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MMATH18-201: Module Theory Lecture Notes: Composition Series

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MMATH18-201: Module Theory Lecture Notes: Composition Series MMATH18-201: Module Theory Lecture Notes: Composition Series 1 Composition Series In this section, we introduce the notion of composition series for modules and give some necessary and sufficient condition for the existence of composition series. As a consequence, we define the length of a module which paves the way for the application of Jordan–H¨older Theorem for modules of finite lengths. Throughout R denotes a commutative ring with unity. The result of this section are from Chapter 6 of [1]. Definition. Let M be an R-module. A chain in M is a finite sequence fMig of submodules of M starting with the module M itself and ending with the zero submodule, such that M = M0 ⊃ M1 ⊃ · · · ⊃ Mn = f0g: The number of inclusions or links (`⊃') is called the length of the chain. If fNig and fMjg are two chains in M, then fNig is called a refinement of fMjg if each Mj equals some Ni. Definition. A chain M = M0 ⊃ M1 ⊃ · · · ⊃ Mn = f0g; of submodules of a module M is called a composition series for M if the quotient modules Mi−1=Mi (1 ≤ i ≤ n) are simple, i.e., has no submodules other than the trivial one and itself. So the chain is maximal in the sense that we cannot insert any submodule between Mi−1 and Mi; for each i 2 f1; 2; : : : ; ng or the chain has no proper refinement. Definition. If a module M has no composition series, then we say that the length of M is infinite. If a module has a composition series, then the least length of a composition series for M is called the length of M and is denoted by by l(M): So for a module M the length l(M) is either infinite or finite depending upon whether M has a composition series or not. If l(M) < 1 then we say that M is of finite length and in this case we prove that l(M) is an invariant for M, i.e., every composition series for M has the same length. For this we use the following lemma. 1 Lemma 1.1. Let N be a proper submoduel of a module M of finite length. Then l(N) < l(M). Proof. Let l(M) = m and M = M0 ⊃ M1 ⊃ · · · ⊃ Mm = f0g; (1) a composition series for M of length m. For i 2 f0; 1; : : : ; mg let Ni = N \ Mi. Then from (1), we have a chain N = N0 ⊇ N1 ⊇ · · · ⊇ Nm = f0g of submodules of N. Consider not the following composition of module homo- morphisms αi−1 πi−1 Ni−1 −−! Mi−1 −−! Mi−1=Mi; for i 2 f1; 2; : : : ; mg; where αi−1 and πi−1 denote respectively the inclusion and canonical projections. The kernel of the αi−1πi−1 is the set of elements of Ni−1 which lie in Mi, i.e., ker αi−1πi−1 = Ni−1 \ Mi = (N \ Mi−1) \ Mi = N \ (Mi−1 \ Mi) = N \ Mi (* Mi ⊂ Mi−1) ) ker αi−1πi−1 = Ni: (2) So there is an induced injective homomorphism (say) φi−1 : Ni−1=Ni −! Mi−1=Mi; (1 ≤ i ≤ n). Since (1) is a composition series, so for each i, Mi−1=Mi is simple therefore φi−1 (Ni−1=Ni) is either the trivial module or whole of Mi−1=Mi; i.e., either Ni−1=Ni is trivial or Ni−1=Ni is simple (* in this case φi−1 is an isomor- phism) which implies that either Ni−1 = Ni or Ni is maximal in Ni−1. Hence by deleting repeated terms (if any) in the series N = N0 ⊃ N1 ⊃ · · · ⊃ Nm = f0g; (3) we obtain a composition series for N of length at most m; i.e., l(N) ≤ l(M). If l(N) = l(M), then Ni−1 6= Ni (1 ≤ i ≤ n) therefore for each i we have that Ni−1=Ni is simple. Now ∼ Mi−1=Mi = Ni−1=Ni = Ni−1= (Ni−1 \ Mi) ; (by (2)) ∼ = (Ni−1 + Mi) =Mi (by Second Isomorphism Theorem) ⊆ Mi−1=Mi; 2 which implies that (Ni−1 + Mi) =Mi = Mi−1=Mi consequently Ni−1 + Mi = Mi−1 for each i 2 f1; 2 : : : ; mg: Since Mm = f0g so for i = m we have Nm−1 = Mm−1: For i = m − 1, Nm−2 + Mm−1 = Mm−2 =) Nm−2 + Nm−1 = Mm−2 (* Nm−1 = Mm−1) i:e:; Nm−2 = Mm−2 (* Nm−1 ⊂ Nm−2): Continuing in this manner we get that Nm−3 = Mm−3;:::;N1 = M1 and finally for i = 1, N0 = M0; i.e., N = M, which contradicts the fact that N is a proper submodule of M. Hence our assumption that l(N) = l(M) is not possible