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

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?

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 $$A/\alpha$$.