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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 $K$?

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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).

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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 $K$.

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Thanks. In an infinite case, $B$ is a direct limit of flat subrings. So I think it is also flat. By the lying-over theorem, $B$ is faithfully flat. –  Makoto Kato Jun 14 '12 at 21:03

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