By S. Raghavan, R. Balwant Singh, R. Sridharan

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Let ¯ = M0 ⊕ M1 ⊕ M2 ⊕ · · · and N ¯ = N ⊕ (M1 ∩ A¯ = A ⊕ a ⊕ a2 ⊕ · · · , M N ) ⊕ (M2 ∩ N ) ⊕ · · ·. Since the filtration on M is a-good, it follows ¯ is a finitely generated A-module. 6. Graded and filtered modules, Artin-Rees Theorem 27 ¯ is a noetherian A¯ A¯ is noetherian. 7, M ¯ ¯ module so that N is finitely generated over A. 38 again, the induced filtration on N is a-good. 40 Let A be a noetherian ring, a an ideal of A, M a finitely generated A-module and N , a submodule of M . Then there exists n0 ∈ Z+ such that for every n ≥ n0 ,we have a(an M ∩ N ) = an+1 M ∩ N .

36 Chapter 2. 11 Let f : M → M ′ be a homomorphism of A-modules and let (P , ε), (P ′ , ε′ ) be projective resolutions of M, M ′ respectively. Then there exists a morphism F : P → P ′ over f . Moreover, if F, G: P → P ′ are morphisms over f then F and G are homotopic. Proof: Existence of F . Consider the diagram P0 fε P0′ ε′ ✲ ❄ M′ ✲ 0 where the row is exact. Since P0 is projective, there exists an Ahomomorphism F0 : P0 → P0′ such that f ε = ε′ F0 . We now define Fn by induction on n, assuming Fm to be defined for m < n.

Ms maximal. We then have an A-monomorphism A/r → i A/mi . 30, that A/r is of finite length. Since A is artinian, we have rj is artinian for every integer j ≥ 0 and hence rj /rj+1 is artinian as an A/r-module. Since A/r is semi-simple, rj /rj+1 is a semi-simple artinian module and is hence of finite length. Considering the sequence A ⊃ r ⊃ r2 ⊃ · · · ⊃ rn = 0, we conclude that A is of finite length. 34 Every artinian ring is noetherian. Proof: In fact, any strictly ascending sequence of ideals can be refined to a Jordan-H¨older series for the ring (as a module over itself).

### Homological methods in commutative algebra by S. Raghavan, R. Balwant Singh, R. Sridharan

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