 Replacing L K by some cyclic subextension of prime order, we may assume that HT (G, CL) G by the class field⊃ theory axioms. Suppose that the set S of primes that do not split completely| in|≥L | | is finite. Given α = (av) IK , using the approximation theorem (c.f. the proof of lemma 2.1.6) we may × ∈ v × v find f K such that f av NL /Kv (L ) for all v S (norm groups of local fields are open of finite ∈ · ∈ ∈ v index by local class field theory). But then f α NL/K IL since L = Kv for all v / S by assumption. × · ∈ × ∈ 0 This proves that K NL/KIL = IK . But since IK /(K NL/KIL) = CK /NL/K CL = HT (G, CL), this cannot happen, proving· (1). · To obtain (2), observe that the prime corresponding to a valuation v of K splits completely in L if and only if Φv = 1. Hence, applying (1) to the fixed field M of all Φv, v / S, we conclude that M = K, that is, the Φ , v / S, generate G, as required. ∈ v ∈ 0 Remark 2.3.4 Observe that the above proof only relies on the fact that HT (G, CL) G . This will be used later in the proof of the so-called second inequality. | | ≥ | |

Theorem 2.3.5 (Haße Principle for CSA) Let L K be a Galois extension of global ﬁelds with G = Gal(L/K). Then ⊃ - - v - 2 - 0 Br(L/K) Br(L /Kv) H (G, CL) 0 Mv is complex that is exact at the ﬁrst two terms. If G is cyclic, it is also exact at the last term. In particular, taking the limit over all L, we have that - - 0 Br(K) Br(Kv) Mv is exact, that is, an element α Br(K) is trivial if and only if it is locally trivial. ∈ Proof Let G = Gal(L/K). From the short exact sequence of G-modules

- × - - - 1 L IL CL 1 we obtain an exact sequence

1 - 2 × - 2 - 2 - 3 × 0= H (G, CL) H (G, L ) H (G, IL) H (G, CL) H (G, L ) which can be rewritten as

- - v - 2 - 3 × 0 Br(L/K) Br(L /Kv) H (G, CL) H (G, L ) Mv using the class ﬁeld axioms. Moreover, if G is cyclic, by Hilbert 90 and the periodicity of cohomology of G, H3(G, L×)= H1(G, L×) = 0 as well, which ﬁnishes the proof.

Corollary 2.3.6 (Haße Norm Theorem) Let L K be a cyclic extension of global ﬁelds. Then an × ⊃ × × element of K is a norm if and only if it is a local norm, that is, a K belongs to NL/K L if and v × ∈ only if a N v (L ) for all v. ∈ L /Kv Proof Just use the previous theorem together with the commutative diagram: K × δχ = H0 (G, L×) ∪ - H2(G, L×) = Br(L/K) × T NL/KL ≈

? ? K × δχ v = H0 (G, (Lv)×) ∪ - H2(G, (Lv)×) = Br(Lv/K ) v × T v NLv/K (L ) v ≈ Here χ H1(G, Q/Z) is a generator of this group, and the rows are isomorphisms by the periodicity of the cohomology∈ of the cyclic group G. The general case 43

Now we can show Lemma 2.3.7 Let K be a number ﬁeld. Any α Br(K) is split by a cyclic extension L of K contained in some cyclotomic extension K(ζ) of K. ∈ Proof From the commutative diagram with exact top row

- v - res - v 0 Br(L /Kv) Br(Kv) Br(L )

inv inv ≈ ≈ ? ? [Lv : K ] Q/Z v- Q/Z

together with the Haße principle for CSA, we reduce the problem to proving the following claim: given an integer n and a finite set of places S of K, there is a cyclic extension L K contained in a cyclotomic v ⊃ extension of K such that n divides the local degrees [L : Kv] for all v S. ∈ r × Clearly, it suffices to prove the claim for K = Q. Let p be a prime. Since Gal(Q(ζpr )/Q) = (Z/p ) , r r−2 Q(ζpr ) Q contains a cyclic subextension F (p ) Q of degree p . Now if l is another prime, we ⊃ r r−1 ⊃ have that [Ql(ζpr ) : Ql] equals φ(l )= l (l 1) if p = l (totally ramified case), and equals the smallest r f − r positive integer f such that p l 1 if p = l (totally unramified case). In either case, [F (p )l : Ql] is a power of p that tends to infinity| as−r grows.6 e1 et Now let n = p1 ...pt be the factorisation of n into powers of distinct primes pi. Choose r large ei r r r enough so that pi [F (pi )l : Ql] for all i = 1, 2,...,t and l S. We may take L = F (p1) ...F (pt ), which is a cyclic extension| of Q with the required properties. ∈

Observe that the above theorem shows that Br(Qmc) = 0 for the maximal cyclotomic extension Qmc of Q, much in the same way that Tsen’s theorem asserts that the Brauer group of the function field of sep mc a curve over Fp is trivial. The field Q (more precisely, a cyclic quotient of it) plays the same role as sep the Fp in the geometric case. We now can finish the proof of the Reciprocity Law:

Proof (of the Reciprocity Law) Let L K be the cyclic extension contained in a cyclotomic extension of K, as the previous theorem. Then⊃ by the explicit computation of theorem 2.2.4 we know that K × ker θ . Hence, by lemma 2.2.3, the sequence ⊂ L/K

invv 0 - Br(L/K) - Br(Lv/K ) v - Q/Z ( ) v P ∗ Mv is a complex. But the previous theorem says that Br(K) is the union of all Br(L/K) as L runs over all cyclic extension contained in a cyclotomic extension of K, hence we obtain a complex

invv 0 - Br(K) - Br(K ) v - Q/Z - 0 ( ) v P ∗∗ Mv

× This, in turn, implies that ( ) is a complex for all L. Finally, we conclude that K ker θL/K for all L by lemma 2.2.3. ∗ ⊂ Now suppose that G is abelian, and observe that

։ θL/K : IK G is surjective since the Frobenius maps Φ generate G (theorem 2.3.3). But ker θ K × N I , v L/K ⊃ · L/K L and since the subgroup K × N I has index at most G in I by corollary 2.3.2, it must equal the · L/K L | | K kernel, and θL/K induces an isomorphism

