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.
