Skip to main content
added 303 characters in body
Source Link
user717
  • 5.2k
  • 2
  • 38
  • 41

If $k$ is algebraically closed and $X$ is a $k$-scheme locally of finite type, then the $k$-rational points are precisely the closed points. (See EGA 1971, Ch. I, Corollaire 6.5.3).

More generally: if $k$ is a field and $X$ is a $k$-scheme locally of finite type, then $X$ is a Jacobson scheme (i.e. it is quasi-isomorphic to its underlying ultrascheme) and the closed points are precisely the points $x \in X$ such that $\kappa(x)|k$ is a finite extension.

You should also confer the appendix of EGA 1971. There it is shown that for any field $k$ the category of $k$-schemes locally of finite type with morphisms locally of finite type is equivalent to the category of $k$-ultraschemes (a $k$-ultrascheme is locally the maximal spectrum of a $k$-algebra).

If $k$ is algebraically closed and $X$ is a $k$-scheme locally of finite type, then the $k$-rational points are precisely the closed points. (See EGA 1971, Ch. I, Corollaire 6.5.3).

More generally: if $k$ is a field and $X$ is a $k$-scheme locally of finite type, then $X$ is a Jacobson scheme (i.e. it is quasi-isomorphic to its underlying ultrascheme) and the closed points are precisely the points $x \in X$ such that $\kappa(x)|k$ is a finite extension.

If $k$ is algebraically closed and $X$ is a $k$-scheme locally of finite type, then the $k$-rational points are precisely the closed points. (See EGA 1971, Ch. I, Corollaire 6.5.3).

More generally: if $k$ is a field and $X$ is a $k$-scheme locally of finite type, then $X$ is a Jacobson scheme (i.e. it is quasi-isomorphic to its underlying ultrascheme) and the closed points are precisely the points $x \in X$ such that $\kappa(x)|k$ is a finite extension.

You should also confer the appendix of EGA 1971. There it is shown that for any field $k$ the category of $k$-schemes locally of finite type with morphisms locally of finite type is equivalent to the category of $k$-ultraschemes (a $k$-ultrascheme is locally the maximal spectrum of a $k$-algebra).

Source Link
user717
  • 5.2k
  • 2
  • 38
  • 41

If $k$ is algebraically closed and $X$ is a $k$-scheme locally of finite type, then the $k$-rational points are precisely the closed points. (See EGA 1971, Ch. I, Corollaire 6.5.3).

More generally: if $k$ is a field and $X$ is a $k$-scheme locally of finite type, then $X$ is a Jacobson scheme (i.e. it is quasi-isomorphic to its underlying ultrascheme) and the closed points are precisely the points $x \in X$ such that $\kappa(x)|k$ is a finite extension.