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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.

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  • $\begingroup$ Why are you saying "Now, let $I$ be a ideal of $R'$, then $I$ is a $R$-submodule of $\sum_j Rb_j.$" It will be $R'$ submodule, I agree, but I do not think it has to $R$ module. $\endgroup$ – Bhaskar Vashishth Feb 26 '15 at 8:13
  • $\begingroup$ Because $R\subseteq R'$, then $I$ is a $R-$ submodule of $\sum_j Rb_j.$. right? $\endgroup$ – user126033 Feb 26 '15 at 8:23
  • $\begingroup$ Ohh ya. Ok. But from here we can only say that $I$ is a noetherian module over $R$. Not that $R'$ is noetherian. Why you concluded this? $\endgroup$ – Bhaskar Vashishth Feb 26 '15 at 8:27
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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.

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