MMATH18-201: Module Theory Lecture Notes: Composition Series

Home , Composition series, Length of a module

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–Hölder 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 {Mi} of submodules of M starting with the module M itself and ending with the zero submodule, such that

M = M0 ⊃ M1 ⊃ · · · ⊃ Mn = {0}.

The number of inclusions or links (‘⊃’) is called the length of the chain. If {Ni} and {Mj} are two chains in M, then {Ni} is called a reﬁnement of {Mj} if each

Mj equals some Ni.

Deﬁnition. A chain

M = M0 ⊃ M1 ⊃ · · · ⊃ Mn = {0}, 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 ∈ {1, 2, . . . , n} 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) < ∞ 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 = {0}, (1) a composition series for M of length m. For i ∈ {0, 1, . . . , m} let Ni = N ∩ Mi. Then from (1), we have a chain

N = N0 ⊇ N1 ⊇ · · · ⊇ Nm = {0} 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 ∈ {1, 2, . . . , m}, 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 isomorphism) 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 = {0}, (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 ∈ {1, 2 . . . , m}. Since Mm = {0} 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 = {0}, (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 = {0}, (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 ∈ {1, 2, . . . , k}, 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–HölderTheorem. Any two composition series {Gi}0≤i≤n and {Gi}0≤i≤n of a group G are equivalent, i.e., there is a one to one correspondence between 0 0 the set of quotients {Gi−1/Gi}0≤i≤n and the set of quotients {Gi−1/Gi}0≤i≤n such that the corresponding quotients are isomorphic. In view of Theorem 1.2, it follows that the Jordan–HölderTheorem 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

{0} = 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 = {0}, then nothing to prove, so suppose that M 6= {0} 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 = {0}, then we have a composition series M = M0 ⊃ M1 = {0} of length 1 and we are done.

Otherwise let M1 6= {0}, 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 = {0}, we obtain a composition series

M = M0 ⊃ M1 ⊂ M2 = {0}, of length 2. Otherwise continuing in this manner, we get a strictly descending chain

M = M0 ⊃ M1 ⊃ M2 ⊃ · · · ⊃ Mi−1 ⊃ Mi ⊃ · · · , of submodules of M such that Mi is a maximal submodule of Mi−1, for i = 1, 2,.... Since M satisﬁes d.c.c., therefore there exists some positive integer n such that Mk = Mn for every k > n, which implies that either Mn = {0} or the set Σn+1 of all proper submodules of Mn has maximal element the trivial submodule, i.e., we obtain a composition series of length n or n + 1. In either case we have that M has a composition series and the result follows.

From the above theorem, we conclude that a module is of ﬁnite length if and only if it satisﬁes both a.c.c. and d.c.c.

Proposition 1.4. The length of a module is an additive function on the class of modules of ﬁnite lengths.

Proof. To prove the result, we need to show that if M 0 and M 00 are two R-modules of ﬁnite length, then l (M 0 ⊕ M 00) = l (M 0) + l (M 00), which will follow (Why?) once we show that if

β 0 −→ M 0 −→α M −→ M 00 −→ 0 (7)

5 is a short exact sequence, then

l(M) = l(M 0) + l(M 00). (8)

Let l(M 0) = r and l(M 00) = s, then M 0 and M 00 have composition series of lengths r and s respectively (say)

0 0 0 0 0 M = M0 ⊃ M1 ⊃ M2 ⊃ · · · ⊃ Mr = {0}, (9) and 00 00 00 00 00 M = M0 ⊃ M1 ⊃ M2 ⊃ · · · ⊃ Ms = {0}. (10)

Since α is injective (∵ (7) is exact), so on applying α to the chain (9) all the inclusions remain strict (Why?), i.e.,

0 0 0 0 0 Img α = (M ) α = (M0) α ⊃ (M1) α ⊃ (M2) α ⊃ · · · ⊃ (Mr) α = {0}. (11)

0 0 Claim I: For each i ∈ {1, 2, . . . , r},(Mi ) α is a maximal submodule of Mi−1 α. Suppose if possible that there is some i ∈ {1, 2, . . . , r} such that

0 0 (Mi ) α ⊂ N ⊂ Mi−1 α, for some submodule N of M, then N ⊆ Img α and since α is injective, we have that ((N 0) α) α−1 = N 0 for every submodule N 0 of M 0, which implies that

0 −1 0 −1 0 0 −1 ((Mi ) α) α = Mi ⊂ (N)α ⊂ Mi−1 = Mi−1 α α .

