# Looking for an example for $R$ valuation ring, $X$ scheme s.t. $\operatorname{Spec}(R)\to X$ image is not affine

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.)

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.
• I guess you mean I could take $Proj(k[x_0,x_1])$. Then there is a generic point whose closure is whole space. – user45765 Sep 19 at 22:23