What is a requirement for an order of algebraic number field $K$ to be integrally closed domain?

Question

Suppose there is an order $O$, a subring, of an algebraic number field $K$. What is needed (necessary and sufficient condition) for $O$ to be integrally closed domain? Or if we need to impose restrictions also to $K$, what is required for $K$ to ensure $O$ to be integrally closed?

Answer 1

In standard terminology, an order is a subring $O$ of $K$ so that $O$ is a free $\mathbb{Z}$ module of rank $[K \colon \mathbb{Q}]$ -- by this $\mathbb{Z}$ is not considered an order of $K$. For this, it's enough to check that $O$ is finitely generated as a $\mathbb{Z}$ module and $\mathbb{Q} \cdot O = K$.

There is only one order that is integrally closed. It is the integral closure of $\mathbb{Z}$ in $K$. This order, also called the maximal order (since it's the largest one in fact!). Usually it's denoted by $O_K$.

You might look at more general things: subrings $O$ of $K$ that are finitely generated $\mathbb{Z}$ modules. If you want them integrally closed then they will be exactly the maximal order in the field of fractions generated by them ( which equals $\mathbb{Q}\cdot O$). Such rings are in $1-1$ correspondence with subfields of $K$.