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Suppose $R$ and $S$ are domains, $S$ is integral over $R$, $R$ is integrally closed, $p_1$ and $p_2$ are primes in the domain $R$, $p_1$ contains $p_2$, $q_1$ is a prime in the domain $S$ and lying over $p_1$, we can prove that $p_2 S_{q_1}\cap R=p_1$,so $p_2S_{q_1}\cap S$ is lying over the prime $p_2$ as we wanted in the proof of the Going-down theorem. Is $p_2S_{q_1}\cap S$ also a prime? I find that many books just play a trick of "ec" to around it,but can we prove it directly?

A generated question: Suppose $R$ and $S$ are domains, $S$ is integral over $R$, $p$ is a prime in $R$, in which situation $pS$ is also a prime in $S$?(As we know, $(2)$ is prime in $\Bbb{Z}$, but not a prime in $\Bbb{Z}[i]$)

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    I don't get anything, sorry. Please edit your question.2012-03-22

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(Is the language mis-representing the question? For $R$ integral over $S$, $R$ is "larger" than $S$, and primes in $R$ lie over primes in $S$. The "generated" question's example is incompatible with the question, to my perception.)

If the "generated" question is construed as asking when primes stay prime in an extension, this has a good answer in the case of rings of algebraic integers, namely, a necessary condition for Galois extensions is that the Galois group be cyclic, since a prime staying prime means that the Galois group of the global extension is isomorphic to the Galois group of the residue field extension. But, even then, we need Dirichlet's theorem to prove that the "expected" proportion of primes do stay prime. (Tchebotarev density addresses an opposite sort of issue, namely, splitting, not "inert" primes.)

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    Thank you for the situation of algebraic integers,but I care about the Going-Down more,a little generated question is that if S is local ring ,does pS also a prime?2011-08-06