THE BRAUER GROUP OF A FIELD IGOR RAPINCHUK This paper is devoted to the construction of the Brauer group of a field and its description in terms of factor sets. Since the elements of the Brauer group are similarity classes of central simple algebras over a given field, we begin by establishing some fundamental theorems for such algebras in §§1 and 2 (this material is contained, for example, in [2], [4] and [6]). In §3, we introduce the Brauer group of a field, and in §4 we describe it using factor sets and crossed products, which leads to an isomorphism between the Brauer group and a certain second cohomology group (this part closely follows the exposition given in [2], Ch. 4). In §5 we specialize to crossed products associated to cyclic Galois extensions. Finally, in §6 we apply the general theory to describe the Brauer group of a local field. (These two sections follow [4], Ch. 15 and 17.) In this paper, all algebras will be associative and finite dimensional. 1. Basic facts about simple algebras Let A be an algebra with identity over a field K. We recall that A is said to be simple if it has no proper two-sided ideals, and central if its center Z(A) coincides with K. We will study algebras by analyzing the structure of modules over them. A (left) A-module M is simple if it contains no proper submodules. The following well-known statement will be used repeatedly. Schur’s Lemma. If M and N are simple A-modules then every nonzero A-module homomorphism f: M → N is an isomorphism. In particular, if M is a simple A-module then End M is a division A ring. Indeed, we have Kerf 6= M, so Kerf = {0}, and f is injective. Similarly, Imf 6= {0}, so Imf = N, making f also surjective, hence an isomorphism. Now, let A be a (finite dimensional) simple K-algebra. By dimension consideration, there exists a minimal nonzero left ideal M ⊂ A. In the sequel, A will denote A considered as a left A-module, and A then M is a simple submodule of A. A Proposition 1. Let A be a finite dimensional simple K-algebra, and M ⊂ A be a nonzero minimal left ideal. Then (1)there exists n > 0 such that A ’ M ⊕···⊕M as A-modules; A | {z } n (2)any A-module is isomorphic to a direct sum of copies of M, in particular M is the only simple A-module; (3)let N and N be A-modules; then N ’ N as A-modules if and only if dim N = dim N 1 2 1 2 K 1 K 2 (we notice that any A-module has the natural structure of a K-vector space). P Proof. (1): SinceM isaleftideal, Maisatwo-sidedideal,hencecoincideswithA.Inparticular, a∈A we can write 1 = m a +···+m a with m ∈ M, a ∈ A, 1 1 n n i i 1 2 IGORRAPINCHUK and then n X (1) A = Ma n i=1 P We can assume that the set {a ,...,a } is minimal with respect to the property A = Ma , and 1 n i then Ma 6= {0} for all i = 1,...,n. Notice that for any a ∈ A, the map f : M → Ma, x 7→ xa, is a i a surjective homomorphism of left A-modules. So, if Ma 6= {0} then arguing as in the proof of Schur’s Lemma, we see that f is injective, hence an isomorphism. Thus, all the Ma ’s in (1) are isomorphic a i to M, and in particular are simple A-modules. It remains to show that the sum (1) is direct. However, if for some j we have \X Ma Ma 6= {0} j i i6=j P then because of the simplicity of Ma we conclude that Ma ⊂ Ma . Then j j i6=j i X A = Ma , i i6=j contradicting the minimality of the set {a ,...,a }. 