# Rings with finitely many prime ideals.

What follows is an argument i found in my textbook which i can't understand.

Let $R$ be a Dedekind domain, with quotient field $K$, $R'$ is the integral closure of $R$ in a finite and separable extension $L$ of $K$. For a maximal ideal $\mathfrak{p}$ of $R$ let $S:=R- \mathfrak{p}$. Then $R_S$ is a DVR and $R_S'$ is a Dedekind domain with only a finite number of maximal ideals, namely those corresponding to the ideals of $R'$ that contain $\mathfrak{p}$.

I can't understand this correspondence. The only fact i know is that, if $R$ is a domain and $S$ a multiplicative subset of $R$, then there is a one-to-one correspondence between prime ideals of $R$ not intersecting $S$ and prime ideals of $R_S$. But in the case $S=R-\mathfrak{p}$, not intersecting $S$ means being contained in $\mathfrak{p}$, which is the opposite of what my textbook states. What am i missing?

## 2 Answers

You are right that a maximal ideal $\mathfrak p'\subset R'$ is disjoint from $S$ if and only its trace on $R$ is included in $\mathfrak p$, i.e. $\mathfrak p'\cap R\subset \mathfrak p$. This is purely set-theoretic .
Is the book then wrong?
No, because we now use that for any integral extension of rings $R\subset R'$ a prime ideal $\mathfrak p'\subset R'$ is maximal if and only if its trace $\mathfrak p'\cap R\subset R$ is maximal in $R$.
Hence if we know that $\mathfrak p'$ is maximal and that $\mathfrak p'\cap R\subset \mathfrak p$, we deduce that $\mathfrak p'\cap R= \mathfrak p$ so that, yes, $\mathfrak p'\supset \mathfrak p$.

Careful! You moved from $R$ to $R'$. In $R'$ we have $\mathfrak{P}\cap S=\emptyset \Longleftrightarrow \mathfrak{P}\cap R\subseteq \mathfrak{p}$. So the maximal ideals of $R'_S$ correspond $1$-$1$ to the prime divisors of $\mathfrak{p}R'$ in $R'$, of which there are at most $[L:K]\lt \infty$.