Subrings of fraction fields

There are lots of examples arising from integral closures. For example, $k[x^2,x^3] \subseteq k[x]$ with field of fractions $k(x)$ is no localization since $k[x^2,x^3]^* = k^* = k[x]^*$.

I've shown here that the ring $S=K[X]+YK(X)[Y]$ is not noetherian. Note that we have $K[X,Y]=R\subset S\subset Q(R)=K(X,Y)$. Since $R$ is noetherian, $S$ can't be a localization of $R$.

Let $U$ be the multiplicative set $\{r\in R\mid \frac 1r\in S\}$, which is the "obvious" (and maximal) candidate. Then clearly $R[U^{-1}]\subseteq S$. But is $S\subseteq R[U^{-1}]$? Can we conclude $\frac1b\in S$ from $\frac ab\in S$? Note how we would e.g. conclude $\frac13\in S$ from $\frac23\in S$: We'd use $2\cdot \frac23-1=\frac13$. Such is not possible in general.

Let $R=\mathbb Z[X]$ and $S=R[\frac X2]$. Then if we assume $S=R[U^{-1}]$, there must be $g\in U$, $f\in R$ such that $\frac X2=\frac fg$, i.e. $2f=Xg$. Since also $\frac 1g\in S\subseteq \mathbb Q[X]$, necessarily $g$ is constant, i.e. $g=2k$ with $k\in\mathbb Z\setminus \{0\}$. but $\frac1{2k}\notin S$ because the image of the homomorphism $S\to\mathbb Q$ induced by $X\mapsto 0$ is in $\mathbb Z$.