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TIFR Notes on Semisimple Rings, Central Simple Algebras, Etc TIFR Notes on Semisimple Rings, Central Simple Algebras, Etc. August 3, 2007 2 Contents 1 Tensor Product 3 2 Semi-Simple Modules and Rings 13 3 Central Simple Algebras 25 4 Brauer Groups of some Fields 39 5 Representations of Finite Groups 47 1 2 CONTENTS Chapter 1 Tensor Product 1. Tensor product of modules and linear § maps. Let M,N and P be abelian groups. A map f : M N P is called bi-additive if × → f(x1 + x2, y)= f(x1, y)+ f(x2, y) and f(x, y1 + y2)= f(x, y1)+ f(x, y2) for x, x1, x2 M and y, y1, y2 N. In other words, f is bi-additive if the maps x f(∈x, y) and y f(x,∈ y) are homomorphisms. It follows that 7→ 7→ f(x, 0) = f(0, y) = 0 and f(x, y)= f( x, y)= f(x, y) − − − for x M and y N. ∈ ∈ Let A be a ring, M a right A-module, N a left A-module and P an abelian group. Then a map f : M N P is called balanced (over A, to be precise) if it is bi-additive and satisfies× → f(xa, y)= f(x, ay) for x M, y N and a A. ∈ ∈ ∈ 3 4 CHAPTER 1. TENSOR PRODUCT A tensor product of M and N is a pair (T,φ), where T is an abelian group and φ is a balanced map from M N into T , which satisfies the following condition: × given any abelian group P and a balanced map f : M N P , there exists a unique homomorphism of groups f˜ : T P such that× the→ diagram → M N × EE EE EE EE f φ EE EE EE EE EE" T / P f˜ is commutative. It is easily seen that if (T ′,φ′) is another tensor product of M and N, then there exists an isomorphism of groups α : T T ′ such that φ′ = α φ. In this sense the tensor product is unique (provided→ that it exists). We now◦ proceed to show that it does exist. Let F denote the free abelian group with the set M N as a basis. Let H denote the subgroup of F generated by the elements× of the form (x + x , y) (x , y) (x , y), 1 2 − 1 − 2 (x, y + y ) (x, y ) (x, y ), 1 2 − 1 − 2 (xa, y) (x, ay), − for x, x1, x2 M; y, y1, y2 N; a A. The natural homomorphism : F F/H, restricted∈ to M N∈, gives∈ a balanced map φ : M N F/H→, and φ(M N) generates× F/H. Now let P be an abelian× group→ and let f : M N× P be a balanced map. Then f extends to a homomorphism × → f : F P . Since f is balanced f vanishes on H and therefore passes down to a homomorphism→ f : F/H P . Evidently the diagram → e M N × DD DD DD DD f φ DD DD DD DD D fe D! F/H / P 5 is commutative. Further, f is unique since φ(M N) generates F/H. Thus (F/H,φ) is a tensor product of M and N. × We shall write M AN for F/H and x y for φ((x, y)), where x M and y N. Since (x,⊗ y) x y is balanced,⊗ it follows that every element∈ ∈ 7→ ⊗ of M AN can be written in the form xi yi for some xi M and ⊗ 1≤i≤n ⊗ ∈ yi N. ∈ P Remark. M AN can be 0 even if neither M nor N is 0. For instance, ⊗ Q ZZ/(n) = 0 for n = 0. ⊗ 6 Let M, M ′ be two right A-modules and N,N ′ two left A-modules. If f : M M ′ and g : N N ′ are A-linear maps, then the composite map → → (f,g) ′ ′ ′ ′ M N M N M AN × −−−→ × −−−→ ⊗ is balanced. Thus there exists a unique homomorphism of groups f g : ′ ′ ⊗ M AN M AN such that the diagram ⊗ → ⊗ (f,g) M N M ′ N ′ × −−−→ × ′ ′ M AN M AN ⊗y −−−→f⊗g ⊗y is commutative. The map f g is called the tensor product of f and g; it is characterized by ⊗ (f g)(x y)= f(x) g(y). ⊗ ⊗ ⊗ The map ′ ′ ′ ′ HomA(M, M ) HomA(N,N ) HomZ(M AN, M AN ) × −→ ⊗ ⊗ given by (f,g) f g is bi-additive. 7→ ⊗ Remarks. 1. 1M 1N = 1M N . ⊗ ⊗ f f ′ g g′ 2. if M M ′ M ′′ (resp. N N ′ N ′′) are A-linear maps, where−→ the M’s−→ (resp. N’s) are right−→ (resp.