# Extension of prime ideals in Dedekind domains

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).

• Do we know that $\mathfrak pS\cap R=\mathfrak p$? If so it would follow from that, no? – Gregory Grant May 19 '15 at 18:14
• @GregoryGrant: In the setting which I described, we do not now that $\mathfrak{p}S\cap R=\mathfrak{p}$. – russoo May 19 '15 at 18:17
• Are you saying it's not true or we just don't know that it's true? – Gregory Grant May 19 '15 at 18:24
• I do not see what would not be true. Every nonzero ideal in a Dedekind doman is a product of prime ideals, this includes the full ring. – quid May 19 '15 at 18:25
• @quid If that were true then we could write $1$ as a product of primes in $\mathbb Z$, since $\mathbb Z$ is, afterall, a dedekind domain. – Gregory Grant May 19 '15 at 18:28

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.

• Maybe that this is a silly question: What do mean with $S_{\mathfrak{p}}$? I am asking since $\mathfrak{p}$ is first of all an ideal in $R$. – russoo May 19 '15 at 20:00
• It's the localisation of $S$ (considered as an $R$-module) at $\mathfrak p$. Didn't you have lectures on modules of fractions? – Bernard May 19 '15 at 20:02
• Rings of fractions yes, but not modules of fractions. But give me some time and I'll just go through the relevant definitions. – russoo May 19 '15 at 20:13
• It's is the same construction and amounts to tensoring with $A_\mathfrak p$. Any way, $S$ is aring, and $R\setminus\mathfrak p$ is a multiplicatively closed subset of $S$. – Bernard May 19 '15 at 20:16
• Ok. I think I see the point: We can simply view $S$ as ring and consruct the ring of fractions $(R\setminus \mathfrak{p})^{-1}S$. Then, we view this ring as an $R_{\mathfrak{p}}$-module and denote it with $S_{\mathfrak{p}}$. Is this what you have in mind? – russoo May 19 '15 at 20:41

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 $I=R$.