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Let $R$ be a commutative ring and $I,J$ ideals of $R$ such that $J$ is finitely generated and the rings $R/I$ and $R/J$ are Noetherian. Are the $R$-modules $R/J$, $J/IJ$, $R/IJ$ Noetherian? Is the ring $R/IJ$ is Noetherian?

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    $\begingroup$ $R/J$ is a noetherian ring, which means that all its ideals are finitely generated. The $R$-submodules of $R/J$ are also $R/J$-submodules which are precisely the ideals of $R/J$. But those are all finitely generated which is equivalent to $R/J$ being noetherian. $\endgroup$
    – Sebastian
    Jan 14, 2011 at 9:42

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Sebastian did the $R/J$ case in the comments. Now $J/IJ$ is $J\otimes R/I$ which, since $J$ is finitely generated and $R/I$ is Noetherian (as an R-module), is a Noetherian R-module. The exact sequence $$ 0\rightarrow J/IJ \rightarrow R/IJ\rightarrow R/J\rightarrow 0$$ and the previous two results show $R/IJ$ is Noetherian.

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