I am trying to prove Corollary 4.6(f) of Hartshorne Chapter 2.4 using the valuative criterion for separated morphisms:
Suppose $f:X\to Y$ is a morphism between Noetherian schemes, and suppose there is an open cover $\{ V_i \}$ of $Y$ such that $f^{-1}(V_i)\to V_i$ is separated for each $i$. Show that $f$ is separated.
Suppose we have a commutative diagram of the form in the valuative criterion, with morphisms $\phi,\psi:\operatorname{Spec}R\to X$, where $R$ is a valuation ring. Let $x_1$ is the point corresponding to the zero ideal in $R$, and $x_0$ the point corresponding to the maximal ideal, where we have $x_0\in\overline{\{x_1\}}$ is a specialization. Suppose $V\subset Y$ is an open affine – is it true that if $f(x_1)\in V$, then $f(x_0)\in V$? If it were true, then the proposition would follow immediately by just considering any of the open subsets in the cover that contain the image of $x_1$, and I have seen a source do this to prove it, but I do not know how or why this is true.
Best Answer
(I write generalization instead of generization because I like it better.)
You have this backwards - the appropriate statement is that if $x$ lies in an open set $U$, then every generalization of $x$ lies in $U$ as well. The reason is just point set topology: if $y$ is a generalization of $x$ not lying in $U$, then $U^c$ is a closed set containing $y$ not containing $x$, so $x\notin\overline{\{y\}}$ by the definition of the closure as the smallest closed set containing $y$. This contradicts the definition of generalization, so $y$ must be in $U$.