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1. Short Exact Sequences 1.1. Definition

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1. Short Exact Sequences 1.1. Definition 1. Short exact sequences ′ ′ f1 f2 ′ f1 ′ f2 ′ 1.1. Definition. Let 0 A1 A2 A3 0 and 0 A1 A2 A3 0 be two short exact sequence of R→-modules.−→ We−→ call these→ short exact→ sequences−→ −→E and E→′. Suppose h : A A′ are R-module homomorphisms such that the following diagram commutes: j j → j f1 f2 0 / A1 / A2 / A3 / 0 (1) h1 h2 h3 f ′ f ′ ′ 1 ′ 2 ′ 0 / A1 / A2 / A3 / 0 ′ Then one says that h = (h1, h2, h3) is a homomorphism from E to E . We say that h is an ′ isomorphism from E to E if each hj is an isomorphism. Let A and C be two R-modules. Let i : A A C be the inclusion and p : A C C be the projection. Then one has an exact sequence→ ⊕ ⊕ → p 0 A i A C C 0. (2) → −→ ⊕ −→ → We say that a short exact sequence is split if it is isomorphic to an exact sequence of the form (2). 1.2. Lemma. Let (1) be a homomorphism of short exact sequences. If h1 and h3 are iso- morphism, then so is h2. j q 1.3. Theorem (Splitting criteria). Let 0 A B C 0 be a short exact sequence of R-modules. Then the following are equivalent:→ −→ −→ → (a) The exact sequence is split. (b) There exists k : B A such that k j = id . → ◦ A (c) There exists r : C B such that q r = idC . → ′ ◦ ′ (d) There exists a submodule C B such that q C′ : C C is an isomorphism. Further, if C′ is any submodule⊆ as in (d), then B| = j(A→) C′. In particular, the short exact sequence above is split, then B A C. ⊕ ≃ ⊕ Proof. (b) = (d) : Define C′ = ker(k) and verify that q ′ : C′ C is an isomorphism. ⇒ |C → (d) = (c) : Define r : C B to be the inverse of the isomorphism q ′ . ⇒ → |C (c) = (a) : Define a homomorphism h2 : A C B by h(a, c) = j(a)+ r(c). One verifies that⇒ the diagram below commutes. ⊕ → i p 0 / A / A C / C / 0 ⊕ id h2 id j q 0 / A / B / C / 0 Lemma 1.2 implies that h2 is an isomorphism. Thus (c) implies (a). Finally (a) = (b) is left as an exercise. ⇒ Let C′ be any submodule as in (d). Let b B. Then q(b rqb)=0, so(b rqb) i(A). So writing b = (b rqb)+ rqb, we see that B∈ = i(A)+ C′.− If b i(A) C′,− then b∈= i(a) for some a A.− so q(b) = qi(a) = 0. Since b C′ and the restriction∈ ∩ of q to C′ is an isomorphism,∈ it follows that b =0. So B = i(A) ∈C′. 1 ⊕ 2. Hom’s and Tensors References: Dummit and Foote (Ch. 10.4 - 10.5), Atiyah Macdonald (Ch. 2). Let R be a commutative ring with 1. 2.1. Exercise: Let M ′,M,M ′′ be R-modules. Given an R-module N and an R-module homomorphism f : M ′ M, one obtains an R-module homomorphism f ∗ : Hom(M, N) Hom(M ′, N) given by f→∗(α)= α f. → Let ◦ f g M ′ M M ′′ 0 (3) −→ −→ → be a sequence of R-modules and homomorphisms. Then the following are equivalent: (a) The sequence (3) is exact. g∗ f ∗ (b) The sequence 0 Hom(M ′′, N) Hom(M, N) Hom(M ′, N) is exact for every R-module N. → −→ −→ 2.2. Definition. Let M,N,P be R-modules. Let α : M N P be a map. If m M, × → ∈ then let αm,• : N P denote that map αm,•(n)= α(m, n). If n N, then let α•,n : M P denote that map →α (m) = α(m, n). Say that α : M N P∈is a R-bilinear map if →α •,n × → m,• and α•,n are R-linear maps for all m M and n N. Let Bilinear(M ,N,P ) denote the set of all R-bilinear maps from M N∈ P . ∈ × × → 2.3. Exercise: Verify that there are isomorphisms i : Bilinear(M ,N,P ) Hom (M, Hom (N,P )) (4) M × ≃ R R given by α αm,•. These isomorphisms are natural in the sense that for all R-module homomorphism7→ φ : M M ′, the following diagram commutes: → iM Bilinear(M N,P ) / HomR(M, HomR(N,P )) O × O ∗ ∗ (φ×idN ) φ ′ ′ Bilinear(M N,P ) / HomR(M , HomR(N,P )) × iM′ Here φ∗ and (φ id )∗ are the maps induced by the functoriality of Hom (see 2.1). × N 2.4. Definition. Let M and N be R-modules. Then there exists a pair (T,g) where T is an