−→ left) A-modules, then (f ′ f) (g′ g)=(f ′ g′) (f g). ◦ ⊗ ◦ ⊗ ◦ ⊗ 6 CHAPTER 1. TENSOR PRODUCT 2. Structure of module on tensor product § We recall that given two rings A and B, an abelian group M is said to be an A-B-bimodule if M is a module over A and B, and if the endomorphism of M effected by any element of A commutes with the endomorphism of M effected by any element of B. In other words the multiplication in M by any element of B is A-linear. If M is an A-B-bimodule on which A operates from the left and B operates from the right, then we shall say that M is an A-B-bimodule of the type AMB. Similar explanations for A,BM and MA,B. Examples. 1. Any two-sided ideal of A is an A-A-bimodule in an obvious manner. 2. If h : A B is a homomorphism of rings, then B can be considered as → an A-B-bimodule of the type ABB or BBA. 3. If A is a commutative ring, then any A-module can be considered as an A-A-bimodule by taking the given A-module structure on M twice. Let N be a left A-module and let M be an A-B-bimodule of the type BMA. We know that the multiplication θb in M by an element b of B is A-linear. This gives rise to an endomorphism θb 1 of the group M AN. From the properties of tensor product of linear maps⊗ mentioned at the⊗ end of 1, it follows that b θb 1 is a homomorphism of rings from B into § 7→ ⊗ EndZ(M AN). Thus M AN acquires the structure of a left B-module which is characterized⊗ by the formula⊗ b (x y)=(bx) y ⊗ ⊗ ′ ′ for b B, x M and y N. If BMA is another bimodule, f : M M linear∈ over A∈and B, and∈g : N N ′ an A-linear map of left A-modules→ then f g is B-linear. → ⊗ A similar discussion applies to the case where M is an A-B-bimodule of the type MA,B or N is an A-B-bimodule (of course A operating on N from the left). Proposition 2.1. If M (resp. N) is a right (resp. left) A-module, then M AA = M as right A-modules (resp. A AN = N as left A-modules). ⊗ ∼ ⊗ ∼ 7 Proof. The map M A M given by (x, a) xa is balanced and gives × → 7→ rise to a homomorphism φ : M AA M with φ(x a)= xa. Clearly φ is A- ⊗ → ⊗ linear. Define an A-linear map ψ : M M AA by ψ(x)= x 1. It is easily checked that φ and ψ are inverses of→ each⊗ other. So φ is an⊗ isomorphism. Similarly A AN is isomorphic to N as left A-modules. ⊗ Proposition 2.2. Let A and B be two rings. Let M be a right A-module, N an A-B-bimodule of the type ANB, and P a left B-module. Then there is an isomorphism of groups ∼ M A(N BP ) (M AN) BP ⊗ ⊗ −→ ⊗ ⊗ which maps x (y z) to (x y) z. ⊗ ⊗ ⊗ ⊗ Proof. For each x M, define a map ∈ αx : N P (M AN) BP × → ⊗ ⊗ by αx((y, z))=(x y) z. The map αx is balanced (over B) and gives rise to ⊗ ⊗ a homomorphism αx : N BP (M AN) BP . Now define a map ⊗ → ⊗ ⊗ β : M (N BP ) (M AN) BP f × ⊗ → ⊗ ⊗ by β((x, w)) = αx(w) for x M and w N BP . Then β is balanced (over ∈ ∈ ⊗ A) and gives rise to a homomorphism f β : M A(N BP ) (M AN) BP ⊗ ⊗ → ⊗ ⊗ such that β(x (y z))=(x y) z. Similarly we have a homomorphism (say ⊗ ⊗ e ⊗ ⊗ γ) in the other direction such that γ((x y) z) = x (y z). The maps β ⊗ ⊗ ⊗ ⊗ and γ aree inverses of each other and hence β is an isomorphism. e e e The above discussion about the structure of module on tensor product e can bee applied in particular to the following two situations: (1) Change of rings. Let h : A B be a homomorphism of rings and → let N be a left A-module. Then B AN is a left B-module in a canonical ⊗ manner (consider B as an A-B-bimodule of the type BBA). We say that the B-module B AN is obtained by a change of base ring (via h). Similarly if ⊗ M is a right A-module, then M AB is canonically a right B-module. ⊗ The first proposition above says that the module obtained by a change of rings via the identity map of the base ring is the same as the original module. 8 CHAPTER 1. TENSOR PRODUCT The second proposition (combined with the first) gives the transitivity of change of rings: if h′ : B C is another homomorphism of rings and N is an A-module, say a left→ module, then the C-module obtained by change of rings through h′ h : A C is canonically isomorphic to the C-module ′ ◦ → obtained via h from the B-module B AN ontained via h from N.
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