But (N)α−1 is a submodule of M 0, so the above (strict) inclusion implies that 0 0 Mi is not maximal in Mi−1 contradicting the fact that (9) is a composition series of M 0. Hence the claim follows. By exactness of (7), we have that β is surjective, so on applying β−1 to the composition series (8) all the inclusion in the resulting chain remains strict (Why?), i.e.,

00 −1 00 −1 00 −1 00 −1 −1 M = (M ) β = (M0 ) β ⊃ (M1 ) β ⊃ · · · ⊃ (Ms ) β = ({0}) β = ker β. (12) 00 −1 00 −1 Claim II: For 1 ≤ j ≤ s, Mj β is a maximal submodule of Mj−1 β . Assume to the contrary that there is some j ∈ {1, 2,...} such that

00 −1 00 −1 Mj β ⊂ N ⊂ Mj−1 β ,

6 for some submodule N of M, then from (12), we have that ker β ⊆ N, which in view of Correspondence Theorem1 implies that

00 −1 00 00 00 −1 Mj β β = Mj ⊂ (N)β ⊂ Mj−1 = Mj−1 β β.

But (N)β is a submodule of M 00, so the above (strict) inclusion implies that 00 00 Mj is not a maximal submodule of Mj−1 contradicting the fact that (10) is a composition series of M 00 and proving the claim. On using Img α = ker β (∵ (7) is exact), together with the chains (11), (12), and the claims I and II, it follows that the following chain

00 −1 00 −1 00 −1 00 −1 −1 M = (M ) β = (M0 ) β ⊃ (M1 ) β ⊃ · · · ⊃ (Ms ) β = ({0}) β = ker β = Img α

0 0 0 0 0 = (M ) α = (M0) α ⊃ (M1) α ⊃ (M2) α ⊃ · · · ⊃ (Mr) α = {0} is a composition series of M of length r + s and hence l(M) = r + s.

The next result shows that finite dimensional vector spaces satisfies both a.c.c. and d.c.c and hence have finite length.

Proposition 1. For a vector space V over a field K, the following statements are equivalent: (i) V is finite dimensional. (ii) V is of finite length. (iii) V satisfies a.c.c. (iv) V satisfies d.c.c.

Proof. (i) =⇒ (ii). Let n be the dimension of V and {e1, e2, . . . , en} a basis of

V . If Wi denote the subspace of V generated by {e1, e2, . . . , ei}, then we have a chain

V = Wn ⊃ Wn−1 ⊃ · · · ⊃ W1 ⊃ W0 = {0}. It can be easily veriﬁed (exercise) that the above chain is a composition series of length n, i.e., V is of ﬁnite length. In fact from above we have that the length of V is same as its dimension. (ii) =⇒ (iii) and (iv) follows immediately from Theorem 1.3. Finally we prove (simultaneously) that (iii) =⇒ (i) and (iv) =⇒ (i). Assume

1Let M be an R-module and N a submodule of M. Then there exists a one to one correspondence between the submodules of M which contain N and the submodules of M/N.

7 to the contrary that V is not a ﬁnite dimensional vector space over K. Then there is a countably inﬁnite linearly independent subset of V (say) {e1, e2,...}.

For each natural number n let Un = he1, e2, . . . , eni and Wn = hen, en+1,...i, then Un and Wn are subspaces of V and we have the following ascending and descending chains

U1 ⊆ U2, ⊆ · · · ⊆ Un ⊆ Un+1 ⊆ · · · , (13)

W1 ⊇ W2, ⊇ · · · ⊇ Wn ⊇ Wn+1 ⊇ · · · , (14) In fact all the inclusions in the above two chains are strict, because if for some n, Un = Un+1, then we have that en+1 ∈ Un = he1, e2, . . . , eni, which implies that en+1 can be expressed as a K-linear combination of e1, e2, . . . , en contradicting the fact that {e1, e2, . . . , en, en+1} is linearly independent. Similarly if Wn = Wn+1 for some n, then en ∈ Wn+1 = hen+1, en+2,...i, i.e., there exist finitely many non- m X zero scalars (say) aj1 , aj2 , . . . , ajm ∈ K such that en = aji en+ji , contradicting i=1 the fact that en, en+j1 , . . . , en+jm are linearly independent over K. Therefore, we have that (13) and (14) are strictly ascending and descending chain of subspaces of V which never stops contradicting the hypothesis that V satisfies a.c.c. and d.c.c. Hence our assumption that V is not finite dimensional is false and thus (i) follows.

8 References

[1] M. F. Athiyah and I. G. Macdonald, Introduction to Commutative Algebra, Addison Wesley, 1969.

[2] D. S. Dummit and R. M. Foote, Abstract Algebra, (Third Edition) Wiley India Pvt. Ltd., 2011.

[3] J. J. Rotman. An introduction to the theory of groups, Springer-Verlag, New York, 1995.