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$R$ is a number ring, $K$ the field of fractions and $R$ is Dedekind.

$\Rightarrow$ exists a set of primes $S$ of $\mathscr{O}_{K}$ and: $R=\bigcap\limits_{\mathbb{p} \notin S}\mathscr{O}_{\mathbb{p}}=\{x\in K:ord_{\mathbb{p}}(x)\geq0\forall\mathbb{p}\notin S\}$

I do not know how to begin here. I started with what this set is: all the $\mathbb{p}$ that are not in $S$ are hence not integral, and thus of the form $\frac{r}{p}$ for some $p$ prime, and hence the inverse to certain primes?

We had a Theorem for number rings that are Dedekind:

$R$ as above, then $\exists \phi: \mathcal{I}(R)\rightarrow \oplus_{\mathbb{p}}$ $\mathbb{Z}$ with $I \mapsto (ord_{\mathbb{p}}(I))_{\mathbb{p}}$, and for all $I = \prod\limits_{\mathbb{p}}\mathbb{p}^{ord_{\mathbb{p}}(I)}$

I'd be very happy about any kind of help :) All the Best, Luca

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When you say "number ring," what do you mean exactly? Presumably $K$ supposed to be a number field? If so, do you just mean a subring of $K$ containing the ring of integers of $K$? –  Keenan Kidwell Oct 13 '12 at 23:58
    
We defined it like this: "A number field is a finite field extension of the field of rational numbers $\mathbb{Q}$, and a number ring is a subring of a number field. This introduction shows how number rings arise naturally when solving equations in ordinary integers." –  Luca Oct 14 '12 at 0:13

1 Answer 1

Lemma 1 Let $A$ be a discrete valuation ring and $K$ be its field of fractions. Let $p$ be its maximal ideal. Let $B$ be a subring of $K$ such that $A \subset B$. Suppose $B$ is a local ring and $p = q \cap A$, where $q$ is its maximal ideal.

Then $A = B$.

Proof: Suppose $A \neq B$. There exists $b \in B - A$. Then $\frac{1}{b} \in p$. Hence $1/b \in q$. Since $b \in B$, $1 = b\frac{1}{b} \in q$. This is a contradiction. QED

Lemma 2 Let $K$ be a finite extension field of $\mathbb{Q}$, $\mathcal{O}$ the integral closure of $\mathbb{Z}$ in $K$. Let $R$ be a discrete valuation ring of $K$. Then there exists a nonzero prime ideal $p$ of $\mathcal{O}$ such that $R = \mathcal{O}_p$.

Proof: Let $m$ be the maximal ideal of $R$. Since $1 \in R$, $\mathbb{Z} \subset R$. Let $\alpha \in \mathcal{O}$. Since $\alpha$ is integral over $\mathbb{Z}$, it is integral over $R$. Since $R$ is a unique factorization domain, $R$ is integrally closed in $K$. Hence $\alpha \in R$. Hence $\mathcal{O} \subset R$. Let $p = m \cap \mathcal{O}$. Then $\mathcal{O}_p \subset R_m = R$. Suppose $p = 0$. Then $\mathcal{O}_p = K \subset R$. This is a contradiction. Hence $\mathcal{O}_p$ is a discrete valuation ring. Clearly $p\mathcal{O}_p = m \cap \mathcal{O}_p$. Hence $\mathcal{O}_p = R$ by Lemma 1. QED

Lemma 3. Let $A$ be an integral domain, $K$ its field of fractions. Then $A = \bigcap_m A_m$, where $m$ runs through all maximal ideals of $A$.

Proof: Clearly $A \subset \bigcap_m A_m$. Hence it suffices to prove the other inclusion.

Let $x \in \bigcap_m A_m$. Suppose $x$ does not belong to $A$. Let $I = \{a \in A\colon ax \in A\}$. Clearly $I$ is an ideal of $A$. Since $x$ does not belong to $A$, $I \neq A$. Hence there exista maximal ideal $m$ such that $I \subset m$. Since $x \in A_m$, there exis $s \in A - m$ such thjat $sx = 0$. This is a contradiction. Hence $x \in A$. Hence $\bigcap_m A_m \subset A$. QED

Theorem Let $K$ be a finite extension field of $\mathbb{Q}$. Let $\mathcal{O}$ be the integral closure of $\mathbb{Z}$ in $K$. Let $R$ be a subring of $K$ such that $K$ is its field of fractions. Suppose $R$ is Dedekind. Then there exists a set $T$ of nonzero prime ideals of $\mathcal{O}$ such that $R = \bigcap_{p\in T} \mathcal{O}_p$.

Proof: Let $m$ be a nonzero prime ideal of $R$. Since $R$ is Dedekind, $R_m$ is a discrete valuation ring. Hence the assertion follows immediately from Lemma 2 and Lemma 3. QED

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