5
$\begingroup$

Siegel's theorem states the following:

Let $C$ be a smooth projective curve over a number field $K$. Let $\tilde C\subset C$ be an open affine subvariety, and $i:\tilde C\hookrightarrow \mathbb{A}^m_K$ be a closed immersion. Then if $i(\tilde C)$ lies over infinitely many $\mathbb{A}^m_{\mathcal{O}_K}(\mathcal{O}_K)$-points, then the genus of $C$ is $0$, and furthermore $|C(\bar{\mathbb{Q}})\smallsetminus \tilde C(\bar{\mathbb{Q}})|\leq 2$.

In order for me to explain what I find confusing about this statement, consider the following definition:

Call a subset $S$ of $C(K)$ Siegel if there exists an open affine $\tilde C\subset C$ and a closed immersion $i:\tilde C\hookrightarrow \mathbb{A}^m_K$ so that the points of $S$ are exactly the $K$-points of $\tilde C$ lying above $\mathcal{O}_K$-points of $\mathbb{A}^m_{\mathcal{O}_K}$.

Question

  1. Can you give a reasonably intuitive characterization of the Siegel subsets of $C(K)$? For example, can you re-phrase the definition of a Siegel set in a manner that avoids reference to affine space, and only deals with $\mathcal{O}_K$-models of $C$ or $\tilde C$?
  2. What can one say about the set of Siegel sets? Is it true that any two maximal Siegel sets are disjoint? Is the set of maximal Siegel sets an equivalence relation?
  3. Can one describe maximal Siegel sets in some nice fashion? Presumably $C(K)$ itself is not a Siegel set in general, because if that were the case then Siegel's theorem would imply Falting's theorem, which is apocryphal.
Improve this question
$\endgroup$
2
  • $\begingroup$ Reformulate Siegel's theorem as follows. Let $C$ be a smooth affine connected curve over $K$ of negative Euler characteristic. Then for each model $\mathscr C$ of $C$ over $\mathscr O_K$, the set $\mathscr C(\mathscr O_K)$ is finite. $\endgroup$
    – ACL
    Commented Dec 16, 2016 at 17:48
  • 1
    $\begingroup$ Presumably by "model" you mean "affine model". Okay. But my questions still stand, as they are about Siegel sets. It sounds like you have an idea for my first question that you used to conclude the content of your comment. Would you mind making it more explicit? $\endgroup$
    – Andrew NC
    Commented Dec 16, 2016 at 18:11

1 Answer 1

Reset to default
3
$\begingroup$

Every finite set is a Siegel set, in your sense for curves of positive genus. Proof: Let $S$ be the set in question, take an affine open subset $U$ of your curve containing $S$ and embed this open set $U$ in affine space. Change variables by clearing denominators such that the points of $S$ are integral with respect to this embedding. Let $T$ be the remaining set of integral points with respect to this embedding, so $T$ is finite by hypothesis. Construct a function $f$ on the curve such that $f(P)=0, P \in S, f(P)=1/2, P \in T$. Now take the open subset of $U$ where $f$ is regular and embed it in affine space of one more dimension by taking $f$ as an extra coordinate in addition to the previous coordinates. Clearly, the set of integral points for this embedding is $S$.

I don't understand your issue with $C(K)$ being Siegel if the genus is at least two. Of course, it's not a direct consequence of Siegel's theorem but it follows from the Mordell conjecture just by clearing denominators.

Not sure what happens in genus zero.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I'm very confused... If for a genus >=2 curve C it is always true that C(K) is Siegel, then Siegel's theorem implies Falting's theorem. But Siegel was known before Falting's theorem, so it can't be that it implies it in a trivial way... $\endgroup$
    – Andrew NC
    Commented Dec 16, 2016 at 20:30
  • 1
    $\begingroup$ @AndrewNC I am not claiming Siegel implies Faltings. I am saying that, as a consequence of Faltings' theorem, I can deduce that $C(K)$ is Siegel. If I don't assume Faltings then I cannot complete my argument, so it's not circular. $\endgroup$
    – Felipe Voloch
    Commented Dec 16, 2016 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .