# Is the trivial ring regular?

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?)

• A ring with one element can't be local. – Matt Samuel Feb 13 at 0:53
• @MattSamuel but it doesn't need to be local, right? – 0x539 Feb 13 at 0:54
• Are you saying it has no localizations? – Matt Samuel Feb 13 at 0:55
• @MattSamuel Yes. It has no localizations at prime ideals. – 0x539 Feb 13 at 0:55
• Is a smooth morphism from a finite scheme into an infinite scheme surjective? – Matt Samuel Feb 13 at 0:56