Let $f : X \rightarrow Y$ be a finite type morphism of Noetherian schemes. The valuative criterion for properness runs as follows. Suppose that for any DVR $R$ with fraction field $K$ that any $K$-valued point of $X$ lying above an $R$-valued point of $Y$ extends uniquely to an $R$-valued point of $X$. Then $f$ is proper.

Does it suffice to check instead that any $K$-valued point of $X$ lying above an $R$-valued point of $Y$ extends to an $R'$-valued point of $X$, where $R'$ is the integral closure of $R$ in a finite extension $K'$ of $K$?


1 Answer 1



I claim that, for any $K'$ point of $X$, if this point extends both to an $R'$ point and an $K$ point, then it extends to a $R$ point. This obviously proves the result.

Let $x'$ be the closed point of $\mathrm{Spec}(R')$. Let $\mathrm{Spec}(A)$ be an affine neighborhood of the image of $x'$. Then the $R'$ point must factor through $\mathrm{Spec}(A)$. The $K$ point also factors through $\mathrm{Spec}(A)$, as it has the same image as the $K'$ point.

We can now transform the question into algebra. The algebraic statement is that we have a map $A \to K'$ which factors through both $K$ and $R'$. But then the image of this map must lie in $K \cap R' = R$. So the map factors through $\mathrm{Spec}(R)$, and our $K$ point extends to an $R$ point.

  • 1
    $\begingroup$ Should be "extends to an R point" in the 2nd paragraph, I think. $\endgroup$
    – TSG
    Nov 8, 2009 at 22:52
  • $\begingroup$ A problem with this argument is that R' is not local in general. Of course this will hold if R is henselian. In general, to make the argument work one probably ought to make an extra hypothesis to ensure that any finite set of points of a fibre of f is contained in an open subset (e.g. that f is quasiprojective). Or one can reduce to the henselian case by the same argument, and make no hypothesis. $\endgroup$
    – Bruno Kahn
    Jun 28, 2022 at 15:46

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