# How did the book conclude specialization of points?

This is related to Ueno's Algebraic Geometry 2, Thm 5.9 Valuative Criterion proof.

Let $$f:X\to Y$$ be a finite type morphism between $$X$$ noetherian scheme and $$Y$$ scheme. Assume $$f$$ is proper and hence $$f$$ is separated.

I want to see for any valuation ring $$R$$ with $$Frac(R)=K$$ s.t. given commutative diagram $$Spec(R)\to Y$$ and $$Spec(K)\to X$$ inducing unique $$Spec(R)\to X$$ lifting s.t. overall triangles of diagram commute.

Suppose $$g,g'$$ are 2 such liftings. Let $$\eta_0,\eta'_0$$ be closed points of $$Spec(R)$$ and $$\zeta_0,\zeta'_0$$ be corresponding images under lift $$g,g'$$ to $$X$$. It is clear that diagonal map induced by $$f:X\to Y$$ for $$X\to X\times_YX$$ is closed by $$f$$ separated. Note that $$Spec(K)\to Spec(R)$$ is identifying the generic point and $$g,g'$$ agrees over image of $$Spec(K)$$ map to $$X$$. Now $$Spec(R)\to X\times_YX$$ will have generic point image lying on diagonal. It suffices to check closed point $$Spec(R)$$ is on diagonal as well. Note that image of $$X$$ under diagonal map is closed. Since $$X$$ is noetherian scheme, $$X\to X\times_YX$$ has closed image iff $$Im(X\to X\times_YX)$$ is closed under specialization. If that is the case, then $$g=g'$$. Denote $$\eta_1$$ as image of $$Spec(K)\to X$$.

$$\textbf{Q:}$$ Why is image of $$(\eta_0,\eta'_0)\in\overline{(\eta_1,\eta_1)}\in X\times_YX$$? In other words, why $$(\eta_0,\eta'_0)$$ is specialization of $$(\eta_1,\eta_1)$$? It seems that the book is using continuity of the map $$Spec(R)\to X\times_YX$$ to conclude $$Im\subset\overline{(\eta_1,\eta_1)}$$. In other words, consider continuous map $$h:X\to Y$$ and say $$x\in X$$ I have $$h(\overline{x})\subset \overline{h(x)}$$

First note that any valuation ring $$R$$ is local, with maximal ideal $$\mathfrak{m} = \{r \in R | v(r) > 0\}$$. So $$\operatorname{Spec}(R)$$ has only one closed point, i.e. $$\eta_0 = \eta_0'$$.
Then $$f^{-1}(\overline{(\eta_1, \eta_1)}) \subset \operatorname{Spec }(R)$$ is a closed set. Any closed set in an affine scheme contains closed points, but there is only one closed point in $$\operatorname{Spec }(R)$$, namely $$\eta_0$$. But $$\eta_0 \in f^{-1}(\overline{(\eta_1, \eta_1)})$$ if and only if $$(\eta_0, \eta_0) = f(\eta_0) \in \overline{(\eta_1, \eta_1)}$$.
• To me, it looks like the book is saying the following. Say $\eta_1$ is generic point of $R$. Then $f(\overline{\eta_1})\subset \overline{f(\eta_1)}=\overline{(\eta_1,\eta_1)}$ where first inclusion is given by continuity which is valid in all topological spaces continuous map sense. – user45765 Sep 21 '19 at 15:19