Inspired by my lectures on the basics of commutative algebra, I tried to investigate whether the noetherian property still holds when taking extensions (here meaning: if $I\lhd R$ and $R/I$ both satisfy the a.c.c., does $R$ as well?). I started to wonder the following:
Let $I,J,K\lhd R$, and $\pi_K: R \to R/K$ the canonical projection. If $I\cap K = J\cap K$ and $\pi_K(I) = \pi_K (J)$, does this already imply $I = J$?
We can represent this graphically via the short exact sequence of $R$-Modules $$ 0 \to K\stackrel{\iota}{\to} R \stackrel{\pi_K}{\to} R/K \to 0 $$ where $\iota$ is just the embedding of $K$ into $R$, and notice that the intersections $I\cap K, J\cap K$ is basically just the preimages under $\iota$.
I tried to consider all these objects as $R$-moduls and play around with isomorphism theorems, but I only got statements for quotiens or intersections of $I, J, I+J$ or similar, but not these objects directly.
Is it even true?
It's not true. For instance, let $R=k[x,y]/(x^2,xy,y^2)$ for some field $k$, and let $K=(x)$, $I=(y)$, and $J=(x+y)$. Then $R/K\cong k[y]/(y^2)$ and $\pi_K(I)=\pi_K(J)=(y)$, and also $I\cap K=J\cap K=0$, but $I\neq J$.
It is true, though, if you know that $I\subseteq J$, which is what you need for proving that Noetherianness is preserved by extensions. Indeed, given $j\in J$, there exists $i\in I$ such that $\pi_K(i)=\pi_K(j)$. Then $i-j\in K$, and also $i-j\in J$ since $i,j\in J$ (here we use the assumption that $I\subseteq J$). Thus $i-j\in J\cap K=I\cap K$ and so $j=i-(i-j)\in I$.