# Show if $R$ is Noetherian, then $R_S$ is Noetherian [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 16 '15 at 12:56
• For $r\in R$, the notation $r/1$ means the image of $r$ in the localized ring (under the usual homomorphism). – Tobias Kildetoft Feb 16 '15 at 10:37