This question already has an answer here:

Show if $R$ is Noetherian, then $R_S$ is Noetherian.

here is what I have read from somewhere else. Suppose $R$ is Noetherian and $J$ is an ideal $R_S$. Then $J=IR_S$ for some ideal $I$ of $R$. Since $R$ is Noetherian, $I$ is finitely generated, say $I=(r_1,r_2,…r_n)$ Thus $J=(r_1/1,r_2/1,…r_n/1)$. Hence every ideal in $R_S$ is finitely generated and so $R_S$ is Noetherian.

My confusion is how to get $J=(r_1/1,r_2/1,…r_n/1)$ from $I=(r_1,r_2,…r_n)$. And I also confuse about the notation of $r_1/1$, what are they, are they suppose just be like $r_1/1=r_1$ ? Thank you.


marked as duplicate by egreg, Martin Sleziak, user26857, Dietrich Burde, Namaste abstract-algebra Feb 16 '15 at 12:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ For $r\in R$, the notation $r/1$ means the image of $r$ in the localized ring (under the usual homomorphism). $\endgroup$ – Tobias Kildetoft Feb 16 '15 at 10:37
  • $\begingroup$ I think this was proved here. $\endgroup$ – Pierre-Guy Plamondon Feb 16 '15 at 11:21