Let $R$ be a Noetherian normal domain with fraction field $K$. Recall that for any ideal $I \subseteq R$, its Nagata transform $T(I)$ is defined as the set of elements $f\in K$ such that $I^n f \subseteq R$ for some $n \in \mathbb{N}$, and that $T(I)$ is a ring between $R$ and $K$. If $P$ is a height one prime ideal of $R$, are there any criteria for when $P T(P) = T(P)$? In particular, if $T(P)$ is not a localization of $R$, can one conclude that $PT(P) \neq T(P)$?
Note that if $P=(x)$ is principal, then $T(P)=R_x$, so that $PT(P) = T(P)$. On the other hand, we do have an example where $PT(P) \neq T(P)$. Namely, let $R = \mathbb{Q}[x,y,u,v] / (xv-yu)$, and let $P=(x,y)$. Then $T(P) = \mathbb{Q}[x,y,\frac{u}{x}]$, whence $PT(P)$ is a height two prime of $T(P)$.
