Let $R$ be an integral domain, and $K$ its field of fractions. It is well known that for a *finitely generated* fractional ideal $I$ of $R$, and $S$ a multiplicative set we have $$(R:\_KI)\_S=(R\_S:\_KI\_S).$$


I suppose that in general the above equation fails, but I don't know such an example. 


The usual non-noetherian ring $k[X\_1,\dots,X\_n,\dots]$ is not an option since it is a Krull domain, or for Krull domains the equality holds for all fractional ideals.



Example of fractional ideal whose inverse does not commute with localization
Let $R = \mathbb{Z}+X\mathbb{Q}[X]$ and $I = X\mathbb{Q}[X] = (X, X/2, X/3, \ldots)$, and let $S = \{1,2,3,\ldots\}$. Then $K = \mathbb{Q}(X)$, $R\_S = \mathbb{Q}[X]$, and $I\_S = X \mathbb{Q}[X]$, whence $(R\_S:\_K I\_S) = (1/X)\mathbb{Q}[X]$. However, $(R:\_K I) = \mathbb{Q}[X] = R\_S$, so that $(R:\_K I)\_S = (R\_S)\_S = R\_S = \mathbb{Q}[X]$.


To see that $(R:\_K I) = \mathbb{Q}[X]$, note that
$y \in (R:\_K I)$ implies $$y = \frac{f\_1}{X} = \frac{f\_2}{X/2} = \frac{f\_3}{X/3} = \cdots$$
for some $f\_1, f\_2, f\_3, \ldots \in R$, which implies that $2, 3, 4, \ldots$ all divide $f\_1$ in $R$, which implies that $f\_1(0) = 0$ and therefore $f\_1 \in X\mathbb{Q}[X]$ and hence $y \in \mathbb{Q}[X]$. Conversely, if $y \in \mathbb{Q}[X]$, then $Iy \subseteq I \subseteq R$, so $y \in (R:\_K I)$.


A similar example that should also work is $R = \operatorname{Int}(\mathbb{Z})$ (the ring of integer-valued polynomials), $I = \{f \in \operatorname{Int}(\mathbb{Z}): f(0) = 0\}$, and $S = \{1,2,3,\ldots\}$.


