Let $R$ be a Noetherian domain with quotient field $K$ and let $b_1,\ldots,b_n\in K$.
Suppose that $R'$ is a integral domain, $R\subseteq R'$ and
$$R'\subseteq \sum_j Rb_j.$$
Remark: It is well know that if $R$ is a Noetherian ring and $M$ is a finitely generated $R$-module then $M$ is Noetherian.
Thus, the $R$-module $\sum_j Rb_j$ is Noetherian.
Now, let $I$ be a ideal of $R'$, then $I$ is a $R$-submodule of $\sum_j Rb_j.$
This implies that $I$ is a finitely generated $R$-submodule, in particular is finitely generated as $R'$-module.
The conclusion is that $R'$ is a Noetherian ring.
Is correct ?
Thank you all.
A minor remark: you should assume that $R'$ is a subring of $K$, containing $R$. This ensures the operations match.
Your proof is correct. Here's how it can be generalized.
First of all, $R'$ is an $R$-algebra, in particular an $R$-module. Since $R'$ is contained in a finitely generated $R$-module, it is noetherian as an $R$-module.
Now we prove that whenever $A$ is an $R$-algebra, finitely generated as $R$-module, it is a noetherian ring. (Of course, we keep the basic hypothesis that $R$ is a noetherian ring).
If $I_0\subseteq I_1\subseteq\dots\subseteq I_n\subseteq \dotsb$ is an ascending chain of ideals of $A$, it is also an ascending chain of $R$-submodules of $A$, so it stabilizes.