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Central Simple Algebras and the Brauer Group

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Central Simple Algebras and the Brauer Group 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<n 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).
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