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Noetherian and Artinian Modules and Rings Artinian and Noetherian Rings Have Some Measure of Finiteness Associated with Them. In

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Noetherian and Artinian Modules and Rings Artinian and Noetherian Rings Have Some Measure of Finiteness Associated with Them. In Noetherian and Artinian Modules and Rings Artinian and Noetherian rings have some measure of finiteness associated with them. In fact, the conditions for Artinian and Noetherian rings, called respectively the descending and ascending chain conditions, are often termed the minimum and maximum conditions. These properties make Artinian and Noetherian rings of interest to an algebraist. Furthermore, these two types of rings are related. Definition 1. Let M be an R-module, and suppose that we have an increasing sequence of submodules M1 M2 M3 . , or a decreasing sequence M1 M2 M3 . We say that the sequence stabilizes if for some t, Mt = Mt+1 = Mt+2 = . The question of stabilization of sequences of submodules appears in a fundamental way in many areas of abstract algebra and its applications. Definition 2. The module M is said to satisfy the ascending chain condition (acc) if every increasing sequence of submodules stabilizes; M satisfies the descending chain condition (dcc) if every decreasing sequence of submodules stabilizes. Definition 3. An R-module M is called Artinian if descending chain conditions (dcc) hold for submodules of M. 1 Definition 4. An R-module M is called Noetherian if ascending chain condition (acc) holds for submodules M. Proposition 1. The following conditions on M are equivalent: (i) M is Artinian; (ii) Every nonempty collection of submodules of M has a minimal element. Proof. (i) (ii) Let F = { Mi | i I } be a non- empty family of submodules of M. Pick any i1 I. M If i1 is a minimal element of F, we are done. If M M not, there exists i2 I such that i2 i1 . If M i is minimal, we are done. Else i I such 2 3 M M that i3 i2 . If F does not contain a minimal element, we obtain an infinite strictly descending M M M ... chain i1 i2 i3 which is a contradiction as M is Artinian. (ii) (i) Consider a descending chain of submodules of M, M1 M2 M3 . Let F = { Mi | i I }, then F has a minimal element say, Mt. Now Mt = Mt+1 = Mt+2 = . Hence M is Artinian. Corollary 1. Every non-zero Artinian module contains a minimal submodule. Proof. Let M be a non-zero Artinian R-module. Let F be a family of all proper submodules of M. Then (0) F implies F . Then F has a minimal 2 element say, N. Clearly N is a minimal submodule of M. Proposition2. The following conditions on an R-module M are equivalent: (1) M is a a Noetherian module; (2) Every nonempty collection of submodules of M has a maximal element (with respect to inclusion). Proof. Assume (1), and let S be a nonempty collection of submodules. Choose M1 S. If M1 is maximal, we are finished; otherwise we have M1 M2 for some M2 S. If we continue inductively, the process must terminate at a maximal element; otherwise the acc would be violated. Conversely, assume (2), and let M1 M2 . The sequence must stabilize; otherwise {M1,M2, . } would be a nonempty collection of submodules with no maximal element. There is another equivalent condition in the Noetherian case. Proposition3. M is Noetherian if and only if every submodule of M is finitely generated. Proof. If the sequence M1 M2 . does N M not stabilize, let r . Then N is a r1 submodule of M, and it cannot be finitely 3 generated. For if x1, . , xs generate N, then for sufficiently large t, all the xi belong to Mt. But then N ⊆ Mt ⊆ Mt+1 ⊆ ·· · ⊆ N, so Mt = Mt+1 = . Conversely, assume that the acc holds, and let N be a submodule of M. If N ≠ 0, choose x1 N. If Rx1 = N, then N is finitely generated. Otherwise, there exists x2 N- Rx1. If x1 and x2 generate N, we are finished. Otherwise, there exists x3 N- (Rx1 + Rx2) . Since Rx1 Rx1 + Rx2 Rx1 + Rx2+ Rx3 …, the acc forces the process to terminate at some stage t, in which case x1, . , xt generate N. Definition5 A ring R is Noetherian [resp. Artinian] if it is Noetherian [resp. Artinian] as a module over itself. If we need to distinguish between R as a left, as opposed to right, R-module, we will refer to a left Noetherian and a right Noetherian ring, and similarly for Artinian rings. Examples 1. Every PID is Noetherian. This follows since every ideal is generated by a single element. 2. Z is Noetherian (a special case of Example 1) but not Artinian. There are many descending chains of ideals that do not stabilize, e.g., Z ⊃ (2) ⊃ (4) ⊃ (8) ⊃ . 