Question related to commutative ring being Noetherian

Question

Let $A$ be a commutative ring with $1$, and $A = (f_1, \ldots, f_n)$. I want to prove the following:

If $A$ is a Noetherian ring, then so is $A_{f_i}$ (which is the ring $A$ localized at $f_i \in A$). If each $A_{f_i}$ is Noetherian, then $A$ is Noetherian.

It says in the exercise to use the hint $A \rightarrow \prod A_{f_i}$ is injective, but I am not sure how this is useful... I would appreciate any help! Thanks!

Answer 1

This is a special case of the following:

Claim: If $A$ is a commutative ring with $A=(f_1,...,f_n)$, then an $A$-module $M$ is finitely generated if and only if each localization $M_{f_i}$ is finitely generated.

In your situation, you can apply this to $M := I$ since in this case $M_{f_i} = I_{f_i}$ is finitely generated over $A_{f_i}$ as a submodule of the - by assumption - Noetherian $A_{f_i}$-module $A_{f_i}$.

Since this is an exercise, do you want to try proving the claim for yourself? If you have trouble, I already typed it and can show it if you want. Use the assumption to find a finite set of elements of $M$ which generates each of the localizations $M_{f_i}$ and deduce that it must already generate $M$.