New PDF release: Commutative Algebra [Lecture notes]

By Wolfram Decker

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6 to the exact sequence 0 → Ker π → Rr → M → 0. 2 Free Resolutions Consider a ring R and an R-module M. Then, as above, M is an epimorphic image of free R-modules: Pick a set of generators {mλ }λ∈Λ and consider a free R-module F0 with free basis {eλ }λ∈Λ and the epimorphism π : F0 → M, eλ → mλ . Apply the same argument to the kernel of π to get a free R-module F1 and an epimorphism F1 → Ker π. Write ϕ for the composite map F1 → Ker π → F0 to get an exact sequence ϕ F1 → F0 → M → 0. 1 Each exact sequence as above is called a free presentation of M.

F1 , . . , fk , also g ∈ I f1 , . . , fk . This contradicts the choice of fk+1 since deg(g) < dk+1 = How do Noetherian rings behave under localization and modulo ideals? 4 Let R be a Noetherian ring. (i) If I is an ideal of R, then R/I is Noetherian. (ii) If U ⊆ R is a multiplicatively closed subset, then R[U −1 ] is Noetherian. Proof. 7: Ideals in R/I correspond to ideals in R containing I and, thus, we can translate the ideal conditions from R to R/I. 5: Let J1 ⊆ J2 ⊆ J3 ⊆ . . be a chain of ideals in R[U −1 ].

The ascending chain condition in Noetherian rings applied to I0 : g ⊆ I0 : g 2 ⊆ . . shows that I0 : g m = I0 : g m+1 for some m ≥ 1. Then I0 = (I0 : g m ) ∩ I0 , g m by Exercise 2e on Sheet 1. Since f √ g ∈ I0 , also f g m ∈ I0 , which means f ∈ I0 : g m . Then I0 I0 : g m since m f I0 . Also, I0 I0 , g since g I0 . It follows that both I0 : g m and I0 , g m are not in Γ , so both ideals have a primary decomposition. But then, I0 must have a primary decomposition as well, combining the PDs of I0 : g m and I0 , g m .

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Commutative Algebra [Lecture notes] by Wolfram Decker

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