IK CK ≈ θL/K: = G K × N I N C → · L/K L L/K L 44 Brauer group of Global Fields

Finally, to show that ( ) is exact, it is enough to show that ∗∗ 1 invv Z 0 - Br(L/K) - Br(Lv/K ) v - [L:K] - 0 ( ) v P Z † Mv is exact for the cyclic extension of the beginning of the proof. Observe that we already know that ( ) is † a complex and that it is exact at the ﬁrst (Haße principle) and last term since the Frobenius maps Φv 1 v generate G, hence invv(Φv δπv)= mod Z generate the last group: here fv = [L : Kv] is the order ∪ fv of Φv and πv is a uniformiser of v. Comparison of the exact sequence in theorem 2.3.5 with the last sequence yields a surjective map 1 Z 2 ։ [L:K] H (G, CL) Z , and we have to show that this map is an isomorphism. For that, just observe that H2(G, C ) is cyclic of order G = [L : K] since by the global Artin reciprocity we have isomorphisms L | |

θL/K 0 ∪δχ- 2 G H (G, CL) H (G, CL) ≈ T ≈

where χ H1(G, Q/Z) is a generator and cup product with δχ H2(G, Z) is the periodicity isomorphism. ∈ ∈

3 Proof of the Class Field Theory Axioms

We are left to compute the cohomology groups of IL and CL. The first computation is fairly straight- 0 1 forward, but verifying that HT (G, CL) = G and HT (G, CL) = 1 is quite delicate, and the proof is broken into two parts, traditionally| called| | “first| and| second inequalities:”| the first inequality is the 0 1 computation of the Herbrand quotient h(G, CL)= HT (G, CL) / HT (G, CL) = G , which already shows that H0 (G, C ) G , while the second inequality| is the opposite| | one. | | | | T L | ≥ | |

3.1 Cohomology of IL Recall that for any set S of valuations of K, we write

IS = K × U K v × v vY∈S vY∈ /S

Theorem 3.1.1 (Cohomology of Idéles) Let L K be a finite Galois extension of global fields with abelian G = Gal(L/K). Then ⊃ p p v × H (G, IL)= H (Gv, (L ) ) Mv 1 Z 1 2 nv v In particular, H (G, IL)=1 and H (G, IL)= v Z where nv = [L : Kv]. L Proof Let S be a finite set of primes containing all archimedean and ramified ones. Using the fact that cohomology commutes with products,

Hp(G, IS )= Hp(G, L× ) Hp(G, U ) L w × w Yw|v Yw|v v∈S v/∈S

By Shapiro’s lemma, Hp(G, IS )= Hp(G , (Lv)×) Hp(G ,U v) L v × v vY∈S vY∈ /S Since every v / S is unramiﬁed, Hp(G ,U v) = 0, and therefore ∈ v

p S p v × H (G, IL)= H (Gv, (L ) ) vY∈S

Taking the injective limit over all ﬁnite sets S yields the result. Proof of the Class Field Theory Axioms 45

From now on we concentrate on the number field case, which is the difficult one. We leave to the reader the task of adapting the proofs to the function field case.

3.2 First inequality Let L K be a cyclic extension of number fields with Galois group G = Gal(L/K). Our goal is to ⊃ compute the Herbrand quotient h(G, CL). Theorem 3.2.1 (First Inequality) Let L K be a Galois extension of number fields with cyclic G = Gal(L/K). Then ⊃ h(G, CL) = [L : K] In particular, H0 (G, C ) [L : K]. | T L |≥ Proof We invoke lemma 2.1.6 and choose a finite set T of valuations of L which 1. includes all archimedean and ramified primes; × T 2. IL = L IL; 3. is G-stable, i.e., w T σ(w) T for all σ G. ∈ ⇒ ∈ ∈ and let S be the set of places of K dividing the places of T . We have

T T IL h(G, IL) h(G, CL)= h G, = IT L× h(G, IT L×) L ∩ L ∩ We already know that

0 T 0 v T HT (G, IL) HT (Gv,L ) h(G, IL)= | 1 T | = | 1 v | = nv H (G, I ) H (Gv,L ) | T L | vY∈S | T | vY∈S df where n = [Lv : K ]. On the other hand, LT = IS L× is just the group of T -units of L. Denote by v v L ∩ df V = Hom(T, R)= Rew wM∈T the T -dimensional real vector space equipped with G-action | | (σf)(w)= f(σ−1w), f V and σ G, w T ∈ ∈ ∈ |T | ′ ′ Here ew denotes the dual of the standard basis of R , that is, ew(w) = 1 and ew(w ) = 0 if w = w. By Dirichlet’s unity theorem, the map of G-modules 6 log u: LT V → u log u e 7→ k kw · w wX∈T has a ﬁnite kernel µ L× and a G-lattice M 0 of rank T 1 as image, contained in the G-invariant hyperplane x⊂e x = 0 by the product| |− formula. We have thus exact sequences of { w∈T w w | w∈T w } G-modules P P log 0 - µ - LT - M 0 - 0 0 - M 0 - M - Z - 0 where M = M 0 Zv and v = (1, 1,..., 1) V . Hence ⊕ ∈ h(G, M)h(G, Z) h(G, LT )= = h(G, M) G h(G, µ) · | |

To compute h(G, M), we may replace M with any G-lattice N such that M Z Q = N Z Q (as G- ⊗ ∼ ⊗ modules) since Herbrand quotients of a ﬁnite G-modules are trivial. Take N =df Hom(T, Z) V . We ⊂ have that M Z R = N Z R M Z Q = N Z Q and ⊗ ∼ ⊗ ⇒ ⊗ ∼ ⊗ G h(G, M)= h(G, N)= h(G, IndGv Z)= h(Gv, Z)= nv vY∈S vY∈S vY∈S Putting everything together, we obtain the desired result. 46 Brauer group of Global Fields

3.3 Second inequality Now we ﬁnish the proof of the class ﬁeld theory axioms by showing

Theorem 3.3.1 (Second Inequality) Let L K be a Galois extension of number ﬁelds with cyclic G = Gal(L/K) of prime order n. Suppose that µ⊃ K ×. Then n ⊂ H0 (G, C ) n | T L |≤

The key idea of the proof is to use Kummer theory in place of the reciprocity law, and a variant of Haße’s norm theorem dealing with n-th powers instead of norms, given by the first lemma below. By Kummer theory, L = K( √n a) for some a K ×. The second inequality follows from ∈ Theorem 3.3.2 Let L K be as above, and let S be a sufficiently large finite set of primes of K such that ⊃ all archimedean primes are in S; • all primes that ramify in L are in S; • all primes dividing a or n are in S; • I = K ×IS • K K Suppose that there is a set T of S 1 primes of K, disjoint from S, satisfying | |− 1. Lv = K for all v T . v ∈ 2. (local power) KS∪T D(S,T ) = (KS∪T )n, where ∩

