Jacobson's Basic Algebra, volume II (second edition) has:
Definition. A domain $D$ is called a Dedekind domain if every $D$-fractional ideal of $F$ (the field of fractions of $D$) is invertible.
Theorem. (Theorem 10.6, page 630) Let $D$ be a domain. The following are equivalent:
- $D$ is Dedekind.
- $D$ is integrally closed, noetherian, and every nonzero prime ideal of $D$ is maximal.
- Every proper integral ideal of $D$ can be written in one and only one way as a product of prime ideals.
- $D$ is noetherian and $D_P$ is a discrete valuation ring for every maximal ideal $P$ in $D$.
The proof that 3 implies 1 is credited to Zariski and Samuel:
First, every invertible prime $P$ of $D$ is maximal in $D$; let $a\in D-P$, and assume that $aD+P\neq D$; then $aD+P$ can be written as a product of primes,
$$aD+P = P_1\cdots P_m$$
and likewise, $a^2D+P\neq D$ so
$$a^2D+P = Q_1\cdots Q_n$$
Since $P$ is property contained in both $aD+P$ and in $a^2D+P$, then $P$ is properly contained in each of the $P_i$ and each of the $Q_j$. Now look at $D/P$, which is also a domain, and in which $\overline{aD} = \overline{P_1}\cdots\overline{P_m}$ and $\overline{a^2D}=\overline{Q_1}\cdots \overline{Q_n}$. The principal ideals $\overline{aD}$ and $\overline{a^2D}$ are invertible, and $\overline{P_i}$ and $\overline{Q_j}$ are prime ideals. Since $\overline{a^2D} = \overline{a}^2\overline{D}=(\overline{aD})^2 = \overline{P_1}^2\cdots\overline{P_n}^2$, and since factorizations for invertible integral ideals are unique, $\{\overline{P_1},\overline{P_1},\ldots,\overline{P_n},\overline{P_n}\}$ is the same sequence as $\{\overline{Q_1},\ldots,\overline{Q_m}\}$, except perhaps for the order. Hence, the corresponding sequences of ideals of $D$ that contain $P$, $\{P_1,P_1,\ldots,P_n,P_n\}$ and $\{Q_1,\ldots,Q_m\}$ are the same up to order. Hence $(aD+P)^2 = a^2D+P$, so
$$P\subseteq (aD+P)^2 = a^2D+aP+P^2 \subseteq aD+P^2.$$
So if $p\in P$, then $p=ax+y$ with $x\in D$ and $y\in P^2$; hence $ax\in P$, and since $a\notin D$, then $x\in P$. So $P\subseteq aP+P^2\subseteq P$, so $P=aP+P^2$. Since we are assuming $P$ is invertible, then $D = P^{-1}P = P^{-1}(aP+P^2) = aD+P$, a contradiction.
So $aD+P = D$ for every $a\notin P$, hence $P$ is maximal.
Now let $P$ be any nonzero prime ideal, and let $b\neq 0$ be an element of $P$. So $bD\subseteq P$ and $bD=P_1P_2\cdots P_m$ where the $P_i$ are prime and are invertible. So the $P_i$ are maximal. Since $P$ is prime, $P_i\subseteq P$ for some $i$; hence $P=P_i$, so $P$ is invertible.
Since every proper ideal is a product of primes, and every prime is invertible, it follows that every integral ideal is invertible.
Given a fractional ideal $I'=aI$, with $I$ integral and $a\neq 0$, we have $I'(I')^{-1}=II^{-1}=D$, so every fractional ideal is invertible. $\Box$
$\newcommand{\cO}{\mathcal{O}}$
$\newcommand{\cS}{\mathcal{S}}$
I'm also not super experienced, so do let me know if something here is unclear or seems wrong!
As you said, we need to show that $\cO_{K,\cS}$ is an integrally closed Noetherian domain of Krull dimension (length of longest chain of prime ideals) one.
- Suppose $\alpha$ is integral over $\cO_{K,\cS}$ and satisfies the relation
$\alpha^d + a_{d-1}\alpha^{d-1} + \dots a_0 = 0$ with coefficients $a_i \in \cO_{K,\cS}$. Then, for any valuation $v$ corresponding to a prime $\mathfrak{p}$,
$$dv(\alpha) = v(a_{d-1}\alpha^{d-1} + \cdots a_0) \geq \min((d-1)v(\alpha)+v(a_{d-1}), \dots , v(a_0))$$
If $\mathfrak{p} \notin \cS$, then $v(a_i) \geq 0$ for all $i$, so it cannot be the case that $v(\alpha) < 0$. Thus, $\alpha \in \cO_{K,\cS}$, meaning that $\cO_{K,\cS}$ is integrally closed in its fraction field $K$.
As TokenToucan pointed out in the comments, the localization of a Noetherian ring is Noetherian, and $\cO_{K,S}$ is the localization with respect to the set of elements outside $\mathfrak{p}$ for all $\mathfrak{p} \notin \cS$ (if you don't see how to go from the valuation condition to ratios of elements check out this answer). Alternately, an overkill way is to observe that the Krull-Akizuki Theorem tells us that subrings of the fraction field of a one-dimensional reduced (no nonzero nilpotents) Noetherian ring are also Noetherian. In our case, $\cO_K$ is one-dimensional and an integral domain and $\cO_{K,\cS}$ contains $\cO_K$ and is contained within $K$, the fraction field of $\cO_K$.
This follows from the Krull-Akizuki theorem, as it actually guarantees that (proper) subrings of $K$ containing $\cO_K$ are Noetherian and one-dimensional.
Putting this all together, unless I messed up, we have the desired result.
Best Answer
Addressing your questions in order:
For knowing that the norm lives in $\mathbb{Z}\cap P$, note that it lives in $P$ since one of its factors lives in $P$, and $P$ is an ideal, and lives in $Z$ since its an algebraic integer, that lives in the image of the norm map, which is $\mathbb{Q}$ (one way to see this is to view the norm via matrices, we are taking the determinant to the underlying field).
The claim that $pS$ is contained in $P$ follows since we know that $P\cap \mathbb{Z}=p\mathbb{Z}$, so in particular, $p$ is in $P$, so since $P$ is an ideal of $S$, $pS$ is in $P$.
Then finally, the claim that $S/pS$ is $n$ dimensional is because we knew $S$ is free of rank $n$ as a $\mathbb{Z}$ module, and we can check that any basis descends to a basis once we reduce mod $p$. Perhaps easier is to convince yourself that any generating set of $S$ over $\mathbb{Z}$ will generate $S/pS$ and $\mathbb{Z}/p$ module, and then worry about the exact dimension of this vector space (since we really only needed finiteness to get the conclusion).
I would reccomend chapter $3$ of Atiyah Macdonald if you want to go over this, this is general module theory rather than anything particularly arithmetic.