Chain conditions. 1. Artinian and noetherian modules. ALGBOOK CHAINS 1.1

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1 CHAINS 1.1 Chai coditios 1. Artiia ad oetheria modules. (1.1) Defiitio. Let A be a rig ad M a A-module. The module M is oetheria if every ascedig chai!!m 1 M 2 of submodules M of M is stable, that is, there is a such that M = M +1 =. The A-module M is artiia if every descedig chai M 1 M 2 of submodules M of M is stable, that is, there is a such that M = M +1 =. (1.2) Example. K be field ad M a fiitely geerated vector space. The M is artiia ad oetheria. I fact, it follows from Remark (MODULES 1.26) that whe L L is a strict iclusio of subspaces of M the dim(l) < dim(l ). Hece there ca oly be ascedig or descedig chais of fiite legth i M. (1.3) Example. The itegers Z is a oetheria Z-module. This is because every ideal of Z is of the form () for some iteger, ad (m) () meas that divides m. The module Z is ot artiia because (2) (2 2 ) (2 3 ) is a ifiite descedig chai of ideals. (1.4) Example. The polyomial rig Z[t] i the variable t over the itegers is ot oetheria as a Z module sice it cotais the ifiite chai Z Z + Zt Z + Zt + Zt 2 of submodules. It is ot artiia either because it cotais the ifiite chai (2) (2 2 ) (2 3 ) of ideals. (1.5) Example. Fix a prime umber p. Let M be the Z-submodule of Q/Z cosistig of all the classes of ratioal umbers m/ such that p q m/ Z for some positive iteger q. That is, the classes of the elemets m/ where is a power of p. We deote by M the submodule of M geerated by the class of 1/p. The M cosists of the p elemets that are the classes i M of the elemets m/p for m = 0, 1,..., p 1. The modules i the chai M 1 M 2 M 3... are the oly proper submodules of M. This is because if L is a proper submodule of M it cotais the class of a elemet m/p with > 1, ad where p does ot divide m. Sice p is a prime umber that does ot divide m there are itegers q ad r such that qp + rm = 1. The rm/p = qp /p + 1/p = q + 1/p. Hece the class of 1/p is i L. It follows that M L, ad that L = M s, where s is the largest iteger such that M s L. Sice we have a chai M 1 M 2 M 3 ad the modules M are the oly proper submodules of M it follows that the Z-module M is artiia but ot oetheria. chais1

2 CHAINS 1.2 (1.6) Lemma. Let A be a rig ad M a A-module. The followig assertios are equivalet: (1) The module M is oetheria. (2) The collectio of submodules of M satisfies the maximum coditio. (3) Every submodule of M is fiitely geerated. Proof. (1) (2) We saw i (TOPOLOGY 1.5) that the assertios (1) ad (2) are equivalet. (2) (3) Assume that M satisfies the maximum coditio ad let L 0 be a submodule of M. Deote by!!l the collectio of o-zero submodules of L that are fiitely geerated. The L is ot empty sice Ax is i L for all x 0 i L. Sice M satisfies the maximum coditio L has a maximal elemet L. We shall prove that L = L. Assume that L L. The there is a elemet x L \ L. We have that Ax+L is a fiitely geerated submodule of L ad L is cotaied properly i Ax+L. This is impossible because L is maximal i L, ad we have obtaied a cotradictio to the assumptio that L L. Hece L = L ad L is fiitely geerated. (3) (1) Assume that every submodule of M is fiitely geerated. Let M 1 M 2 be a chai of submodules of M. The L = =0M is a submodule of M ad thus fiitely geerated. Each fiite set of geerators of L must be cotaied i some M. The L = M = M +1 =. That is, the chai M 1 M 2 is stable. (1.7) Propositio. Let A be a rig ad let 0 M u M v M 0 be a exact sequece of A-modules. The: (1) The module M is oetheria if ad oly if the modules M ad M are oetheria. (2) The module M is artiia if ad oly if the modules M ad M are artiia. Proof. (1) Assume that M is oetheria. Every chai M 1 M 2 of submodules of M gives rise to a chai v 1 (M 1 ) v 1 (M 2 ) of submodules of M. Sice M is oetheria we have that v 1 (M ) = v 1 (M +1) = for some positive iteger, ad thus M = M +1 =. Every chai M 1 M 2 of submodules of M gives rise to a chai u(m 1) u(m 2) i M. Sice M is oetheria u(m ) = u(m +1) = for some positive iteger, ad thus M = M 2 =. Coversely assume that M ad M are oetheria, ad let M 1 M 2 be a chai of submodules of M. The v(m 1 ) v(m 2 ) is a chai of submodules of M, ad u 1 (M 1 ) u 1 (M 2 ) is a chai of submodules of M. Sice M ad M are oetheria there is a positive iteger such that v(m ) = v(m +1 ) = ad u 1 (M ) = u 1 (M +1 ) =. However, the M = M +1 = sice M i is completely determied by v(m i ) ad u 1 (M i ). I fact a elemet x of M is i M i if ad oly if v(x) v(m i ) ad there is a x u 1 (M i ) such that x u(x ) M i. (2) The proof of the secod part is aalogous to the proof of the first part, with descedig chais istead of ascedig chais.

