Cohomological interpretation of Brauer Groups 20172252 Seewoo Lee

Abstract

In this note, we introduce a centeral simple algebra and a Brauer group, which classifies central division algebra over a given field. We prove that there exists a canonical isomorphism between a relative Brauer group Br(L/k) and the second group cohomology H2(Gal(L/k),L×), which shows that the cohomology of group can be used to classify central division algebras over a given field. We also present some examples and computations.

1 Introduction

Let k be a field. A central simple algebra A over k is a finite dimensional associative algebra which is simple and center is k. For example, the quoternions H is a 4-dimensional central simple algebra over R. Our basic question is the following: Classify the central simple algebras over a given field k. This seems very hard problem at first glance. However, we will show that the Artin-Wedderburn theorem claims that every central simple algebra over k is isomorphic to a matrix algebra over some central division algebra over k which is determined unique up to isomorphism. Hence the theorem reduces the problem as Classify the central division algebras over a given field k. However, this isn’t an easy problem. To do this, we will define the Brauer group Br(k) of a given field, which is a group that classifies central division algebras over k. More precisely, each elements of Br(k) corresponds to the isomorphism class of the central division algebras over k. Computation of the Brauer group can be done by the computation of the second cohomology group, by the following main theorem we will prove: Br(L/k) ' H2(Gal(L/k),L×).

Using this, we will compute Brauer groups over various fields such as R, Fp, Qp, Q. Also, we will show that there exists a noncommutative division algebra of any given positive characteristic p > 0. In Appendix, we will introduce another interpretation of the second cohomology group H2(G, A) as certain group extensions of G by A.

Before we start, we will give some basic definitions. Definition 1. Let k be a field and A be a finite dimensional k-vector space. (1) A is k-algebra if A has a ring structure and k is contained in the center of A. (2) k-subalgebra of A is a subring of A containing k. (3) k-algebra homomorphism ϕ : A → B is a ring homomorphism with ϕ(a) = a for all a ∈ k. (4) Oppsite Aopp is the k-algebra A with same addition but with reversed multiplication α·β := βα. l (5) For a basis e1, . . . , en for A as a k-vector space, the constants {aij} that satisfies n X l eiej = aijel l=1 are called the structure constants of A relative to the basis (ei)i.

2 Simple Algebras; Semisimple Modules

In this section, we define CSA, central simple algebras over a field k, and study some basic properties. At last, we will show that classification of CSA’s can be reduced to the classification of central division algebras over k (Artin-Wedderburn theorem), and the second one can be done by computing so called Brauer groups.

2.1 Semisimple Modules

In this article, A-modules are finitely generated left A-modules. Let V be an A-module. If V is a such A-module and A is finite dimensional k-vector space, then V is also a finite dimensional k-vector space with a k-algebra homomorphism A → Endk(V ) (representation of A on V ).

Definition 2. (1) V is a faithful A-module if the map A → Endk(V ) is injective. (2) V is a simple A-module if V 6= 0 and the only A-submodule of V are 0 and B itself. (3) V is a semisimple A-module if it is isomorphic to a direct sum of simple A-modules. (4) V is indecomposable if it can not be written as a direct sum of two nonzero A-modules. Theorem 1. Every A-module V admits a filtration

V = V0 ⊃ · · · ⊃ Vr = 0 whose quotients Vi/Vi+1 are simple A-modules. If

V = W0 ⊃ · · · ⊃ Ws = 0 is a second such filtration, then r = s and there is a permutation σ of {0, . . . , r − 1} such that Vi/Vi+1 ' Wσ(i)/Wσ(i)+1 for all i.

Proof. Existence follows from finite dimensionality of V . Uniqueness can be proved as in the Jordan- H¨oldertheorem.

Corollary 1. If V ' V1 ⊕ · · · ⊕ Vr ' W1 ⊕ · · · ⊕ Ws with all A-modules Vi,Wj’s are simple, then r = s and there exists σ ∈ Sr such that Vi ' Wσ(i) for all i. P Proposition 1. Let V be an A-module. Assume that V = i∈I Si for some simple submodules Si L of V . Then for every submodule W of V , there exists a subset J if I such that V = W ⊕ i∈J Si.

Proof. Use Zorn’s lemma; choose maximal J among the subsets of I such that the sum SJ = P j∈J Sj is direct and W ∪ SJ = 0. Then we can prove that W ⊕ SJ = V . Corollary 2. TFAE: (a) V is semisimple; (b) V is a sum of simple submodules; (c) every submodule of V has a complement. Corollary 3. Sums, submodules and quotient modules of semisimple modules are semisimple.

2 2.2 Simple k-algebras

Definition 3. A k-algebra A is simple if it contains no proper two-sided ideals other than 0.

Example 1. 1. Mn(k), the ring of n × n matrices over k, is a simple k-algebra. 2. A is a division algebra (skew field) if every nonzero element has an inverse. Every division algebra is simple. Also, matrix ring over a division algebra is also simple. 3. For a, b ∈ k×, define a quaternion algebra H(a, b) be the k-algebra with basis 1, i, j, ij (as a k-vector space) and with the multiplication determined by i2 = a, j2 = b, , ij = −ji.

One can show that H(a, b) is either a division algebra or isomorphic to M2(k), hence simple.

2.3 Centralizers

Definition 4. Let A be a k-subalgebra of B. Then the centralizer of A in B is

CB(A) = {b ∈ B : ba = ab ∀a ∈ A} which is also a k-subalgebra of B.

Example 2. 1. We have CMn(k)(k · In) = Mn(k) and CMn(k)(Mn(k)) = k · In.

2. If A ⊂ Mn(k) is a set of diagonal matrices in Mn(k), then CMn(k)(A) = A.

For both cases, we have C(C(A)) = A. This can be generalized for any faithful semisimple A- modules. Theorem 2 (Double centralizer theorem). Let A be a k-algebra and V be a faithful semisimple A-module. Then C(C(A)) = A (centralizers taken in Endk(V )).

Proof. Let D = B(A) and B = C(D). Clearly A ⊂ B and the reverse inclusion follows from the next lemma when we take v1, . . . , vn to generate V as a k-vector space.

Lemma 1. For any v1, . . . , vn ∈ B and b ∈ B, there exists a ∈ A s.t. avi = bvi holds for 1 ≤ i ≤ n.

Proof. We will prove this for n = 1 first. Since Av1 is an A-submodule of V , by the Proposition 1, we can find W ≤ V s.t. V = Av1⊕W . Consider the projection π : V → V defined as (av1, w) 7→ (av1, 0). Since π is A-linear, π ∈ D and π(v) = v if and only if v ∈ Av1. Now

π(bv1) = b(πv1) = bv1 so bv1 ∈ Av1.

For general n, let W be the direct sum of n copies of V with A acting diagonally, i.e.

a(v1, . . . , vn) = (av1, . . . , avn), a ∈ A, vi ∈ V.

Then W is a semisimple A-module and we can check that CEndk(W )(A) = Mn(D) = Mn(CEndk(V )(A)) and CEndk(W )(Mn(D)) = B · In. Now apply the case n = 1 to A, W, b and (v1, . . . , vn) ∈ W .

