I have a curious question about an argument/hint given in following thread:


The OP asked if there an irreducible smooth proper one-dimensional $\mathbb{Z}$-scheme that is not isomorphic to the projective line.

In the answer Ariyan Javanpeykar showed that the generic fiber $C := X_{\mathbb{Q}}:= X \times_{Spec\mathbb{Z}} Spec(\mathbb{Q})$ of the morphism$X \to Spec(\mathbb{Z}) $ is a smooth proper geometrically connected curve of genus zero.

It stays to show that this implies $C_{\mathbb{Q}} =\mathbb{P}^1_{\mathbb{Q}}$ from what we can deduce $X = \mathbb{P}^1_{\mathbb{Z}}$. According to the hint one have to use that the sets of $\mathbb{F}_p$-valued points $X(\mathbb{F}_p)$ are non empty for all $p$.

Could anybody explain in what way this hint with non empty $X(\mathbb{F}_p)$ has to be used to acquire desired result?

My considerations base essensially on Ravi Vakhil's https://math.stanford.edu/~vakil/0708-216/216class41.pdf

using the "curve to projective" extension theorem (2.1; p. 5) and the embedding theorem for non singular curves (Thm 0.2; p. 7):

I think that for $C_{\mathbb{Q}} =\mathbb{P}^1_{\mathbb{Q}}$ as well for $X = \mathbb{P}^1_{\mathbb{Z}}$ we have essential problem is to find "comparing" morphisms $f: C \to \mathbb{P}^1$; the candidates for an isomorphism.

Firslty regarding $C_{\mathbb{Q}} =\mathbb{P}^1_{\mathbb{Q}}$:

Thm. 0.2 provides an open immersion $f: C_{\mathbb{Q}} \to \mathbb{P}^1_{\mathbb{Q}}$ which is proper, so closed and since both irreducible, a homeomorphism. I think that I can use a genus argument (interpreting it as dimension of first cohomology of $C$) on the resulting sequnce of the induced morphisms between structure sheaves to verify that $f$ is already an isomorphism of schemes.

I think that similar combination of thms 0.2 and 2.1 would provide an immersion $g:X \to \mathbb{P}^1_{\mathbb{Z}}$ and the verification this this is an isomorphism on the level of structure sheaves should work similar as in case above using genus property on resulting sheaf cohomology.


The point that I really curious about is what argument is meant by the observation that the $\mathbb{F}_p$-valued points $X(\mathbb{F}_p)$ aren't empty?

How does it help here? Where does it flow in for the argumentation?

  • $\begingroup$ @AknazarKazhymurat: I don't see where in the proof from the given reference is explicitely used that $X(\mathbb{F}_p)$ are non empty as input $\endgroup$ – Karl_Peter Mar 9 at 17:43
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    $\begingroup$ @AknazarKazhymurat To prove that every Brauer-Severi scheme $X$ over $\mathbb{Z}$ is trivial, you use that $X(\mathbb{F}_p)$ is non-empty. See my following comment for more. $\endgroup$ – Ariyan Javanpeykar Mar 11 at 11:10
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    $\begingroup$ @Karl_Peter Let $X$ be a smooth proper geometrically connected scheme whose fibres are curves of genus zero. Note that $X(\mathbb{F}_p)$ is non-empty for every $p$. You can prove this using Hasse-Weil's theorem or a more direct argument à la Chevalley-Warning. Since $X\to \mathrm{Spec} \ \mathbb{Z}$ is smooth and proper, the non-emptyness of $X(\mathbb{F}_p)$ implies that $X(\mathbb{Q}_p)$ is non-empty. Then, $X$ is a smooth conic in $\mathbb{P}^2_{\mathbb{Q}}$ with a $\mathbb{Q}_p$-point for every prime $p$. $\endgroup$ – Ariyan Javanpeykar Mar 11 at 11:12
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    $\begingroup$ ...We conclude that $X(\mathbb{Q})$ is non-empty by Hasse-Minkowski's theorem, so that $X\cong \mathbb{P}^1_{\mathbb{Q}}$. (Recall that a smooth proper geometrically connected curve $X$ of genus zero over a field $k$ is isomorphic to $\mathbb{P}^1_k$ if and only if $X(k)$ is non-empty.) $\endgroup$ – Ariyan Javanpeykar Mar 11 at 11:13
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    $\begingroup$ @Karl_Peter This is an application of the infinitesimal lifting criterion for (formally) smooth morphisms. You lift a $\mathbb{F}_p$-point to a $\mathbb{Z}/p^2$-point. Then you lift this $\mathbb{Z}/p^2$-point to a $\mathbb{Z}/p^3$-point, etc... This gives you a compatible system of $\mathbb{Z}/p^i$-points. The latter gives you a $\mathbb{Z}_p$-point. See stacks.math.columbia.edu/tag/02GZ for a discussion of formally smooth morphisms. $\endgroup$ – Ariyan Javanpeykar Mar 13 at 14:47

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