Timeline for An invariant subspace under $G_Q$ action , and BSD-rank
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
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| May 7, 2021 at 11:50 | comment | added | Yuan Yang | @Chris Wuthrich Yes, you are right... In this case if an element $P$ in $E(\bar{Q})\otimes Q$ is stable under $G_Q$ action, then there is an integer $n$ such that $nP$ in $E(\bar{Q})$ is stable under $G_Q$ action. So they are indeed equal. But generally, are tensor functor and taking invariant functor always commute with each other? | |
| May 7, 2021 at 8:37 | comment | added | Chris Wuthrich | Isn't your space just $E(\bar{\mathbb{Q}})\otimes_{\mathbb{Z}} \mathbb{Q}$ as the map from $E(\bar{\mathbb{Q}})$ to it is surjective? Then it is clear that the Galois-invariant part is $E(\mathbb{Q})\otimes \mathbb{Q}$, isn't it? | |
| May 7, 2021 at 7:43 | history | asked | Yuan Yang | CC BY-SA 4.0 |