R-module and g : M N T is an R-bilinear map satisfying the following universal property: Given any R-module× →P and an R-bilinear map α : M N P , there exists a unique R-module homomorphism α¯ : T P such that α =α ¯ g. × → One verifies that such a pair (T,g) is necessarily→ unique upto unique◦ isomorphism. We write T =(M R N) and call it the tensor product of M and N over R. We write M N instead of M ⊗N, when there is no chance of confusion. ⊗ ⊗R 2.5. Construction of the tensor product: Let F be the free module on the set M N with a basis indexed by all ordered pairs (m, n), with m M and n N. Let S be× the submodule of F generated by all expressions of the following∈ form: ∈ (m + m , n) (m , n) (m , n), r(m, n) (mr, n), 1 2 − 1 − 2 − (m, n + n ) (m, n ) (m, n ), r(m, n) (m, rn). 1 2 − 1 − 2 − 2 Define M N = F/S. The image of (m, n) F in M N is denoted by m n. So ⊗R ∈ ⊗R ⊗ by construction, M R N consists of finite expressions of the form j rj(mj nj) and the following properties⊗ hold: P ⊗ (m + m ) n = m n + m n, m (n + n )= m n + m n , 1 2 ⊗ 1 ⊗ 2 ⊗ ⊗ 1 2 ⊗ 1 ⊗ 2 r(m n)=(mr) n = m (rn). ⊗ ⊗ ⊗ 2.6. Lemma. Let φ : M M ′ and ψ : N N ′ be R-module homomorphisms. Then we get a map φ ψ : M N →M ′ N ′. Let (M→ N,g) and (M ′ N ′,g′) be the tensor products. Then there× exists× a unique→ R×-module homomorphism⊗ φ ψ⊗: M M ′ N N ′ such that g′ (φ ψ)=(φ ψ) g. ⊗ ⊗ → ⊗ ◦Let ×be the category⊗ ◦ of R-modules. The construction (M, N) (M N) and (φ, ψ) (φ ψC) defines a functor from the category to . 7→ ⊗ 7→ ⊗ C × C C 2.7. Lemma. Let M,N,P be R-modules: Show that there are the following natural isomor- phisms: (a) R R M M. (b) (M⊗ N)≃ P M (N P ). (c) M ⊗(N ⊗P ) ≃ (M⊗ N)⊗ (M P ). ⊗ ⊕ ≃ ⊗ ⊕ ⊗ 2.8. Remark. The above lemma says that the category of R-modules with the operations and behave like a ring where R plays the role of anC unit. ⊕ ⊗ 2.9. Lemma. There are natural isomorphisms M N N M induced by (m n) (n m). ⊗ ≃ ⊗ × 7→ ⊗ 2.10. Exercise: Let (M N,g) be the tensor product of the R-modules M and N. Show that the universal property⊗ of (M N,g) implies that we have isomorphisms ⊗ g∗ : Hom (M N,P ) Bilinear(M N,P ) R ⊗ ≃ × given by g∗(¯α)=α ¯ g which are natural in the following sense. Given any R-module homomorphisms φ : M◦ M ′ and ψ : N N ′, the following diagram commutes: → → g∗ HomR(M N,P ) / Bilinear(M N,P ) O ⊗ O × (φ⊗ψ)∗ (φ×ψ)∗ ′ ′ ′ ′ HomR(M N ,P ) / Bilinear(M N ,P ). ⊗ (g′)∗ × Hint: 2.6. 2.11. Lemma (Hom- adjunction). Let M,N,P be R-modules.There are isomorphisms ⊗ γ : Hom (M N,P ) Hom (M, Hom (N,P )). M R ⊗R ≃ R R which are natural in the following sense. Given any R-module homomorphism φ : M M ′, the following diagram commutes: → γM HomR(M R N,P ) / HomR(M, HomR(N,P )) O ⊗ O ∗ ∗ (φ⊗idN ) φ ′ ′ HomR(M R N,P ) / HomR(M , HomR(N,P )) ⊗ γM′ 3 Proof. Use the commutative diagrams in 2.3 and 2.10 with ψ = idN . f g 2.12. Lemma (Exactness property of ). Let M ′ M M ′′ 0 be an exact sequence of R-modules. Then the sequence ⊗ −→ −→ → f⊗id g⊗id M ′ N N M N N M ′′ N 0 ⊗ −−−−→ ⊗ −−−→ ⊗ → is exact for all R-module N. Proof. Let N,P be any two R-modules. Then the implication (a) = (b) in 2.1 tells us that ⇒ g∗ f ∗ 0 Hom (M ′′, Hom (N,P )) Hom (M, Hom (N,P )) Hom (M ′, Hom (N,P )) → R R −→ R R −→ R R is exact. Now the 2.11 tells us that the induced sequence (g⊗id )∗ (f⊗id )∗ 0 Hom (M ′′ N,P ) N Hom (M N,P ) N Hom (M ′′ N,P ) → R ⊗ −−−−−→ R ⊗ −−−−−→ R ⊗ is exact. Note that this uses the naturality in 2.11. Now the implication (b) = (a) in 2.1 f⊗id g⊗id ⇒ tells us that M ′ N N M N N M ′′ N 0 is exact too. ⊗ −−−−→ ⊗ −−−→ ⊗ → .Lemma. Let M be an R-module and I be an ideal in R. Let i : I ֒ R be the inclusion .2.13 i⊗idM → (a) Show that I M IM and I M R M is simply the inclusion of IM ⊗R ≃ ⊗R −−−→ ⊗R in M once we identify R R M with M.
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