Let $R$ be a commutative ring, and suppose that $U$ is an $R$-module with local endomorphism ring. (In particular, note that $U$ is indecomposable.) Consider the ring morphism $f:R\to\operatorname{End}_RU$ taking $r\in R$ to the multiplication-by-$r$ map, $f_r$. The maximal ideal of $\operatorname{End}_RU$ is the set of non-automorphisms in $\operatorname{End}_RU$, and so its contraction under $f$ is the prime ideal $$P:=\{r\in R:f_r\notin\operatorname{Aut}_R U\}\in\operatorname{Spec}R.$$ Now $U$ admits unique division by all elements of $R$ outside $P$, and so can be considered as an $R_P$-module in the natural way. An immediate corollary of this is that $U$ can also be considered as an $R_M$ module for some maximal ideal $M<R$; just take any maximal ideal containing $P$. However, $P$ itself does not have to be maximal: for instance, consider $\mathbb{Q}$ as a $\mathbb{Z}$-module. This has endomorphism ring isomorphic to $\mathbb{Q}$, and so $P$ in this case is $\{0\}$, which is of course not maximal in $\mathbb{Z}$.
Question: Under what circumstances is $P$ maximal? Are there any "nice" characterizations of when this is the case, perhaps with some additional constraints on $U$ or $R$?
Context: By a theorem of Ziegler, indecomposable and pure-injective modules have local endomorphism rings, so the fact above shows that classifying those modules over a commutative ring is equivalent to classifying them over the localizations at all maximal ideals. (One also needs to show that an $R_M$-module is pure-injective as an $R_M$-module if and only if it is pure-injective as an $R$-module, but this is not hard.) This was the context that prompted the question for me, so I'm especially interested in the special case when $U$ not only has local endomorphism ring but is also pure-injective. Can anything be said about the question about in that case?