D(S,T )= (K ×)n K × U v × v × v vY∈S vY∈T v∈ /YS∪T

and KS∪T denotes the group of (S T )-units. ∪ Then H0 (G, C ) divides n. | T L | Proof Since D(S,T ) N I (use corollary II.2.2), it is enough to prove that IK divides n. ⊂ L/K L | K×D(S,T ) | Using the relation [AC : BC] = [A : B] [A C : B C] · ∩ ∩ for subgroups A, B, C of an abelian group, and condition 2, we have that

S∪T IK × × n I K ×IS∪T D(S,T ) [K : (K ) ] K = K = = v∈S v v × × × IS∪T S∪T S∪T n K D(S,T ) K D(S,T ) K ∩ K Q[K : (K ) ] K ×∩D(S,T )

But by Kummer theory and the structure theorem of the multiplicative group of a local ﬁeld, we have that [K × : (K ×)n]= n2/ n (which also works for archimedean v in view of the assumption µ K ×). v v k kv n ⊂ On the other hand, since µ K ×, Dirichlet’s unit theorem implies that [KS∪T : (KS∪T )n]= n|S|+|T | = n ⊂ n2|S|−1. Hence 2 n 2|S| I v∈S n K = knkv = = n K ×D(S,T ) Qn2|S|−1 n2|S|−1

using the product formula.

We now construct T . First we make the following reduction:

Lemma 3.3.3 Let L K and S be as above. Suppose T is a set of places of K, disjoint from S, such that the map ⊃ S ։ n K Uv/Uv vY∈T is surjective. Then KS∪T D(S,T ) = (KS∪T )n. ∩ Proof of the Class Field Theory Axioms 47

Proof Observe that x K is an n-th power if and only if the ﬁeld M = K( √n x) is equal to K, and by the ﬁrst inequality (or rather∈ theorem 2.3.3) this happens if and only if

H0 (Gal(M/K), C )=1 I = K ×N I T M ⇐⇒ K M/K M

S∪T × Hence we have to show that if x K D(S,T ) then IK = K NM/KIM , so we need to ﬁnd many norms. Notice that ∈ ∩ N I E =df K × U n U M/K M ⊃ v × v × v vY∈S vY∈T v∈ /YS∪T

× n S∪T since x D(S,T ), hence x (Kv ) , and x K , hence M is unramified outside S T and norm maps∈ are surjective by∈ corollary II.2.2. But∈ now given α IS there exists f KS such∪ that ∈ K ∈ fα E N I by assumption, and therefore I = K ×IS = K ×N I , as required. ∈ ⊂ M/K M K K M/K M The next lemma finishes the proof of the second inequality. Lemma 3.3.4 With the above notation and hypotheses, there is a set T of S 1 primes in K such that | |− 1. Lv = K for all v T . v ∈ S ։ n 2. the map K v∈T Uv/Uv is surjective. Q Proof Let s = S , and H be the kernel of the map in (2). We claim that it suffices to choose T so n | | that L = K( √H) (by the Kummer correspondence). In fact, (1) is clear, while in order to show (2), it S n suffices to show that K /H and v∈T Uv/Uv have the same cardinality. But Q [KS : (KS)n] ns KS/H = = = ns−1 | | [H : (KS)n] [L : K] by Dirichlet’s unit theorem and Kummer theory, while

U /U n = n/ n = ns−1 | v v | k kv vY∈T vY∈T by the structure theorem of units of local ﬁelds and the product formula. S Now choose representatives a1 = a,a2,...,as K of a basis of the s-dimensional Z/n-vector space KS/(KS)n. Let ∈

n S n n n n M = K( √K ) and Mi = K( √a1,..., √ai−1, √ai+1 ..., √as)

Then by Kummer theory we have that L = M , and each M M is a cyclic extension of 2≤i≤s i ⊃ i degree n. Hence by theorem 2.3.3 we may chooseT places wi of Mi such that their restrictions vi to K are all distinct and lie outside S, and such that wi does not split completely in M. Hence Mi is the decomposition ﬁeld of the unique place of M lying over wi, and therefore L K is the maximal ⊃ × n S subextension of M K where all places vi split completely. Let T = v2,...,vs . Now H = (L ) K since, for x KS, we⊃ have { } ∩ ∈ x (L×)n K( √n x) L ∈ ⇐⇒ ⊂ K ( √n x)= K for i =2,...,s (since each v splits completely) ⇐⇒ vi vi i x U n for i =2,...,s ⇐⇒ ∈ vi x H ⇐⇒ ∈ and by Kummer theory we have that L = K( √n H), as required.

Chapter 4

Appendix: Galoiscohomology

0.1 Definitions In this subsection, G will denote a finite group. Definition 0.1.1 A G-module M is just a left Z[G]-module, where Z[G] is the group ring of G with integer coefficients. In other words, M is an abelian group together with a G-action, that is, a map G M M sending (σ, m) to an element σ m M such that, for all m,m′ M, and σ, σ′ G, × → · ∈ ∈ ∈ 1. 1 m = m · 2. σ (m + m′)= σ m + σ m′ · · · 3. (σσ′) m = σ (σ′ m) · · · A morphism of G-modules f: M N is a map of left Z[G]-modules, i.e., a group morphism such that f(σm)= σf(m) for all σ G and→ m M. ∈ ∈ If M is a G-module, we write M G for the subgroup of G-fixed points:

M G =df m M σ m = m for all σ G { ∈ | · ∈ } Example 0.1.2 Examples of G-modules arising in nature are 1. M = Z, M = Q, or M = Q/Z where G operates trivially: σm = m for all σ G and m M; ∈ ∈ 2. M = Z[G] or M = IG, where G operates by left multiplication. Here IG is the so-called augmentation ideal of Z[G], deﬁned as the kernel of the augmentation map ǫ: Z[G] Z given by → n σ n σ 7→ σ σX∈G σX∈G

Observe that IG is a free Z-module with basis σ 1, σ G, σ = 1. If Z is given the trivial action as above, then the augmentation map is a morphism− ∈ of G-modules;6

+ × 3. if L K is a ﬁnite Galois extension with G = Gal(L/K), then M = L , M = L and M = µL are all⊃ examples of G-modules where G operates via the Galois action. Here L+ and L× are the × additive and multiplicative groups of L and µL L is the subgroup of roots of 1 contained in L. ⊂ 4. for any G-module M we have a morphism of G-modules N : M M G given by G → N (m) =df σm, m M G ∈ σX∈G

