Is the trivial ring regular?

Question

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?)

Is a smooth morphism from a finite scheme into an infinite scheme surjective? — Matt Samuel, Feb 13 '19 at 00:56
No, smooth morphisms need not be surjective. Every open immersion is smooth. That is, if $X$ is a variety, $U\subset X$ an open set, then the inclusion $U\to X$ is smooth. — Mohan, Feb 13 '19 at 01:02

score 6 · Accepted Answer · answered Feb 13 '19 at 08:52

6

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.)

answered Feb 13 '19 at 08:52

Fred Rohrer

521

Dr. Rohrer, see comments here https://math.stackexchange.com/a/3215492/690882 – Jul 27 '19 at 11:57

Is the trivial ring regular?

1 Answers1