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

https://math.stackexchange.com/questions/3062986/irreducible-smooth-proper-one-dimensional-mathbbz-schemes

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.

QUESTION:

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?

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  • $\begingroup$ @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 $\endgroup$
    – user267839
    Commented Mar 9, 2019 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
    Commented Mar 11, 2019 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
    Commented Mar 11, 2019 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
    Commented Mar 11, 2019 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
    Commented Mar 13, 2019 at 14:47

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