Finite morphisms over $\operatorname{Spec} O_K$

Question

the following question is closely related to this one.

Suppose that $X$ is an integral, Noetherian scheme with a finite and dominant morphism $$f: X\to \operatorname{Spec} O_K$$ where $K$ is a number field and $O_K$ is its ring of integers.

Is there any classification theorem for $X$? My conjecture is the following:

1. If $X$ is regular, then $X=\operatorname{Spec} O_L$ for a finite extension $L|K$
2. If $X$ is not regular then it is a spectrum of an order (not maximal) of $L$. Here $L$ is still a finite extension of $K$.

Is it true?

In the linked question the OP accepted an incomplete answer, and now I'm curious.

Edit: As suggested in the comments, the question can be rephrased in the following way. Let $A$ be a finite $O_K$-algebra wich is also an integral domain and such that $\operatorname{Frac}(A)=L$. Is it true that $A$ is an order of $L$? Of course we have that $A$ is integral over $\mathbb Z$, so $A\subseteq O_L$. It remains to show that $A$ contains an integral $\mathbb Z$-basis of length $[L:\mathbb Q]$.

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Let's recall the definition of order:

An order of $L$ is a subring of $O_L$ containing a $\mathbb Z$-basis of length $[L:\mathbb Q]$.

Answer 1

As you have remarked yourself already, there is no need to use the language of schemes to formulate this problem. Finite morphisms are affine, so $A = \mathcal O_X(X)$ is a finite $\mathcal O_K$-algebra. Since you assumed that $f$ is dominant, the map $O_K\to A$ is injective (use that $O_K$ is a reduced ring).

You seem to be aware of the fact that $O_K$ is a free $\mathbb Z$-module of finite rank. The composition of two finite ring homomorphisms is still finite, so $\mathbb Z\to A$ is finite (in particular $A$ is a noetherian ring, so there is no need to put this in the hypothesis). Since the underlying abelian group of $A$ cannot contain non-trivial torsion elements (it is an integral domain by your assumption), it is free of finite rank (say $n$).

Now is easy to see that $L:=\mathrm{Frac}\, A$ is a number field of dimension $n$ over $\mathbb Q$, and $A$ is an order in $L$. One could observe for example that the natural map $\mathbb Q\otimes_{\mathbb Z} A \to L$ is an isomorphism.

[The ring $\mathbb Q\otimes_{\mathbb Z} A$ is a localization of $A$ and its dimension over $\mathbb Q$ is equal to $n$. Since $A$ is integral this localization can be identified with a subring of its fraction field; but any finite-dimensional domain is a field so this subring is in fact equal to the fraction field.]

Among all orders in $L$ there is a unique one that is integrally closed, it is $O_L$ (in fact, for any subring $R$ of a number field $L$ its normalization is given by $RO_L$). Considering the fact that, in this situation, being regular is equivalent with being normal, you will see directly that both of your statements are true (in general being regular is a stronger condition than being normal, but $X = \mathrm{Spec}\, A$ is 1-dimensional).