Is the following proposition true?

Proposition? Let $A$ be an integral domain, $K$ its the field of fractions. Let $B$ be the integral closure of $A$ in $K$. Suppose $B$ is finitely generated $A$-module. Let $I = \{a \in A; aB \subset A\}$. Let $P$ be a prime ideal of $A$. Then $A_P$ is not integrally closed if and only if $I ⊂ P$.

EDIT I'm particularly interested in the case when $K$ is an algebraic number field and $A$ is its order(a subring of K which is a finitely generated $\mathbb{Z}$-module and contains a $\mathbb{Q}$-basis of $K$). In this case, the above condition gives a necessary and sufficient condition for $A_P$ to be a discrete valuation ring.

  • $\begingroup$ Context? Partial work? Thoughts? $\endgroup$ – rschwieb Jun 15 '12 at 1:32
  • $\begingroup$ I added my motivation. $\endgroup$ – Makoto Kato Jun 15 '12 at 7:06

Yes, your proposition is true.

The main point to know is that normalisation commutes with localization. That is to say, if $S$ is a multiplicative part of $A$, then $(S^{-1} A)'$ equals $S^{-1} A'$, where the “prime” denote the integral closure.

In your case, set $S = A\setminus P$, so that $(A_P)' = S^{-1} A'$. And you're asking when $S^{-1} A' \subset S^{-1} A$, the reverse inclusion being obvious.

The intuition is the following : $A'$ is generated over $A$ by some fractions, and if $S$ contains the denominators of these fractions, then $A'$ is a subset of $S^{-1} A'$, and thus we have the equality.

Lets go for a proof. Let $I$ be the conductor ideal $[A:A'] = \{ a \in A \mid aA' \subset A \}$. This is a kind of a l.c.m. of the denominators. We shall prove that $[S^{-1} A : S^{-1} A'] = S^{-1} I$ :

  1. Let $a/s$ be an element of $S^{-1}I$, with $a\in I$. Then $aS^{-1} A' = S^{-1}(aA') \subset S^{-1} A$.

  2. Let $a$ be an element of $[S^{-1} A: S^{-1} A']$. Since $A'$ is finite over $A$, let we can write $A' = A[f_1,\dotsc,f_n]$, for some fractions $f_i$'s. By definition of the conductor ideal $a f_i \in S^{-1} A$, let write it as $a_i/s_i$. Let $s$ be the product $\prod_i s_i$. It is clear that for all $i$ we have $saf_i\in A$, and thus $sa A'\subset A$, i.e. $sa\in I$, hence $a\in S^{-1} A$.

Since $S^{-1} A' = S^{-1} A$ if and only if the corresponding conductor ideal contains $1$, the problem is almost solved. Indeed $S^{-1} I$ contains $1$ is and only if $S$ and $I$ have non zero intersection.

Hence, we have proved the following :

A localisation $A_P$ at a prime ideal $P$ (assuming that $A'$ is finitely generated over $A$) is integrally closed if and only if $I$ is not included in $P$.

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  • $\begingroup$ Thanks. A few minor points: The conductor $[A' : A]$ is usally written as $(A : A')$. Let $a$ be an element of $[S^{-1} A' : S^{-1} A]$. I prefer using $\alpha$ for example instead of $a$ to distinguish $\alpha$ from an element of $A$. hence $a\in S^{-1} A$. should be hence $a\in S^{-1} I$. $\endgroup$ – Makoto Kato Jun 15 '12 at 18:49

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