Answer to Question #144113 in Abstract Algebra for anjali g

Question

If R is noetherian and I is a ideal show that R/I is noetherian.

Answer 1

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

Proof. Let $J_1 ⊂ J_2 ⊆ ··· ⊆ J_i ⊆ J_{i+1} ⊆ ···$ be an ascending chain of ideals in $R/I$. By the Correspondence Theorem, $R$ has ideals $K_i (i = 1, 2, . . . )$  such that $I ⊆ K_i$ and $K_i/I = J_i$ and $K_i ⊆ K_{i+1}$ for all $i$.  Then $I ⊆ K_1 ⊆ K_2 ⊆ ··· ⊆ K_i ⊆ K_{i+1} ⊆ ···$ is an ascending chain of ideals of $R$. Since $R$ is Noetherian, there is some $N$ such that $K_i = K_N$ whenever $i \geqslant N,$ so $J_i = K_i/I = K_N/I = J_N$ for $i \geqslant N$. Hence $R/I$ is Noetherian.