3 2.4 Classification of simple k-algebras

In this subsection, we prove the Artin-Wedderburn theorem, which claims that the classification of central simple k-algebras can be reduced to the classification of central division algebras.

Lemma 2 (Schur’s lemma). If the A-module S is simple, then EndA(S) is a division algebra.

Proof. Let γ : S → S be an A-linear map. Then ker(γ) is an A-submodule of a simple A-module S, so it is either S or 0. In the first case, γ = 0, and in the second case it is an isomorphism, i.e., it has an inverse that is also A-linear.

Theorem 3 (Artin-Wedderburn). Every simple k-algebra A is isomorphic to Mn(D) for some n and some division k-algebra D.

Proof. Choose any simple A-module S, for example, any minimal left ideal of A. Then A acts faithfully on S, and let D = CEndk(S)(A). By the double centralizer theorem, CEndk(S)(D) = A, so A ' EndD(S). By Schur’s lemma, D ' EndA(S) is a division algebra, so S is a free D-module and n opp EndD(S) ' EndD(D ) ' Mn(D ).

2.5 Modules over simple k-algebras

Proposition 2. For every simple k-algebra A, the simple submodules of AA are the minimal left ideals in A; any two such ideals are isomorphic as left A-modules, and A is a direct sum of its minimal left ideals.

Proof. By the Artin-Wedderburn theorem, we can assume A = Mn(D) for some division algebra D. Note that every minimal left ideals of Mn(D) has a form of L({j}) where L(I), I ⊂ {1, 2, . . . , n} is a set of matrices whose jth columns are zero for j 6∈ I.

Theorem 4. Let A be a simple k-algebra, and let S-be a simple A-module. Then every A-module V is isomorphic to a direct sum of copies of S. In particular, it is semisimple.

Proof. Let S0 be a minimal left ideal of A. Let e1, . . . , er be a set of generators for V as an A-module. The map r X (a1, . . . , ar) 7→ aier i=1 realizes V as a quotient of a sum of copies of AA, and hence as a quotient of a sum of copies of S0. Thus, V is a sum of simple submodules, each isomorphic to S0, and the Proposition 1 shows that V is isomorphic to a direct sum of copies of S0. Corollary 4. Let A be a simple k-algebra. Then any two simple A-modules are isomorphic, and any two A-modules having the same dimension over k are isomorphic.

Proof. Obvious from the previous Theorem.

4 Corollary 5. The integer n in the Artin-Wedderburn theorem is uniquely determined by A, and D is uniquely determined up to isomorphism.

Proof. If A ' Mn(D), then D ' EndA(S) for any simple A-module S. Thus the statement follows from the Corollary 4.

3 Definition of the Brauer Group

Now we study some properties of tensor products and define the most important object in this paper, Brauer groups.

3.1 Tensor products of algebras

Let A, B be k-algebras and A ⊗k B be the tensor product of A and B as k-vector spaces. There is a unique k-bilinear multiplication on A ⊗k B, (a ⊗ b)(a0 ⊗ b0) = aa0 ⊗ bb0, a, a0 ∈ A, b, b0 ∈ B and this gives a k-algebra structure on A ⊗k B. Proposition 3. (1) For any k-algebras A, B, C, we have canonical isomorphisms;

A ⊗k B ' B ⊗k A, A ⊗k (B ⊗k C) ' (A ⊗k B) ⊗k C. (2) For any k-algebras A, B, we have

A ⊗k Mn(B) ' Mn(A ⊗k B)

(3) Mn(k) ⊗ Mm(k) ' Mmn(k).

3.2 Centralizers in tensor products

The next proposition shows that taking centralizers preserves tensor products. Proposition 4. Let A, A0 be k-algebras with subalgebras B and B0. Then 0 0 0 0 CA⊗kA (B ⊗k B ) ' CA(B) ⊗k CA (B ).

In particular, if Z(A) is the center of A, then Z(A ⊗k B) ' Z(A) ⊗k Z(B).

0 0 0 Proof. Clearly C(B ⊗k B ) ⊃ C(B) ⊗k C(B ). Let (fi)i be a basis for A as k-vector space. Then 0 0 (1 ⊗ fi)i is a basis for A ⊗k A as an A-module, and so an element α of A ⊗k A can be written P uniquely in the form α = i αifi, αi ∈ A. Let β ∈ B. Then α commutes with β ⊗ 1 if and only 0 0 if βαi = αiβ for all i. Therefore, the centralizer of B ⊗ 1 in A ⊗ A is C(B) ⊗ A . Similarly, the centralizer of 1 ⊗ B0 in C(B) ⊗ A0 is C(B) ⊗ C(B0). Certainly, C(B ⊗ B0) ⊂ C(B ⊗ 1), and so C(B ⊗ B0) is contained in C(B) ⊗ A0, and, in fact, is contained in the centralizer of 1 ⊗ B0 in C(B) ⊗ A0, which is C(B) ⊗ C(B0).

5 In particular, we have Z(A ⊗k B) = Z(A) ⊗k Z(B). Corollary 6. The center of a simple k-algebra is a field.

Proof. We have Z(Mn(D)) ' Z(Mn(k)) ⊗k Z(D) ' k ⊗k Z(D) ' Z(D), where Z(D) is a field. Definition 5. A k-algebra is central if Z(A) = k.

The corollary shows that every simple k-algebra is central simple over a finite extension of k. We will call central simple algebra as CSA.

3.3 Primordial elements

Definition 6. Let V be a k-vector space (possibly infinite dimensional) and let (ei)i∈I be a basis P for V . For v ∈ V , define J(v) = {i ∈ I | ai 6= 0} where v = i aiei.(J(v) is a finite subset of I.) For a subspace W of V , a nonzero element w ∈ W is primordial relative to the basis (ei)i∈I if 0 0 J(w) is minimal among the sets J(w ) for 0 6= w ∈ W and ai = 1 for some i in the expression P w = i aiei. Proposition 5. (a) Let w be a nonzero element of W such that J(w) is minimal, and let w0 be a second nonzero element of W . Then J(w0) ⊂ J(w) if and only if w0 = cw for some c ∈ k, in which case J(w) = J(w0). (b) The set of primordial elements of W spans it.

0 0 P Proof. (a) Clearly, if w = cw and c 6= 0, then J(w ) = J(w). For the converse, let w = i∈J(w) aiei 0 P 0 0 and w = i∈J(w0) biei be nonzero elements of W with J(w ) ⊂ J(w). If we choose j ∈ J(w ), then −1 0 −1 0 J(w − ajbj w ) ⊂ J(w)\{j}, and so w = ajbj w by the minimality of j(w). P (b) Use induction on the size of |J(w)|. Let w = i∈J(w) aiei ∈ W and assume w 6= 0. Among nonzero elements w0 of W with J(w0) ⊂ J(w), choose one s.t. J(w0) has the fewest possible number of elements. Then w = cw0 will be primordial for some c 6= 0, say w = P b e with b = 1. 0 0 i∈J(w0) i i j Now w = ajw0 + (w − ajw0) with w − ajw0 ∈ W and |J(w − ajw0)| < |J(w)|. Hence by inductive hypothesis, w − ajw0 is a linear combination of primordial elements, and so is w.

The above results also hold when k is a division algebra.

3.4 Simplicity of tensor products

To define Brauer group, we have to show that the tensor product of two CSAs over k is again a CSA over k. For this, we need a Lemma.