1 n (2): Let N be a (nonzero) left A-module. Then N is a quotient of a free A-module which in combination with part (1) shows that there is a surjective homomorphism M f: M −→ N i i∈I where each M is isomorphic to M. Set N = f(M ). We can discard those i for which N = {0}. Then i i i i clearly f gives an isomorphism between M and N , and in particular, N is simple. Furthermore, i i i P N = N , and it remains to find a subset I ⊂ I such that i∈I i 0 M (2) N = N . i i∈I0 P For this we consider the collection J of all subset J ⊂ I for which the sum N is direct. Clearly, i∈J i all one-element subsets of I belong to J, in particular, J =6 ∅. We can order J by inclusion, and then it is easy to see that J satisfies Zorn’s Lemma. Let I ∈ J be a maximal element provided by 0 P the latter. Then by our construction the sum N is direct, and we only need to show that it i∈I0 i P coincides with N. Assume the contrary. Then in view of N = N , there exists i ∈ I such that i∈I i 0 P P N 6⊂ N . Since N is simple, this actually means that N ∩ N = {0}, implying that the i0 P i∈I0 i i0 i0 i∈I0 i sum N is also direct. This contradicts the maximality of I and proves (2). i∈I0∪{i0} i 0 (3): We embed K ,→ A by x 7→ x·1 , so any A-module can indeed be considered as a vector space A over K. By part (2), we have N ’ Mα1 and N ’ Mα2 1 2 for some cardinal number numbers α and α . Then 1 2 dim N = (dim M)α , K i K i and since dim M is finite, we see that K dim N = dim N ⇔ α = α , K 1 K 2 1 2 and our claim follows. (cid:3) Part (1) of Proposition 1 will enable us to prove Wedderburn’s Theorem (see Theorem 1) which describes the structure of finite dimensional simple algebras. The argument will require the following. BRAUER GROUP 3 Lemma 1. Let A be an arbitrary ring, M be a left A-module, and E = End (M). Then for any A n > 1, there exists a ring isomorphism (3) End (Mn) ’ M (E), A n the ring of n×n-matrices over the ring E. Furthermore, if A is a K-algebra with identity then E, End (Mn), and M (E) have the natural structures of a K-algebra for which (3) is an isomorphism A n of K-algebras. Proof. Define ε : M → Mn and π : Mn → M by i i ε : m 7→ (0,...,m,...,0) and π : (m ,...,m ) 7→ m . i i 1 n i Then n X ε π = id and π ◦ε = id if k = j and 0 if k 6= j. k k Mn k j M k=1 Given f ∈ End (Mn), we let f = π ◦f◦ε ∈ E for i,j = 1,...,n. We claim that the correspondence A ij i j End (Mn) 3 f 7→ϕ (f ) ∈ M (E) A ij n yields the required isomorphism (3). Indeed, for f,g ∈ End (Mn) we have A ϕ(f +g) = (π ◦(f +g)◦ε ) = (π ◦f ◦ε +π ◦g◦ε ) = (f )+(g ) = ϕ(f)+ϕ(g), i j i j i j ij ij and n ! n n X X X ϕ(fg) = π ◦f ◦ ε π ◦g◦ε = (π ◦f ◦ε )(π ◦g◦ε ) = f g = (ϕ(f)ϕ(g)) ij i k k j i k k j ik kj ij k=1 k=1 k=1 for all i,j, so ϕ(fg) = ϕ(f)ϕ(g). Thus, ϕ is a ring homomorphism. Given (f ) ∈ M (E), we define ij n f: Mn → Mn by n n ! X X f(m) = f (π (m)),..., f (π (m)) . 