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I have this exercise:

Let $R$ be a Noetherian ring. Let $S\subseteq R$ be a multiplicatively closed set. Prove that $S^{-1}R$ is a Noetherian ring.

First off we claim that given a subset $I\subset R$, $$S^{-1}I \lhd S^{-1}R \quad \Rightarrow \quad I\lhd R\ \land \ I\ \cap S=\emptyset $$ That's true because $$ S^{-1}I \lhd S^{-1}R \quad \Rightarrow \quad (\ \frac {i_k} s \in S^{-1}I \Rightarrow \exists \ d_k\ \mathrm {maximal} :d_k|i_k,d_k\notin S)$$ So if we take the ideal in $S^{-1}R\ $generated by the elements $(\frac {d_1} 1,\frac {d_2} 1,\dots ,\frac {d_k} 1,\dots)$, clearly we obtain $S^{-1}I$, plus $I=(d_1,d_2,\dots,d_k,\dots)\lhd R$. Since every ideal on $R$ is finitely generated, it follows that ideals are finitely generated on $S^{-1}R$ too.

Is this correct? Do I need to be more exhaustive or am I wrong with something? Thank you :)

user26857
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Dr. Scotti
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  • When I say that $d_k$ is maximal I mean that there is no $g\in R$ s.t. $g|i_k, g\notin S$ and $d_k|g$, except for $g=d_k$ – Dr. Scotti Jun 27 '19 at 13:46
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    Please do not ask two completely different questions in a single post. It is better for users to tackle a single, or at most a few highly related problems, rather than distinct ones. Also, please search for your questions first. In this case both are duplicates. Compare your solution to existing ones, and if you think you still dont' know if your approach is valid, post it as a new solution. It will bump the question into view and people will take a look. – rschwieb Jun 27 '19 at 13:51

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