Lemma 3. Let A be a k-algebra, and let D be a division algebra with center k. Then every two-sided ideal A in A ⊗k D is generated as a left D-module by a := A ∩ (A ⊗ 1).

6 Proof. Define a left D-module structure on A ⊗k D as

δ(α ⊗ δ0) := α ⊗ δδ0, α ∈ A, δ, δ0 ∈ D.

The ideal A of A ⊗k D is a D-submodule of A ⊗k D. Let (ei) be a basis for A as a k-vector space. Then (ei ⊗ 1) is a basis for A ⊗k D as a left D-module. Let α ∈ A be primordial with respect to this basis, say X X α = δi(ei ⊗ 1) = ei ⊗ δi. i∈J(α) i∈J(α) P For any nonzero δ ∈ D, αδ ∈ A and αδ = i∈I (δiδ)(ei ⊗ 1). In particular, J(αδ) = J(α), and so 0 0 0 αδ = δ α for some δ ∈ D by the Proposition 5 (a). As some δj = 1, this implies that δ = δ , so each δi commutes with every δ ∈ D and δi ∈ Z(D) = k, so α ∈ A ⊗ 1. This implies that every primordial element of A is in A ⊗ 1, which completes the proof by the Proposition 5 (b).

Proposition 6. The tensor product of two simple k-algebras, at leas one of which is central, is again simple.

Proof. By Artin-Wedderburn’s theorem, we can assume that one of the algebras is Mn(D) for some division algebra D with Z(D) = k. If A is the second simple k-algebra, then the previous Lemma 0 0 shows that A ⊗k D is simple, so A ⊗k D ' Mm(D ) for some m and a division algebra D . This gives 0 0 A ⊗k Mn(D) ' Mn(A ⊗k D) ' Mn(Mm(D )) ' Mmn(D ) and this is simple.

Corollary 7. The tensor product of two central simple k-algebras is again central simple.

If A is a central simple algebra over k, we have a natural homomorphism

opp 0 0 A ⊗k A → Endk(V ), a ⊗ a 7→ (v 7→ ava ).

opp The map is injective since A ⊗k A is simple and kernel of the map does not contain 1. Since both have k-dimension [A : k]2 = (dim V )2, the map is an isomorphism.

Corollary 8. opp A ⊗k A ' Endk(A) ' Mn(k), n = [A : k].

Here we use the notation [A : k] = dimk A.

3.5 The Noether-Skolem Theorem

The following theorem is very useful.

Theorem 5 (Noether-Skolem). Lef f, g : A → B be homomorphisms from k-algebra A to B. If A is simple and B is central simple, then there exists b ∈ B× such that f(a) = b · g(a) · b−1 for all a ∈ A, i.e. f and g differ by an inner automorphism of B.

7 n n Proof. If B = Mn(k) = Endk(k ), the homomorphisms give A-module structures on k , and two modules should be isomorphic since they have the same dimension. Then such isomorphism is given by an element b ∈ Mn(k) satisfying f(a) · b = b · g(a). For general case, consider the homomorphisms

opp opp f ⊗ 1, g ⊗ 1 : A ⊗k B → B ⊗k B ' Mn(k).

opp opp Since B⊗k B ' Mn(k) for some n, the first part of the proof shows that there exists b ∈ B⊗k B s.t. (f ⊗ 1)(a ⊗ b0) = b · (g ⊗ 1)(a ⊗ b0) · b−1 for all a ∈ A, b0 ∈ Bopp. Take a = 1 and then (1 ⊗ b0) = b · (1 ⊗ b0) · b−1 for all b ∈ Bopp, so opp 0 opp b ∈ CB⊗kB (k ⊗k B ) = B ⊗k k, i.e. b = b0 ⊗ 1 for some b0 ∈ B. Take b = 1 and we find

−1 f(a) ⊗ 1 = (b0 · g(a) · b0 ) ⊗ 1 for all A ∈ A.

Corollary 9. Let A be a central simple algebra over k and B1,B2 ⊂ A be a simple sub k-algebra. −1 If h : B1 → B2 is an k-algebra isomorphism, then it extends to eh : A → A s.t. eh(b) = aba .

Proof. Lef f : B1 ,→ A be an inclusion map and g : B1 → B2 ,→ A be a composition of h with an inclusion map. By Noether-Skolem theorem, there exists b ∈ A× s.t. g(a) = b · f(a) · b−1 for all −1 a ∈ B1, which is same as h(a) = bab . Corollary 10. All automorphisms of a central simple k-algebra are inner.

3.6 Definition of the Brauer group

Now we can define the Brauer group of given field (and a relative Brauer group of given field extension).

Definition 7. Let A, B be central simple algebra over k. (1) A and B are similar (A ∼ B) if A⊗k Mn(k) ' B⊗k Mm(k) for some m, n. (This is an equivalence relation). (2) Define the Brauer group Br(k) as

Br(k) := {[A] | A is a central simple algebra over k.} with a group structure given by [A][B] := [A ⊗k B]. −1 opp Note that the operation is well-defined, [Mn(k)] is an identity element and [A] = [A ]. Also, Br(k) is an abelian group.

If [A] = [B], then A and B have same division ring parts. So we can think Brauer group Br(k) as a classifying object that classifies division algebra over given field k. Here are some examples that we will see again.

8 Example 3. (1) If k is algebraically closed, Br(k) = 0. More generally, Br(k) = 0 if k is quasi- algebraically closed. (A field k is quasi-algebraically closed if every non-constant homogeneous poly- nomial P over k has a non-trivial zero provided the number of its variables is more than its degree.) (2) Br(R) = {[R], [H]}' Z /2 Z. (3) By Wedderburn’s theorem on finite division algebra, Br(k) = 0 whenever k is a finite field. (4) Brauer group of a nonarchimedean local field is canonically isomorphic to Q / Z. (5) (Albert, Brauer, Hasse, Noether) For a number field K, there is an exact sequence

M Σ 0 → Br(K) → Br(Kv) −→ Q / Z → 0. v

3.7 Extension of the base field

Proposition 7. Let A be a central simple algebra over k and K ⊃ k is another field. Then A ⊗k K is a central simple algebra over K.

Proof. By the Proposition 4, we have Z(A⊗k K) = Z(A)⊗k K = k ⊗k K ' K. Also, by the Lemma 3, every two-sided ideal in A ⊗k K is generated as an A-module by its intersection with K, and therefore is 0 or A ⊗k K. In the general case, A ' Mn(D) and so

A ⊗k K ' Mn(D) ⊗k K

' (Mn(k) ⊗k D) ⊗k K

' Mn(k) ⊗k (D ⊗k K)

' Mn(D ⊗k K)

' Mn(K) ⊗K (D ⊗k K) which is simple.

Corollary 11. For every central simple algebra A over k,[A : k] is a square.

Proof. [A : k] = [A ⊗k k : k] and A ⊗k k ' Mn(k).