1k k nk k k=1 k=1 Clearly, f ∈ End (Mn). Furthermore, for any i,j we have A n X (π ◦f ◦ε )(m) = f ((π ◦ε )(m)) = f (m), i j ik k j ij k=1 showing that the correspondence (f ) 7→ f is inverse to ϕ and thus making ϕ a ring isomorphism. ij AsweobservedintheproofProposition1,ifAisaK-algebra,anyA-moduleN becomesaK-vector space. Moreover, since K is containedin the centerof A, End (N)becomes a K-algebraforthescalar A multiplication (af)(x) = f(ax) = af(x) for a ∈ K, f ∈ End (N), x ∈ N. A Since ε and π are A-module homomorphisms, we have i j (af) = π ◦(af)◦ε = a(π ◦f ◦ε ) = af , ij i j i j ij which shows that (3) is an isomorphism of K-algebras. (cid:3) The following theorem is the main result of this section. Theorem 1. (Wedderburn) Let A be a finite dimensional simple algebra over a field K. Then A ’ M (D) for a unique n > 1 and a unique up to isomorphism division K-algebra D. Conversely, any n algebra of the form M (D), where D is a division algebra, is simple. n 4 IGORRAPINCHUK Proof. We recall that the opposite algebra Aop is obtained by giving the same K-vector space A a new product defined by a∗b = ba where ba is the product in the original algebra A. First, we notice that End ( A) ’ Aop. Indeed, if ϕ ∈ End ( A) then ϕ(x) = xϕ(1) for all x ∈ A, and then then the A A A A correspondence ϕ 7→ ϕ(1) yields the required isomorphism. On the other hand, by Proposition 1(1), for some n > 1, there is an isomorphism of left A-modules: A ’ Mn, where M is a minimal nonzero A left ideal of A. Then by Lemma 1, End ( A) ’ M (E), where E = End (M). Since M is simple as A A n A A-module, E is a division algebra. Thus, Aop ’ End ( A) ’ M (E). A A n It remains to observe that the map a = (a ) 7→ ta = (a ) gives an isomorphism M (E)op ’ M (Eop). ij ji n n So,weeventuallyobtainthatA ’ M (D)withD = Eop (noticethatthealgebraoppositetoadivision n algebra is itself a division algebra). For the uniqueness of n and D, we need the following lemma. Lemma 2. Let A = M (D), where D is a division ring, and let V = Dn be the space of n-columns on n which A acts by matrix multiplication on the left. Then V is a simple A-module and End (V) ’ Dop. A Proof. Given any nonzero v,w ∈ V, there exists a ∈ A such that av = w, and the simplicity of V     1 d  0   ∗  follows. Now, let f ∈ EndA(V). Let v0 =  .. , and suppose that f(v0) =  .. . We claim that  .   .  0 ∗   a 1  a2  f(v) = vd for all v ∈ V. Indeed, let v =  . . Then  .   .  a n       a 0 ... 0 a 0 ... 0 d 1 1 . . . . . . . f(v) = f .. .. .. v0 =  .. .. ..  ..  = vd. a 0 ... 0 a 0 ... 0 ∗ n n Then the map f 7→ d gives the required isomorphism End (V) ’ Dop. (cid:3) A Now, suppose A ’ M (D ) and A ’ M (D ). Let V = Dn1 and V = Dn2. Then both V and V n1 1 n2 2 1 1 2 2 1 2 can be considered as A-modules. It follows form Lemma 2 that they are simple A-modules, and then by Proposition 1(2), they are isomorphic as A-modules. Using Lemma 2, we obtain Dop ’ End (V ) ’ End (V ) ’ Dop, 1 A 1 A 2 2 so D ’ D as K-algebras. Furthermore, 1 2 dim A = n2dim D = n2dim D , K 1 K 1 2 K 2 so n = n . 