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MMATH18-201: Module Theory Lecture Notes: Chain Conditions

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MMATH18-201: Module Theory Lecture Notes: Chain Conditions MMATH18-201: Module Theory Lecture Notes: Chain Conditions Contents 1 Noetherian and Artinian Modules1 2 Composition Series 13 3 Hilbert Basis Theorem 20 4 Exercises 23 References 24 In the previous chapter we saw that a submodule of a finitely generated module need not be finitely generatd, in these notes we characterize such module. Throughout R denotes a commutative ring with unity. The result of this section are from Chapter 6 and 7 of [1]. 1 Noetherian and Artinian Modules In this section we introduce the notion of Noetherian and Artinian modules and give some equivalent definitions of the same. Definition. Let M be an R-module. A sequence fMig of submodules of M is called (i) ascending (or increasing) if M1 ⊆ M2 ⊆ · · · ; (ii) descending (or decreasing) if M1 ⊇ M2 ⊇ · · · and (iii) stationary if for ascending or descending chain there exists some integer n > 0 such that Mn = Mn+1 = ··· : Definition. An R-module M is said to satisfy ascending chain condition ( or a.c.c. for short) if every ascending chain of submodules of M is stationary. Definition. An R-module M is said to satisfy maximal condition if every non- empty subset of submodules of M has a maximal element1. The next result show that for an R-module M; the above two definition are equivalent. Theorem 1.1. For an R-module M the following are equivalent: (i) M satisfies a.c.c. (ii) M satisfies maximal condition. Proof. (i) =) (ii). Let Σ be a non-empty set of submodules of M. Assume to the contrary that Σ has no maximal element. Since Σ is non-empty, there exists a submodule (say) M1 in Σ: As Σ has no maximal element so we can find some submodule M2 in Σ such that M1 ⊂ M2. Again M2 cannot be maximal, so there exists some M3 in Σ such that M2 ⊂ M3: Continuing this way, we obtain a strictly ascending chain of submodules of M, M1 ⊂ M2 ⊂ M3 ⊂ · · · ; 1An element a of a partially ordered set (X; ≤) is called a maximal element of X if there is no element in X strictly greater than a; i.e., if for some x 2 X such that a ≤ x; then a = x. 1 which is not stationary. This contradicts (i) hence (ii) holds. (ii) =) (i). Let M1 ⊆ M2 ⊆ · · · ; (1) be any ascending chain of submodules of M and let Σ = fMi j i = 1; 2;:::g: Then Σ 6= φ so by our hypothesis, Σ has a maximal element (say) Mn for some positive integer n. Now in view of (1), we have that Mn ⊆ Mk; for every k > n: (2) But Mn is a maximal element of Σ, therefore (2) implies that Mn = Mk for every k > n, i.e., the sequence (1) is stationary which proves (ii). Definition. An R-module M is called Noetherian if it satisfies the ascending chain condition or equivalently it satisfies the maximal condition. The following result gives another definition of Noetherian modules. Theorem 1.2. An R-module M is Noetherian if and only if every submodule of M is finitely generated. Proof. Suppose first that M is Noetherian, that is it satisfies a.c.c., and let N be any submodule of M. If N = f0g, then it is finitely generated and we are done. So assume that N is a non-trivial submodule of M. Then there exist some non-zero element in N (say) x1 and let N1 = hx1i be the submodule of generated by x1. If N = N1, then N is cyclic and we are done otherwise N1 ( N; i.e., there exists some x2 in N n N1 and let N2 = hx1; x2i. Again if N = N2, then we have that N is finitely generated otherwise continuing in this manner, we obtain an ascending chain of submodules of M N1 ( N2 ( ··· : (3) Since M satisfies a.c.c., therefore the chain (3) is stationary, i.e., there exists some positive integer k such that Nk = Nk+1 = ··· ; which in view of the construction of Ni's implies that N = Nk = hx1; x2; : : : ; xki and hence N is finitely generated. Conversely assume that every submodule of M is finitely generated, we need to show that M satisfies a.c.c. Let M1 ⊆ M2 ⊆ · · · ; (4) 2 [ be any ascending chain of submodules of M and let N = Mi, verify that N i2N is a submodule of M: Then in view of the hypothesis N is finitely generated [ (say) by x1; x2; : : : ; xn; i.e., hx1; x2; : : : ; xni = N = Mi; which implies that i2N there exist positive integers