Let $R$ be a DVR and $K$ its fraction field. Let $X$ be a proper integral separated scheme of finite type over $Spec(K)$. Does it always exists a proper integral separated scheme $Y$ of finite type over $Spec(R)$, such that its generic fiber is isomorphic to $X$?

Of course, if the variety is projective, the answer is yes. But I would like to know if there are proper non-projective varieties that do not admit a proper integral model as before.

  • 1
    $\begingroup$ proper includes separated and finite-type already. Clearly we can do this for affine schemes. I'm trying to figure out if we can glue these together. $\endgroup$
    – Will Sawin
    Commented Jun 5, 2012 at 3:01
  • $\begingroup$ @Will: for proper replaced by affine, you can just take $Y=X$. $\endgroup$
    – user2035
    Commented Jun 5, 2012 at 7:12

1 Answer 1


From Nagata's compactification theorem, we can find a proper map $\overline{X} \to \mathop{\rm Spec}R$, containing $X$ as an open dense subscheme; then it is immediate to check that the generic fiber must be $X$.


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