REPRESENTATION THEORYWEEK 91. Jordan-H¨older theorem and indecomposable modulesLet M be a module satisfying ascending and descending chain conditions (ACCand DCC). In other words every increasing sequence submodules M1⊂ M2⊂ . . . andany decreasing sequence M1⊃ M2⊃ . . . are finite. Then it is easy to see that thereexists a finit e sequenceM = M0⊃ M1⊃ · · · ⊃ Mk= 0such that Mi/Mi+1is a simple module. Such a sequence is called a Jordan-H¨olderseries. We say that two Jordan H¨older seriesM = M0⊃ M1⊃ · · · ⊃ Mk= 0, M = N0⊃ N1⊃ · · · ⊃ Nl= 0are equivalent if k = l and for some permutation s Mi/Mi+1∼=Ns(i)/Ns(i)+1.Theorem 1.1. Any two Jordan-H¨older series are equivalent.Proof. We will prove that if the statement is true for any submodule of M then itis tr ue for M. (If M is simple, the stateme nt is trivial.) If M1= N1, then thestatement is obvious. Otherwise, M1+ N1= M, hence M/M1∼=N1/ (M1∩ N1) andM/N1∼=M1/ (M1∩ N1). Consider the seriesM = M0⊃ M1⊃ M1∩N1⊃ K1⊃ · · · ⊃ Ks= 0, M = N0⊃ N1⊃ N1∩M1⊃ K1⊃ · · · ⊃ Ks= 0.They are obviously equivalent, and by induction assumption the first series is equiv-alent to M = M0⊃ M1⊃ · · · ⊃ Mk= 0, and the second one is equivalent toM = N0⊃ N1⊃ · · · ⊃ Nl= {0}. Hence they are equivalent. Thus, we can define a length l (M) of a m odule M satisfying ACC and DCC, andif M is a proper submodule of N, t hen l (M) < l (N).A module M is indecomposable if M = M1⊕ M2implies M1= 0 or M2= 0.Lemma 1.2. Let M and N be indecomposable, α ∈ HomR(M, N), β ∈ H omR(N, M)be such that β ◦ α is an isomorphism. Then α and β are isomorphisms.Proof. We claim that N = I m α⊕ Ker β. Indeed, Im α∩Ker β = 0 and for any x ∈ None can write x = y + z, where y = α ◦ ( β ◦ α)−1◦ β (x), z = x − y. Then since N isindecomposable, Im α = N, Ker β = 0 and N∼=M. Date: November 7, 2005.12 REPRESENTATION THEORY WEEK 9Lemma 1.3. Let M be indecomposable module of finite le ngth and ϕ ∈ EndR(M),then either ϕ is an isomorphism or ϕ is nil potent.Proof. There is n > 0 such that Ker ϕn= Ker ϕn+1, Im ϕn= Im ϕn+1. In this caseKer ϕn∩ Im ϕn= 0 and hence M∼=Ker ϕn⊕ Im ϕn. Either Ker ϕn= 0, Im ϕn= Mor Ker ϕn= M. Hence the le mma. Lemma 1.4. Let M be as in Lemma 1.3 and ϕ, ϕ1, ϕ2∈ EndR(M), ϕ = ϕ1+ ϕ2. Ifϕ is an isomorphism then at least one of ϕ1, ϕ2is also an isomorphism.Proof. Without loss of generality we may assume that ϕ = id. But in this case ϕ1and ϕ2commute. If both ϕ1and ϕ2are nilpotent, then ϕ1+ ϕ2is nilpotent, but thisis impossible as ϕ1+ ϕ2= id. Corollary 1.5. Let M be as in Lemma 1.3. Let ϕ = ϕ1+ · · · + ϕk∈ EndR(M). Ifϕ is an isomorphism then ϕiis an isomorphism at least for one i.It is obvious that if M satisfies ACC and DCC then M has a decompositionM = M1⊕ · · · ⊕ Mk,where all Miare indecomposable.Theorem 1.6. (Krull-Schmidt) Let M be a module of finite length andM = M1⊕ · · · ⊕ Mk= N1⊕ · · · ⊕ Nlfor some indecomposable Miand Nj. Then k = l and the r e exists a permutation ssuch that Mi∼=Ns(j).Proof. Let pi: M1→ Nibe the restriction to