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By definition, a Zariski closed subset of $\operatorname{Spec}A$ is a set of the form $V(I) = \{P \in \operatorname{Spec}A \mid I \subset P \}$. What if we work in a ZF model where AC is violated? (See below the note for the question)

Note: previous edits were confused by my troubles with the terms closed point and generic point, so I removed them. Sorry :(

$\mathrm{Edit}^2$: What a non-closed point that does not contain closed points looks like? Or, in non-Zariski language, if a ring $A$ has a proper ideal $I$ that is not contained in any maximal ideal, what does $\operatorname{Spec} A$ look like near the corresponding point?

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    What do you mean by "non-generic points"? Do you mean that the closed set consists entirely of points which are dense in it (in other words, has the indiscrete topology)? Presumably you want it to have at least $2$ points.2012-09-27
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    I think that you meant $I\subseteq P$ in the definition of $V(I)$, no? Also if you write the definition of a non-generic point, I might be able to give you an answer.2012-09-27
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    @Asaf A generic point in a scheme is a point whose closure is an irreducible component.2012-09-27
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    @Keenan: So a non-generic point would be a [prime?] ideal such that what exactly? Its closure is reducible, I suppose. Not being fluent in the language of Zariski topology, what would that mean in the language of ideals?2012-09-27
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    @Asaf, I'm sorry, I should have said the generic points of $\mathrm{Spec}(A)$ are exactly the minimal prime ideals. This is because irreducible components are (by definition) maximal irreducible subsets (which exist by Zorn!), and the assignment $\mathfrak{p}\mapsto V(\mathfrak{p})$ is an inclusion-reversing bijection between primes and irreducible closed subsets of $\mathrm{Spec}(A)$ ($V(\mathfrak{p})$ is exactly the closure of the point $\mathfrak{p}$ in $\mathrm{Spec}(A)$). So a prime is not a generic point if it properly contains some other prime.2012-09-27
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    The closure of an irreducible subset $Y$ of a space $X$ is always irreducible, so in particular the closure of a point is irreducible. Schemes are "sober" topological spaces, meaning every irreducible closed subset has a unique point whose closure is that subset.2012-09-27
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    @Keenan: Thank you very much for the detailed explanation. Sadly, I'm still not following the exact meaning of the question. Perhaps due to the double-negation, we want some $V(I)\neq\varnothing$ such that there is no non-generic points in $V(I)$. So $V(I)$ should be pretty much a single point? I don't see how the axiom of choice comes in here. Or am I reading the question wrong?2012-09-27
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    @Asaf, I'm not sure I understand it either. Issues involving AC are not really my thing. It seems to me the OP is asking whether, when AC doesn't hold, there is a closed subset $V(I)$ of a scheme such that every point in $V(I)$ is dense in $V(I)$. If the proof that schemes are sober works without AC (which I think it does, but again, I don't really think about that), then this is equivalent to $V(I)$ being a point. It seems to me the $\mathrm{Spec}(k)$ works with $k$ a field, but perhaps I misunderstand.2012-09-27
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    Dear Alexei, in very standard algebraic geometry the spectrum of any field (with itself as closed subset) has the property you require...2012-09-27
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    @Georges: Isn't that a trivial spectrum (i.e. a single point)?2012-09-27
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    Dear @Asaf, yes exactly: so that point is obviously the generic point and the closed set does not contain non generic points since there are no other points at all!2012-09-27
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    @Georges: Thank you. I have to say that despite not understanding a whole lot about algebraic geometry, I still fail to come up with a reason why the axiom of choice should be involved here. I can understand why it might be involved in the existence of a generic point, but in the existence of non-generics? Sounds strange!2012-09-27
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    @Asaf: I now see that Keenan had written that already. I just reacted to the question without reading all the comments: my apologies to Keenan ( whose comments are remarkably lucid and informative) and to all users. I agree with you that the axiom of choice, while relevant in proving the existence of minimal primes, doesn't seem to have much to do with non-generic points. And you are also right that in the definition of $V(I)$ it should be $I\subset P$.2012-09-27
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    @Georges: Thank you very much for your reply. I tried to answer the question to the extent of my knowledge in the Zariski topology based on the comments by yourself and Keenan.2012-09-27
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    @QiaochuYuan By non-generic points I meant points which are themselves Zariski closed subsets.2012-09-27
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    Dear @Georges, Thank you for the kind words, and no need to apologize! I didn't initially have the example of $\mathrm{Spec}(k)$ in my comment, and I think I was editing it in while you were composing your comment.2012-09-27
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    @Alexei, This is not the standard usage of generic\non-generic points. What you're talking about are simply "closed points." So I guess what you're asking for is an example of a scheme without closed points (you needn't restrict to closed subsets because every closed subset of a scheme can be given the structure of a scheme whose underlying topology is unchanged). There are examples, but they have to be non-quasi-compact, at least if you allow AC (because, with Zorn, you can prove that every non-empty quasi-compact scheme has a closed point). For what happens without AC, see Asaf's answer.2012-09-27
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    @KeenanKidwell Thanks!2012-09-27
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    Alexei, could you (or anyone else) translate the second edit to a non-Zariski language? And please don't remove edits, add them to previous edits.2012-09-27

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