Question #144113
If R is noetherian and I is a ideal show that R/I is noetherian.
1
Expert's answer
2020-11-16T07:08:52-0500

Theorem. Let RR be a Noetherian ring. If II is an ideal of RR then R/IR/I is Noetherian.

Proof. Let J1J2JiJi+1J_1 ⊂ J_2 ⊆ ··· ⊆ J_i ⊆ J_{i+1} ⊆ ··· be an ascending chain of ideals in R/IR/I. By the Correspondence Theorem, RR has ideals Ki(i=1,2,...)K_i (i = 1, 2, . . . )  such that IKiI ⊆ K_i and Ki/I=JiK_i/I = J_i and KiKi+1K_i ⊆ K_{i+1} for all ii.  Then IK1K2KiKi+1I ⊆ K_1 ⊆ K_2 ⊆ ··· ⊆ K_i ⊆ K_{i+1} ⊆ ··· is an ascending chain of ideals of RR. Since RR is Noetherian, there is some NN such that Ki=KNK_i = K_N whenever iN,i \geqslant N, so Ji=Ki/I=KN/I=JNJ_i = K_i/I = K_N/I = J_N for iNi \geqslant N. Hence R/IR/I is Noetherian.


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