# Showing that a ring is Noetherian

I show the following:

Let $R$ be a commutative ring with unity and $I \subseteq R$ an ideal. Prove: if $R/I$ is a Noetherian ring and $I/I^2$ is a finitely-generated $R$-module, then $R/I^n$ is a Noetherian ring.

My work: I show that $I/I^n$ and $R/I$ are Noetherian $R/I^n$-modules. I can show they are $R/I^n$-modules, but I cannot show they are Noetherian. E.g., why is $I/I^2$ a Noetherian $R/I^2$-module? It is only finitely-generated $R$-module, so it should be a finitely-generated $R/I^2$-module.

The exact sequence $0 \to I/I^n \to R/I^n \to R/I \to 0$ shows that it suffices to prove that $I/I^n$ and $R/I$ are noetherian $R/I^n$-modules, as you have already observed.
Since $R/I$ is noetherian over $R/I$, it is also noetherian over $R$ and over $R/I^n$ (the submodules with respect to all these rings are just the same (as sets)).
To show that $I/I^n$ is noetherian (say over $R$) for $n \geq 1$, induct on $n$, and use the exact sequence $0 \to I^{n-1}/I^n \to I/I^n \to I/I^{n-1} \to 0$, to reduce the claim to $I^{n-1}/I^n$. Now use another exact sequence and induction to reduce the claim to the cases $n=1,2$, i.e. $R/I$ and $I/I^2$. But $R/I$ is noetherian by assumption, and $I/I^2$ is a finitely generated module over the noetherian ring $R/I$, hence also noetherian.
• How can I show that $I^2/I^3$ is Noetherian R module given $I/I^2$ is Noetherian R module..? – Via Nov 20 '14 at 12:54