Dedekind Domains
A Dedekind domain is an integral domain $R$ such that

Every ideal is finitely generated.

Every nonzero prime ideal is a maximal ideal.

$R$ is integrally closed in its field of fractions.
The last condition means that if $\alpha/\beta \in K$ is a root of a monic polynomial over $R$, then $\alpha/\beta \in R$, that is, $\beta  \alpha$ in $R$.
The first condition is equivalent to both of the following:

Every increasing sequence of ideals is eventually constant, that is, given $I_1 \subset I_2 \subset I_3 ...$, there exists some $k$ for which $I_k = I_n$ for all $n \ge k$.

Every nonempty set $S$ of ideals has a maximal member, that is, there exists $M \in S$ such that $M\subset I \in S \implies M = I$.
A ring satisfying these conditions is called a Noetherian ring.
It is not difficult to show that these three conditions are equivalent. First suppose we are given some sequence $I_1 \subset I_2 \subset I_3 ...$. Then consider the ideal generated by all the ideals in the sequence. If it is finitely generated, then for some $k$, $I_k$ contains all the generators, and thus $I_k = I_n$ for all $n \ge k$.
Now suppose we are given a nonempty set $S$ of ideals. Then take any ideal $I_1 \in S$, and look for an ideal $I_2$ that strictly contains $I_1$. Iterating this procedure produces a sequence $I_1 \subset I_2 \subset ...$. If there exists some $k$ for which $n \ge k$ implies $I_k = I_n$, then $I_k$ is a maximal ideal in $S$.
Next suppose we are given some ideal $I$. Consider the set $S$ of subideals of $I$ that are finitely generated. If it has a maximal element $I'$, then we must have $I = I'$.
Theorem: Every number ring is a Dedekind domain.
Proof: Since a number ring is a free abelian group of finite rank, any ideal must also be a free abelian group of finite rank (because it is a additive subgroup) thus every ideal is finitely generated.
For the second condition, it suffices to show that for every nonzero prime ideal $P$, the integral domain $R/P$ is a field. We do so by showing $R/P$ is finite, which implies it is a field because every finite integral domain is a field (because for any element $a$, we must have $a^n = 1$ for some $n$, thus $a^{n1}$ is the inverse of $a$).
We shall show that $R/I$ is finite for any nonzero ideal $I$. Take any nonzero $\alpha\in I$, and let $m = N^K(\alpha)$ where $K$ is the number field of $R$. So $m = \alpha\beta$ where $\beta$ is the product of the conjugates of $\alpha$. Hence $\beta$ is also be an algebraic integer. Now $m$ is a nonzero integer, so $\beta= m /\alpha \in K$, so we deduce $\beta\in R$ which means $m \in I$. Now $R/(m)$ is finite (it contains $m^n$ elements), hence $R/I$ is finite (its order divides $m^n$).
Lastly, suppose $\alpha$ is a root of a monic polynomial $f(x) = a_0 +...+ a_{n1}x^{n1} +x^n$ over $R$. Now $R=\mathbb{Z}[a_0,...,a_{n1}]$ is clearly finitely generated, and since $\mathbb{Z}[a_0,...,a_{n1},\alpha]$ is also finitely generated since any expression containing $\alpha^m$ for $m \ge n$ can be rewritten using smaller powers of $\alpha$. Hence $R$ is integrally closed.