7
$\begingroup$

The valuative criterion for separatedness (resp. properness) says that a morphism of schemes (resp. a quasi-compact morphism of schemes) f:X→Y is separated (resp. proper) if and only if it satisfies the following criterion:

For any valuation ring R (with K=Frac(R)) and any morphisms Spec(R)→Y and Spec(K)→X making the following square commute

Spec(K) ---> X
  |          |
  |          | f
  v          v
Spec(R) ---> Y

there exists at most one (resp. exactly one) morphism Spec(R)→X filling in the diagram.

But if Y is locally noetherian and f:X→Y is of finite type, then this condition only needs to be verified for discrete valuation rings. Does anybody know of an example where it is not sufficient to use DVRs? In other words, does there exist a morphism of schemes f:X→Y which is not separated (resp. proper), but does satisfy the valuative criterion for DVRs?

Reference: EGA II, Proposition 7.2.3 and Theorem 7.3.8

$\endgroup$
10
$\begingroup$

You can probably just take Y to be the spectrum of a valuation ring A which is not a DVR, for example the integral closure of C[[t]] in an algebraic closure of C((t)). In this case any homomorphism from A to a DVR R has to factor through the quotient field or the residue field of A.

For an explcit example, Let X be two copies of Spec(A) glued along the complement of the closed point and let X --> Y be the map which is the identity on both copies. The fibres of this map are both proper so it satisfies the valuative criterion using only DVRs by the observation above but the map itself fails to be proper.

$\endgroup$
  • $\begingroup$ Very nice. The thing I didn't realize is that a map from a non-discrete valuation ring to a discrete valuation ring has to factor through the residue field. $\endgroup$ – Anton Geraschenko Oct 19 '09 at 5:56
  • $\begingroup$ unknown -- you should log in and claim all that reputation you're collecting! $\endgroup$ – Scott Morrison Oct 26 '09 at 16:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.