If $A$ is a commutative, unitary ring and $I$ an ideal of $A$ such that $I$ and $A/I$ are Noetherian rings, then $A$ is Noetherian?

Question

If $A$ is a commutative, unitary ring and $I$ an ideal of $A$ such that $I$ and $A/I$ are Noetherian rings, then $A$ is Noetherian ?

I know just that if $A$ is Noetherian then $A/I$ is Noetherian.

Answer 1

Yes. Let $J$ be an ideal in $A$.

Clearly $J \cap I \subseteq I$ is finitely generated as $I$ is Noetherian.

Also $J/(J \cap I) \cong (I + J)/I$ which is a submodule of $A/I$, hence it's finitely generated.

Hence, you can conclude that $J$ is finitely generated which implies that $A$ is Noetherian, as one of the characterizations of being a Noetherian ring is that every ideal is finitely generated.

The reason you can conclude this is by taking a finite generating system for $I \cap J$, pushing it forward, taking a finite generating system for $J/(J \cap I)$ and pulling it back, then showing the result is a finite generating system for $J$.