1 2 Finally, we need to show that A = M (D), where D is a division algebra, is simple. Let e be the n ij standard basis of A. Suppose a ⊂ A is a nonzero two-sided ideal, and pick a nonzero a = (a ) ∈ a ij where, say, a 6= 0. It is easy to check that i0j0 e = e (a−1 a)e , ij ii0 i0j0 j0j so e ∈ a for all i,j, and therefore a = A. (cid:3) ij Corollary 1. Suppose K is an algebraically closed field. If A is a finite dimensional simple algebra over K then A ’ M (K) for some n. n BRAUER GROUP 5 Indeed, it is enough to show that if D is a finite dimensional division algebra over K then D = K. Assume the contrary, and pick a ∈ D\K. Then K(a)/K is a nontrivial finite field extension, which cannot exist because K is algebraically closed. Thus, D = K. The following statement is well-known. Lemma 3. Let A = M (D). Then the center Z(A) is naturally isomorphic to the center Z(D). n Indeed, if a ∈ Z(A) then using the fact that a commutes with all elements of the standard basis e , ij we immediately see that a is a scalar matrix. Furthermore, if α is its diagonal element then α ∈ Z(D). Conversely, any such scalar matrix is in Z(A). 2. Fundamental theorems for simple algebras The following simple facts will be used repeatedly. Lemma 4. Let V and W be vector spaces over a field K, and suppose w ,...,w ∈ W are linearly 1 n independent over K. If a ,...,a ∈ V are such that 1 n a ⊗w +···+a ⊗w = 0 in V ⊗ W 1 1 n n K then a = ··· = a = 0. 1 n Proof. Being linearly independent, w ,...,w can be included in a basis w ,...,w ,... of W. Let 1 n 1 n P v ,...,v ,... be a basis of V. We can write a = α v with α ∈ K, and then 1 m i j ij j ij   X X X 0 = a1⊗w1+···+an⊗wn =  αijvj⊗wi = αij(vj ⊗wi). i j i,j But it is well-known that the elements v ⊗w form a basis of V ⊗ W. So, all α = 0, and therefore j i K ij a = ··· = a = 0. (cid:3) 1 n If A and B are K-algebras then the tensor product of vector spaces A ⊗ B can be given a K multiplication satisfying (a ⊗b )(a ⊗b ) = a a ⊗b b , 1 1 2 2 1 2 1 2 and this multiplication makes A ⊗ B into a K-algebra. Furthermore, A and B can be identified K with subalgebras of A⊗ B by the maps a 7→ a⊗ 1 and b 7→ 1 ⊗b, and then A and B commute K K B A inside A⊗ B. It is not difficult to see that A⊗ B can in fact be characterized by the following K K universal property: given algebra homomorphisms f: A → C and g: B → C such that f(A) and g(B) commute inside C then there exists a unique algebra homomorphism F: A⊗ B → C such that K F(a⊗b) = f(a)g(b). Proposition 2. For any two K-algebras A and B we have Z(A⊗ B) = Z(A)⊗ Z(B). K K In particular, if A and B are central over K then so is A⊗ B. K Proof. The inclusion ⊃ is obvious. To prove the opposite inclusion, take any z ∈ Z(A⊗ B) and pick K a shortest presentation of the form n X (4) z = a ⊗b . i i i=1 Then the systems a ,...,a and b ,...,b are linearly independent over K. Indeed, if b ,...,b are 1 n 1 n 1 n linearly dependent then one of them, say, b , is a linear combination of others: 1 b = β b +···+β b . 1 2 2 n n 6 IGORRAPINCHUK Then z = a ⊗(β b +···+β b )+a ⊗b +···+a ⊗b = (β a +a )⊗b +···+(β a +a )⊗b 1 2 2 n n 2 2 n n 2 1 2 2 n 1 n n isashorterpresentation,acontradiction. Now,weclaimthatin(4),a ,...,a ∈ Z(A)andb ,...,b ∈ 1 n 1 n Z(B). Indeed, for any a ∈ A we have n X 0 = (a⊗1)z−z(a⊗1) = (aa −a a)⊗b . i i i i=1 Since the b ’s are linearly independent, by Lemma 4, we have aa −a a = 0 for all i = 1,...,n. Since i i i a ∈ A was arbitrary, we conclude that a ∈ Z(A) for all i. The argument for b ,...,b is similar. (cid:3) i 1 n The definition of the product on the Brauer group, which we will discuss in the