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Consider $R$ a valuation ring and $X$ a scheme and a morphism $f:\operatorname{Spec}(R)\to X$. $k(\eta)$ denotes residue field at a point $\eta$. This $f$ is equivalent to prescription of $\eta\in X$ image of generic point of $\operatorname{Spec}(R)$ and $m\in X$ image of maximal ideal of $R$ s.t. $k(\eta)\subset \operatorname{Frac}(R)$ and $\overline{\eta}$ reduced scheme along with $O_{X,m}\to R_m$ where $R_m$ dominates over $O_{X,m}$.

$\textbf{Q:}$ There is no reason to expect $\overline{\eta}$(closure of $\eta$ in $X$) affine. I am looking for an example $\overline{\eta}$ being not affine. The examples I could imagine are almost all affine.(e.g. $k[x_1,\dots, x_n]\to k[[x_1]]$ by evaluation at $x_i=0$ for $i\neq 1$ and $x_1\to x_1$. This will show image of valuation ring spectrum is sitting in $V(x_2,\dots, x_n)$ which is already affine.)

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Let $X$ be a proper irreducible reduced scheme over a field $k$. Let $\eta_X$ be the generic point of $X$ and $k(\eta_X)$ be residue field corresponding to the generic point. Let $R = k(\eta_X)$, then there is an obvious map $Spec(R) \rightarrow X$. Clearly the generic point of $Spec(R)$ gets mapped to $\eta_X$ by construction and hence $\overline{\eta_X} = X$. Now, just observe that $R$ is a field and hence a valuation ring.

In this example one can also replace the field $k(\eta_X)$ by a valuation subring say $R$. Then since $X$ is proper there exists a map $Spec(R) \rightarrow X$, extending the natural map $Spec(k(\eta_X)) \rightarrow X$. Clearly the generic point of $Spec(R)$ is mapped to $\eta_X$ and $\overline{\eta_X} = X$ again by definition.

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  • $\begingroup$ I guess you mean I could take $Proj(k[x_0,x_1])$. Then there is a generic point whose closure is whole space. $\endgroup$ – user45765 Sep 19 at 22:23
  • $\begingroup$ @user45765 Yes. That is definitely a special case of the example we saw in the answer above. $\endgroup$ – random123 Sep 20 at 2:08

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