0
$\begingroup$

Let $X$ be any irreducible scheme with the property that the generic point $\eta$ of $X$ is an set open wrt underlying Zariski topology.

What can we say about the structure of such schemes? Topologically, algebraically,...

(note, that I'm not assuming $X$ to be Jacobson, what would imply that every open subset contains a closed point of $X$ ( compare to the proof in 33.20.3 (3) in https://stacks.math.columbia.edu/tag/0A21), since in this case this would trivially imply that $X$ is zero dimensional)

Edit#1: Let follow Martin Brandenburg's suggestion to assume for sake of simplicity $X=\operatorname{Spec}R$ to be affine.

#Edit#2: If we moreover pass to reduced ring, the question becomes about the structure of rings which become fields after beeing localized at a single element.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Have you looked at the affine case already? This is also more or less sufficient, an open singleton is affine. $\endgroup$
    – Martin Brandenburg
    Commented Oct 10, 2023 at 13:58
  • $\begingroup$ @MartinBrandenburg: say $X$ is affine wrt to ring $R$, then if the generic pt of open, then we can deduce that there exist a ring element $r\to R$ such that $R$ localized at $r$ is local Artin ring, namely the localization at the generic point. Can we conclude more from that? Especially about "global" topology of $X$? $\endgroup$
    – user267839
    Commented Oct 10, 2023 at 14:23
  • $\begingroup$ The problem is that I'm not sure if there exist a way to say something interesting about the comparison of dimensions of $R$ and $R_r$ except the trivial inequality estimation, if we assume $R$ really to be commutative with one, but else arbitrary $\endgroup$
    – user267839
    Commented Oct 10, 2023 at 14:28
  • $\begingroup$ By the way you may also replace $X$ by $X_{red}$ (your question is purely topological) and hence assume that $X$ is reduced, meaning that $R$ is an integral domain. $\endgroup$
    – Martin Brandenburg
    Commented Oct 10, 2023 at 14:30
  • $\begingroup$ #typo: above of course I meant $r\in R$, not the "arrow"... $\endgroup$
    – user267839
    Commented Oct 10, 2023 at 14:30

1 Answer 1

Reset to default
3
$\begingroup$

Let's assume $X = \mathrm{Spec}(R)$ is affine. Since $X$ can be replaced by $X_{\mathrm{red}}$, let's also assume that $X$ is reduced. Then $R$ is an integral domain. The condition that $\{ \eta \}$ is open means that there is some $f \in R$ with $\{ \eta\} = D_f$. So $D_f$ is an integral affine scheme with exactly one point, so it must correspond to a field. But then $Q(R) = R[f^{-1}]$. Conversely, if there is some $f \neq 0$ such that $Q(R) = R[f^{-1}]$, then $D(f) = \{ \eta\}$.

The condition $Q(R) = R[f^{-1}]$ is equivalent to: every non-zero element of $R$ divides some power of $f$. If $R$ is factorial, this means that $R$ has at most one prime element (up to a unit). I am not sure what else to say. Of course, the most "famous" example is that of a DVR.

The same question has been asked at MSE:

open generic points of affine scheme?

Improve this answer
$\endgroup$
4
  • $\begingroup$ If $R$ is noetherian, the condition is equivalent to having a finite number of height $1$ primes. This clearly generalizes to noetherian schemes. $\endgroup$
    – Giulio Bresciani
    Commented Oct 10, 2023 at 14:56
  • 2
    $\begingroup$ Let me explain this. Assume $R$ noetherian. If $Q(R)=R(f^{-1})$, then by Krull's principal ideal theorem a prime has height $1$ if and only if it is a minimal prime containing $f$, and thus there is a finite number of them. On the other hand, if the number of height $1$ primes is finite, by taking a suitable product we can find a non-nilpotent $f$ contained in all of them, so the generic point is open. $\endgroup$
    – Giulio Bresciani
    Commented Oct 10, 2023 at 15:04
  • $\begingroup$ @GiulioBresciani: and that there is a finite number of them follows from Noetherness, right? Otherwise one could construct from infinitely many of them a non terminating chain... $\endgroup$
    – user267839
    Commented Oct 10, 2023 at 15:11
  • $\begingroup$ Yes, it is noetherianity. In general, every noetherian ring has a finite number of minimal primes, and we apply this to $R/(f)$. $\endgroup$
    – Giulio Bresciani
    Commented Oct 10, 2023 at 15:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .