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How to construct connected scheme $X$ over $\text{Spec}\,\mathbb{Z}$ such that for $p\neq0$ the fiber $X_p=X\times_{\text{Spec}\,\mathbb{Z}}k(p)$ over the prime ideal $(p)$ contains precisely $p$ points over $\mathbb{F}_p$?

The same question for $p+1$ and $p-1$.

miguels
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1 Answers1

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Take the canonical morphism $X=\mathbb A^1_{\mathbb Z}=\operatorname {Spec} (\mathbb Z[T])\to \operatorname {Spec} \mathbb Z$, which has fiber $\mathbb A^1_{\mathbb F_p}=\operatorname {Spec} (\mathbb F_p[T])$ over $(p)\in \mathbb Z$.
This fiber has exactly $p$ points rational over $\mathbb F_p$ .

Edit
The canonical morphism $Y=\mathbb P^1_{\mathbb Z}=\operatorname {Proj} \mathbb Z[T,U]\to \operatorname {Spec} \mathbb Z$ has as fiber over $(p)\in \operatorname {Spec} \mathbb Z$ the projective line $\mathbb P^1_{\mathbb F_p}=\operatorname {Proj} \mathbb F_p[T,U]$, and that fiber has exactly $p+1$ points rational over $\mathbb F_p$-points.
This answers miguels's supplementary question in his comment.

Georges Elencwajg
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  • Thanks. What about $p+1$ points? – miguels Mar 21 '15 at 19:43
  • Dear miguels, I have answered this supplementary question in an Edit. – Georges Elencwajg Mar 21 '15 at 20:10
  • thank you for your answers. Can you give an example with $p-1$ points in fiber? – miguels Apr 27 '15 at 23:36
  • @miguels The affine line and the projective line over $\mathbb Z$ answered your first two questions. To construct schemes with less points just remove points from these schemes. For instance, you can remove the zero section of $\mathbb A^1_{\mathbb Z}$ to get a scheme with $p-1$ rational points in its fiber over $(p)$. – Ariyan Javanpeykar Apr 28 '15 at 07:56