Skip to main content
added 632 characters in body; edited body
Source Link
Pete L. Clark
  • 65.4k
  • 12
  • 241
  • 381

Does every smooth proper morphism $X \to \operatorname{Spec} \mathbf{Z}$ with $X$ nonempty have a section?

EDIT [Bjorn gave additional information in a comment below, which I am recopying here. -- Pete L. Clark]

Here are some special cases, according to the relative dimension $d$. If $d=0$, a positive answer follows from Minkowski's theorem that every nontrivial finite extension of $\mathbf{Q}$ ramifies at at least one prime. If $d=1$, it is a consequence (via taking the Jacobian) of the theorem of Abrashkin and Fontaine that there is no nonzero abelian scheme over $\mathbf{Z}$, together with (for the genus $0$ case) the fact that a quaternion algebra over $\mathbf{Q}$ split at every finite place is trivial.

Does every smooth proper morphism $X \to \operatorname{Spec} \mathbf{Z}$ with $X$ nonempty have a section?

Does every smooth proper morphism $X \to \operatorname{Spec} \mathbf{Z}$ with $X$ nonempty have a section?

EDIT [Bjorn gave additional information in a comment below, which I am recopying here. -- Pete L. Clark]

Here are some special cases, according to the relative dimension $d$. If $d=0$, a positive answer follows from Minkowski's theorem that every nontrivial finite extension of $\mathbf{Q}$ ramifies at at least one prime. If $d=1$, it is a consequence (via taking the Jacobian) of the theorem of Abrashkin and Fontaine that there is no nonzero abelian scheme over $\mathbf{Z}$, together with (for the genus $0$ case) the fact that a quaternion algebra over $\mathbf{Q}$ split at every finite place is trivial.

Apparently it's not an open problem!
Link
Greg Kuperberg
  • 56.6k
  • 10
  • 203
  • 282
Source Link
Bjorn Poonen
  • 23.8k
  • 7
  • 90
  • 109

Smooth proper scheme over Z

Does every smooth proper morphism $X \to \operatorname{Spec} \mathbf{Z}$ with $X$ nonempty have a section?