× called norm map. In the previous example, when M = L the norm map NG coincides with the usual norm of ﬁelds N : L K. When M = L+ then it becomes the trace T : L K. L/K → L/K → From now on, unless otherwise stated the abelian groups M of the example will always be given the G-actions above. With this convention, for any G-module M we have an isomorphisms of abelian groups

G HomZ[G](Z,M)= M

and M Z Z M = ⊗ [G] I M G · 50 Appendix: Galois cohomology

since Z[G]/IG = Z (induced by augmentation). Now we show how to build new modules from old ones. Let M and N be two G-modules. Then the set HomG(M,N) of all morphisms of G-modules between M and N can be made into a G-module by “conjugation” df −1 (σf)(m) = σf(σ m) for f HomG(M,N) and m M. Similarly, the tensor product of M and N over Z ∈ ∈ M N ⊗ can be made into a G-module by “diagonal action” σ(m n) =df σ(m) σ(n) for m M and n N. The next deﬁnition shows how to “lift” modules from subgroup⊗ s: ⊗ ∈ ∈ Deﬁnition 0.1.3 Let H G be a subgroup and N be an H-module. The induced module from N is the G-module ≤ G df IndH (N) = HomH (G, N) of all H-linear functions f: G N (i.e. f(h g) = h f(g) for all h H and g G) and where the → · · ∈ ∈ G-action is given by (σf)(g) = f(gσ) for all σ, g G. When H = 1 we simply write IndG(N). A ∈ G-module M is called induced if M = IndG(N) for some abelian group N.

Another way to “lift” an H-module is via the “base change” Z[G] Z[H] N, where the G-action is given by multiplication on the left component (in the tensor product, the⊗ left component is viewed as a right Z[H]-module via m σ =df σ−1 m for all σ H and m Z[G]). Base change is related to induced modules via the isomorphism· of G-modules· ∈ ∈

IndG(N) = Z[G] N ∼ ⊗ given by φ g φ(g−1). Here N is any abelian group with trivial G-action. Observe that an 7→ g∈G ⊗ induced G-moduleP M = Z[G] N is also induced as an H-module: if Hσ1,...,Hσn are right cosets of H then M = Z[H] N ′ where⊗N ′ = σ N. ⊗ 1≤i≤n i L Example 0.1.4 Let L K be a ﬁnite Galois extension with G = Gal(L/K). Then L+ is an induced G-module. In fact, by the⊃ normal basis theorem there exists ω L such that σ(ω) σ G is a basis ∈ { | ∈ } of L over K. Then we have a G-isomorphism K[G] = L+ given by a σ a σ(ω). Since ∼ σ∈G σ 7→ σ∈G σ K[G]= Z[G] K is induced by the above, so is L+. P P ⊗ The following lemma will be useful in arguments involving dimension shifting (which we will see later).

Lemma 0.1.5 Let M be an arbitrary G-module and denote by M0 the underlying abelian group. Then there is an injective map of G-modules ( M ֒ IndG(M → 0 sending m to the function φm: G M0 given by φm(σ) = σ m. There is also a surjective map of G-modules → · G Ind (M0) ։ M given by φ σφ(σ−1). 7→ σ∈G P Definition 0.1.6 Let M be a G-module and i 0. The i-th cohomology group of M is defined as ≥ i df i H (G, M) = ExtZ[G](Z,M) while the i-th homology group of M is defined as

df Z[G] Hi(G, M) = Tori (Z,M)

We briefly recall the definitions of Ext and Tor below, but we warn you that except for low degrees i =0, 1 the computations of these groups are not done directly from their definitions but rather via their functorial properties and some “vanishing theorems” that tell you sufficient conditions under which these groups are trivial. Proof of the Class Field Theory Axioms 51

A G-module I is called injective if the functor Hom ( , I) is exact. A G-module P is called G − projective if the functor HomG(P, ). For instance, free Z[G]-modules are projective. It can be shown that any G-module M can be embedded− into an injective module and it can also be written as a quotient of a projective module. Hence we can inductively construct an injective resolution of M, namely an exact sequence 0 M I0 I1 I2 → → → → →··· where the G-modules Ii are all injective, and similarly one may construct a projective resolution

P 2 P 1 P 0 M 0 ···→ → → → → i Now we can define ExtZ[G](Z,M) as follows. Choose an injective resolution of M as above and apply the G-fixed point functor Hom (Z, ) = ( )G to it. We obtain a complex G − − 0 M G (I0)G (I1)G (I2)G → → → → →··· Then ker (Ii)G (Ii+1)G i df → ExtZ[G](Z,M) = im (Ii−1)G (Ii)G → (Here we interpret I−1 = 0). Now an easy but rather tedious computation shows that the groups thus obtained are independent of the choice of the injective resolution of M. Alternatively, one can choose a projective resolution of Z P 2 P 1 P 0 Z 0 ···→ → → → → and apply the functor Hom ( ,M) to it, defining G −

i i+1 ker HomG(P ,M) HomG(P ,M) i df → ExtZ[G](Z,M) = i−1 i im HomG(P ,M) HomG(P ,M) → (Here we interpret P −1 = 0) Again, this is independent of the chosen projective resolution of Z, and it can be shown that either procedure, via projective resolutions of Z or injective resolutions of M, yield isomorphic groups. Z[G] For Tori (Z,M), the procedure is similar, except that Hom is replaced by the tensor product and the we use projective resolutions for both entries (in the tensor product M Z N we view M as a right ⊗ [G] Z[G]-module via m σ =df σ−1 m for σ G and m M). For instance, if P M 0 is a projective · · ∈ ∈ • → → resolution of M, applying the functor Z Z = /I we obtain ⊗ [G] − − G ·−

ker Pi/IG Pi Pi−1/IG Pi−1 Z[G] · → · Tori (Z,M)= im Pi+1/IG Pi+1 Pi/IG Pi · → · Example 0.1.7 (Cyclic groups) Let G be a cyclic group of order n and let σ be a generator of G. Then we have a projective resolution of Z

I N I N I ǫ - Z[G] - Z[G] - Z[G] - Z[G] - Z[G] - Z - 0 · · · where ǫ is the augmentation map, and I and N denote multiplication by σ 1 and 1+ σ + + σn−1, respectively. Hence we obtain − · · ·

