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In algebraic geometry if $f: X \to Y$ is locally of finite presentation (where $X, Y$ are schemes) then smoothness of $f$ implies that for all $y \in Y$ the "geometric fiber" $\DeclareMathOperator{\Spec}{Spec} \Spec{\overline{\kappa(y)}} \times_{Y} X$ is regular.

If $y$ has no preimage under $f$ then the geometric fiber should be empty, so it is the trivial ring $\{0\}$. Is this ring considered to be regular? I guess technically all of its localizations are regular local.

(In other words, must every smooth morphism of Schemes be surjective?)

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  • $\begingroup$ A ring with one element can't be local. $\endgroup$
    – Matt Samuel
    Feb 13, 2019 at 0:53
  • $\begingroup$ @MattSamuel but it doesn't need to be local, right? $\endgroup$
    – Carlos Esparza
    Feb 13, 2019 at 0:54
  • $\begingroup$ Are you saying it has no localizations? $\endgroup$
    – Matt Samuel
    Feb 13, 2019 at 0:55
  • $\begingroup$ @MattSamuel Yes. It has no localizations at prime ideals. $\endgroup$
    – Carlos Esparza
    Feb 13, 2019 at 0:55
  • $\begingroup$ Is a smooth morphism from a finite scheme into an infinite scheme surjective? $\endgroup$
    – Matt Samuel
    Feb 13, 2019 at 0:56

1 Answer 1

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The empty scheme is regular according to EGA IV.5.8.2. Accordingly, the zero ring is regular according to Bourbaki's AC.VIII.5 Exercice 6. (Of course, the point is that regularity is defined as some property for every point or for every prime ideal, resp.)

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