What properties $R \subseteq S$ should have in order that every prime ideal of $S$ is extended?

My question is almost the same as In what conditions every ideal is an extension ideal?; I allow myself to ask this question, since there is no answer to the above question.

My question:

Given commutative rings with 1, $R \subseteq S$, when every prime ideal $q$ of $S$ is an extension of some prime ideal $p$ of $R$? namely, there exists a prime ideal $p$ of $R$ such that $q=pS$. Is it easier to answer if I only demand that there exists a (not necessarily prime) ideal $I$ of $R$ such that $q=IS$? (maybe this is an equivalent question?)

Remarks:

(1) I have already asked a similar (more general) question concerning modules in MO; however, I hope to get a more precise answer if I restrict to the case of (prime) ideals. Notice that this paper was mentioned in MO, and I am looking for something similar to Proposition 3.3/Theorem 3.4 in that paper (preferably with less conditions; for example, without assuming locality).

(2) A plausible answer to my question: Every ideal of the localization is an extended ideal. However, I wish to know if there are other known cases which answer my question.

(3) Two slightly relevant questions are: faithfully flat ring extensions where primes extend to primes and In a faithfully flat ring extension, is $\operatorname{ht}I=\operatorname{ht}IS$ right?. (I suspect that demanding that $R \subseteq S$ is faithfully flat is not enough for a positive answer to my question?).

Thank you very much!

• The Cohen-Seidenberg "going up" and "going down" theorems might help you. – Dominik B. Jul 21 '15 at 16:41
• Thanks!! Please if @DominikB. or someone else can help me with the following: Assume $R \subseteq S$ is an integral extension, $q$ is a prime ideal of $S$, and I am looking for a (prime) ideal $I$ of $R$ such that $q=IS$. Now, $p:=q \cap R$ is a prime ideal of $R$. If I am not wrong, an integral extension $A \subseteq B$ (and a faithfully flat extension) satisfies: $aB \cap A=a$, for every ideal $a$ of $A$. Therefore, here: $pS \cap R=p$. However, I am not sure if this helps. – user237522 Jul 21 '15 at 18:24
• It is clear that $pS=(q \cap R)S \subseteq q$, but it seems (even when $R \subseteq S$ is integral) that the other inclusion is not necessarily true, so it is not necessarily true that $pS=q$. (Moreover, $pS$ may not be a prime ideal of $S$?). Am I missing something trivial? – user237522 Jul 21 '15 at 21:20