4.3 Composition Series
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1 4.3 Composition Series Let M be an A-module. A series for M is a strictly decreasing sequence of submodules M = M 0 M 1... M n = {0} beginning with M and finishing with {0 }. The length of this series is n. 768
2 A composition series is a series in which no further submodule can be inserted which, for the above, is equivalent to saying each composition factor M i /M i+1 is simple, that is, each M i /M i+1 is nontrivial and has no submodule except for itself and the trivial submodule. 769
3 Example: The Z-module Z 30 has the following lattice of submodules: Z 30 = {0} = 0 770
4 Any path from the top to the bottom will yield a composition series: e.g , with composition factors: 1 / 2 = Z 30 /2Z 30 = (Z/30Z)/(2Z/30Z) = Z/2Z = Z 2, 771
5 2 / 6 = 2Z/6Z = Z3, 6 / 0 = 6Z/30Z = Z5 In fact, all composition series for Z 30 produce composition factors Z 2, Z 3, Z 5 (in accordance with the Jordan-Holder Theorem below). 772
6 Notation: If N is a module, let l(n) denote the least length of a composition series of N, if one exists, and put l(n) = if no composition series for N exists. Theorem: Suppose M has a composition series of length n. Then every composition series of M has length n, and every series can be refined (that is, submodules can be inserted) to yield a composition series. 773
7 Proof: Here l(m) n. Consider a submodule N of M where N M. We first show l(n) < l(m). ( ) Put l = l(m) and let M = M 0 M 1... M l = {0} be a composition series of length l. Then N = N M 0 N M 1... N M l = {0} is a chain of submodules from N to {0}. 774
8 By a module isomorphism theorem, for each i, N M i N M i+1 = N M i (N M i ) M i+1 = (N M i) + M i+1 M i+1, the last of which is a submodule of the simple module M i / M i+1. Hence N M i N M i+1 is trivial or simple. 775
9 Thus, deleting repetitions from the above chain from N to {0} must yield a composition series for N, which proves l(n) l(m). Suppose l(n) = l. Then, no repetitions occurred in the previous process, so, by the earlier isomorphism, (N M i ) + M i+1 = M i ( i). ( ) 776
10 We observe, by induction, that N M i = M i ( i), since, clearly N M l = N {0} = {0} = M l, which starts the induction, and, for i l 1, using ( ) and an inductive 777
11 hypothesis, N M i = (N M i ) + (N M i+1 ) = (N M i ) + M i+1 = M i. In particular, M = M 0 = N M 0 = N M = N, which contradicts that N M. Hence l(n) < l and ( ) is proved. 778
12 Now consider any series M = M 0 M 1... M k = {0} ( ) of length k. By ( ), l = l(m) > l(m 1) >... > l(m k) = 0, so l k. If ( ) is a composition series, then l k, by definition of l, so l = k. 779
13 This proves all composition series of M have the same length n. If ( ) is not a composition series, then k < n, because if k = n, then we can insert another module somewhere to get another series of length n + 1 l = n, which is nonsense. Hence any series which is not a composition series can be successively refined until its length is n, in 780
14 which case it becomes a composition series, and the Theorem is proved. Corollary: A module M has a composition series iff M satisfies the a.c.c. and d.c.c. Proof: (= ) If M has a composition series of length n, then, by the previous Theorem, all series have length n, 781
15 so all ascending and descending chains must be stationary, that is, M satisfies both the a.c.c. and the d.c.c. ( =) Suppose M satisfies both the a.c.c. and the d.c.c. If M = {0} then certainly M has a composition series. Suppose M {0}. 782
16 Since M satisfies the maximal condition (equivalent to the a.c.c), the set of proper submodules of M has a maximal element M 1, so M/M 1 is simple. Suppose we have found submodules M = M 0 M 1... M k (k 1) where M i /M i+1 is simple for i = 0,...,k
17 If M k = {0} then we have a composition series for M. If M k {0}, then, again since M satisfies the maximal condition, we can find a submodule M k+1 of M k such that M k /M k+1 is simple. Either M has a composition series, or by induction we have an infinite strictly descending chain M = M 0 M 1... M k
18 The latter is excluded because M satisfies the d.c.c. Hence M has a composition series and the Corollary is proved. Say that a module M has finite length if it has a composition series (equivalently satisfies both the a.c.c. and d.c.c.) in which case all composition series have the same length l(m), called the length of M. 785
