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I am trying to prove that if $R$ is Noetherian, and $I$ is an ideal of $R$, then $R/I$ is Noetherian.

I know that both $R$ and $I$ are finitely generated since they are both ideals of $R$. I’m having trouble seeing why the quotient $R/I$ must be finitely generated. I know that elements of $R/I$ take the form $r+I$,but why must this set be finitely generated? Is it because if each $a\in I$ is finitely generated, then $r+I$ must necessarily be a set of finitely generated elements. But then the generator has to belong to this set so it must be finitely generated. Do I have the right idea?

user873295
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I would use Correspondence here. Namely, if $\{J_\alpha\}_{\alpha\in\Lambda}$ is a non-terminating chain of nested ideals in $R/I$ (for some index set $\Lambda$ with $\alpha < \beta \Rightarrow J_\alpha \subsetneq J_\beta$), then $\{\pi^{-1}(J_\alpha)\}_{\alpha\in\Lambda}$ is a non-terminating chain of nested ideals in $R$ containing $I$ (where $\pi:R\to R/I$ is defined by $\pi(r)=r\pmod{I}$). But $R$ is Noetherian, so every ascending chain terminates, giving a contradiction.

IAAW
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