Timeline for Jacobians of pointed curves

6 events

when toggle format	what		by	license	comment
Jan 8, 2019 at 13:39	comment	added	Roman Fedorov		@AriyanJavanpeykar: Of course, you are right.
Jan 8, 2019 at 13:39	comment	added	Roman Fedorov		@Stanley Yao Xiao: you say "this is an exact statement over an algebraically closed field" -- this is not unless you work with polarized abelian varieties. From the above discussion it seems to follow that there exists a polarized Jacobian abelian variety $A/K$ such that whenever $Jac(Y)$ is isomorphic to $A$ as polarized abelian variety, $Y$ has no rational points.
Dec 23, 2018 at 23:26	comment	added	Ariyan Javanpeykar		@RomanFedorov What about a higher genus curve $Y$ over $\mathbb{Q}$ with no non-trivial automorphisms and $Y(\mathbb{Q}) = \emptyset$? Such a curve has no twists. (Does this really answer the question though? It seems to me that there could still be a curve $Y'$ with $Y'(\mathbb{Q})\neq \emptyset$ such that $Jac(Y') \cong Jac(Y)$. This isomorphism won't respect the theta divisor.)
Dec 14, 2018 at 14:13	comment	added	Roman Fedorov		If you can find a curve $Y/K$ with the following property: for every $Y'/K$ such that $Y$ and $Y'$ are isomorphic over $\bar K$, $Y'(K)=\emptyset$. Then $Jac(Y)$ is such an example.
Dec 12, 2018 at 20:01	history	edited	Stanley Yao Xiao	CC BY-SA 4.0	added 166 characters in body
Dec 12, 2018 at 19:56	history	asked	Stanley Yao Xiao	CC BY-SA 4.0