4
$\begingroup$

This question is partially inspired by this question: Does every non-empty set admit a group structure (in ZF)?

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):

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 $S$?

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, 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:

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 $S$?

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.

$\endgroup$
11
$\begingroup$

Take a field of arbitrary infinite cardinality; 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.

$\endgroup$
  • $\begingroup$ Thanks! I suspected something like that would work. I don't know why I didn't just think to look for "fields of arbitrary cardinality" (instead of try to construct them using groups!). I will hold off accepting for now to see if the second question gets any interesting responses. (+1) $\endgroup$ – Sean Lawton May 6 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.