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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.