Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 1 / b

Let $n_1,\dots ,n_r$ generate $N$, and let the images of $m_1,\dots ,m_s$ generate $M/N$. For any $m\in M$, subtracting a suitable linear combination of the $m_j$ leaves an element of $N$, which is a combination of the $n_i$. Thus

If $M$ is Noetherian, every submodule of $N$ is a submodule of $M$, and submodules of $M/N$ correspond to submodules of $M$ containing $N$; hence both are Noetherian.

Conversely, suppose $N$ and $M/N$ are Noetherian. For any $P\subseteq M$, the intersection $P\cap N$ is finitely generated and the image $(P+N)/N$ is finitely generated. Lifting generators of the image and applying part i toshows that $P$ is finitely generated. Thus $M$ is Noetherian.

The exact sequenceand part ii show that a direct sum of two modules is Noetherian exactly when both summands are. Induction proves the assertion for every finite direct sum.

Each $R/I_j$ is Noetherian as an $R$-module because its $R$-submodules are precisely its ideals as a Noetherian ring. The diagonal maphas kernel ${\bigcap }_j{I}_j=0$. Hence $R$ is an $R$-submodule of a finite direct sum of Noetherian modules, so $R$ is a Noetherian $R$-module; equivalently, $R$ is a Noetherian ring.