M G if i =0

 ker NG  if i is odd Hi(G, M)=  I M  G ·  M G if i> 0 is even  NG(M)   52 Appendix: Galois cohomology

G where NG: M M is the norm map and IG is the augmentation ideal of Z[G]. Using the same projective resolution→ of Z, for homology we obtain M if i =0 I M  G ·  G  M Hi(G, M)=  if i is odd  NG(M) ker N  G if i> 0 is even  I M  G  · The main functorial property of Tor and Ext, and thus of the homology and cohomology groups, are their long exact sequences. For any short exact sequence of G-modules

0 A B C 0 → → → → one has long exact sequences 0 - H0(G, A)= AG - H0(G, B)= BG - H0(G, C)= CG δ0 - H1(G, A) - H1(G, B) - H1(G, C) δ1 δ2 - H2(G, A) - H2(G, B) - H2(G, C) - · · · and δ -3 H (G, A) - H (G, B) - H (G, C) · · · 2 2 2 δ-2 - - H1(G, A) H1(G, B) H1(G, C) δ A B C -1 H (G, A)= - H (G, B)= - H (G, C)= - 0 0 I A 0 I B 0 I C G · G · G · i for cohomology and homology respectively. The maps δ and δi are called connecting morphisms. The other maps are the natural ones induced by the maps A B and B C. → → Think of the short exact sequence as a way to “decompose” B into simpler modules, a submodule A and a quotient module C. If we know the homology/cohomology of A and C then the long exact sequence allows us to ﬁnd out the homology/cohomology of B. Example 0.1.8 Let ǫ be the augmentation map. From the exact sequence of G-modules - - ǫ- - 0 IG Z[G] Z 0 and the fact that Z[G] is free (and thus has trivial homology), we conclude that the connecting morphisms give isomorphisms H (G, Z)= H (G, I ) for all p 1. In particular, for p = 1 we have that p p−1 G ≥ IG ab H1(G, Z)= H0(G, IG)= 2 = G IG

ab df ab 2 where G = G/[G : G] is the maximal abelian quotient of G. The isomorphism G IG/IG is given by σ [G : G] (σ 1) I2 . ≈ · 7→ − · G The main vanishing theorem is Shapiro’s lemma. Lemma 0.1.9 (Shapiro) Let H G be a subgroup and N be an H-module. Then for all i 0 we have isomorphisms ⊂ ≥

i i G G H (H,N)= H (G, IndH (N)) and Hi(H,N)= Hi(G, IndH (N)) In particular, any induced G-module has trivial cohomology and homology. Proof (Sketch) For any any G-module M and any H-module N we have a canonical isomorphism G G HomG(M, IndH N) = HomH (M,N) and the functor IndH ( ) is exact. From these two properties it G − • follows that IndH ( ) preserves injectives, hence given an injective resolution 0 N I of N, we − G G • G → → obtain an injective resolution 0 IndH (N) IndH (I ) of IndH (N). The result follows by applying → → G HomG(Z, ) and using the fact that HomG(Z, IndH N) = HomH (Z,N). The proof for homology is similar. − Proof of the Class Field Theory Axioms 53

Corollary 0.1.10 If L K is a finite Galois extension with G = Gal(L/K) then H0(G, L+) = K+ and Hp(G, L+)=0 for p⊃ 1. ≥ Definition 0.1.11 The p-th Tate cohomology group is defined by

Hp(G, M) if p 1 ≥  M G  if p =0 p  N M H (G, M)=  G T  ker N  G if p = 1 IG M −  ·  H−p−1(G, M) if p 2  ≤−  The importance of the Tate cohomology groups is that it allows us to splice the long exact sequences of homology and cohomology into a single very long one. Given a short exact sequence of G-modules

0 A B C 0 → → → → we have a commutative diagram with exact rows - - - - H1(G, C) H0(G, A) H0(G, B) H0(G, C) 0

NG NG NG ? ? ? 0 - H0(G, A) - H0(G, B) - H0(G, C) - H1(G, A)

where the vertical arrows are induce by the norm maps. Hence we obtain an exact sequence

H−1(G, A) H−1(G, B) H−1(G, C) ···→ T → T → T H0 (G, A) H0 (G, B) H0 (G, C) → T → T → T H1 (G, A) H1 (G, B) H1 (G, C) → T → T → T →··· Example 0.1.12 (Periodicity of Tate cohomology for cyclic groups) If G is cyclic we have that

M G if p is even p  N M H (G, M)= G T  ker N  G if p is odd I M  G ·  0.2 Explicit Resolutions Here we show how to deﬁne homology and cohomology via an explicit projective resolution of Z: Deﬁnition 0.2.1 The standard resolution of Z is the projective resolution

d d d d -3 C˜ -2 C˜ -1 C˜ -0 Z - 0 · · · 2 1 0 where C˜ = Z[Gp+1] is the free Z-module with basis (σ , σ ,...,σ ) Gp+1 = G G (p + 1 times) p 0 1 p ∈ ×···× with “diagonal” G-action s (σ ,...,σ ) = (s σ ,...,s σ ), and where d : C˜ C˜ is given by · 0 p · 0 · p p+1 p+1 → p d(σ , σ ,...,σ )= ( 1)k(σ , σ ,...,σ , σ ,...,σ ) 0 1 p+1 − 0 1 k−1 k+1 p+1 0≤kX≤p+1

Applying Hom ( ,M) to the standard resolution, we obtain a complex of abelian groups G − d˜0 d˜1 d˜2 0 - C˜0(G, M) - C˜1(G, M) - C˜2(G, M) - · · · where ˜ ˜ ˜p df ˜ p+1 f(s σ0,...,s σp) = s f(σ0,...,σp) for all C (G, M) = functions f: G M · · · p+1 n → s G and (σ0,...,σp) G o ∈ ∈