Let L be a (possibly infinite) field extension of k. Then the map

Br(k) → Br(L),A 7→ A ⊗k L defines a homomorphism since

0 0 Mn(k) ⊗k L ' Mn(L), (A ⊗k L) ⊗L (A ⊗k L) ' (A ⊗k A ) ⊗k L Let’s denote the kernel of the homomorphism by Br(L/k), which is a group of similarity class represented by central simple k-algebras A such that A ⊗k L ' Mn(L) for some n. We call A is split by L and L is a splitting field of A, if [A] ∈ Br(L/k). Proposition 8. [ Br(k) = Br(K/k) K⊂k [K:k]<∞

9 Proof. We have to show that every CSA A over k is split by a finite extension of k. From A ⊗k k = Mn(k), there exists a basis (eij)1≤i,j≤n for A ⊗k k s.t. eijelm = δjleim for all i, j, l, m. Since A ⊗k k = ∪[L:k]<∞A ⊗k L, there exists a finite extension field K over k s.t. all eij lie in A ⊗k K, and it follows that A ⊗k K ' Mn(K).

4 The Brauer Group and Cohomology

Now we prove the main theorem: there exists a natural isomorphism

H2(L/k) = H2(Gal(L/k),L×) ' Br(L/k).

Using this, we will compute Brauer groups of some fields by computing cohomology groups.

4.1 Maximal subfields

Theorem 6. Let B ⊂ A be central k-algebras. Then C = CA(B) is simple and CA(C) = B. Moreover, [B : k][C : k] = [A : k]. (This is a generalized version of the double centralizer theorem.)

Proof. Let V denote B regarded as a k-vector space. Then B and Bopp act on V , by right and left multiplication respectively, and each is the centralizer of the other.

Consider the CSA A ⊗k Endk(V ). Proposition 4 shows that CA⊗kEndk(V )(B ⊗ 1) = C ⊗k Endk(V ) opp and CA⊗kEndk(V )(1 ⊗ B) = A ⊗k B . By applying Nother-Skolem theorem to b 7→ b ⊗ 1 and b 7→ 1 ⊗ b (both are maps from B to A ⊗k Endk(V )), we obtain an invertible element u of this k-algebra s.t. b ⊗ 1 = u(1 ⊗ b)u−1 for all b ∈ B. Clearly then

C(B ⊗ 1) = u · C(1 ⊗ B) · u−1

opp so the centralizers are isomorphic. Therefore C ⊗k Endk(V ) is simple because A ⊗k B is simple, and this implies that C itself is simple because, for every ideal a of C, a ⊗k Endk(V ) is an ideal in 2 C ⊗k Endk(V ). As Endk(V ) has degree [B : k] over k,

2 [C ⊗k Endk(V ): k] = [C : k][B : k] , and obviously opp [A ⊗k B : k] = [A : k][B : k]. On comparing these equalities, we find that

[A : k] = [B : k][C : k].

If B0 denotes the centralizer of C in A, then B0 ⊃ B. But after above, [A : k] = [C : k][B0 : k], so [B : k] = [B0 : k] and B = B0.

Corollary 12. If in the statement of the Theorem 6, B has center k, then so also does C, and the canonical homomorphism B ⊗k C → A is an isomorphism.

10 Proof. The centers of B and C both equal B ∩ C, and so B central implies C central. Therefore the k-algebra B ⊗k C is central simple, which implies that B ⊗k C → A is injective. It is surjective because the algebras have the same dimension over k.

Corollary 13. Let A be a central simple algebra over k, and L be a subfield of A containing k. TFAE:

1. L = CA(L) 2.[ A : k] = [L : k]2 3. L is a maximal commutative k-subalgebra of A.

Proof. (a) ⇔ (b). Because L is commutative, it is contained in C(L), but

[A : k] = [L : k][C(L): k] and so C(L) = L if and only if [A : k] = [L : k]2. (b) ⇒ (c). Let L0 be a maximal commutative subalgebra containing L. Then L0 ⊂ C(L), and so

[A : k] ≥ [L : k][L0 : k] ≥ [L : k]2.

Thus (b) implies that [L0 : k] = [L : k] and L0 = L. (c)⇒(a). If L 6= C(L), then there exists γ ∈ C(L)\L. Now L[γ] is a commutative subalgebra of A, contradicting the maximality of L.

Corollary 14. The maximal subfields containing k of a central division k-algebra D are exactly those with degree p[D : k] over k.

Proof. Any commutative k-subalgebra of D is an integral domain of finite degree over k, and hence is a field.

Corollary 15. Let A be a CSA over k. A field L of finite degree over k splits A if and only if there exists an algebra B similarly to A containing L and [B : k] = [L : k]2. In particular, every subfield L of A of degree [A : k]1/2 over k splits A.

opp opp Proof. Suppose L splits A. Then L also splits A , say, A ⊗k L = Endk(V ). This equality opp opp states that CEndk(V )(L) = A ⊗k L and so CEndk(L)(A ⊗k L) = L by Theorem 6. Let B = opp opp CEndk(V )(A ). Then L ⊂ B and by Corollary 12, B is central simple and that B ⊗k A ' opp Endk(V ). On tensoring both sides with A and using that A ⊗k A is a matrix algebra, we find that B ∼ A.

For the converse, it suffices to show that L splits B. Because L is commutative, L = Lopp ⊂ Bopp, and because [L : k] = p[B : k], L is equal to its centralizer in Bopp. Therefore the centralizer of opp opp 1 ⊗ L in B ⊗k B is B ⊗k L. With an identification B ⊗k B ' Endk(B), the centralizer of L becomes identified with EndL(B). Corollary 16. Let D be a central division algebra of degree n2 over k, and let L be a field of degree n over k. then L splits D if and only if L can be embedded in D (i.e. there exists a homomorphism of k-algebra L → D).

11 Proof. If L embeds into D, then it splits D by the previous corollary. Conversely, if L splits D, then there exists a CSA B over k containing L, similar to D, and of degree [L : k]2. But B ∼ D implies B ' Mm(D) for some m and the condition on the degrees implies m = 1. Proposition 9. Every central division algebra over k contains a maximal subfield separable over k.

Proof. Let D be a central division algebra over k, and let L be a subfield that is maximal among the separable subfields of D. If L is not a maximal subfield of D, then its centralizer D0 is a central division algebra over L not equal to L. The next lemma shows that D0 contains a separable subfield properly containing L, which contradicts to the definition of L.

Lemma 4. Every central division algebra over k and not equal to it contains a subfield separable over k and not equal to it.

Proof. Let D be a central division algebra over k. Choose a basis (ei)1≤i≤n2 for D with e1 = 1, and P m P let x = i aiei be an element of D. For every integer m, x = i Pi(m; a1, . . . , an2 )ei where Pi are polynomials in the aj’s with coefficients in k that depend only on the structure constants for D. Therefore, if K is a field containing k and X x = ai(ei ⊗ 1), ai ∈ K, i P is an element of D ⊗k K, then i Pi(m; a1, . . . , an2 )ei ⊗ 1 with the same polynomials Pi.

When k is perfect we know the lemma, and so we may assume that k has characteristic p 6= 0 and is infinite. Suppose that k[a] is purely inseparable over k for every a ∈ D. Then apr ∈ k for some r > 0, and one sees easily that there exists an r that works for every element of D. For this particular r r r and i 6= 1, Pi(p ; a1, . . . , an2 ) = 0 for all a1, . . . , an2 ∈ k, which implies that Pi(p ; X1,...,Xn2 ) is pr the zero polynomial. Therefore, a ∈ k for every a ∈ D ⊗k k. But D ⊗k k ' Mn(k), and so

r diag(1, 0,..., 0) = diag(1, 0,..., 0)p ∈ k, which implies that n = 1.

