Is ring of $S$-integers a Dedekind domain? For $\mathcal{O}_{K}$, the integer ring of a global field, we denote $S$ to be any set of primes of a global field $K.$ Let
 $$\mathcal{O}_{K,S}:=\{x\in K\mid v_{\mathfrak{p}}\geq 0\text{ for }\mathfrak{p}\notin S\}$$ be the ring of $S$-integers of $K$ (see Neukirch, Schmidt, Wingberg Cohomology of Number Fields, Ch. VIII, § 3).
A ring $R$ is Dedekind iff $R$ is a Noetherian integrally closed domain where the height of its prime ideals is $1.$
The Prop.1 in Bourbaki Commutative Algebra, Ch.VII.1 tells us that if $$S=\{\text{all primes of }K\}-\{\text{finitely many primes}\},$$ then $\mathcal{O}_{K,S}$ is Dedekind.
I guess $\mathcal{O}_{K,S}$ is not a Dedekind domain except for some special $K$ if $S$ is finite.
But I failed in a clear proof that $\mathcal{O}_{K,S}$ is Dedekind or not for $S$ to be finite. 
Due to my poor knowledge in commutative algebra, it will great if there will be an explicit proof.
Please forgive me for the long description.
 A: $\newcommand{\cO}{\mathcal{O}}$
$\newcommand{\cS}{\mathcal{S}}$
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.
