Timeline for Does bounded-degree base extension yield Zariski-dense Mordell-Weil group?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Sep 3, 2015 at 22:42 | comment | added | Vesselin Dimitrov | Indeed. I missed this point, but the remarks of grghxy and Laurent Moret-Bailly solve the problem. $B$ depends only on $n$. | |
| Sep 3, 2015 at 22:02 | comment | added | Laurent Moret-Bailly | By a theorem of Zarhin, if $A'$ is the dual of $A$, then $(A\times A')^4$ admits a principal polarization. So the principally polarized case (in dimension $8n$) implies the general case. | |
| Sep 3, 2015 at 22:02 | comment | added | grghxy | @VesselinDimitrov: For abelian varieties $A$ and $A'$, and subsets $S \subset A(L)$ and $S' \subset A'(L)$, the closure of $S \times S'$ in $(A \times A')_L$ is $\overline{S} \times \overline{S}'$. So to ensure $A(L)$ is Zariski-dense in $A_L$ (surely that is what is meant, not "Zariski-dense in $A$", right?) it suffices that $L$ works the same way for $(A \times A^{\vee})^4$. Hence, by Zarhin's trick (and allowing any dimension) it seems we can assume there is a principal polarization. Am I overlooking something? | |
| Sep 3, 2015 at 19:19 | comment | added | Vesselin Dimitrov | Yes, this argument only gives a $B$ that depends on $n$ and on the minimum degree of a polarization of $A$ over $K$. | |
| Sep 3, 2015 at 19:00 | comment | added | Vesselin Dimitrov | @O-RenIshii: I would have to think about that, maybe I would need to restrict the abelian variety $A/K$ to be principally polarized (or admit a polarization of a fixed degree). But if this is the case, and if $\mathcal{O}(D)$ defines a principal polarization, then the very ample linear series $|3D|$ embeds $A$ as a degree-$n! 3^n$ subvariety of $\mathbb{P}_K^{3^n - 1}$. Then use a linear projection. | |
| Sep 3, 2015 at 18:40 | comment | added | O-Ren Ishii | @VesselinDimitrov: How do you carry out step 2.? | |
| Sep 3, 2015 at 18:33 | comment | added | Vesselin Dimitrov | I think we could follow the same construction as for elliptic curves to yield in fact a $B$ depending only on $n$: 1. Reduce to the case that $A$ is simple; 2. Present $A$ as a branched cover $\pi: A \to \mathbb{P}_K^n$ with degree bounded only in terms of $n$; 3. Observe that $\pi^{-1}(\mathbb{P}^n(K))$ contains a non-torsion point $P$; 4. Then $[K(P):K] \leq \deg{\pi}$, and the group $\langle P \rangle \subset A(K(P))$ is Zariski-dense. | |
| Sep 3, 2015 at 18:06 | history | edited | Michael Zieve | CC BY-SA 3.0 |
added 9 characters in body
|
| Sep 3, 2015 at 18:00 | history | edited | Jeremy Rouse | CC BY-SA 3.0 |
deleted 92 characters in body
|
| S Sep 3, 2015 at 18:00 | history | suggested | Jesper Petersen | CC BY-SA 3.0 |
added hyperlink
|
| Sep 3, 2015 at 17:50 | review | Suggested edits | |||
| S Sep 3, 2015 at 18:00 | |||||
| Sep 3, 2015 at 17:18 | history | asked | Michael Zieve | CC BY-SA 3.0 |