next section, relies on the following statement. Theorem 2. Let A be a central simple K-algebra, and B be an arbitrary K-algebra. Then any two- sided ideal a ⊂ A⊗ B is of the form a = A⊗ b for some two-sided ideal b of B. In particular, if K K B is also simple (but not necessarily central), then A⊗ B is simple. K Proof. We may assume that a 6= {0}. First, we will show that \ (5) a B 6= {0}. For this we pick a nonzero x ∈ a which has a presentation of the form n X x = a ⊗b i i i=1 with the smallest possible n. Then a ,...,a and b ,...,b are linearly independent. In particular, 1 n 1 n a 6= 0, so, since A is simple, we have Aa A = A, i.e. there exist c ,...,c ,d ,...,d ∈ A such that 1 1 1 ‘ 1 ‘ c a d +···+c a d = 1. 1 1 1 ‘ 1 ‘ Consider x˜ = (c ⊗1)x(d ⊗1)+···+(c ⊗1)x(d ⊗1) = (c a d +···+c a d )⊗b +···+(c a d +···+c a d ) = 1 1 ‘ ‘ 1 1 1 ‘ 1 ‘ 1 1 n 1 ‘ n ‘ = 1⊗b +a˜ ⊗b +···+a˜ ⊗b . 1 2 2 n n Clearly x˜ ∈ a, x˜ 6= 0 and x˜ has length 6 n. So, we may assume from the very beginning that a = 1. 1 We now claim that actually n = 1. Indeed, suppose n > 2. Since a ,...,a are linearly independent 1 n over K, we have a ∈/ K = Z(A). So, there exists a ∈ A such that aa 6= a a. Then 2 2 2 y = (a⊗1)x−x(a⊗1) = (aa −a a)⊗b +···+(aa −a a)⊗b 2 2 2 n n n is a nonzero element in a having length < n, a contradiction. So, n = 1, and x = 1⊗b ∈ a, and (5) 1 follows. Thus, b := a∩B is a nonzero two-sided ideal of B. We claim that a = A⊗ b. In any case, A⊗ b K K is a two-sided ideal of A⊗ B contained in a. Then one can consider the canonical homomorphism K ϕ: A⊗ B −→ (A⊗ B)/(A⊗ b) ’ A⊗ B/b K K K K with Ker ϕ = A ⊗ b ⊂ a. If a 6= A ⊗ b then ϕ(a) is a nonzero two-sided ideal of A ⊗ B/b. K K K Applying (5) to the latter algebra, we obtain that ϕ(a)∩B/b 6= {0}. Taking pullbacks, we obtain that for a = ϕ−1(ϕ(a)) one has a∩B % b, which contradicts our construction. (cid:3) The proof of the following corollary requires one general remark: if A is a K-algebra then for any field extension L/K, the algebra A := A⊗ L can be considered as an algebra over L for the scalar L K multiplication ‘·(a⊗b) = a⊗‘b, and dim A = dim A . K L L BRAUER GROUP 7 Corollary 2. Let A be a finite dimensional central simple algebra over a field K. Then dim A is a K perfect square. Proof. Let K¯ be an algebraic closure of K. Consider B := A⊗ K¯ as a K¯-algebra. It follows from K Theorem 2 that B is simple, and then by Corollary 1 we have B ’ M (K¯) for some n > 1. Thus, n dim A = dim B = n2. K K¯ (cid:3) The following theorem will enable us to construct the inverses of elements in the Brauer group. Theorem 3. Let A be a central simple algebra over a field K, dim A = n2. Then K A⊗ Aop ’ End (A) ’ M (K). K K n2 Proof. For a ∈ A, define λ : A → A by λ (x) = ax. Clearly, λ ∈ End (A), and the correspondence a a a K L: A → End (A), a 7→ λ , is an algebra homomorphism. Similarly, for b ∈ A, we define ρ : A → A K a b by ρ (x) = xb. Again, ρ ∈ End (A), and the correspondence b 7→ ρ defines an algebra homomor- b b K b phism R: Aop → End (A). (The homomorphisms L and R are called the left and the right regular K representations of A, respectively.) For any a,b,x ∈ A we have (λ ◦ρ )(x) = a(xb) = (ax)b = (ρ ◦λ )(x), a b b a i.e. λ and ρ commute in End (A). Thus, there exists a homomorphism F: A⊗ Aop → End (A) a b K K K P which takes a ⊗ b to the endomorphism that acts as follows x 7→ axb (then an element a ⊗ b i i corresponds to the endomorphism x 7→ Pa xb ). By Theorem 2, the algebra A⊗ Aop is simple, so i i K since F is not the zero homomorphism, we have KerF = {0}, i.e. F is injective. On the other hand, dim A⊗ Aop = (n2)2 = dim End (A), K K K K which implies that F is also surjective, hence an isomorphism. (cid:3) The following two theorems are the most important results about simple algebras. Theorem 4. (Skolem-Noether) Let A and B be finite dimensional simple K-algebras, with B central. If f,g: A → B are two K-algebra homomorphisms then there exists b ∈ B∗ such that g(a) = bf(a)b−1 for all a ∈ A. Proof. Consider C = A⊗ Bop. Since B is central, Bop is also central, so it follows from Theorem 2 K that C is simple. Associated with every homomorphism f: A → B, one has a C-module structure on B given by (a⊗b) ·x = f(a)xb. f We will use B to denote B endowed with this structure. For our two homomorphisms f,g: A → B, f we obviously have dim B = dim B , so by Proposition 1(3) we have B ’ B as C-modules. Let K f K g f g ϕ: B → B be a C-module isomorphism. Set b = ϕ(1). Then for any x ∈ B we have f g ϕ(x) = ϕ((1⊗x) ·1) = (1⊗x) ·ϕ(1) = bx. f g Applying the same argument to ψ = ϕ−1: B → B , we see that ψ(x) = b0x where b0 = ψ(1). Then g f x = (ϕ◦ψ)(x) = bb0x, so substituting x = 1, we get bb0 = 1. Similarly, b0b = 1, i.e. b ∈ B∗. Furthermore, for any a ∈ A, we have bf(a) = ϕ(f(a)) = ϕ((a⊗1) ·1) = (a⊗1) ·ϕ(1) = g(a)b, f g yielding g(a) = bf(a)b−1, as required. (cid:3) 8 IGORRAPINCHUK Corollary 3. Let A be a central simple algebra over K. Then every K-algebra automorphism of A is inner. Indeed, given a K-algebra automorphism g: A → A, our claim follows from the theorem applied to f = id . (A different proof based on Theorem 3 is given in [6], Ch. XI, Prop. 4.) A Theorem 5. (theDoubleCentralizerTheorem) Let A be a central simple algebra over K of dimension dim A = n, and let B ⊂ A be a simple subalgebra of dimension dim B = m. Denote K K Z (B) = {x ∈ A | xb = bx for all b ∈ B}. A Then (1) Z (B)⊗ M (K) ’ A⊗Bop; A K m (2) Z (B) is a simple subalgebra of A of dimension dim Z (B) = n/m; A K A (3) Z (Z (B)) = B. A A Proof. The proof is based on two simple observations that slightly generalize our previous construc- tions: • In Proposition 2 we proved that for any K-algebras A and B one has Z(A⊗ B) = Z(A)⊗ Z(B). K K The same argument shows that for any K-algebras A and B and any subalgebras A0 ⊂ A and B0 ⊂ B one has Z (A0⊗ B0) = Z (A0)⊗ Z (B0). A⊗KB K A K B • In the proof of Theorem 3, we constructed the representations L: A → End (A), a 7→ λ , and K a R: Aop → End (A), b 7→ ρ , and observed that L(A) and R(Aop) commute inside End (A). In fact, K b K Z (L(A)) = R(Aop). EndK(A) Indeed,iff ∈ Z (L(A))thenf(ax) = af(x)foralla,x ∈ A.Lettingx = 1,wegetf(a) = af(1), EndK(A) i.e. f = ρ . f(1) To prove the theorem, we consider two embeddings f,g: B → A⊗ End (B) = A⊗ M (K) K K K m given by f(b) = b⊗id and g(b) = 1⊗λ . B b We have Z(A⊗ M (K)) = Z(A)⊗ Z(M (K)) = K ⊗ K = K, K m K m K which means that A ⊗ M (K) is