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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. – Qiaochu Yuan Sep 27 '12 at 16:27
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. – Asaf Karagila Sep 27 '12 at 16:30
@Asaf A generic point in a scheme is a point whose closure is an irreducible component. – Keenan Kidwell Sep 27 '12 at 16:44
@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? – Asaf Karagila Sep 27 '12 at 16:58
@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. – Keenan Kidwell Sep 27 '12 at 17:00

To sum up the comments by Keenan and Georges, generic points are minimal prime ideals. It seems that you are asking for a closed set $V(I)$ such that every prime ideal in $V(I)$ is minimal.

Since no two points in $V(I)$ are comparable in $\subseteq$ (otherwise we contradict minimality) we have to have that every point is maximal. Since the intersection of two prime ideals containing $I$ is also a prime ideal containing $I$ then we have to have that $V(I)$ is a singleton to begin with.

Assuming, however, that you confused the double-negations a bit let me answer a slightly different question. Without the axiom of choice there exists a [commutative unital] ring $R$ and a decreasing chain of prime ideals in $R$ without a minimal element.

To see a wonderfully vivid proof that the existence of a minimal prime ideal implies the axiom of choice see this MathOverflow answer.

Therefore assuming choice fails there is such a ring that has a decreasing chain of prime ideals without a minimal element, as promised.

To the edit: much like the part above, the existence of maximal ideals in commutative unital rings is also equivalent to the axiom of choice (see this). Namely without the axiom of choice there is a ring with a chain of ideals without a maximal element.

I am unsure, however if you can find a "maximal" prime which is not maximal and not contained in any real maximal ideal.

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I tried to use my [quite] limited knowledge about alg. geometry, I'd be glad to hear some feedback. – Asaf Karagila Sep 27 '12 at 18:36
Your edit seems to answer the (apparently) intended question of whether or not there is a scheme without closed points, because in an affine scheme, closed points are just maximal ideals. – Keenan Kidwell Sep 27 '12 at 20:05
@Keenan: Ah, very good. When I return home I will add a linked reference. – Asaf Karagila Sep 27 '12 at 20:09

I'm guessing, but my instinct is that without the axiom of choice, you don't want to use the prime spectrum. Instead of a topological space, you want a locale. This is because the description of open sets in terms of ring elements should be well-behaved, despite the notion of using prime ideals as points being ill-behaved.

Before you scoff, I want to point out this really isn't a silly thing to do: remember that one of the main reasons you were using the prime spectrum in the first place is because it is better behaved than the maximal spectrum.

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