Intersection of ideals is zero and quotients are noetherian 
Assume $R$ is a ring and $I_1,...,I_n\subseteq R$ ideals s.t $R/I_i$ is noetherian for every $i$, and $\bigcap_{i=1}^nI_i=\{0\}$, then $R$ is noetherian.

My attempt: I defined the natural homomorphism $\phi:R\rightarrow R/I_1\times\dots\times R/I_n: \phi(r)=(r+I_1,\dots,r+I_n)$. Since the intersection is the zero ideal, I then can say that: $R\simeq\{(r+I_1,\dots,r+I_n)|r\in R\}\subsetneq R/I_1\times\dots\times R/I_n$. I haven't learned modules - so the existing solution to this problem on this site doesn't help me.
How do I show that the right side of the isomorphism is noetherian? Any hint would be appreciated.
 A: I show the case $n=2$, the general case is very similar.
Let $R$ be a ring with two ideals $I,J$ with $I \cap J = 0$ such that $R/I$ and $R/J$ are Noetherian. We want to show directly that $R$ is Noetherian. Since it was asked for a proof without module theory, the proof does not use the standard arguments from module theory (see e.g. here). But the arguments are basically the same, just written down in this special case here, which makes the proof somewhat clumsy in my opinion.
Let $\pi_I : R \to R/I$, $\pi_J : R \to R/J$ denote the canonical projections, and let $\pi := (\pi_I,\pi_J) : R \to R/I \times R/J$.
Consider an ascending chain of ideals $(K_n)$ in $R$. This yields an ascending chain of ideals $\pi_I(K_n)$ in $R/I$. Since $R/I$ is Noetherian, we may assume (after shifting the indices) $\pi_I(K_n)=\pi_I(K_0)$ for all $n$.
Now consider $L_n := \{y \in R/J : (0,y) \in \pi(K_m)\}$. Then $(L_n)$ is an ascending chain of ideals in $R/J$ (check this!). Since $R/J$ is Noetherian, we may assume (after shifting the indices) that $L_n = L_0$ for all $n$.
I claim $\pi(K_n)=\pi(K_0)$. In fact, let $k \in K_n$, and $\pi(k) = (x,y)$. Then $x \in \pi_I(K_n)=\pi_I(K_0)$, hence $\pi_I(k')=x$ for some $k' \in K_0$. Let $y' := \pi_J(k')$, so that $\pi(k')=(x,y')$. Therefore, $\pi(k-k')=(0,y-y')$. Hence, $y-y' \in L_n=L_0$, so that $(0,y-y')=\pi(k'')$ for some $k'' \in K_0$. Hence, $\pi(k) = \pi(k-k')+\pi(k')=\pi(k''+k')$, and we are done.
Since $\pi$ is injective, it follows that $K_n=K_0$ for all $n$.
