Timeline for Q(\sqrt{-l_0}) satisfies Heegner hypothesis for an Elliptic curve of conductor C implies C is a square
Current License: CC BY-SA 3.0
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| Feb 27, 2015 at 0:59 | comment | added | tracing | In the context of GZ, the root number over $K$ is always $-1$. This is how one knowns that the $L$-function vanishes at $s = 1$, and why it makes sense to try to find a formula for its derivative (which is then given by the GZ formula). | |
| Feb 26, 2015 at 7:55 | history | edited | George Turcas | CC BY-SA 3.0 |
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| Feb 26, 2015 at 7:45 | comment | added | George Turcas | Thank you very much for your response, but as far as I understand, GZ just tells us that the $L'(E/K,1) \neq 0$. To be able to say that $L(E/K,s)$ vanishes at $s=1$, we need to look at something else. Is there something I miss? Thank you again | |
| Feb 26, 2015 at 1:40 | answer | added | Kestutis Cesnavicius | timeline score: 1 | |
| Feb 26, 2015 at 0:42 | comment | added | tracing | ... you can just choose $l_0$ such that $C$ splits in $K$. | |
| Feb 26, 2015 at 0:41 | comment | added | tracing | As far as I can tell, the argument uses Gross--Zagier: over $K$ one deduces from the fact that Heegner point has infinite order that the $L$-function has a simple zero (this is Gross--Zagier), thus the the sign in the fun'l equation over $K$ is $-1$. This is the product of the signs for each of $E$ and its twist. Earlier in the argument the sign for $E$ was shown to be $1$, and so the sign for the twist must be $-1$. As to why the root number over $K$ is always $-1$, you can look in GZ for a proof. Why should the Heegner hypothesis imply that $C$ is a square? E.g. if $C$ is prime, ... | |
| Feb 25, 2015 at 23:42 | review | First posts | |||
| Feb 25, 2015 at 23:44 | |||||
| Feb 25, 2015 at 23:42 | history | asked | George Turcas | CC BY-SA 3.0 |