19 We now prove a uniqueness result concerning the composition factors: Jordan-Holder Theorem: There is a one-one correspondence between the composition factors of any two composition series of a module of finite length such that corresponding factors are isomorphic. Proof: Let M be a module of finite length l. 786
20 If l = 0 then the set of composition factors is always empty, so the result is vacuously true, which starts an induction. Suppose l > 0 and let M = M 0 M 1... M l = {0} M = M 0 M 1... M l = {0} be two composition series. 787
21 If M 1 = M 1 then, by an inductive hypothesis (since l(m 1 ) = l 1 ), there is an appropriate correspondence between { M i /M i+1 i = 1,...,l 1 } and so, since { M i/m i+1 i = 1,...,l 1 }, M 0 /M 1 = M 0/M 1, 788
22 there is an appropriate correspondence between { M i /M i+1 i = 0,...,l 1 } and { M i/m i+1 i = 0,...,l 1 }, and we are done. Suppose then M 1 M 1, so M 1 + M 1 = M, since M 1 is maximal in M. 789
23 M = M 1 + M 1 M 1 M 1 M 1 M 1 790
24 By module isomorphism theorems, and But M/M 1 = M 1 + M 1 / M 1 = M 1 / M 1 M 1 M/M 1 = M 1 + M 1 / M 1 = M 1 / M 1 M 1. l(m 1 M 1) < l(m 1 ), l(m 2 ) < l so we can apply an inductive hypothesis to composition series of these modules. 791
25 Let F 1, F 2, F 1 be the collections of composition factors for M 1, M 1 M 1, M 1 respectively. Then there are appropriate correspondences between F 1 { M/M 1 } and F 2 { M 1 /M 1 M 1, M/M 1 } and F 2 { M/M 1, M 1/M 1 M 1 } and F 1 { M/M 1 } which proves the Theorem. 792
26 Theorem: The length l(m) of a module M defines an additive function on the class of all A-modules of finite length. Proof: Suppose 0 M α M β M 0 is exact, where all modules have finite length. We need to show l(m) = l(m ) + l(m ). 793
27 Let M = M 0 M 1... M l(m ) = {0} M = M 0 M 1... M l(m ) = {0} be composition series for M and M respectively. Since α is injective, α(m ) = α(m 0 ) α(m 1 )... α(m l(m ) ) = {0} is a composition series for α(m ) = kerβ. 794
28 Since β is onto, M/ker β = β 1 (M 0)/ kerβ β 1 (M 1)/ ker β... β 1 (M l(m ))/ker β = {ker β} is a composition series for M/ ker β. Combining these two series produces the following series for M : 795
29 M = β 1 (M 0) β 1 (M 1)... β 1 (M l(m ) ) = ker β {0} = α(m l(m ) )... α(m 0) = α(m ) But this is a composition series, because, by another isomorphism theorem, for i = 0,...,l(M ) 1, 796
30 β 1 (M i ) / β 1 (M i+1) = ( ) β 1 (M i ) / kerβ / ( ) β 1 (M i+1) / kerβ, which is simple. Hence l(m) = l(m ) + l(m ), and the Theorem is proved. 797
31 Interpreting the theory for vector spaces yields: Proposition: Let V be a vector space over a field F. TFAE (i) V is finite dimensional. (ii) V has finite length. (iii) V satisfies the a.c.c. (iv) V satisfies the d.c.c. If any of these hold, then dimension equals length. 798
32 Proof: (i) = (ii) If { x 1,..., x n } is a basis for V then V = x 1,..., x n x 1,..., x n 1 is a composition series of length n.... x 1 {0} (ii) = (iii), (ii) = (iv): follow from the earlier Corollary on page
33 (iii) = (i), (iv) = (i): Suppose (i) is false, so V contains an infinite sequence x 1,..., x n,... of linearly independent vectors. For each n 1 put U n = x 1,..., x n, Then V n = x n+1, x n+2,
34 {0} U 1 U 2... U n... V V 1 V 2... V n... are strictly ascending and descending chains respectively, so that both (iii) and (iv) fail to hold. 801
35 Corollary: Let A be a ring in which M 1 M 2... M N = {0} for some (not necessarily distinct) maximal ideals M 1,...,M n. Then A is Noetherian iff A is Artinian. 802
36 Proof: First note that if I A, M A, M maximal then I / IM is an A-module, so also I/IM is an A/M-module (since M Ann (I/IM) ), that is, I/IM is a vector space over the field A/M, 803
37 so, by the previous Proposition, iff I/IM satisfies the a.c.c. on subspaces (ideals) I/IM satisfies the d.c.c. on subspaces (ideals). Consider the chain A M 1 M 1 M 2... M 1... M n = {0}. Then, by repeated application of an earlier Theorem about exactness (on page 756), 804
38 A satisfies the a.c.c. on ideals iff each factor A/M 1, M 1 /M 1 M 2,..., M 1...M n 1 / M 1...M n satisfies the a.c.c. on ideals iff each factor satisfies the d.c.c. on ideals iff A satisfies the d.c.c. on ideals, and the Corollary is proved. 805
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