22 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESQuestion: Is it possible for two different ideals I, J to give isomorphiccyclic modulesR/I∼=R/J?If so, we need to describe all the ideals J which give the same quotientup to isomorphism.Cyclic modules can be visualized as a diagram with one peak:•!!!!!!!"""""""•!!!!!!!•"""""""•"""""""• •!!!!!!!•!!!!!!!•"""""""For example, this could be K[X, Y ]/(X3, X2Y2, Y3). A finitely gener-ated module could be visualized as a union of overlapping pictures ofthis kind.[If someone can create the picture, I will insert it here.]MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 237. Products and coproductsToday we talked about products and coproducts (sums) and some ofthe consequences of these concepts. We also talked about thedifference. In particular, infinite products can be finitely generatedbut nontrivial infinite sums cannot.(1) products(2) sums(3) projective modules(4) finite generation7.1. products. If Mα, α ∈ I, is a family of R-modules then we canform the Cartesian product!α∈IMα= {(xα) | xα∈ Mα}This is an R-module with addition and action of R givencoordinate-wise. We have projection maps onto each coordinate:pα:!α∈IMα→ MαThese are R-module homomorphisms.Theorem 7.1. The Cartesian product"α∈IMαis the product in thecategory of R-modules.Proof. The statement is that, given any module L and homomorphismsfα: L → Mα, there exists a unique homomorphism f ="fα: L →"Mαmaking the following diagram commute:"Mαpα!!MαLfα""##########∃!f##This is obviously true: f must send x ∈ L to the unique element of"Mαwhose α coordinate is fα(x). !7.2. direct sums. The direct sum is equal to the weak product:#α∈IMα=!α∈I#Mαwhere the weak product is the subset of the product consisting ofelements where only finitely many coordinates are nonzero:!#Mα=$(xα) ∈!Mα| xα= 0 ∀∀α%24 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESThere are inclusion mapsjα: Mα→#Mα.Every element of&Mαis a sum of elements in the images of theseinclusion maps.Theorem 7.2. The direct sum is the coproduct in the category of R-modules.Proof. Again, the main point is to understand the statement. Theproof is trivial. This theorem says that, given any module L and ho-momorphisms fα: Mα→ L, there is a unique morphism f from thedirect sum to L making the following diagram commute.Mαjα!!fα$$$$$$$$$$$$&Mα∃!f%%LThe mapping f takes the element (xα) to the sum'fα(xα). This isdefined since only finitely many of the co ordinates xαare nonzero. !When the index set I is finite, the weak product is equal to theproduct. So, we get the following.Corollary 7.3. Given a finite collection of modules M1, · · · , Mnthedirect sum is equal to the direct product:n#i=1Mi=n!i=1Mi.The argument that proves that finite sums and products agree worksin any preadditive category. Recall that a category C is preadditive ifthe Hom sets are additive groups and composition is biadditive.Theorem 7.4. If M1, · · · , Mnare objects of any preadditive category,the finite product"Miexists in the category if and only if the finitesum (coproduct)&Miexists in the category. Furthermore, they agreewhen they exist.Proof. Suppose the product"Miexists. Then, for each i, we have theidentity map idMi: Mi→ Miand the zero maps 0 : Mi→ Mjwheni %= j. By the universal property of the product, this give a morphismji= (0, 0, · · · , 0, idMi, 0, · · · , 0) : Mi→!Miwhich has the property thatpj◦ ji= δij.MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 25I.e., this is the identity if i = j and zero if i %= j. Given any object L ofthe category and morphisms fi: Mi→ L, let f :"Mi→ L be givenbyf =n(i=1fi◦ pi.Note that this uses only categorical properties: we can compose mapsand we can add maps. Objects do not have elements! Since composi-tion is biadditive we can make the following computation.f ◦ ji=n(j=1fj◦ pj◦ ji=n(j=1δijfj= fiThus f makes the diagram in the definition of the coproduct commute.To prove the uniqueness of f we need the following equation whichholds in the endomorphism group of"Mi.n(i=1ji◦ pi= id.To prove this, let h be the endomorphism given on the left. Thenpj◦ h =n(i=1pj◦ ji◦ pi=n(i=1δijpi= pj.By the universal property of the product, this forces h = id.Returning to the uniqueness of f . Suppose that g :"Mi→ L isanother morphism so that g ◦ ji= fi. Then(f − g) ◦ ji= fi− fi= 0.So,0 = (f − g) ◦n(i=1ji◦ pi= (f − g) ◦ id = f − gwhich implies that f = g. !Definition 7.5. An additive category is a preadditive category whichhas finite products and sums including the empty sum which is the zeroobject. (Recall that a zero object is an object which is both initial andterminal.)Thus, the corollary can be rephrased to say:Corollary 7.6. The category of R-modules is an additive category.26 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES7.3. pr ojective modules.Definition 7.7. A module M is called projective if, for any epimor-phism p : A → B and any morphism f : M → B, there is a morphism˜f : M → A so that p ◦˜f = f. I.e., the following diagram commutes.Ap%%%%M∃˜f&&%%%%f!!B(1) Free modules are projective.(2) What happens when B is projective?(3) Partial converse for (1).(4) Example.7.3.1. free ⇒ projective.Theorem 7.8. Free modules are projective.Proof. Suppose that M = F (X) is the free R -module on the set X.We have a homomorphism f : F = M → B and we want to lift it to ahomomorphism˜f : F → A.We need the following adjunction property.HomR(F (X), A)∼=HomEns(X, A)Any homomorphism F (X) → A gives a set map X → A by restrictionand any set map g : X → A extends uniquely to a homomorphismg#: F (X) → A by the formulag#)(rixi*=(rig(xi).In this case, g : X → A is given by the fact that p is surjective: Forany xi∈ X, f(xi) = bi∈ B comes from some element ai∈ A. Letg(xi) = ai. Then the lifting˜f : F → A is given by˜f)(rixi*=(rig(xi) =(riai.This is a lifting of f sincef)(rixi*=(rif(xi) =(ribi= p)(riai*= p˜f)(rixi*!MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 277.3.2. What if B were projective?Theorem 7.9. If p : A → B is onto and B is projective thenA∼=B ⊕ ker p.Proof. What students immediately realized is that there is a mappings : B → A. This follows from the definition of a projective moduleapplied to the following diagram.Ap%%%%B∃s&&!!!!!!!idB!!BThe morphism s : B → A is called a section of p because it satisfies thecondition p ◦ s = idB. So, we