3 CHAINS 1.3 (1.8) Corollary. Let M 1, M 2,..., M be A-modules. (1) If the modules M 1, M 2,..., M are oetheria, the the direct sum i=1 M i is oetheria. (2) If the modules M 1, M 2,..., M are artiia, the the direct sum i=1 M i is artiia. Proof. (1) We shall prove the first assertio of the Corollary by iductio o. It holds trivially for = 1. Assume that it holds for 1. We clearly have a short exact sequece 0 M i=1 M i 1 i=1 M i 0. It follows from the iductio hypothesis that 1 i=1 M i is oetheria. The first part of the Corollary hece follows from the Propositio. (2) The proof of the secod part of the Corollary is similar to the proof of the first part. (1.9) Defiitio. Let!!0 = M 0 M 1 M = M be a chai of submodules of M. We call the legth of the chai. A chai is a refiemet of aother chai if we obtai the secod by addig modules to the first. We call a chai 0 = M 0 M 1 M = M a compositio series if the modules M 1 /M 0, M 2 /M 1,..., M /M 1 have o proper submodules. (1.10) Theorem. (The Jorda Theorem) Let A be a rig ad M a A-module that has a compositio series. The all compositio series of M have the same legth, ad every chai ca be refied to a compositio series. Proof. For every submodule L of M that has a compositio series we let!l(l)! be the smallest legth of a compositio series of L. Let l(m) = ad let {0} = M 0 M 1 M = M be a compositio series for M. We shall first show that every submodule L of M has a compositio series ad that l(l) < l(m) for all proper submodules L of M. To see this we cosider the chai {0} = L 0 = L M 0 L 1 = L M 1 L = L M = L of submodules of L. We have that L i /L i 1 has o proper submodules sice, by Lemma (MODULES 1.13), we have a ijective map L i /L i 1 M i /M i 1. Sice M i /M i 1 has o proper submodules either L i = L i 1 or L i /L i 1 M i /M i 1 is a isomorphism. Hece, removig terms where L i 1 = L i from the chai {0} = L 0 L 1 L = L we obtai a compositio series for L. It follows that l(l) l(m). We shall show by iductio o i that if l(l) = l(m) the M i = L i for i = 1, 2,...,. Whe l(l) = l(m) all the maps L i /L i 1 M i /M i 1 are isomorphisms. I particular M 1 = L 1. Assume that L i 1 = M i 1. Sice L i /L i 1 M i /M i 1 is a isomorphism we have for each x i M i that there is a elemet x 1 i M i 1 ad a elemet y i L i such that x + x 1 = y. The x = y x 1 is i L i sice x 1 is i M i 1 = L i 1 L i. Hece we have that L i = M i. I particular we have that L = M. We have thus show that whe l(l) = l(m) the L = M. Hece whe L is properly cotaied i M we have that l(l) < l(m). We shall ow prove that all compositio series of M have the sema legth. Let K 1 K 2 K m = M be a chai i M. The m l(m) because l(k 1 ) <

4 CHAINS 1.4 < l(k m ) = l(m). I particular, if {0} K 1 K 2 K m = M is a compositio series we must have that m = l(m) sice l(m) is the legth of the shortest compositio series of M. Hece we have proved that all compositio series have the same legth. If m < l(m), the chai K 1 K 2 K m ca ot be a compositio series. Cosequetly at least oe of the residue modules K 1 /{0}, K 2 /K 1,..., K m /K m 1 cotais a proper submodule differet from (0). Hece we may add oe more term to the chai to obtai a chai of legth m + 1. I this way we ca add groups i the chai util the chai has legth i which case it is a compositio series. Hece every chai ca be refied to a compositio series. (1.11) Defiitio. Let A be a rig ad M a A-module. We say that M has fiite legth if it has a compositio series. The legth of M is the commo legth l(m) = l A (M) of all compositio series of M. (1.12) Example. Let K be a field. A fiitely dimesioal K-vector space M has fiite legth ad dim K (M) = l K (M). This is because K has o proper K- submodules, ad thus, if x 1, x 2,..., x is a basis for M, the Kx 1 Kx 1 + Kx 2 Kx 1 + Kx Kx is a compositio series. (1.13) Example. The rig Z/6Z has the compositio series {0} 2Z/6Z Z/6Z. Clearly (Z/6Z)/(2Z/6Z) is isomorphic to Z/2Z. There is aother compositio series {0} 3Z/6Z Z/6Z. (1.14) Propositio. Let A be a rig ad M a A-module. The legth of M is fiite if ad oly if M is a artiia ad oetheria A-module. Proof. If M is of fiite legth it follows from Theorem (1.10) that all chais i M are of fiite legth. Hece M is artiia ad oetheria. Coversely, assume that M is artiia ad oetheria. Deote by!!l be the collectio of submodules L (0) of M such that there is a chai L = M M 1 M 1 = M for some positive iteger where M i 1 /M i has o proper submodules for each i = 2, 3,...,. The L is ot empty because M belogs to L. Sice M is artiia there is a miimal elemet L i L. If L has o proper submodules we have foud a compositio series (0) L = M M 1 M 1 = M of M ad we have proved the Propositio. We shall show that L ca ot have proper submodules. Assume to the cotrary that L has proper submodules ad let!!l be the collectio of proper submodules of L. Sice M is oetheria there is a maximal proper submodule M +1 of L. The L /M +1 has o proper submodules, ad thus M +1 M M 1 = M is a chai such that M i 1 /M i has o proper submodule for i = 2, 3,..., + 1. Sice M +1 L this is impossible sice L is miimal i L. This cotradicts the assumptio that L has proper submodules ad we have proved the Propositio. (1.15) Propositio. Let A be a rig ad 0 M u M v M 0 a exact

