(All rings assumed to be commutative and unital)


Given a ring $R$, a prime ideal $\mathfrak{p}$ of $R$, and an extension ring $S$ (the algebra map $R\to S$ is injective), are there any nontrivial sufficient conditions for $rad(\mathfrak{p}S)$ to be prime? How about when $S$ is integral over $R$? How about when it is finitely presented as an $R$-module?


EDIT: Qing Liu's comment means that I should be even more specific.


Assume the third question above ($S$ is finitely presented as an $R$-module.).


Suppose further that there exists an element $f \in R$ such that $R\_f$=$S\_f$. Then even further, suppose $f \in \mathfrak{p}$.


EDIT 2: It is now open season for adding any extra conditions on $S$ that you want. Please don't add stupid conditions like S=R. Note that the condition about $f$ is actually a condition on $S$.



When is the radical of the extension of a prime ideal prime?
Probably Harry has already thought of it like this, but rephrased in terms of algebraic geometry, this question is basically "under what conditions is the preimage of an irreducible variety under a map be irreducible."


When you phrase it like that, it sounds like you might be able to get a statement like "when all maximal prime ideals have prime (up to taking radical) extension" (I think this really does follow over an algebraically closed field by the Nullstellensatz, and some easy topology) but I don't think you'll really do any better than that. This at least gives you some geometric intuition (lots of maps, like smooth one with connected fibers satisfy this condition). 


I have no idea how the condition that $f$ exists with $R\_f=S\_f$ is supposed to help. This says that the map is an isomorphism away from the 0 set of f, but $V(\mathfrak{p})$ is in that zero set, so you haven't gained any control.


