$A/\alpha$ is Noetherian as an $A$-module but how to get it is Noetherian as an $A/\alpha$-module?

Question

I meet a problem in commutative algebra of Atiyah.

Pprposition $6.6.$ Let $A$ be Noetherian (resp. Artinian), $\alpha$ an ideal of $A$. Then $A/\alpha$ is a Noetherian (resp. Artinian) ring.

Proof. By $(6.3)$ $A/\alpha$ is Noetherian (resp. Artinian) as an $A$-module, hence also as an $A/\alpha$-module.

What puzzles me is the last sentence: $A/\alpha$ is Noetherian as an $A$-module , but how to get that it is an Noetherian module as an $A/\alpha$-module?

Answer 1

The structure of $A/\alpha$ as an $A$-module is virtually the same as its structure as an $A/\alpha$-module; just think of how the action of $A$ resp. $A/\alpha$ on $A/\alpha$ are related.

For this specific situation, it suffices to notice that the subgroups of $A/\alpha$ which are $A$-submodules coincide with the subgroups which are $A/\alpha$-submodules. Hence $A/\alpha$ is Noetherian over $A$ if and only if it is Noetherian over $A/\alpha$.