# Integral and prime ideal in Dedekind domain

Let $A$ be an Dedekind domain, $K$ its quotient field, $L$ a finite separable extension of $K$, and $B$ the integral closure of $A$ in $L$.

If $p$ is a prime ideal of $A$, then $pB$ has a factorization $pB=P_1^{e_1}\cdots P_r^{e_r}$ into primes of $B$. Now let $B_{P_i}$ be the localization of $B$ in $P_i$, i.e., $S_i^{-1}B$ where $S_i$ is the complement of $P_i$ in $B$.

Assume $A$ is a local ring (so $p$ and $P_i$ are principal). Then $B_{P_i}$ is integral over $A_p$ iff there exists only one prime ideal $P'$ in $B$ lying over $p$.

How to prove the iff statement above? I read it in Algebraic Number Theory by Serge Lang, but those kind of things are far deep away in my memory, so I can't come up with any useful strategy.

• In your case, $A$ is local, so already $A_p = A$. – D_S Feb 29 '16 at 21:18

All the lemmas I mentioned are pretty standard facts, or you can do the proofs yourself without too much trouble.

Lemma 1: Let $A$ be a subring of $K$, and $B$ the integral closure of $A$ in $L$. If $S$ is a multiplicative set, then $S^{-1}B$ is the integral closure of $S^{-1}A$ in $L$.

In particular, any subring of $L$ containing $S^{-1}B$ as a proper subset will not be integral over $S^{-1}A$.

Lemma 2: Let $R$ be a Dedekind domain (or more generally an integral domain where every prime ideal is maximal), and $P$ a nonzero prime ideal of $R$. Then the inclusion $R \subseteq R_P$ is an equality if and only if $R$ is a local ring, i.e. $P$ is the only nonzero prime ideal.

Lemma 3: If $R$ is a Dedekind domain, and $S$ is a multiplicative set in $R$, then $S^{-1}R$ is also a Dedekind domain, and $P \mapsto S^{-1}P$ is a bijection between the nonzero prime ideals of $R$ which are disjoint from $S$, and the nonzero prime ideals of $S^{-1}R$.

Proposition: Let $A$ be a Dedekind domain with quotient field $K$, and $B$ the integral closure of $A$ in a finite separable extension $L$ of $K$. Let $\mathfrak p$ be a prime of $A$, and let $\mathfrak P_1, ... , \mathfrak P_g$ be the primes of $B$ lying over $\mathfrak p$. Then $B_{\mathfrak P_i}$ is integral over $A_{\mathfrak p}$ for every $i$, if and only if $g = 1$ (if and only if $B_{\mathfrak P} = B_{\mathfrak p}$..

Proof: By Lemma 1, $B_{\mathfrak p}$ is the integral closure of $A_{\mathfrak p}$ in $L$, and by Lemma 3, the primes of $B_{\mathfrak p}$ are $\mathfrak P_i B_{\mathfrak p}, i = 1, ... , g$. Also, localization is transitive: $B_{\mathfrak P_i}$ is the localization of $B_{\mathfrak p}$ at the prime $\mathfrak P_i B_{\mathfrak p}$.

So by Lemma 2, all the inclusions $B_{\mathfrak p} \subset B_{\mathfrak P_i}, i = 1, ... , g$ are proper if and only if $g > 1$. But to say that $B_{\mathfrak p} \subsetneq B_{\mathfrak P_i}$ is equivalent to saying that there exists an $x \in B_{\mathfrak P_i}$ is not integral over $A_{\mathfrak p}$.