3
$\begingroup$

Consider Siegel's theorem. It says that for a smooth affine algebraic curve $C$ over $\mathbb{Q}$ such that $g(C)>0$ any model $\mathcal{C}$ of $C$ over $\mathbb{Z}$ has finitely many $\mathbb{Z}$-points.

Take $C$ to be an elliptic curve with infinitely many rational points. Let $C'$ be the smooth compactification of $C$. Then for any proper model $\mathcal{C}'$ of $C'$ over $\mathbb{Z}$ the valuative criterion implies that $\mathcal{C}'(\mathbb{Z})=C'(\mathbb{Q})$.

To my embarassment I was confused by this situation because it seem that one can embed $C'$ into the 3-dimensional space and then cover it by its intersections with the standard affine opens. Each of them has finitely many integral points so how can the compactification have infinitely many? The point seems to be that not every $\mathbb{Z}$-point of $\mathcal{C}'$ is contained in a standard affine open (consider $(2, 3)$ in $\mathbb{P}^1_{\mathbb{Z}}$). In general there seems to be no finite affine open cover of $\mathcal{C}'$ such that each $\mathbb{Z}$-points is contained in one of the opens. What does seem to be true is that each $\mathbb{Z}$-point is contained in some affine open but these affine opens are different for different points.

There are 2 points here which are not clear to me in a general situation. First, is it true that if $X\to Spec(\mathbb{Z}$) is proper then any $\mathbb{Z}$-point on $X$ is contained in an affine open? This fails in equal characteristic i.e. a $k[t]$-point on a variety over $k$ need not be contained in an affine open. Second, are there positive-dimensional situations where the reasoning in the previous paragraph can be made to work, i.e. a finite affine open cover can be found such that every integral point lies entirely in one of the opens? This is meaningless as written since there are plenty of varieties with no rational points whatsoever. One can demand that there exist infinitely many rational points to make the search more interesting.

$\endgroup$
6
  • 1
    $\begingroup$ Note that Siegel's theorem assumes $g(C)>0$ ... $\endgroup$
    – abx
    Oct 31, 2019 at 5:12
  • $\begingroup$ If $X$ is projective, then it can be embedded in $\mathbb{P}^n$ over $\mathbb{Z}$ as a closed subscheme. So your first question reduces to the case of projective space, in which case the answer is obvious. I am not sure what happens if $X$ is not projective, but only proper. $\endgroup$
    – naf
    Oct 31, 2019 at 6:37
  • $\begingroup$ I assume by $C$ you mean an affine open subset of an elliptic curve, rather than an elliptic curve, since elliptic curves are by definition proper. You should think of an integral point of $C$ as being a rational point $x$ of $C'$ such that $x \bmod p \in C$ for all primes $p$. This might clear up some of your confusion. $\endgroup$ Oct 31, 2019 at 11:09
  • 1
    $\begingroup$ "This fails in equal characteristic": The analogous statement is not what you claim, but the fact that if $f:X\to\mathrm{Spec}(k[t])$ is projective then every section of $f$ lands into an affine open of $X$. This is true, even with any affine scheme $S=\mathrm{Spec}(R)$ in place of $\mathrm{Spec}(k[t])$. Proof: you may assume $X=\mathbb{P}^n_S$. A section corresponds to an invertible quotient $L$ of $R^n$. But $L$ is a projective $R$-module, so $R^n=L\oplus E$. This $E$ determines a relative hyperplane disjoint from the section. QED. $\endgroup$ Oct 31, 2019 at 17:04
  • $\begingroup$ @LaurentMoret-Bailly you are completely right. Do you know what happens if $f$ is proper but not projective? $\endgroup$
    – user145520
    Oct 31, 2019 at 18:41

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.