2
$\begingroup$

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.

  • 0
    Indeed, thanks for pointing that out. I'll change it immediately2012-08-31

1 Answers 1

2

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).

  • 0
    @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