Corollary 17. For every field k, [ Br(k) = Br(L/k). k⊂L⊂ksep [L:k]<∞

Proof. The proposition shows that every element of Br(k) is split by a finite separable extension, and therefore by a finite Galois extension.

12 4.2 Proof of the main theorem

Now we will prove the main theorem. We construct a map from Br(L/k) to H2(L/k), and show that the map is an isomorphism. To do this, we will construct bijections

[A]7→[γ(A)] Br(L/k) H2(L/k) [A(ϕ)]←[ϕ] [

A(L/k)/ ' where A(L/k)/ ∼ is a set of isomorphism classes of certain CSA over k that we will define. The horizontal maps are desired isomorphisms between the Brauer group and the second cohomology group.

Let A(L/k) be a class of CSA over k containing L s.t. [A : k] = [L : k]2. By the Corollary 13, L = CA(L) and L is a maximal subfield of A containing k. Let G = Gal(L/k). For each σ ∈ G, × −1 there exists eσ ∈ A s.t. σ(x) = eσ · x · eσ for all x ∈ L, by the Noether-Skolem therem. Also, if fσ × −1 −1 is another element of A satisfying the same property, then fσ eσ · x = x · fσ eσ for all x ∈ L, i.e. −1 × fσ eσ ∈ CA(L) = L. Hence the choice of eσ is unique up to multiplication by an element of L .

Now we can rewrite the previous equation about eσ as

eσ · x = σ(x) · eσ. (1)

Since both eσeτ and eστ satisfies the above equation (we have eσeτ · x = σ(τ(x)) · eσeτ and eστ · x = × σ(τ(x)) · eστ ), there exists ϕ(σ, τ) ∈ L s.t.

eσeτ = ϕ(σ, τ)eστ . (2)

Since A is associative, we should have (eρeσ)eτ = eρ(eσeτ ) for all ρ, σ, τ ∈ G. This implies

(eρeσ)eτ = ϕ(ρ, σ)eρσeτ = ϕ(ρ, σ) · ϕ(ρσ, τ)eρστ

= eρ(eσeτ ) = eρ(ϕ(σ, τ)eστ ) = ρ(ϕ(σ, τ))ϕ(ρ, στ)eρστ so we have ϕ(ρ, σ)ϕ(ρσ, τ) = ρ(ϕ(σ, τ)) · ϕ(ρ, στ), which implies that ϕ : G × G → L× is a 2-cocycle. We can check that the cohomology class of ϕ, 2 0 × [ϕ] ∈ H (L/k), is independent of the choice of eσ. If eσ = uσeσ is another choice with uσ ∈ L , then the corresponding 2-cocycle ϕ0 should satisfy

0 0 0 0 0 −1 eσeτ = ϕ (σ, τ)eστ ⇔ ϕ (σ, τ) = uσ · σ(uτ ) · uστ · ϕ(σ, τ).

× 0 If we consider {uσ}σ∈G as u : G → L , then the above equation implies ϕ = (du) · ϕ (du is a 2-coboundary) and [ϕ0] = [ϕ]. From this, we have a map γ : A(L/k) → H2(L/k).

Proposition 10. γ is surjective, and its fibres are the isomorphism classes.

Proof. We need a lemma.

13 Lemma 5. The set (eσ)σ associated to the algebra A ∈ A(L/k) forms a basis of A as a (left) L-vector space.

Proof. Since [A : L] = [A : k]/[L : k] = n, it suffices to show that eσ are linearly independent. Suppose not, and let (eσ)σ∈J be a maximal linearly independent set. If τ 6∈ J, then X eτ = aσeσ σ∈J for some aσ ∈ L. For a ∈ L, we have X eτ a = τ(a) · eτ = τ(a) · aσeσ σ∈J and X X eτ a = aσeσ · a = aσ · σ(a) · eσ. σ∈J σ∈J

From this, we have τ(a) · aσ = σ(a) · aσ for all σ ∈ J. For at least one σ ∈ J, aσ 6= 0, and then the equation shows that τ = σ, which contradicts to τ 6∈ J. Hence J = G.

Now A is uniquely determined by the following properties: A ⊃ L;(eσ)σ∈G is a basis of A as an L-vector space; multiplication in A satisfies (1) and (2).

First, assume that A, A0 ∈ A(L/k) are isomorphic elements with an isomorphism f : A → A0. By the Corollary 9, we can choose f so that f(L) = L and f|L = idL. If eσ satisfies the condition (1) 0 0 for A, then f(eσ) also satisfies the condition (1) for A . With the choices (eσ) and (f(eσ)), A and A define the same cocycle. Hence γ induces a map A(L/k)/ ≈→ H2(L/k). By abusing the notation, we will denote this map as γ again.

It is easy to show that this γ is injective. Let A, A0 ∈ A(L/k) s.t. γ(A) = γ(A0), i.e. there exists 0 0 P P 0 (eσ) ⊂ A and (eσ) ⊂ A s.t. both defines a same 2-cocycle. Then the map σ aσeσ 7→ σ aσeσ defines an isomorphism from A to A0.

For the surjectivity, we will construct an inverse map of γ. Choose any [ϕ] ∈ H2(L/k). Note that we can always choose a normalized cocycle as a representative, i.e. ϕ(σ, 1) = ϕ(1, σ) = 1 for all σ ∈ G. (If ϕ : G × G → L× is an arbitrary cocycle, define ϕ0 : G × G → L× as ϕ0(σ, τ) := σ(ϕ(1, 1)), then ϕ0 is a coboundary and ϕ(ϕ0)−1 is a normalized cocycle which is cohomologous to ϕ.) With such choice, define an algebra A as an L-vector space with basis (eσ)σ∈G, i.e. M A(ϕ) = L · eσ σ∈G with a multiplication given by eσeτ = ϕ(σ, τ)eστ . Now we show that A is a CSA over k which lies in A(L/k), and the map ϕ 7→ A(ϕ) only depends on the cohomology class of ϕ.

1. Associativity: As before, A(ϕ) is associative since ϕ is a 2-cocycle.

14 2. Identity: Since ϕ is normalized, we have eσe1 = ϕ(σ, 1)eσ = eσ and e1eσ = eσ similarly. Hence e1 became an identity of A(ϕ). P 3. Center of A(ϕ) is k: Assume that x = σ∈G aσeσ ∈ Z(A(ϕ)). Then for any β ∈ L, we have ! ! X X X X aσβeσ = β aσeσ = βx = xβ = aσeσ β = aσσ(β)eσ, σ∈G σ∈G σ∈G σ∈G

so σ(β) = β if aσ 6= 0. However, if σ 6= 1, then we can always choose β s.t. σ(β) 6= β. So aσ = 0 for all σ 6= 1 and x = a1e1. Now xeτ = eτ x implies τ(a1) = a1 for any τ, hence a1 ∈ k and Z(A(ϕ)) = k · e1 ' k.

