This question is partially inspired by this question: *[Does every non-empty set admit a group structure (in ZF)?](https://mathoverflow.net/q/12973/12218)*


It was also inspired by my desire to explain the importance of quotient morphisms when discussing algebraic quotients. I wanted to point out to a student that the mere existence of a scheme structure on the orbit *set* was insufficient to understand the motivation for a categorical quotient. I then realized I did not know how to prove my first desired phrase: "Since every set can be the scheme of some ring,..."


So I figured someone here might have thought through this before. Here is my question (as in the title):



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> Given a set $S$, does there exist a commutative ring $R$ with identity so that the underlying set of $\mathrm{Spec}(R)$ is bijective to $S$?
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Every finite set is bijective to the zeros of a polynomial and so that should handle finite sets. $\mathrm{Spec}(\mathbb{Z})$ handles the countable case, and affine space over $\mathbb{C}$ gives us an example that is uncountable. After that I am out of my set theory depth, unfortunately. 


I feel like I should be able to use the fact that every set (in ZFC) admits a group structure, to some avail, but I don't see how exactly. Maybe [turn your set into an abelian group](https://mathoverflow.net/q/89877/12218), then group ring, then quotient field and then consider affine space with respect to that field?


Of course maybe this general set-theoretic statement answers the question in the affirmative abstractly: <https://mathoverflow.net/a/89878/12218>. However, I would personally feel better with explicit examples/constructions.


Regardless, I suspect the answer is yes and not very interesting. But perhaps a more interesting question is this:



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> Given a set $S$, does there exist an *interesting* commutative ring $R$ with identity so that the underlying set of $\mathrm{Spec}(R)$ is bijective to $S$?
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If my suspicions are right, then "interesting" here should not mean smooth or normal, but should mean something like "with as bad of a singularity as you want" (some kind of "Murphy's Law" condition). But to keep the second question more flexible, let's leave it vague for now.



Does every non-empty set admit an (affine) scheme structure (in ZFC)?
Take a [field of arbitrary infinite cardinality](https://math.stackexchange.com/q/1296889/29145); let $K$ be its algebraic closure (which has the same cardinality) and consider Spec(K[X]), which has the same cardinality plus one, i.e. the same cardinality.