54 Appendix: Galois cohomology

and d˜p: C˜p(G, M) C˜p+1(G, M) is given by → (d˜pf˜)(σ ,...,σ )= ( 1)kf˜(σ ,...,σ , σ ,...,σ ) 0 p+1 − 0 k−1 k+1 p+1 0≤kX≤p+1 We have an isomorphism between C˜p(G, M) and the abelian group Cp(G, M) of all functions from Gp to M: it takes f C˜p(M) to the function ∈ (σ , σ ,...,σ ) f˜(1, σ , σ σ , σ σ σ ,...,σ σ σ ) 1 2 p 7→ 1 1 2 1 2 3 1 2 · · · p The above complex is therefore isomorphic to the following “inhomogeneous” one d0 d1 d2 0 - C0(G, M) - C1(G, M) - C2(G, M) - · · · where dp: Cp(G, M) Cp+1(G, M) is given by → (dpf)(σ ,...,σ )= σ f(σ ,...,σ ) 1 p+1 1 · 2 p+1 + ( 1)kf(σ ,...,σ , σ σ , σ ,...,σ ) − 1 k−1 k · k+1 k+2 p+1 1≤Xk≤p + ( 1)p+1f(σ ,...,σ ) − 1 p Hence we obtain an explicit formula ker dp Hp(G, M)= im dp+1 An element of ker dp is called a p-cocycle while an element of im dp−1 is called a p-coboundary. Let 0 A B C 0 → → → → be a short exact sequence of G-modules. In terms of cocycles and coboundaries, one has a very explicit description of the connecting morphism δ: Hp(G, C) Hp+1(G, A) → as follows: given a p-cocycle f: Gp C representing an element ϕ = [f] Hp(G, C), we may lift it → ∈ to a function fˆ: Gp B. Then dpfˆ is in the image of the map Cp+1(G, A) Cp+1(G, B) induced by → → A B, so we may view it as a p-cocycle dpfˆ: Gp+1 A, and δ(ϕ) = [dpfˆ] Hp+1(G, A). → → ∈ Similarly, for homology one obtains an inhomogeneous complex d d d -3 C (G, M) -2 C (G, M) -1 C (G, M) - 0 · · · 2 1 0 where C (G, M)= Cp(G, M) and d : C (G, M) C (G, M) is given now by p p p → p (d f)(σ ,...,σ )= σ−1 f(σ, σ ,...,σ ) p 1 p−1 · 1 p−1 σX∈G + ( 1)k f(σ ,...,σ , σ σ, σ−1, σ ,...,σ ) − 1 k−1 k · k+1 p−1 1≤kX≤p−1 σX∈G + ( 1)p+1 f(σ ,...,σ , σ) − 1 p−1 σX∈G Example 0.2.2 We have that a 1-cocycle is a function f: G M such that f(στ)= σf(τ)+ f(σ). It is a coboundary if and only if it has the form f(σ) = σm →m for some m M. In particular, if the G-action on M is trivial, then all 1-coboundaries are trivial− and a 1-cocycle∈ is just a group morphism: H1(G, M) = Hom(G, M) in this case. Example 0.2.3 Let G be a cyclic group of order n generated by σ. Then one has an exact sequence of G-modules 0 Z Q Q/Z 0 → → → → As we shall see later, Hi(G, Q) = 0 for all i 1 and hence the connecting maps induce isomorphisms ≈ ≥ δ: Hi(G, Q/Z) Hi+1(G, Z) for all i 0. Then H1(G, Q/Z) = Hom(G, Q/Z) is a group of order n, → ≥ 1 2 generated by a morphism f: G Q/Z given by f(σ) = n mod Z, and hence H (G, Z) is also cyclic of order n, generated by the class→ of the 2-cocycle i j 1 if i + j n δf(σ , σ )= ≥ for 0 i,j

Theorem 0.2.4 (Hilbert’s Satz 90) Let L K be a ﬁnite Galois extension with Galois group G = Gal(L/K). Then ⊃ H1(G, L×)=0

Proof Let f: G L× be a 1-cocycle, and consider the element → m =df f(σ) σ(a) · σX∈G where a L× is chosen so that m = 0, which is possible by Dedekind’s independency of characters. Then for∈ every τ G we have that 6 ∈ m = f(τσ) τσ(a)= τ f(σ) f(τ) τσ(a)= τ(m) f(τ) · · · · σX∈G σX∈G

That is, f(τ)−1 = τ(m)/m, showing that f is a coboundary.

0.3 Dimension shifting; Inﬂation, Restriction, Corestriction We have seen how changing modules alters the cohomology groups via the long exact sequence. Now we show how changing the group alters the cohomology. Let f: G′ G be a group morphism and A be a G-module. We may view A as a G′-module as well via →

σ′ a =df f(σ′) a for a A, σ′ G′ · · ∈ ∈ ′ We denote this G′-module by f ∗A. Clearly AG is a subgroup of (f ∗A)G , hence the inclusion map deﬁnes a functorial map H0(G, A) H0(G′,f ∗A). By general properties of derived functors, this map in degree 0 extends uniquely for all →p 0 to a functorial map Hp(G, A) Hp(G′,f ∗A), compatible with the connecting maps. We recall the≥ proof of this fact since it relies on→ a recurrent technique in group homology/cohomology known as dimension shifting. First write an exact sequence of G-modules

0 A I B 0 → → → → with I injective (cohomologically trivial would do, for instance an induced module). Since Hp(G, I)=0 for all p > 0 we have that the connecting maps give isomorphisms δ: Hp−1(G, B) ≈ Hp(G, A) for all p> 1. Now, for p> 1, if we already know the map Hp−1(G, ) Hp−1(G′,f ∗ ) in→ dimension p 1 we may deﬁne it in dimension p via the composition − → − −

δ−1 ∂ Hp(G, A) - Hp−1(G, B) - Hp−1(G′,f ∗B) - Hp(G′,f ∗A) ≈ where ∂ denotes the connecting map with respect to the exact sequence of G′-modules

0 f ∗A f ∗I f ∗B 0 → → → → For the “base case” p = 1 the procedure is similar but we need to use the commutative diagram instead

0 - AG - IG - BG - H1(G, A) - 0 ∩ ∩ ∩

? ? ? ? ′ ′ ′ 0 - (f ∗A)G - (f ∗I)G - (f ∗B)G - H1(G′,f ∗A) - · · · Similar procedures work for homology and also for Tate cohomology groups. One may consider the more general case where A′ is a G′-module and g: A A′ is a group morphism which is compatible with f in the sense that g(f(σ′) a)= σ′ g(a) for all a →A and σ′ G′. Then the degree 0 map induced by g extends to a map Hp (G, A· ) Hp·(G′, A′) for all∈p. ∈ T → T 56 Appendix: Galois cohomology

Deﬁnition 0.3.1 Let H G be a subgroup and M be a G-module. We deﬁne the restriction map ≤

res: Hp (G, M) Hp (H,M) T → T to be the map induced by the inclusion map M G ֒ M H in degree 0 (take f: H ֒ G to be the inclusion map). → → Now suppose that H ⊳ G is normal. We deﬁne the inﬂation

inf: Hp (G/H,M H ) Hp (G, M) T → T to be the map induced by the identity (M H )G/H = M G in degree 0 (take f: G ։ G/H to be the quotient .(map and g: M H ֒ M to be the inclusion → For p 0, the restriction and inﬂation maps have a very simple description in terms of the standard resolution.≥ Given a p-cocycle f: Gp M, res([f]) is represented by the restriction f: Hp M of f to Hp. On the other hand, for a p-cocycle→ f: (G/H)p M H we have that inf([f]) is given by→ the p-cocycle → f˜: Gp M given by the composition →

f Gp ։ (G/H)p - M H ⊂ - M

where the unlabelled maps are the natural ones.