4. Simplicity: Let I be a nonzero ideal. Let x = aσ1 eσ1 + ··· + aσm eσm be a nonzero element of I with minimal number of terms (i.e. minimal m). If m > 1, we can find β ∈ L× s.t. −1 σm(β) 6= σm−1(β). Then y := x − σm(β)xβ ∈ I and coefficient of eσm in y is zero, which contradicts to the minimality of x. Hence m = 1 and x = aeσ, then x is a unit with an inverse −1 −1 −1 x = σ (a )eσ−1 .

5. A(ϕ) ∈ A(L/k): L ' L · e1 ⊂ A(ϕ). Also, from [A : L] = | Gal(L/k)| = [L : k], we have [A : k] = [A : L][L : k] = [L : k]2.

6. ϕ 7→ A(ϕ) only depends on the cohomology class: Let [ϕ0] = [ϕ], i.e.

0 −1 ϕ (σ, τ) = ϕ(σ, τ) · (σuτ · uστ · uσ)

× 0 0 for some u : G → L . Define a map F : A(ϕ ) → A(ϕ) as L-linear map with F (eσ) := uσeσ. Then

0 0 F (eσ)F (eτ ) = uσeσuτ eτ = uσσ(uτ ) · eσeτ = uσσ(uτ )ϕ(σ, τ) · eστ 0 0 0 0 = ϕ (σ, τ)uστ eστ = F (ϕ (σ, τ)eστ ) = F (eσeτ )

so F is an L-algebra homomorphism between A(ϕ0) and A(ϕ).

We have a bijection between A(L/k)/ ' and H2(L/k). Using this, we show our main theorem.

Theorem 7. For every finite Galois extension L/k, the map [ϕ] 7→ [A(ϕ)] defines an isomorphism of abelian groups H2(L/k) → Br(L/k).

Proof. To show that this map is bijective, by the Proposition 10, it suffices to prove that A 7→ [A]: A(L/k)/ '→ Br(L/k) is bijective.

0 0 If A ∼ A , then A ' Mn(D) and A ' Mm(D) for some central division algebra D over k. Then [A : k] = [L : k]2 = [A0 : k] gives n = m and so A ' A0. This proves that the map A(L/k)/ '→ Br(L/k) is injective, and the Corollary 15 proves that the map is surjective.

To show that the map is a homomorphism, we have to prove the following.

15 0 0 0 Lemma 6. For any two 2-cocycles ϕ and ϕ , A(ϕ + ϕ ) ∼ A(ϕ) ⊗k A(ϕ ).

0 0 0 00 Proof. Let A = A(ϕ),B = A(ϕ ),C = A(ϕ + ϕ ) and (eσ), (eσ), (eσ) be standard bases for A, B, C respectively. Regard A, B as left L-vector spaces, and define V = A ⊗L B. Note that

V ' (A ⊗k B)/hla ⊗ b − a ⊗ lb : a ∈ A, b ∈ B, l ∈ Li.

V has a unique right A ⊗k B-module structure

0 0 0 0 0 0 (a ⊗L b )(a ⊗k b) = a a ⊗L b b, a , a ∈ A, b , b ∈ B and a unique left C-module structure

00 (leσ)(a ⊗L b) = leσa ⊗L eσb, l ∈ L, σ ∈ G, a ∈ A, b ∈ B.

The two actions commute, and so the right action of A ⊗k B on V defines a homomorphism of k-algebras opp f :(A ⊗k B) → EndC (V ) which is injective because A ⊗k B (and hence its opposite) is simple. By dimension counting, both opp 4 4 (A ⊗k B) and EndC (V ) have degree n = [L : k] over k, so f is an isomorphism. By the n n opp Corollary 4, V ' C as C-module and EndC (V ) ' EndC (C ) ' Mn(C ). From this, we have an isomorphism of k-algebras opp opp (A ⊗k B) → Mn(C) which is same as A ⊗k B → Mn(C).

As a corollary, we also have an isomorphism between Br(k) and H2(k/k). Corollary 18. For every separable algebraic closure k of k, there is a canonical isomorphism Br(k) → H2(k/k).

Proof. For every tower of fields E ⊃ L ⊃ k with E,L finite Galois over k, the diagram

H2(L/k) Inf H2(E/k)

Br(L/k) Br(E/k) commutes. (Here Inf : H2(L/k) → H2(E/k) is so-called inflation map.) Now the claim follows from the Corollary 17 and [ H2(k/k) = H2(L/k). k⊂L⊂k L/k fin. Galois

16 4.3 Examples

Here we present some examples of Brauer groups, using the results of group cohomology. First, we need a concept of Tate cohomology.

Definition 8. Let G be a finite group and M be a G-module. Define a norm map NmG : M → M P n as m 7→ g∈G gm. The Tate cohomology group HT (G, M) is defined as

 n H (G, M) n > 0  M G/ Nm (M) n = 0 Hn (G, M) := G T ker(Nm )/I M n = −1  G G  H−n−1(G, M) n < −1 where IG is an augmentation ideal, kernel of the augmentation map Z[G] → Z, g 7→ 1.

Tate cohomology group is a special kind of cohomology group which connects the original group co- homology with group homology. When G is a (finite) cyclic group, we have the following proposition which reduces the computation of cohomology group to n = 0, 1.

Proposition 11. Let G be a finite cyclic group. A choice of generator for G determines isomor- phisms n n+2 HT (G, M) ' HT (G, M) for all G-modules and for all n ∈ Z.

Proof. See the Proposition 3.4. in Chapter II of [3].

Hilbert’s theorem 90 implies that the first cohomology group vanishes in case of Galois cohomology.

Theorem 8 (Hilbert’s theorem 90). Let L/k be a Galois extension. Then

H1(L/k) := H1(Gal(L/k),L×) = 0.

Proof. See the Proposition 1.22. in Chapter II of [3].

Definition 9. Assume that G is a finite cyclic group and M be a G-module. If the cohomology r groups HT (G, M) are finite, define Herbrand quotient as

0 #HT (G, M) h(M) := 1 . #HT (G, M) Proposition 12. If M is a finite module, then h(M) = 1.

Proof. See the Proposition 3.8. in Chapter II of [3].

Also, it is known that cohomology of finite groups are always a torsion group.

17 Proposition 13. Let G be a finite group and M be a G-module. Then |G| · Hr(G, M) = 0 for all r > 0.

Proof. Se the Corollary 1.31. in Chapter II of [3].

As a Corollary, we obtain the following. Corollary 19. (a) For any finite field extension L/k, Br(L/k) is a torsion group. Also, Br(k) is a torsion group for any field k. (b) For any CSA A over k, there exists m, n > 0 s.t. A ⊗k A ⊗k · · · ⊗k A ' Mn(k), where the LHS is tensor product of A’s m times.

Proof. (a) This follows from the previous Proposition, Corollary 17 and the main theorem. (b) directly follows from (a).

4.3.1 Real field (Archimedean local field)

Let G = Gal(C/R) = {1, τ} where τ : z 7→ z is a conjugation. By the periodicity of the (Tate) cohomology group of cyclic group, we have 2 × 0 × × × × × H (G, C ) ' HT (G, C ) = R /NmC/R(C ) = R /R+ = {±} From this, we have the following theorem: Proposition 14. Br(R) = Z/2Z. Here the isomorphism is given by [R] 7→ 0 and [H] 7→ 1.

