Timeline for Jacobians defined over smaller fields
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 7, 2012 at 8:44 | answer | added | inkspot | timeline score: 3 | |
| Oct 7, 2012 at 4:32 | answer | added | user27056 | timeline score: 8 | |
| Oct 5, 2012 at 5:17 | comment | added | Ben Wieland | Here is an approach to $g=1$ that similar to grp's. It's not so explicit as grp's, but it is well-organized by some machinery; and thus motivates the machinery. Given an elliptic curve $E$ over a local field $K$, the group of genus $1$ curves with $J(C)=E$ is approximately $K/O_K$. If $K$ is degree $d$ over $\mathbb Q_p$, then $K/O_K=(\mathbb Q_p/\mathbb Z_p)^d$ .... (nb $\mathbb Q_p/\mathbb Z_p=$"$\mathbb Z/p^\infty$" the Prüfer group). Thus there are lots of torsors defined over $L$ that aren't defined over $K$. | |
| Oct 5, 2012 at 4:56 | comment | added | Ben Wieland | Even if the jacobin is ppav over $K$, I do not think Torelli is straight-forward to apply to this problem, for away from the hyperelliptic locus it is morally 2:1, not injective. I expect that sometimes those two fibers form a field extension. Tangentially, people claim that $J(C)=J(C')$ implies that $C=C'$, but it seems to me that they use $Pic^1(C)=Pic^1(C')$, which seems like a stronger assumption. | |
| Oct 5, 2012 at 0:06 | comment | added | grp | @Piotr: It sounds like you ask just that the p.p. does not descend to $K$ respecting the given $K$-structure on the abelian surface. In principle, it might happen that the p.p. abelian surface over $L$ admits another $K$-descent as a p.p. abelian surface (i.e., with a $K$-structure different from the given one on the abelian surface), so the associated curve over $L$ would then be defined over $K$. So to make an example in this way one needs a stronger "does not descend to $K$" property. Perhaps you were already aware of this, in which case all I'm doing is clarifying your comment. | |
| Oct 4, 2012 at 23:31 | comment | added | Piotr Achinger | My guess is that this is impossible if the polarization of $J(X)$ is also defined over $K$, due to Torelli. So I would look for an abelian surface over $K$ with a principal polarization defined over $L$ that does not descend to $K$ (I think that any p.p. abelian surface is a Jacobian of a genus 2 curve). | |
| Oct 4, 2012 at 23:28 | comment | added | grp | The genus-1 curve $ax^3 + by^3 + cz^3 = 0$ has Jacobian (away from characteristics 2 and 3) depending only on $abc$, so you can probably make some explicit examples based on that. | |
| Oct 4, 2012 at 22:44 | comment | added | Tyler Lawson | A not-very-interesting example would be a curve $X$ of genus zero over $L$ which does not lift to $K$. | |
| Oct 4, 2012 at 22:06 | history | asked | Harry | CC BY-SA 3.0 |