I struggle proving the following statement:
Let $A$ be a ring. If for any $a\in A\setminus\{0\}$, $A/aA$ is noetherian, then $A$ is a noetherian ring.
I first noticed that because $0\to aA\to A\to A/aA\to 0$ is a short exact sequence, $A$ is noetherian if and only if $aA$ and $A/aA$ are so, but $aA$ doesn't clearly appear being a noetherian ring, am I wrong?
Then, I was trying to find $n\in\mathbb{N}$ and $a_1\ldots,a_n\in A\setminus\{0\}$ such that $\bigcap_{i=1}^na_iA=\{0\}$, thus $A\cong\prod_{i=1}^nA/a_iA$ and $A$ would be noetherian since isomorphic to a finite product of noetherian rings, but finding such elements depends strongly of the structure of $A$, in a principal ring (for instance) it is impossible. Indeed, $\bigcap_{i=1}^na_iA=\{0\}$ would be the ideal generated by a lcm of $a_1,\ldots,a_n$ and never be $\{0\}$.
Can you provide me some help solving this problem?
Yours sincerely,
Sheol
Let $I$ be a non trivial ideal of $A$ and $a\in I, a\neq 0$, let $p:A\rightarrow A/aA$ the projection. $p(I)$ is finitely generated since $A/aA$ is noetherian. Let $p(x_1),...,p(x_p), x_i\in I$ generators of $p(I)$. Then $a,x_1,...,x_p$ are generators of $I$ since $p^{-1}(p(I))=I$.