Theorem 0.3.2 (Inﬂation-restriction sequence) Let M be a G-module, and H be a normal subgroup of G. Suppose that Hp(H,M)=0 for p =1,...,q 1. Then −

inf res 0 - Hq(G/H,M H ) - Hq(G, M) - Hq(H,M)

is exact.

Proof The result follows easily for p = 1 by direct computation with cocycles. The general case follows by dimension shifting. Suppose that q 2 and consider an exact sequence of G-modules (see lemma 0.1.5) ≥

0 M IndG(M ) N 0 ( ) → → 0 → → ∗

Note that the middle term is induced also as an H-module. Since H1(H,M) = 0 by hypothesis we have that the sequence of G/H-modules

H 0 M H IndG(M ) N H 0 ( ) → → 0 → → ∗∗ G H G/H is still exact. The middle term Ind (M0) = Ind (M0) is induced as a G/H-module, and we have a commutative diagram

inf res 0 - Hq(G/H,M H ) - Hq(G, M) - Hq(H,M) 6 6 6

≈ ≈ ≈ inf res 0 - Hq−1(G/H,N H ) - Hq−1(G, N) - Hq−1(H,N)

where the vertical arrows are the connecting maps associated to ( ) and ( ), which are isomorphisms since the middle terms of ( ) and ( ) are cohomologically trivial.∗ Since H∗∗p(H,N)= Hp+1(H,M)=0 for p =1, 2,...,q 2 we have∗ that the∗∗ bottom row is exact by induction on q. Hence the top row is also exact. − Proof of the Class Field Theory Axioms 57

Remark 0.3.3 For the expert (but then you shouldn’t be reading this appendix!): the previous result follows directly from the Hochschild-Serre spectral sequence

Hp(G/H, Hq(H,M)) Hp+q(G, M) ⇒ Remark 0.3.4 Using the inflation map one can define cohomology for profinite groups as well. A group is profinite if it is the projective limit of finite groups. For instance, for any field k, its absolute Galois df group Gk = Gal(ksep/k) is profinite. If G = lim Gi, giving the discrete topology to the finite groups ←−i∈I Gi, G is made into a topological space as well, namely the projective limit of the topological spaces Gi. Since the product of discrete groups is compact by Tychonoff’s theorem and G is a closed subgroup of

i∈I Gi we have that G is compact. Now let M a continuous G-module, namely a G-module for which theQ action G M M is continuous. This means that for every m M the orbit G m is ﬁnite. Then we can deﬁne× the p→-th cohomology group as ∈ ·

Hp(G, M) =df lim Hp(G/H,M H ) ←−H where H runs over all open normal subgroups of G and the transition maps are given by inflation. Let us go back to the finite case. Let H G be a subgroup and M be a G-module. We define the H G ≤ norm map NG/H : M M via → df NG/H (m) = σ(m) σX∈S where S is a set of left cosets representatives of G/H. Clearly this does not depend on the choice of S and the above sum is G-invariant. Now dimension shifting allows us to extend this map to all other dimensions using an exact sequence of G-modules (see lemma 0.1.5)

0 M IndG(M ) N 0 → → 0 → → and the fact that the middle term is also induced as an H-module. Deﬁnition 0.3.5 The map cor: Hp (H,M) Hp (G, M) T → T induced by the norm map N : M H M G in degree 0 is called corestriction. G/H → Theorem 0.3.6 (Restriction-Corestriction) Let H G be a subgroup and M be a G-module. Then the composition ≤ p res- p cor- p HT (G, M) HT (H,M) HT (G, M) equals multiplication by [G : H]. In particular, Hp (G, M) is killed by G . T | | Proof For p = 0 we have that the above composition is

G H G M - M NG/H- M NGM NH M NGM where the ﬁrst map is the natural one. This composition is clearly multiplication by [G : H]. The general case now follows easily by dimension shifting.

Example 0.3.7 Let G be any p-Sylow subgroup of G. Then res: Hp(G, M) Hp(G ,M) is injective p → p on the p-primary components of these groups. In fact, cor res is multiplication by [G : Gp], which is prime to p, and thus is an automorphism on these p-primary◦ components.

p Example 0.3.8 We have that HT (G, Q) = 0 for all p. In fact, multiplication by G is an automorphism of Q. Since Hp(G, ) is a functor, this implies that multiplication by G is also| | an automorphism of p − | | HT (G, Q) which must then be zero by the theorem. 58 Appendix: Galois cohomology

0.4 Cup product Deﬁnition 0.4.1 There exists a unique family of maps

∪ Hp (G, A) Hq (G, B) - Hp+q(G, A B) T ⊗ T T ⊗ (tensor products over Z) called cup products, which are characterised by the following properties: 1. the cup product is a morphism of bifunctors (a.k.a natural transformations) in the pair (A, B); 2. for p = q = 0 the cup product is induced by the natural map AG BG (A B)G; ⊗ → ⊗ 3. the cup product is compatible with connection morphisms: if

0 A′ A A′′ 0 → → → → is an exact sequence of G-modules and B is a G-module such that

0 A′ B A B A′′ B 0 → ⊗ → ⊗ → ⊗ → is exact then the diagram

Hp (G, A′′) Hq (G, B) ∪- Hp+q(G, A′′ B) T ⊗ T T ⊗

δp id δp+q ⊗ ? ? Hp+1(G, A′) Hq (G, B) -∪ Hp+q+1(G, A′ B) T ⊗ T T ⊗ commutes. On the other hand, if

