Number Ring, Dedekind = Intersection Ring of Integers by inverting primes $R$ 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
 A: 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