so l(N) < l(M): We now prove that for a module of finite length its length is independent of the choice of composition series. Theorem 1.2. If a module M has a composition series of length n; then every composition series of M has length n and every chain can be extended to a composition series. Proof. Since M has a composition series of length n, so l(M) ≤ n. Let M = M0 ⊃ M1 ⊃ · · · ⊃ Mk−1 ⊃ Mk = f0g; (4) be any chain in M. Since the inclusions in the above chain are strict so by Lemma 1.1, we have that l(M) = l(M0) > l(M1) > ··· > l(Mk−1) > l(Mk) = 0: (5) As k > k − 1 > ··· > 2 > 1 > 0 and l(Mi) are non-negative integers so from (5) it follows that l(Mk) = l(M) ≥ k: Since (4) was an arbitrary chain in M, we have that length of every chain in M is atmost l(M). But every composition series is also a chain and M has a composition series of length n therefore we have that n ≤ l(M), which in view of l(M) ≤ n implies that l(M) = n. Hence every composition series in M has length n. Now let M = N0 ⊃ N1 ⊃ · · · ⊃ Nk−1 ⊃ Nk = f0g; (6) 3 be any chain in M, then from above we have that k ≤ n = l(M): If k = n, then the chain (6) is a already a composition series because otherwise for some i, Ni−1=Ni is not simple, i.e., we can insert a submodule between Ni−1 and Ni; resulting in a chain of length more than n which is not possible. So assume that k < n: Since all composition series of M have same length, we have that (6) cannot be a composition series, i.e., for some i 2 f1; 2; : : : ; kg, Ni is not 0 a maximal submodule of Ni−1: So there exist a submodule (say) Ni such that 0 Ni−1 ⊃ Ni ⊃ Ni and we obtain a chain of length k + 1. If k + 1 = n, then this chain becomes a composition series and we are done otherwise we repeat this procedure until we obtain a chain of length n, which must be a composition series. We now recall the Jordan–H¨olderTheorem for finite groups. For proof see Theorem 22, Chapter 3 of [2] or Theorem 5.12 of [3]. 0 Jordan–H¨olderTheorem. Any two composition series fGig0≤i≤n and fGig0≤i≤n of a group G are equivalent, i.e., there is a one to one correspondence between 0 0 the set of quotients fGi−1=Gig0≤i≤n and the set of quotients fGi−1=Gig0≤i≤n such that the corresponding quotients are isomorphic. In view of Theorem 1.2, it follows that the Jordan–H¨olderTheorem is also applicable to modules of finite length. The proof is the same as that for finite groups. The following result give some necessary and sufficient condition for an R- module to have a composition series. Theorem 1.3. An R-module M has a composition series if and only if it satisfies both chain conditions (a.c.c., and d.c.c). Proof. Suppose first that M has a composition series of length n. Then in view of Theorem 1.2, l(M) = n and every chain in M has length at most n. Therefore we cannot have a strictly ascending or descending chain of submodules of M of length more than n. Because if N1 ⊂ N2 ⊂ · · · ; is a strictly ascending chain of submodules of M consisting of more than n submodules, then on adding the trivial submodule and the module itself, we 4 obtain a chain f0g = N0 ⊆ N1 ⊂ N2 ⊂ · · · ⊂ Nn+1 ⊆ M; of M of length atleast n + 1 which contradicts the fact that length of M is n. So M satisfies a.c.c. Similarly (verify) we can show that M also satisfies d.c.c. Conversely assume that M satisfies both a.c.c., and d.c.c., we need to show that M has a composition series. If M = f0g, then nothing to prove, so suppose that M 6= f0g and let Σ0 be the set of all proper submodules of M. Since M is Noetherian (* it satisfies a.c.c.), so Σ0 has a maximal element (say) M1. By definition of Σ0, M1 is a maximal submodule of M: Now if M1 = f0g; then we have a composition series M = M0 ⊃ M1 = f0g of length 1 and we are done. Otherwise let M1 6= f0g, then it is also Noetherian and the set Σ1 of all proper submodules of M1 has a maximal element (say) M2, which must be a maximal submodule of M1. Again if M2 = f0g, we obtain a composition series M = M0 ⊃ M1 ⊂ M2 = f0g; of length 2.
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