$\textbf{Q:}$ Suppose $R$ is noetherian domain and $K=\operatorname{Frac}(R)$. If $S$ is a subring of $K$ s.t. $R\subset S\subset K$, is $S$ noetherian? I suspect this will be the case in general. One can consider $(R:_R I)$ with $S$-ideal $I\subset S$. There is no good reason that $(R:_R I)\neq 0$. Furthermore, there is no good reason that $S$ being localization of $R$ in general unless $R$ is Bézout.
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Is it not true that every intermediate ring is obtained by a localization? Say, by localizing $R$ at the set of units of $S$? Maybe it is not true. But if it were true, then $S$ would be Noetherian (as a localization of $R$) . Maybe $S$ can be weird so that the localization at its units is properly contained in $S$. – rschwieb Nov 26 '18 at 21:43
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1@rschwieb I think I asked a question on this before and someone gave a concrete example that an intermediate ring is not localization. However, not being localization does not say anything about noetherian property of a subring that is not localization. I think there is a loop hole in my logic somewhere. I was expecting the quotient ideal non-trivial but I cannot see any good reason that it is not trivial. – user45765 Nov 26 '18 at 21:47
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5The answer to both claims (every subring of field of fractions is a localization and every intermediate ring is noetherian) is no: https://math.stackexchange.com/questions/287245/subrings-of-fraction-fields#answer-287262 – freakish Nov 26 '18 at 21:51
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@freakish glad I asked: thanks for the link – rschwieb Nov 27 '18 at 01:04