$R$ is a Noetherian ring if and only if both $I$ and $J$ are Noetherian $R$-modules, where $I,J$ are distinct maximal ideals

Question

Problem. Let $R$ be a commutative ring with unity, and $I, J\subset R$ be maximal ideals such that $I \neq J$. Show that $R$ is a Noetherian ring if and only if both $I$ and $J$ are Noetherian $R$-modules.

My attempt: Suppose $R$ is Noetherian. Then every ideal of $R$ is finitely generated, so every submodule of $I$, which is an ideal of $R$, is finitely generated. Thus $I$ is Noetherian, and similarly $J$ is also Noetherian.

But I can't see where to start the opposite direction. Any hints?

Answer 1

Hint:

If $I$ and $J$ are distinct maximal ideals, you have a surjective homorphism $$I\oplus J\longrightarrow I+J=R.$$