Tag 01RC
Chapter 28: Morphisms of Schemes > Section 28.7: Scheme theoretic closure and density
Example 28.7.2. Here is an example where scheme theoretic closure being $X$ does not imply dense for the underlying topological spaces. Let $k$ be a field. Set $A = k[x, z_1, z_2, \ldots]/(x^n z_n)$ Set $I = (z_1, z_2, \ldots) \subset A$. Consider the affine scheme $X = \mathop{\mathrm{Spec}}(A)$ and the open subscheme $U = X \setminus V(I)$. Since $A \to \prod_n A_{z_n}$ is injective we see that the scheme theoretic closure of $U$ is $X$. Consider the morphism $X \to \mathop{\mathrm{Spec}}(k[x])$. This morphism is surjective (set all $z_n = 0$ to see this). But the restriction of this morphism to $U$ is not surjective because it maps to the point $x = 0$. Hence $U$ cannot be topologically dense in $X$.
The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 1022–1038 (see updates for more information).
\begin{example}
\label{example-scheme-theretically-dense-not-dense}
Here is an example where scheme theoretic closure being $X$ does not
imply dense for the underlying topological spaces.
Let $k$ be a field.
Set $A = k[x, z_1, z_2, \ldots]/(x^n z_n)$
Set $I = (z_1, z_2, \ldots) \subset A$.
Consider the affine scheme $X = \Spec(A)$ and the
open subscheme $U = X \setminus V(I)$.
Since $A \to \prod_n A_{z_n}$ is injective we see that the scheme theoretic
closure of $U$ is $X$. Consider the morphism
$X \to \Spec(k[x])$. This morphism is surjective
(set all $z_n = 0$ to see this). But the restriction
of this morphism to $U$ is not surjective because it maps
to the point $x = 0$. Hence $U$ cannot be topologically dense
in $X$.
\end{example}Comments (0)
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