Is the ring of integers in a relative algebraic number field faithfully flat over a ground ring?

Let $L$ be a finite extension of an algebraic number field $K$. Let $A$ and $B$ be the rings of integers in $K$ and $L$ respectively. Is $B$ faithfully flat over $A$? What if $L$ is an infinite algebraic extension of $K$?

Solution 1:

A module over a Dedekind domain is flat if and only if it is torsion-free. So any extension of rings $A\rightarrow B$ with $A$ Dedekind and $B$ a domain is flat. In particular, if the extension is integral, it is faithfully flat (by lying-over, as you point out in your comment).

Solution 2:

Yes, because $B$ is flat over $A$ (since it is locally free) and every prime ideal of $A$ is the contraction of a prime ideal of $B$.

I don't know what happens if $L$ is not finite over $K$.