M1of the natural projec t ion M → Ni,and qj: Nj→ M1be the restriction to Njof the natural projection M → M1. Thenobv iously q1p1+ · · · + qlpl= id, and by Corollary 1.5 there exists i such that qipiisan isomorphism. Lemma 1.2 implies that M1∼=Ni. Now one can easily finish theproof by induction on k. 2. Some facts from homological algebraThe comple x is the graded abel ian group C·= ⊕i≥0Ci. We wil l assume later thatall Ciare R-modules for some ring R. A differential is an R-morphism of degree −1such that d2= 0. U sually we realize C·by the pictured−→ · · · → C1d−→ C0→ 0.We also consider d of degree 1, in this case the superindex C·and0 → C0d−→ C1d−→ . . .All the proofs are similar for these two cases.Homology group is Hi(C) = (Ker d ∩ Ci) /dCi+1.REPRESENTATION THEORY WEEK 9 3Given two complexes (C·, d) and (C′·, d′). A morphism f : C·→ C′·preservi nggrading and satisfying f ◦ d = d′◦ f is called a morphism of complexes. A morphismof complex es induces the morphism f∗: H·(C) → H·(C′).Theorem 2.1. (Long exact sequence). Let0 → C·g−→ C′·f−→ C′′·→ 0be a short e xact seque nce, then the long exacts sequenceδ−→ Hi(C)g∗−→ Hi(C′)f∗−→ Hi(C′′)δ−→ Hi−1(C)g∗−→ . . .where δ = g−1◦ d′◦ f−1, is exact.Let f, g : C·→ C′·be two morphisms of complexes. We say that f and g arehomotopically equivalent if there exists h : C·→ C′·(+1) (the morphism of degree 1)such that f − g = h ◦ d + d′◦ h.Lemma 2.2. If f and g are homotopically e quivalent then f∗= g∗.Proof. Let φ = f − g, x ∈ Ciand dx = 0. Thenφ (x) = h (dx) + d′(hx) = d′(hx) ∈ Im d′.Hence f∗− g∗= 0. We say that complexes C·and C′·are homotopically equivalent if there exist f :C·→ C′·and g : C′·→ C·such that f ◦ g is homotopically equivalent to idC′and g ◦ fis homotopically equivalent to idC. Lemma 2.2 impli es that homotopically equivalentcomplexes have isomorphic homology. The f ol lowing Lemma is straightforward.Lemma 2.3. If C·and C′·are homotopically e quivalent then the compl exes HomR(C·, B)and HomR(C′·) are also homotopically equivalent.Note that the differential in HomR(C·, B) has degree 1.3. Projective modulesAn R-mo dule P is projective if for any surjective morphism φ : M → N and anyψ : P → N there exists f : P → M such that ψ = φ ◦ f.Example. A free module is projective. Indeed, let {ei}i∈Ibe the set of freegenerators of a free module F , i.e. F = ⊕i∈IRei. Define f : F → M by f (ei) =φ−1(ψ (ei)).Lemma 3.1. The following conditions on a module P are equivalent(1) P is projective;(2) There exists a free module F such that F∼=P ⊕ P′;(3) Any exact seq uence 0 → N → M → P → 0 splits.4 REPRESENTATION THEORY WEEK 9Proof. (1) ⇒ (3) Consider the exact sequence0 → Nϕ−→ Mψ−→ P → 0,then sinc e ψ is surjective, there exists f : P → M such that ψ ◦ f = idP.(3) ⇒ (2) Every module i s a quotient of a free module. Therefore we just have toapply (3) to the exact sequence0 → N → F → P → 0for a fr ee module F .(2) ⇒ (1) Let φ : M → N be surjective and ψ : P → N. Choose a free module Fso that F = P ⊕P′. …