Exercise 15 of section 2-1 of Kaplansky's Commutative Rings is to show that if $T$ is a Noetherian ring and is finitely generated module over a subring $R$ of $T$, then $R$ is Noetherian. Kaplansky says that the problem can be reduced to the case where $T$ is a domain and $T/J$ is a Noetherian $R$-module for every nonzero ideal $J$, using the following:
If $T$ is a ring with a subring $R$, and $I$ is an ideal in $T$ maximal with respect to the property that $T/I$ is not a Noetherian $R$-module, then $I$ is a prime ideal of $T$.
Now, since $T_{0}=T/I$ is a domain and $T_{0}/J_{0}$ is Noetherian for every nonzero ideal $J_{0}$ of $T_{0}$, we have that $R/R\cap I$ is Noetherian, assuming what Kaplansky said above. But I still don't know how to show that $R$ is Noetherian from that. Am I missing something?
Although I don't understand very well your question, let me try to sketch the proof of the theorem (as it is given in Kaplansky's book):
Reduce the proof to the case $T$ domain and $T/J$ noetherian $R$-module for any $J\subset T$ non-zero ideal: if $T$ is a noetherian $R$-module, then $R$ is a noetherian ring since a ring which has a noetherian faithful module is noetherian (why?). Then suppose that $T$ is not a noetherian $R$-module and take an ideal $K$ in $T$ maximal with the property that $T/K$ is not a noetherian $R$-module. Prove that $K$ is prime.
Consider a non-zero ideal $I$ of $R$. From a previous exercise one knows that there is $J\subset T$ non-zero ideal such that $J\cap R\subseteq I$.
Since $T/J$ is a noetherian $R$-module, then it is also a noetherian $(R/J\cap R)$-module. But a ring which has a noetherian faithful module is noetherian (why?). In our case the module is $T/J$ and the ring is $R/J\cap R$. We deduce that $R/J\cap R$ is noetherian, and therefore $R/I$ is noetherian.