central. Then by the Skolem-Noether Theorem, f and g are K m conjugate, i.e. there exists x ∈ (A⊗ End (B))∗ such that K K f(b) = xg(b)x−1 for all b ∈ B. This implies that Z (f(B)) = xZ (g(B))x−1, A⊗KEndK(B) A⊗KEndK(B) in particular, these centralizers are isomorphic. But Z (f(B)) = Z (B⊗ K) = Z (B)⊗ End (B) A⊗KEndK(B) A⊗KEndK(B) K A K K and Z (g(B)) = Z (K ⊗ L(B)) = A⊗ R(Bop). A⊗KEndK(B) A⊗KEndK(B) K K Thus, Z (B)⊗ End (B) ’ A⊗ Bop, A K K K proving (1). BRAUER GROUP 9 (2): By Theorem 2, the algebra A⊗ Bop is simple. So, the isomorphism in part (1) implies that K Z (B)⊗ End (B) is simple, and therefore Z (B) is simple. Counting dimensions, we obtain A K K A dim Z (B)·m2 = (dim A)·(dim B) = nm. K A K K So, dim Z (B) = n/m (in particular, m divides n). K A (3): Obviously, B ⊂ Z (Z (B)). Applying part (2) to Z (B) (which is simple), we obtain A A A n n dim Z (Z (B)) = = = m. K A A dim Z (B) n/m K A So, B = Z (Z (B)) by dimension considerations. (cid:3) A A Corollary 4. Let A be a central simple algebra over K of dimension dim A = d2. If L is a field K extension of K of degree ‘ then ‘ divides d and Z (L) is a central simple algebra over L of dimension A dim Z (L) = (d/‘)2. In particular, if ‘ = d then Z (L) = L, and consequently, L is a maximal L A A subfield of A. Proof. Since L is commutative, L ⊂ Z (L). Then A dim Z (L) = d2/‘ = (dim Z (L))·‘, K A L A so dim Z (L) = (d/‘)2. Since L A Z(Z (L)) ⊂ Z (Z (L)) = L, A A A we obtain that Z (L) is central over L. (cid:3) A Corollary 5. Let D be a central division algebra over K of dimension dim D = d2. If P ⊂ D is a K maximal subfield then dim P = d. K NoticethateverymaximalsubfieldP ⊂ DnecessarilycontainsK asotherwisethesubringgenerated by P and K would be a subfield of D strictly containing P. Furthermore, since D is finite dimensional, maximal subfields obviously exist. Now, let P ⊂ D be a maximal subfield. Then P = Z (P). Indeed, D if a ∈ Z (P)\P then P[a] would be a subfield strictly containing P. Applying the previous corollary, D we obtain dim P = d. (In this argument we used the obvious fact that any subalgebra of a finite K dimensional division algebra is itself a division algebra.) The following proposition is needed to give a cohomological interpretation of the Brauer group. Proposition 3. Let D be a central division algebra over a field K. Then D contains a maximal subfield P which is a separable extension of K. Proof. Of course, there is nothing to prove if K has characteristic zero or is finite. So, we can assume that K is an infinite field of characteristic p > 0. Next, it is enough to show that there always exists an element a ∈ D\K which is separable over K. Indeed, given this fact, we can complete the argument by induction on dim D = d2. Indeed, if ‘ = [K(a) : K] > 1 then by Corollary 4, the centralizer K Z (K(a)) is a central division algebra over K(a) such that dim Z (K(a)) = (d/‘)2 < dim D. D K(a) D K Thenbyinductionhypothesis,Z (K(a))containsamaximalsubfieldP whichisaseparableextension D of K(a). Then P is a separable extension of K and by Corollary 5, [P : K(a)] = d/‘, implying that [P : K] = d. Then by Corollary 4, P is a maximal subfield of D. An element a ∈ D\K separable over K can be found