i1; i2; : : : ; in such that xk 2 Mik for 1 ≤ k ≤ n. Let m = maxfi1; i2; : : : ; ing; then in view of (4), we have that Mik ⊆ Mm, i.e., x1; x2; : : : ; xn 2 Mm consequently N = hx1; x2; : : : ; xni ⊆ Mn: The reverse [ inclusion Mm ⊆ N follows from the fact that N = Mi, hence we have that i2 [ N N = Mm: Now Mi = N = Mm together with (4) implies that Mi = Mm for i2N every i ≥ m. Thus ths chain (4) is stationary, i.e., M satisfies a.c.c. Exercise 1.3. Show that the above result can also be proved if instead of a.c.c., we use maximal condition as the definition of a Noetherian module. We also have some similar result for descending chain of submodules of an R-module M: Definition. An R-module M is said to satisfy descending chain condition (or d.c.c. for short) if every descending chain of submodules of M is stationary. Definition. An R-module M is said to satisfy minimal condition if every non- empty subset of submodules of M has a minimal element2. The following result shows that the above two definitions are equivalent, whose proof is on the similar lines as that of Theorem 1.1 and left as an exercise. Theorem 1.4. For an R-module M the following are equivalent: (i) M satisfies d.c.c. (ii) M satisfies minimal condition. Definition. An R-module M is called Artinian if it satisfies the descending chain condition or equivalently it satisfies the minimal condition. The next result give another characterization of Noetherian (respectively Ar- tinian) modules in terms of exact sequences. 2An element a of a partially ordered set (X; ≤) is called a minimal element of X if there is no element in X strictly smaller than a, i.e., if for some x 2 X such that x ≤ a; then x = a. 3 Theorem 1.5. Let β 0 −! M 0 −!α M −! M 00 −! 0 (5) be a short exact sequence of R-modules and R-homomorphisms. Then M is Noetherian (respectively Artinian) if and only if both M 0 and M 00 are Noetherian (respectively Artinian). Proof. Suppose first that M is Noetherian and let 0 0 M1 ⊆ M2 ⊆ · · · ; (6) be any ascending chain of submodules of M 0. Then on applying the R-homomorphism α to (6), we obtain an ascending chain of submodules of M 0 0 (M1) α ⊆ (M2) α ⊆ · · · ; (7) 0 −1 0 00 because if x 2 (M1) α; then (x)α 2 M1 ⊆ M2 which implies that x = −1 00 ((x)α ) α 2 (M2 ) α: Since M is Noetherian, therefore the chain (7) is stationary, i.e., there exists some positive integer n such that 0 0 (Mn) α = Mn+1 α = ··· : (8) As the given sequence (5) is exact, so α is injective and hence an isomorphism when restricted to the image, which implies that ((N 0) α) α−1 = N 0 for every submodule N 0 of M 0. Consequently on applying α−1 to (8), we get that 0 0 Mn = Mn+1 = ··· ; i.e., the chain (6) is stationary. Therefore M 0 satisfies a.c.c., and hence is a Noetherian module. Let 00 00 M1 ⊆ M2 ⊆ · · · ; (9) be any ascending chain of submodules of M 00, then on applying β−1 we get 00 −1 00 −1 (M1 ) β ⊆ (M2 ) β ⊆ · · · (10) 00 −1 00 is an ascending chain of submodules of M (* if x 2 (M1 ) β , then (x)β 2 M1 ⊆ 00 00 −1 M2 , which implies that x 2 (M2 ) β ). As M is Noetherian therefore 00 −1 00 −1 (Mm) β = Mm+1 β = ··· (11) 4 for some positive integer m. Since β is surjective (* (5) is exact), so we have that ((N 00) β−1) β = N 00 for every submodule N 00 of M 00 (Why?). Therefore on applying β to (11), we get that 00 00 Mm = Mm+1 = ··· ; i.e., the chain (9) is stationary. Hence M 00 satisfies d.c.c., and is a Noetherian module. Conversely assume that both M 0 and M 00 are Noetherian, we need to show that M is also Noetherian. Let M1 ⊆ M2 ⊆ · · · ; (12) be any ascending chain of submodules of M, then on applying α−1 and β to (12), we obtain −1 −1 (M1) α ⊆ (M2) α ⊆ · · · ; and (M1) β ⊆ (M2) β ⊆ · · · ; ascending chain of submodules of M 0 and M 00 respectively. Since both M 0 and M 00 are Noetherian the above chains must be stationary, i.e., there exist positive integers m and n such that −1 −1 (Mm) α = (Mm+1) α = ··· and (Mn) β = (Mn+1) β = ··· : We can assume without loss of generality that n ≥ m then −1 −1 (Mn) α = (Mn+1) α = ··· (13) and (Mn) β = (Mn+1) β = ··· : (14) Now in view of (12), Mn ⊆ Mk for every k > n; so to prove that the ascending chain (12) is stationary it is enough to show that Mk ⊆ Mn for every k > n: Let k > n and x 2 Mk.
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