0 B′ B B′′ 0 → → → → is an exact sequence of G-modules and A is a G-module such that

0 A B′ A B A B′′ 0 → ⊗ → ⊗ → ⊗ → is exact then the diagram

Hp (G, A) Hq (G, B′′) ∪- Hp+q(G, A B′′) T ⊗ T T ⊗

id δq ( 1)p δp+q ⊗ − · ? ? Hp (G, A) Hq+1(G, B′) -∪ Hp+q+1(G, A B′) T ⊗ T T ⊗ commutes. Uniqueness of this family is easily proven by dimension shifting, while existence is given by the explicit formulas in cocycles. First we extend the standard resolution by setting C˜ = C˜∗ for p 1, −p p−1 ≥ ˜∗ ˜ p where Cp−1 is the dual of Cp−1 = Z[G ], namely the free Z-module with basis given by functions (σ∗,...,σ∗), which send (σ ,...,σ ) to 1 Z and every other basis element of C˜ to 0 Z. The 1 p 1 p ∈ p−1 ∈ boundary map d : C˜ C˜ is given by −p −p → −p−1 d (σ∗,...,σ∗)= ( 1)i(σ∗,...,σ∗,s∗, σ∗ ,...,σ∗) −p 1 p − 1 i i+1 p sX∈G 0≤Xi≤q and d : C˜ C˜ is given by d (σ )= (s∗). One may then compute the Tate cohomology groups 0 0 → −1 0 0 s∈G by applying HomG( ,M) to this sequenceP and computing the cohomology of the resulting complex. − Now deﬁne φ : C˜ C˜ C˜ as follows: p,q p+q → p ⊗ q Proof of the Class Field Theory Axioms 59

For p, q 0 • ≥ φ (σ ,...,σ ) = (σ ,...,σ ) (σ ,...,σ ) p,q 0 p+q 0 p ⊗ p p+q For p, q 1 • ≥ φ (σ∗,...,σ∗ ) = (σ∗,...,σ∗) (σ∗ ,...,σ∗ ) −p,−q 1 p+q 1 p ⊗ p+1 p+q For p 0, q 1 • ≥ ≥ ∗ ∗ ∗ ∗ ∗ ∗ φp,−p−q(σ1 ,...,σq )= (σ1,s1,...,sp) (sp,...,s1, σ1 ,...,σq ) X ⊗ ∗ ∗ ∗ ∗ ∗ ∗ φ−p−q,p(σ1 ,...,σq )= (σ1 ,...,σq ,s1,...,sp) (sp,...,s1, σq) X ⊗ φ (σ ,...,σ )= (σ ,...,σ ,s ,...,s ) (s∗,...,s∗) p+q,−q 0 p 0 p 1 q ⊗ q 1 X ∗ ∗ φ−q,p+q (σ0,...,σp)= (s1,...,sq ) (sq,...,s1, σ0,...,σp) X ⊗ where the sum runs over all (s ,...,s ) Gp. 1 p ∈ Then a straightforward but long check shows that (ǫ ǫ) φ0,0 = ǫ (here ǫ: C˜0 Z denotes the augmentation map) and that ⊗ ◦ → φ d = (d 1) φ + ( 1)p(1 d) φ p,q ◦ ⊗ ◦ p+1,q − ⊗ ◦ p,q+1 for all p, q Z (we omitted the indices of the coboundary maps d for notational clarity). Now given ∈ f Hom (C˜ , A) and g Hom (C˜ ,B) we deﬁne f g Hom (C˜ , A B) by ∈ G p ∈ G q ∪ ∈ G p+q ⊗ f g = (f g) φ ∪ ⊗ ◦ p,q and now it is easy to check that if f and g are cocycles, so is f g, and that its class depends only on the classes of f and g. A lengthy but easy check shows that the∪ pairing so deﬁned has the desired properties of the cup product. ∪ The following lemma can be easily proved by dimension shifting, and is left as an exercise for the reader since I’m running out of time and energy: Lemma 0.4.2 Let H G be a subgroup. We have ≤ 1. (a b) c = a (b c) in Hp+q+r(G, A B C) for a Hp (G, A), b Hq (G, B), c Hr (G, C); ∪ ∪ ∪ ∪ T ⊗ ⊗ ∈ T ∈ T ∈ T 2. a b = ( 1)pq b a under the isomorphism A B = B A for a Hp (G, A), b Hq (G, B); ∪ − · ∪ ⊗ ⊗ ∈ T ∈ T 3. res(a b) = res a res b Hp+q(H, A B) for a Hp (G, A) and b Hq (G, B); ∪ ∪ ∈ T ⊗ ∈ T ∈ T 4. cor a res b = cor a b for a Hp(H, A) and b Hq(G, B). ∪ ∪ ∈ ∈ Our last theorem shows that how the cup product enters in the periodicity of the cohomology of cyclic groups: Theorem 0.4.3 Let G be a cyclic group of order n and σ be a generator. Consider the 2-cocycle f: G G Z given by × → i j 1 if i + j n f(σ , σ )= ≥ for 0 i,j

Chapter 5

Bibliography

1. E. Artin and J. Tate, Class field Theory, Addison-Wesley. 2. Z. I. Borevich, I. R. Shafarevich, Number Theory, Academic Press. 3. J. W. S. Cassels, Local Fields, London Mathematical Society Student Text 3, Cambridge Uni- versity Press. 4. J. W. S. Cassels and A. Fröhlich, Algebraic Number Theory, Academic Press. 5. P. Gille and T. Szamuely, Central Simple Algebras and Galois Cohomology, Cambridge Univer- sity Press. 6. J. S. Milne, Algebraic Number Theory, course notes, available at http://www.jmilne.org/math/ 7. J. S. Milne, Class Field Theory, course notes, available at http://www.jmilne.org/math/ 8. J. S. Milne, Abelian Varieties, course notes, available at http://www.jmilne.org/math/ 9. J. Neukirch, Algebraic Number Theory, Grundlehren der mathematischen Wissenschaften 322, Springer-Verlag. 10. J. Neukirch, A. Schmidt, and K. Wingberg, Cohomology of Number Fields, Grundlehren der Mathematischen Wissenschaften 323, Springer-Verlag. 11. R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag. 12. W. Scharlau, Quadratic and Hermitian forms, Grundlehren der Mathematischen Wissenschaften, vol. 270, Springer-Verlag. 13. J-P. Serre, A Course in Arithmetic, Graduate Texts in Mathematics 7, Springer-Verlag 14. J-P. Serre, Local Fields, Graduate Texts in Mathematics 67, Springer-Verlag 15. S. S. Shatz, Profinite groups, Arithmetic, and Geometry, Annals of Mathematical Studies 67, Princeton University Press.