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 :)

  • $\begingroup$ 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 $g=d_k$ $\endgroup$ – Dorian Jun 27 at 13:46
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    $\begingroup$ 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. $\endgroup$ – rschwieb Jun 27 at 13:51
  • $\begingroup$ @rschwieb Isn't better to ask here for a verification of the proof instead of posting it as a "new" answer (when, in fact, there is a similar answer posted long time ago) in another thread? $\endgroup$ – user26857 Jun 27 at 22:14
  • $\begingroup$ @user26857 No... That seems like a very useless thing to keep, a question version of a solution that already exists. $\endgroup$ – rschwieb Jun 28 at 0:57
  • $\begingroup$ I don't get for whom and for what is useful to post an answer similar to an existing one and asking if it is correct. And many questions are useful only for the ones who asked. This won't be the first or the last. Moreover, when post this as an answer people can only say something in comments, while here an extended explanation if necessary can be posted as an answer. $\endgroup$ – user26857 Jun 28 at 6:51

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