Extension of prime ideals in Dedekind domains

Question

In various textbooks and lecture notes on algebraic number theory, I have found the following claim without proof:

Let $R$ be a Dedekind domain with field of fractions $F$ and let $S$ be its integral closure in some finite separable extension $E/F$. If $\mathfrak{p}$ is a prime ideal in $R$, then $\mathfrak{p}S$ has a prime factorization in $S$.

My problem is here that the authors seem to assume implicitly that $\mathfrak{p}S$ is a proper ideal in $S$. (Otherwise, I think the claim is false.) Up to now, I have never seen a proof that $\mathfrak{p}S$ is indeed proper in $S$. So, why is this true? For me, this is not obvious.

$\textbf{Edit:}$ To clarify the references, I have found this claim for example in the lecture notes "Algebraic Number Theory" by John Milne, ch. 3, section "Factorization in extensions" (see here: http://www.jmilne.org/math/CourseNotes/ANT.pdf).

Answer 1

If $\mathfrak p S=S$, then $\mathfrak p S_\mathfrak p=S_\mathfrak p$, and, as $S_\mathfrak p$ is a finitely generated $R_\mathfrak p$-module, there results by Nakayama's lemma that $S_\mathfrak p=0$, which contradicts the fact that $R_\mathfrak p\to S_\mathfrak p$ is injective.

Answer 2

Your claim holds for any integral ring extension $R\subset S$. If $p$ is a prime ideal of $R$, then $pS\ne S$ for there is a prime ideal $P$ of $S$ such that $P\cap R=p$.

Moreover, the following holds:

If $R\subset S$ is an integral ring extension, $I\subseteq R$ an ideal, and $x\in IS$, then there is $n\ge1$ and $a_r\in I^r$ for $r=1,\dots,n$ such that $$x^n+a_1x^{n-1}+\cdots+a_n=0.$$

This proves that $IS\ne S$ unless $I=R$.