2 Proof. It is enough to show that nonzero element of H (C/R) corresponds to [H]. Such element is × given by the 2-cocycle ϕ : G × G → C defined as ( −1 ρ = σ = τ ϕ(ρ, σ) := 1 otherwise

2 × Such 2-cocycle defines a nonzero element in H (C/R): if ϕ = du for some 1-coclyce u : G → C , then −1 ϕ(ρ, σ) = ρ(uσ)uρσ uρ for all ρ, σ ∈ G. For ρ = τ and σ = 1, −1 1 = ϕ(τ, 1) = τ(u1)uτ uτ = u1 so u1 = 1. For ρ = σ = τ, −1 2 −1 = ϕ(τ, τ) = τ(uτ )u1 uτ = |uτ | , which gives contradiction. Now we can show that the map A(ϕ) → H defined by xσ 7→ j is an R-algebra isomorphism, where A(ϕ) is a central simple R-algebra corresponds to ϕ which is defined in the proof of the main theorem.

Corollary 20. Up to isomorphism, there are two central division R-algebras, R and H.

Note that H ⊗R H ' M4(R).

18 4.3.2 Finite fields

We can show that H2(Gal(l/k), l×) = 0 for any finite extension l/k of finite fields. (This requires Hilbert’s theorem 90 and the fact that Herbrand quotient h(l×) is 1. In fact, Hn(l/k) = 0 for any n.) Hence Br(l/k) = 0 and Br(k) = 0 for all finite field k. This gives a simple and cohomological proof of Wedderburn’s (little) theorem. Theorem 9 (Wedderburn). Every finite division algebra is a field.

Proof. Let D be a finite division algebra over k, and let l be a center of k, which is a finite extension of k. We know that D is a central simple algebra over l. Since Brauer group Br(l) is trivial, we have D ' Mn(l) for some n. Then we should have n = 1 since Mn(l) is not a division algebra for n ≥ 1. (We can find zero divisors easily.) Hence D ' M1(l) ' l and D is a field.

4.3.3 non-Archimedean Local fields

Let K be a non-Archimedean local field, i.e. finite extension of Qp or Fq((T )). We can show that 2 al H (K /K) ' Q/Z by using homological algebra and some properties of local fields. Such isomor- 2 al phism is called an invariant map invK :H (K /K) ' Q/Z, and this is a main ingredient in the local class field theory. However, it is possible to describe such invariant map more directly, as a map invK : Br(K) ' Q/Z.

We know that any element of Br(K) is represented by a central division algebra D over K, which is unique up to isomorphism. One can show that there exists a maximal subfield L of D that is unramified over K. Let σ ∈ Gal(L/K) be a Frobenius element, i.e. generator of the Galois group. (Note that Galois group of unramifield extension of local field is cyclic.) By Noether-Skolem theorem, there exists α ∈ D s.t. σ(x) = αxα−1 for all x ∈ L. If α0 also has this property, then α0 = cα for some c ∈ L, hence

0 ord(α ) = ord(α) + ord(c) ≡ ord(α) (mod Z) where ord : D → Q is a unique extension of ord : k → Z. Now we define

invK (D) = ord(α) (mod Z). Then this gives a desired map.

Theorem 10. The map invK : Br(K) → Q/Z is a bijection.

For example, let L be the unramifield extension of K of degree n (which is unique up to isomorphism) and let σ be the Frobenius automorphism of L/K, so that G = Gal(L/K) = hσi. Let ϕ be the 2-cocycle ( 1 if i + j ≤ n − 1 ϕ(σi, σj) := , π if i + j > n − 1 where π is a uniformizer of K (an element of K s.t. ordk(π) = 1). Then the crossed-product algebra A(ϕ) equals ⊕0≤i≤n−1L · ei with the multiplication determined by i ei · a = σ (a) · ei, a ∈ L

19 and ( ei+j if i + j ≤ n − 1 eiej = . π · ei+j−n if i + j > n − 1 −1 We identify L is L·e0 ⊂ A(ϕ). Since e1ae1 = σ(a) for a ∈ L, we can use e1 to compute invK (A(ϕ)). n We have e1 = en−1e1 = π · e0 = π, so 1 1 1 inv (A(ϕ)) = ord(e ) = ord(en) = ord(π) = . K 1 n 1 n n

4.3.4 Global fields

For the Brauer groups of global fields, we introduce the Albert-Brauer-Hasse-Noether theorem and Hasse’s theorem, which is about local-global property (Hasse principle) of central simple algebra over number fields. Theorem 11 (Albert-Brauer-Hasse-Noether). Let A be a central simple algebra over a number field K. Suppose that A splits over a local field Kp for any prime p in K, i.e.

A ⊗K Kp ' Md(Kp) for some fixed d > 0. Then A ' Md(K), i.e. A splits over K.

This theorem also can be stated as follows: for each prime p of K, we have a homomorphism Br(K) → Br(Kp) defined as [A] 7→ [A ⊗K Kp]. One can show that for any [A] ∈ Br(K), its image in Br(Kp) is zero for all but finitely many p. Hence we have a well-defined map Σ M Br(K) −→ Br(Kp) p and Albert-Brauer-Hasse-Noether theorem theorem shows that this map is injective. Hasse proved more stronger result: Theorem 12 (Hasse). We have an exact sequence Σ M inv 1 → Br(K) −→ Br(Kp) −−→ Q/Z → 0 p where inv is the Hasse invariant map.

This theorem can be proved by Class field theory. (Hasse invariant map plays the central role of local class field theory.) Using this and the previous results, we can describe Br(K) in terms of subgroup of countable direct sum of Q/Z. For example, if K = Q, we have Br(Qp) ' Q/Z and 1 Br(R) ' Z/2Z ' 2 Z/Z. Hence Hasse’s theorem gives ! 1 M inv 1 → Br( ) → / ⊕ / −−→ / → 0 Q 2Z Z Q Z Q Z p<∞ where the map inv is just a componentwise identity map. Hence we have ( ) M X Br(Q) ' (a, x): a ∈ {0, 1/2}, x = (xp)p ∈ Q/Z, a + xp = 0 p<∞ p<∞ Other number fields can be described in the similar way.

20 5 Noncommutative division algebra of characteristic p > 0

As an application of the main theorem, we will show that there exists a noncommutative division algebra of characteristic p > 0 for any prime p. First, let’s define the norm map.

Definition 10. Let L/k be a finiite Galois extension with Galois group G = Gal(L/k). The norm × × map NL/k : L → k is given by Y NL/k(u) := σ(u). σ∈G

Note that this norm map coincides with the norm map that we defined in the previous section.

Proposition 15. Let k be a field. If there is a cyclic Galois extension L/k s.t. the norm map × × NL/k : L → k is not surjective, then there exists a noncommutative division algebra over k.

Proof. Since G = Gal(L/k) is a finite cyclic group, by the periodicity of the Tate cohomology group (Proposition 11), we have

2 0 × G × Br(L/k) ' H (L/k) ' HT (L/k) = (L ) / im(NL/k) = k / im(NL/k).

Therefore, Br(L/k) 6= 0 if NL/k is not surjective, and so Br(k) 6= 0.

