Timeline for Specialization Map of family of abelian varieties
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2014 at 5:55 | answer | added | user27920 | timeline score: 4 | |
| Jun 13, 2014 at 14:29 | history | edited | wongpin101 | CC BY-SA 3.0 |
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| Jun 13, 2014 at 13:22 | comment | added | wongpin101 | @ACL: Yes, I agree that $S$ has to be normal. In that case, a morphism $Spec(F) \rightarrow X$ can be extended to $Spec(S) \rightarrow X$ by Weil's Extension Theorem. | |
| Jun 13, 2014 at 5:33 | comment | added | ACL | @wongpin101 (followed). This continues to work if $S$ is the spectrum of a Dedekind ring. In the general case, the morphism of groups $X(S)\to X_F(F)$ is injective but has no reason to be an isomorphism. This is already the case if $S$ is not normal. And if $S$ is normal, a morphism $Spec(F)\to X$ only extends outside of a codimension 2 subset in general. | |
| Jun 13, 2014 at 5:31 | comment | added | ACL | @wongpin101: I think you are missing a point in the definition of the specialization map. Let $X/S$ be an abelian scheme, where $S$ is integral, with field of fractions $F$. First assume that $S$ is the spectrum of a DVR. Then the valuative criterion of properness furnishes an isomorphism of groups $X(S) \to X_F(F)$. On the other hand, if $s$ is the special point of $S$, one has a morphism of groups (functoriality) $X(S)\to X_s(\kappa(s))$. These two properties give the specialization morphism $X_F(F)\to X_s(\kappa(s))$. | |
| Jun 12, 2014 at 21:23 | comment | added | wongpin101 | I think the point is: what is the group structure on sections $Mor_{Y_0}({Y_0},X)$? If it's defined by $\mu(s_P,s_Q):=\mu \circ (s_P \times s_Q) \circ \delta_{Y_0}$ where $\delta_{Y_0}$ is the diagonal map, in this case, $s_P(y)+s_Q(y)=\mu(s_P,s_Q)(y)$ is by definition and $\mu(s_P,s_Q)=s_{P+Q}$ is not obvious. However, if the definition is $\mu(s_P,s_Q)=s_{P+Q}$, then $s_P(y)+s_Q(y)=\mu(s_P,s_Q)(y)$ is not obvious. | |
| Jun 12, 2014 at 21:22 | comment | added | wongpin101 | To Felipe: I believe user52824 gave the answer to 1 by Weil's Extension Theorem if X is an abelian scheme over smooth scheme $Y_0$. Thanks for your answer for 2. But could you explain "$\mu(s_p,s_Q)=s_R$ by definition". | |
| Jun 12, 2014 at 19:11 | history | edited | wongpin101 | CC BY-SA 3.0 |
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| Jun 12, 2014 at 17:56 | comment | added | Felipe Voloch | If $X$ is an abelian scheme over $Y_0$, then the group law is given by a map $\mu: X\times_Y X \to X$ and if $P+Q=R$ in $X_F(F)$ then $\mu(s_P,s_Q)=s_R$ by definition. So, as long as the specializations at $y$ make sense, $s_P(y)+s_Q(y)=\mu(s_P,s_Q)(y)=s_R(y)$. This answers 2. contingent to 1. | |
| Jun 12, 2014 at 17:45 | comment | added | wongpin101 | The case I'm interested in is $X_F$ is an algebraic variety over $F$. Let $Y_0$ be the Zariski open subset of $Y$ such that each fiber $X_y$ is an abelian variety for $y \in Y_0$. | |
| Jun 12, 2014 at 17:36 | history | edited | wongpin101 | CC BY-SA 3.0 |
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| Jun 12, 2014 at 17:23 | comment | added | user27920 | Are you perhaps just starting with an abelian variety over $F$ and then "spreading out" to an abelian scheme over a smooth dense open subset of $Y$? The algebraic geometry technique in Lang's book is poorly-suited for relative questions in algebraic geometry (though Lang et al. certainly got some mileage out of it before the whole framework was abandoned in the late 1950's). | |
| Jun 12, 2014 at 17:21 | comment | added | user27920 | It is not true that $s_P$ is a genuine section in general, so the answer to question 1 is "they're not". But if $X$ is an abelian variety then Weil's Extension Theorem ensures that any rational map to $X$ from a $k$-variety is a morphism. But question 2 makes no sense at all since $X_F$ isn't an $F$-group in general. Perhaps if you provide some context/motivation for your question then it will become apparent how to give useful answers to these questions that are somehow each not well-posed. | |
| Jun 12, 2014 at 15:34 | history | asked | wongpin101 | CC BY-SA 3.0 |