in any maximal subfield P of D if d is not a power of p because in this case P/K cannot be purely inseparable (recall that the degree of a purely inseparable extension must be a power of p). So, we only need to consider the case where d = pα. AssumethatD\K doesnotcontainanyelementsseparableoverK.Thenalltheseelementsarepurely inseparable, and since the degree of any element over K divides pα, we obtain that apα ∈ K for all 10 IGORRAPINCHUK a ∈ D. Now, pick a basis e = 1,e ,...,e of D over K, and let t ,...,t be variables. Then there 1 2 d2 1 d2 exist polynomials f ,...,f ∈ K[t ,...,t ] such that 1 d2 1 d2 (t e +···+t e )pα = f (t ,...,t )e +···f (t ,...,t )e . 1 1 d2 d2 1 1 d2 1 d2 1 d2 d2 Since apα ∈ K for all a ∈ D, we have (6) f (a ,...,a ) = ··· = f (a ,...,a ) = 0 2 1 d2 d2 1 d2 for all (a ,...,a ) ∈ Kd2. Then, because K is infinite, we conclude that f = ··· = f = 0, and 1 d2 2 d2 therefore (6) for all (a ,...,a ) ∈ K¯d2. This means that apα ∈ K¯ for all a ∈ D ⊗ K¯. But by 1 d2 K Corollary 1, D⊗ K¯ ’ M (K¯), and for the element e of the standard basis we have epα = e ∈/ K¯, K d 11 11 11 a contradiction, proving the existence of separable elements. (cid:3) 3. The Brauer group of a field Two central simple algebras A and A are called similar (written A ∼ A ) if the division algebras 1 2 1 2 D and D such that A ’ M (D ) and A ’ M (D ), are isomorphic. 1 2 1 n1 1 2 n2 2 Lemma 5. (1) For any K-algebra R, R⊗ M (K) ’ M (R); K n n (2) M (K)⊗ M (K) ’ M (K); m K n mn (3) A ∼ A if and only if there exist m and m such that A ⊗ M (K) ’ A ⊗ M (K); 1 2 1 2 1 K m1 2 K m2 (4) similarity is an equivalence relation. Proof. (1): There is an algebra homomorphism R⊗ M (K) → M (R) such that r⊗x 7→ rx. The K n n P inverse homomorphism is given by (r ) 7→ r ⊗e , where e is the standard basis of M . ij i,j ij ij ij n (2): We have a natural homomorphism End (Km)⊗ End (Kn) → End (Km⊗Kn) = End (Kmn). K K K K K It is injective because it is nonzero and the algebra in the left-hand side is simple (Theorem 2), and it is then surjective by dimension count. (3): Suppose A ’ M (D ). If A ∼ A then D ’ D so using (1) and (2) we obtain i ni i 1 2 1 2 A ⊗ M (K) ’ D ⊗ M (K)⊗ M (K) ’ M (D ) ’ M (D ) ’ A ⊗ M (K). 1 K n2 1 K n1 K n2 n1n2 1 n1n2 2 2 K n1 Conversely, suppose A ⊗ M (K) ’ A ⊗ M (K). As above, we see that 1 K m1 2 K m2 A ⊗ M (K) ’ M (D ) for i = 1,2. i K mi mini i So, by the uniqueness part of Theorem 1 we obtain that D ’ D , and A ∼ A . 1 2 1 2 (4): Follows immediately from the definitions. (cid:3) For a (finite dimensional) central simple algebra A over a field K, we let [A] denote the equivalence class of algebras similar to A. As a set, the Brauer group of K (denoted Br(K)) is the collection of all such classes (thus, the elements of Br(K) bijectively correspond to the isomorphism classes of central division algebras over K). We introduce a product on Br(K) by using tensor product of algebras: (7) [A][B] = [A⊗ B]. K WenoticethatthealgebraA⊗ B iscentralbyProposition2andsimplebyTheorem2, so[A⊗ B] ∈ K K Br(K). If A ∼ A0 and B ∼ B0 then A⊗ M (K) ’ A0⊗ M (K) and B⊗ M (K) ’ B0⊗ M (K) K m K m0 K n K m0 for some integers m,m0,n,n0, and then (A⊗ B)⊗ M (K) ’ (A0⊗ B0)⊗ M (K), K K mn K K m0n0