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If $X$ is a smooth proper one dimensional scheme over $\text{Spec} \mathbb{Z}$ such that $X_{\mathbb{Q}}\cong \mathbb{P}_{\mathbb{Q}}^1$, then how to see $X(\mathbb{F}_p)$ is non-empty for every $p$ and $X\cong \mathbb{P}_Z^1$?

Thanks!

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    Because $X_{\mathbb{F}_p}$ is a genus $0$ curve and those always have points over $\mathbb{F}_p$. Use Riemann-Roch to show that it at least can be embedded as a quadric in $\mathbb{P}^2$ and use a counting argument to show that $ax^2+by^2+cz^2$ has a non-zero solution if $a,b,c\ne 0$ over $\mathbb{F}_p$. For the second part see Theorem 4 of my blog post here https://ayoucis.wordpress.com/2016/04/29/around-abelian-schemes-over-the-integers/ – Alex Youcis Oct 08 '19 at 17:15
  • @AlexYoucis Thanks very much! –  Oct 09 '19 at 02:53
  • @AlexYoucis Would you mind answer this question about a proof in your blog? Thanks! –  Nov 16 '19 at 09:10

1 Answers1

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Just to get this off the unanswered list I will make my comment an answer.

Since $X$ is a genus $0$ smooth proper curve we know that $X_{\mathbb{F}_p}$ is also a smooth proper one-dimensional curve of genus $0$. There are several ways to show that such curves have rational points, but the most elementary is as follows. Use Riemann-Roch to show that you can embed $X_{\mathbb{F}_p}$ in to $\mathbb{P}^2_{\mathbb{F}_p}$ as $V(ax^2+by^2+cz^2)$ where $a,b,c\ne 0$. Note then that we need to show that $ax^2+by^2+cz^2=0$ has a non-zero solution. But, note that $\{ax^2+by^2\}$ and $\{-cz^2\}$ both contain at least $\frac{p-1}{2}+1$ elements and thus they must have at least two members in common for counting reasons.

For the second question see this: https://ayoucis.wordpress.com/2016/04/29/around-abelian-schemes-over-the-integers/

Alex Youcis
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