5 CHAINS 1.5 sequece of A-modules. (1) The A-module M is of fiite legth if ad oly if the A-modules M ad M are of fiite legth. (2) Whe all the modules are of fiite legth we have that l(m) = l(m )+l(m ). Proof. (1) It follows from Propositio (1.7) ad Propositio (1.14) that the first part of the Propositio holds. (2) To prove the secod part of the Propositio we take a compositio series 0 = M 0 M 1 M = M of M ad a compositio series 0 = M 0 M 1 M = M of M. The 0 = u(m 0) u(m 1) u(m ) = u(m ) = v 1 (0) = v 1 (M 0 ) v 1 (M 1 ) v 1 (M ) = v 1 (M ) = M is a compositio series of legth + for M. I fact the homomorphism u clearly iduces a isomorphism M i /M i 1 u(m i )/u(m i 1 ), ad the homomorphism v clearly iduces a isomorphism v 1 (M i )/v 1 (M i 1 ) M i /M i 1. (1.16) Remark. Let ϕ : A B be a surjectio of rigs ad let N be a B-module of fiite legth. The we have that l B (N) = l A (N [ϕ] ). I fact every A-submodule of N is also a B-submodule. (1.17) Theorem. (The Hölder Theorem) Let A be a rig ad let M be a A module of fiite leght. Two compositio series ad {0} = M 0 M 1 M = M (1.13.1) {0} = M 0 M 1 M = M (1.13.2) of M are equivalet, that is, there is a permutatio σ of the umbers {1, 2,..., } such that M σ(i) /M σ(i) 1 is isomorphic to M i /M i 1 for i = 1, 2,...,. Proof. We show the Theorem by iductio o l(m). It is trivially true whe l(m) = 1. Assume that it holds whe l(m) = 1. If M 1 = M 1 the Theorem holds by the iductio hypothesis. Assume that M 1 M 1. We choose a compositio series {0} = L 0 L 1 L 2 = M 1 M 1 of M 1 M 1. The we obtai two compositio series {0} = L 0 L 1 L 2 = M 1 M 1 M 1 M = M (1.13.3) {0} = L 0 L 1 L 2 = M 1 M 1 M 1 M = M (1.13.4) for the module M. Sice M/M 1 ad M/M 1 have o proper submodules ad M 1 M 1 it follows from Lemma (MODULES 1.13) that the iclusios M 1 M ad M 1 M iduce isomorphisms M 1 /M 1 M 1 M/M 1 ad M 1/M 1 M 1 M/M 1. Hece (1.13.3) ad (1.13.4) are equivalet compositio series. The compositio series (1.13.1) ad (1.13.3) have M 1 i commo. Hece it follows from the iductio hypothesis used to the module M 1 that the compositio series are equivalet. Similarly the compositio series (1.13.2) ad (1.13.4) are equivalet sice they have M 1 i commo. It follows that (1.13.1) ad (1.13.2) are equivalet compositio series, as we wated to prove.

6 CHAINS 1.6 (1.18) Exercises. 1. Let L be the submodule of the Z-module Z Z geerated by the elemets (2, 3). (1) Is the residue module (Z Z)/L oetheria? (2) Is the residue module (Z Z)/L artiia? 2. Fid all compositio series for the module Z/12Z. 3. Let K[t] be the polyomial rig i the variable t over the field K. Fid the compositio series of the K[t]-module K[t]/(t 3 (t + 1) 2 ). 4. Let A be a rig ad 0 M 0 M 1 M 0 be a exact sequece of A-modules of fiite legth. Show that i=0 ( 1)i l(m i ) = Let 0 u 1 M 0 legth. u o u 1 u 1 M1 M u 0 be a complex of modules of fiite (1) Show that the A-modules H i = Ker(u i )/ Im(u i 1 ) is of fiite legth for i = 0, 1,.... (2) Show that i=0 ( 1)i l(m i ) = i=0 ( 1)i l(h i ). 6. Let A be a rig ad m a maximal ideal. Moreover let M be a fiitely geerated A-module. (1) Show that the A-module M/mM is of fiite legth. (2) Give a example of a rig A, a prime ideal p of A, ad a fiitely geerated A-module M such that M/pM is ot of fiite legth. 7. Let M be a oetheria A-module, ad let u : M M be a A-liear surjective map. Show that u is a isomorphism. 8. Let M be a artiia A-module, ad let u : M M be a A-liear ijective homomorphism. Show that u is a isomorphism.