Let $A$ be a ring e $\mathfrak{a}$ and $\mathfrak{b}$ two ideals, such that $\mathfrak{a} \cap \mathfrak{b} =(0)$. Prove that $A$ is Noetherian if and only if $A/\mathfrak{a}$ and $A/\mathfrak{b}$ are both Noetherian.
I was thinking about the following sequence:
$0 \rightarrow A/\mathfrak{a} \overset{f}\rightarrow A \overset{v}\rightarrow A/\mathfrak{b} \rightarrow 0$.
$v$ is a quotient map, so it is surjective. Is it possible to find an injective function $f$ which makes this sequence exact? This way I could easily conclude.
Otherwise is there another way to solve this problem? Thank you for you help.
Hint: Take two ideals $I\subseteq J$ in $A$ (such as in an ascending chain of ideals), and show that $I+(\mathfrak{a} \cap \mathfrak{b})= J+(\mathfrak{a} \cap \mathfrak{b})$ iff both $I+\mathfrak a= J+\mathfrak a$ and $I+\mathfrak b= J+\mathfrak b$.