× × For example, if L/k is a finite extension between finite fields, then the norm map NL/k : L → k is surjective since Br(L/k) = 0 in this case. Using this, we can prove existence of noncommutative division algebra of positive characteristic.

Theorem 13. For any prime p, there exists a noncommutative division algebra of characteristic p.

Proof. (This proof is slightly different from the proof in [4].) By the previous proposition, it is enough to show find a cyclic extension L/k s.t. the norm map NL/k is not surjective.

Assume that p is an odd prime. Let k = (x). Then t2 − x ∈ k[t] is a separable irreducible √ Fp polynomial and L = k( x) is a degree 2 Galois extension of k. Any elements u ∈ L has a form √ u = (a + b x)/c for some a, b, c ∈ Fp[x] and its norm is given by a2 − b2x N (u) = . L/k c2

2 × Now our claim is that if α ∈ Fp is not a square (such α always exists), then x − αx 6∈ NL/k(L ). Otherwise, we have

a2 − b2x = c2(x2 − αx) ⇔ a2 − c2x2 = (b2 − αc2)x.

If LHS and RHS are nonzero, then LHS has even degree and RHS has odd degree, which is a contradiction. So we should have a2 = c2x2 ⇔ a = ±cx and b2 − αc2 = 0 ⇔ b2 = αc2. By 2 comparing the coefficients of the highest degree, the coefficients β, γ ∈ Fp satisfies (β/γ) = α, contradiction. Hence x2 − αx can’t be a norm.

21 2 For p = 2, let k = F2(x) and L = k(α), where α is a root of f(t) = t + t + x + 1 ∈ k[t]. (f(t) is irreducible and separable, where the other root is α + 1.) As before, each u ∈ L can be written as a form u = (a + bα)/c where a, b, c ∈ F2[x]. We may assume that x is not a divisor of all three polynomials. Moreover,

(a + bα)(a + bα + b) a2 + ab + b2(x + 1) N (u) = = . L/k c2 c2 We claim that x is not a norm. Otherwise, we have

a2 + ab + b2(x + 1) = c2x.

If we put x = 0, then a(0)2 + a(0)b(0) + b(0)2 = 0 and the only possible case is a(0) = 0 = b(0). This implise x|a and x|b, which gives x2|c2x ⇒ x|c2 ⇒ x|c, contradiction.

If we follow the isomorphisms, it would be possible to construct a noncommutative division algebra of characteristic p explicitly.

6 Appendix - H2(G, A) and a group extension

It is possible to interpret the second cohomology group H2(G, A) as a group extension, which is similar to the description of Ext group. Note that such bijection depends on the G-action on A.

Proposition 16. Let A be a G-module with a fixed action. There exists a bijection between H2(G, A) and the set of equivalence classes of group extensions of G by A, i.e. the set of equivalence classes of the short exact sequences

0 → A → E → G → 1 where the G-action on A is a conjugation action in E, i.e. ga = s(g) · a · s(g)−1 for some section s : G → E of π : E → G (as a multiplicative notation. Such action doesn’t depend on the choice of section s). Also, two extensions 0 → A → E → G → 1 and 0 → A → E0 → G → 1 are equivalent if and only if there exists an isomorphism f : E → E0 s.t. the following diagram commutes:

0 A E G 1 f 0 A E0 G 1

For example, if G acts on A trivially, then H2(G, A) classifies central extensions, i.e. the extensions where A is contained in the center of E. However, for other G-actions, H2(G, A) classifies some other extensions.

Proof. We will follow the proof in [1]. First, choose any 2-cocycle ϕ : G × G → A. Define the group structure on E = A × G as

(a, g)(a0, g0) := (a + ga0 + ϕ(g, g0), gg0).

22 We can check that such multiplication is associative if and only if ϕ satisfies

g · ϕ(g0, g00) − ϕ(gg0, g00) + ϕ(g, g0g00) − ϕ(g, g0) = 0 which is true because ϕ is a 2-cocycle. Also, one can check that (−ϕ(1, 1), 1) is a 2-sided identity and (−g−1a − ϕ(g−1, g) − c(1, 1), g−1) is a 2-sided inverse of (a, g), by using the cocycle condition again. At last, the map A → E defined by a 7→ (a − ϕ(1, 1), 1) and E → G defined by (a, g) 7→ g are group homomorphisms which makes 0 → A → E → G → 1 exact. Now assume that ϕ0 and ϕ are cohomologous, i.e. represents the same cohomology class and let E0 = A × G (as a set) be a corresponding extension of ϕ0. Then ϕ0 = ϕ + dh for some h : G → M. Then we can check that the map E → E0 defined as (a, g) 7→ (a+h(g), g) is an isomorphism which makes 0 → A → E → G → 1 and 0 → A → E0 → G → 1 equivalent.

Conversely, let 0 → A → E → G → 1 be an arbitrary extension. We can identify E = A × G as a set: choose any set-theoretic section s : G → E of π : E → G and then we have a E = A · s(g) = A × G g∈G where A·s(g) = π−1(g). To describe the group structure on E, we note that the subgroup structure on A has been specified but s(1) may not equal to the identity of E.

−1 Now consider s(g1)s(g2) and s(g1g2). Both are elements of π (g1g2) (since π is a group homo- morphism), there exists a unique ϕ(g1, g2) ∈ M s.t. s(g1)s(g2) = ϕ(g1, g2)s(g1g2). (Note that ϕ(1, 1) = s(1).) Thus there exists a function ϕ : G × G → A such that the group law on E = A × G is defined by (a, g)(a0, g0) = (a + g · a0 + ϕ(g, g0), gg0). Since E is associative, we should have

g · ϕ(g0, g00) − ϕ(gg0, g00) + ϕ(g, g0g00) − ϕ(g, g0) = 0 which claims that ϕ is a 2-cocycle and this relation forces ϕ(1, g) = ϕ(1, 1) and g · ϕ(1, 1) = ϕ(g, 1) for all g ∈ G. If we choose another section s0 : G → E, by replacing s as s0 = h · s for some h : G → A then ϕ is replaced by ϕ0 = ϕ + dh. Hence the class [ϕ] ∈ H2(G, A) is independent of s and so depends only on the isomorphism class of the given extension structure E of G by A.

There’s also a connection of this description with a Brauer group. Let A be a CSA over k of degree n2 which is in A(L/k), i.e. contains a field L which is Galois of degree n over k. Let −1 −1 E := {α ∈ A : αLα = L}. Then each α ∈ E defines an element σα : x 7→ αxα of Gal(L/k), and the Noether-Skolem theorem implies that every element of Gal(L/k) arises from α ∈ E. Since p [L : k] = [A : k], CA(L) = L and so the sequence

1 → L× → E× → Gal(L/k) → 1 is exact. Such process gives a map from A(L/k)/ ' to the set of isomorphism classes of extensions of Gal(L/k) by L×, and hence to H2(L/k).

23 References

[1] B. Conrad, Group cohomology and Group extensions, Stanford Math 210B lecture note.

[2] Evan, A Crash Course in Central Simple Algebras, Online note.

[3] J. S. Milne, Class Field Theory, Ver 4.02, 2013.

[4] J. J. Rotman, An Introduction to Homological Algebra, Universitext, Springer.

[5] J. P. Serre, Local Fields, Graduate Texts in Mathematics, Springer.

24