What kind of points are there in a finite type $k$-scheme?

Question

Let $k$ be an arbitrary field and $X$ a $k$-scheme of finite type (i.e. a scheme with a finite cover of spectra of finitely generated $k$-algebras).

How can I think of the points $x\in X$? What does it mean intuitively for a point to be open, closed, or neither of both? What is a characterization of the points $x\in X$?

I think I found out a characterization of closed points. Please correct me, if there is something wrong.

• The closed points $x\in X$ can be characterized as

  • for the special case $X=\operatorname{Spec}(A)$ exactly the maximal ideals of $A$,
  • the $x\in X$ for which the canonical morphism $\operatorname{Spec}(k(x))\to \operatorname{Spec}(\mathcal{O}_{X,x})\to X$ is a closed immersion.
  • the $x\in X$ for which $k(x)$ is a finite field extension of $k$ (Zariki's Lemma).

My question is also about the open points $x\in X$ and the points which are neither closed nor open. I guess that a point $x\in X$ is open iff $\operatorname{Spec}(k(x))\to \operatorname{Spec}(\mathcal{O}_{X,x})\to X$ is an open immersion.

Of course, if $X=\operatorname{Spec(A)}$ is affine, the open points $x\in X$ should be the complements of $V(I)$ containing one element for some ideal $I\subseteq A$.

Do non-closed points correspond to certain subschemes of $X$ which are in a certain sense more that zero-dimensional? If yes, what does the (transcendence) degree of the field extension $k(x)/k$ say about the point $x$ and the subscheme? How is it reflected in the subvariety if the point is open? Short: How can I think of non-closed points?

I have the impression that $X$ may have at most one open point which is the (unique) generic point of $X$, if is exists, for example if $X$ is integral. Perhaps the neither closed nor open points $x\in X$ are (exactly?) the generic points of the irreducible closed subschemes of $X$ which are not $X$ itself? What does the (transcendence) degree of the field extension $k(x)/k$ say about the point $x$? If my guess about the irreducible closed subschemes corresponding to the points was right, is this transcendence degree the Krull dimension of that subvariety?

Answer 1

Yes, you essentially understand the situation : here is the classification of the points of a scheme $X$ of finite type over a (completely arbitrary) field $k$.

0) The closed points are the points $x\in X$ whose residue field $\kappa(x)$ is a finite extension of $k$, i.e. $[\kappa(x):k]\lt \infty$

1) The other points $y\in Y$ correspond bijectively to the integral subschemes $Y\subset X$ of positive dimension.
In this correspondence $Y$ is the closure of $\{y\}$ and $y$ is the unique generic point of $Y$.
Edit Note the subtle and potentially confusing point that if $X$ is irreducible but not reduced, it nevertheless has a generic point: the point $\eta$ corresponding to the integral subscheme $X_{\operatorname {red}}\subset X$ !

2) In the above correspondence the dimension of $Y$ is equal to the transcendence degree of the extension $\kappa(y)/k$ : $$ \operatorname {dim}(Y)=\operatorname {tr.deg. } (\kappa(y)/k) $$
3) If the scheme $X$ is connected of positive dimension it has no open points.
The only way to obtain open points in $X$ is to artificially take disjoint unions of schemes without open points like above with spectra $\operatorname {Spec}(L)$ of finite extensions $L$ of $k$.
Edit For example for any $k$-scheme $Y$ the scheme $X=Y\sqcup \operatorname {Spec}(k)$ has $\operatorname {Spec}(k)$ as an open point.