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In Milne Prop 2.29, it is said that the integral closure $B$ of a PID $A$ in a separable finite extension of its fraction field is a free $A$-module. On the other hand, I have read here that if the base ring is a complete DVR, $\mathrm{Frac}(B)$ need not be separable over $\mathrm{Frac}(A)$ for $B$ to be finitely generated over $A$ (although I would very much like to see a reference for this), but my question is : is it still a free $A$-module ?

The question in the title is a little more restrictive (although not much), but is what I'm really interested in.

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    What is "separated"? Is it the same thing as a separable extension? And just to be clear, you mean $B$ is a free $A$-module?2012-08-31
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    Indeed, thanks for pointing that out. I'll change it immediately2012-08-31

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By the theory of Nagata rings (or Japanese rings), the integral closure of $k[[t]]$ in a finite extension of $k((t))$ will be f.g. It will also be torsion-free (pretty obviously) and so will be free (since $k[[t]]$ is a DVR, and f.g. torsion-free modules over a DVR are always free).

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    Thanks for this answer, it helps very much ! Could you give me a reference for freeness of f.g. torsion-free modules over a DVR (I haven't done much commutative algebra before)?2012-08-31
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    In any (good) elementary book in abstract algebra you can find the following theorem: Over a principal ideal domain a finitely-generated torsion-free module is free. If you are too lazy to check one, then look at http://www.math.umn.edu/~garrett/m/algebra/notes/11.pdf2012-08-31
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    @zozoens: Dear zozoens, This is a direct consequence of the classification of f.g. modules over PIDs, which you can find in most algebra texts, probably on wikipedia, etc. (or see the link in the comment above). Regards,2012-08-31