4 3. If F is a field, then the polynomial ring F[X] is Noetherian (another special case of Example 1) but not Artinian. A descending chain of ideals that does not stabilize is (X) ⊃ (X2) ⊃ (X3) ⊃ . 4. The ring F[X1,X2, . ] of polynomials over F in infinitely many variables is neither Artinian nor Noetherian. A descending chain of ideals that does not stabilize is constructed as in Example 3, and an ascending chain of ideals that does not stabilize is (X1) ⊂ (X1,X2) ⊂ (X1,X2,X3) ⊂ . 5. A module with only finitely many submodules is Artinian and Noetherian. In particular, finite abelian groups are both Artinian and Noetherian over Z. 6. Let V be a finite dimensional vector space over a field F, say, dim V = n. Then V is both Artinian as well as Noetherian F-module. For, if W is a proper subspace of V , then dimW < n. Thus any proper ascending or descending chain of subspace of V can have at most n + 1 terms. However, infinite dimensional vector spaces are not Artinian. For, suppose V is infinite dimensional vector space over the field K with basis B. Let ui B, i N be distinct elements in B. Define Un => un, un+1, · · · < as a subspace of V . Then U1 ⊃ U2 ⊃ · · · is an infinite descending chain of subspaces. 5 7- Q has only two ideals, (0) and Q. Clearly, Q⊃ (0). This finite chain of ideals clearly bottoms out. This is the only possible chain of proper ideal inclusions in Q since Q only has two ideals. Thus, Q is Artinian. The field of rational numbers provides us with an example of an Artinian ring. It should be noted that by the same logic as in the preceding paragraph, Q is also Noetherian. Remark The following observations will be useful in deriving properties of Noetherian and Artinian modules. If N M, then a submodule L of M that contains N can always be written in the form K+N for some submodule K. (K = L is one possibility.) By the correspondence theorem, (K1 + N)/N = (K2 + N)/N implies K1 + N = K2 + N and (K1 + N)/N (K2 + N)/N implies K1 + N K2 + N. Proposition4. Submodules Noetherian [resp. Artinian] modules are Noetherian [resp. Artinian]. Proof. Assume M is Noetherian and let N be its submodule. Then Any submodule of N is also a submodule of M. Hence acc holds for submodules of N ( similarly any descending chain of submodules of N is a chain of submodules of M). Hence the result follows. Proposition5. If N is a submodule of M, then M is Noetherian [resp. Artinian] if and only if N and M/N are Noetherian [resp. Artinian]. 6 Proof. Assume M is Noetherian. Then N is Noetherian. An ascending chain of submodules of M/N looks like (M1 +N)/N (M2 +N)/N ·· · . But then the Mi +N form an ascending sequence of submodules of M, which must stabilize. Consequently, the sequence (Mi + N)/N, i = 1, 2, . , must stabilize. Conversely, assume that N and M/N are Noetherian, and let M1 M2 ··· be an increasing sequence of submodules of M. Take i large enough so that both sequences {Mi ∩N} and {Mi+N} have stabilized. If x Mi+1, then x+N Mi+1+N = Mi+N, so x = y + z where y Mi and z N. Thus x − y Mi+1 ∩ N = Mi ∩ N, and since y Mi we have x Mi as well. Consequently, Mi = Mi+1 and the sequence of Mi’s has stabilized. The Artinian case is handled by reversing inequalities (and interchanging indices i and i + 1 in the second half of the proof). Corollary If M1, . , Mn are Noetherian [resp. Artinian] R- modules, then so is M1 ⊕M2 ⊕·· ·⊕Mn. Proof. It suffices to consider n = 2 (induction will take care of higher values of n). The submodule N = M1 of M = M1 ⊕ M2 is Noetherian by hypothesis, and M/N M2 (apply the first isomorphism theorem to the natural projection of M onto M2, i.e. f : M M2, where ( ) is also Noetherian. 7 By Proposition5, M is Noetherian. The Artinian case is done the same way. Proposition. In an artinian ring all the prime ideals are maximal. 8 9 11 11 Proposition . A quotient ring of an Artinian (Noetherian) ring is Artinian (Noetherian). Proof. We prove for Artinian rings. The Noetherian case follows on similar lines. Let R be an Artinian ring and I be an ideal of R. Then R/I is an Artinian R-module (as seen in the module case). Now any left ideal of R/I is of the form J/I, where J is a left ideal of R. Hence a descending chain of left ideals of R/I gives descending chain of left ideals of R. Hence R/I is an Artinian ring. Remark . Subring of Artinian or Noetherian may not be Artinian or Noetherian. For example, the subring Z of the Artinian ring Q is not Artinian.
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