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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?

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    $\begingroup$ The generic point is usually not open. Think of an algebraic curve, for example. $\endgroup$
    – Zhen Lin
    Jul 20 '14 at 22:51
  • $\begingroup$ Dear @ZhenLin, KeenanKidwell, thank you both for the clarification. $\endgroup$ Jul 20 '14 at 22:57
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    $\begingroup$ Dear @Keenan: but yes, the scheme morphism $\operatorname {Spec}(k(x))\to X$ is a closed immersion if $x$ is closed. Indeed it is closed, injective and $\mathcal O_{X,x}\to k(x)$ is surjective, cf. EGA I, Proposition (4.2.2.b). Am I misunderstanding you? $\endgroup$ Jul 20 '14 at 23:15
  • $\begingroup$ Dear @Georges, No, you're not misunderstanding me. I'm just wrong! Thank you for the correction! $\endgroup$ Jul 20 '14 at 23:17
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    $\begingroup$ Dear @jeffrey, As Georges kindly points out, my original comment was completely wrong. The morphism is a closed immersion for a closed point. My apologies! $\endgroup$ Jul 20 '14 at 23:19
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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.

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  • $\begingroup$ Dear @GeorgesElencwajg, thank you for the clarifying answer. Does it follow from (1), that an irreducible but non-reduced $k$-scheme of finite type does not have a generic point? Concerning (2): Your answer lets me guess that open points are some trivial exceptions. I don't quite get your example of an open point. You take $X=Spec(L\times L)$ for $L$ a finite extension of $k$ and...? Thank you $\endgroup$ Jul 21 '14 at 0:00
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    $\begingroup$ Dear jeffrey: 1) Ha, but yes every irreducible scheme does have a generic point, namely the one corresponding to the integral subscheme $Y_{\operatorname {red}}\subset Y$. This is a bit confusing and I'll write an Edit about this. 2) A typical scheme with open point is $X=Y\sqcup \operatorname {Spec}(k)$ where $Y$ is an arbitrary $k$-scheme . $\endgroup$ Jul 21 '14 at 0:41
  • $\begingroup$ Dear @GeorgesElencwajg, thank you very much. I certainly understand the situation much better now. $\endgroup$ Jul 21 '14 at 8:56
  • $\begingroup$ Dear jeffrey, you are welcome. $\endgroup$ Jul 21 '14 at 9:10
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    $\begingroup$ Dear jeffrey, only the first definition is correct. Indeed, every point satisfies the second definition: this is the content of part 1) of my answer! (The third definition is the characterization of a closed point: exactly the opposite of generic...) $\endgroup$ Jul 21 '14 at 11:16

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