Exercise A0.V.1.01

Proposition

Let $R$ be a Noetherian ring with an ideal $I.$ Then the ring $R/I$ is Noetherian.

Proof

Take the quotient map $\pi : R \to R/I.$ Let $J$ be an arbitrary ideal of $R/I.$ Since the ideals of $R/I$ are in bijection with the ideals of $R$ containing $I$, there exists an ideal $J'$ of $R$ such that $\pi(J') = J.$ Since $R$ is Noetherian, $J'$ is finitely generated. The image of the generating set of $J'$ under $\pi$ gives a finite generating set of $J.$

$\blacksquare$