Timeline for About isogeny theorem for elliptic curves
Current License: CC BY-SA 2.5
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| Oct 13, 2010 at 4:16 | history | edited | Torsten Ekedahl | CC BY-SA 2.5 |
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| Oct 13, 2010 at 1:06 | comment | added | natura | @Emerton. Yes, you are right, it's all tensored with $\mathbb{Q}$. Thank you! | |
| Oct 12, 2010 at 23:54 | comment | added | Emerton | (In the example of $X_0(11)$, the ground field should be taken to be $\mathbb Q$.) | |
| Oct 12, 2010 at 23:53 | comment | added | Emerton | Dear Dror, This has nothing to do with primes of bad reduction. It has to do with those primes $p$ for which $E$ admits a $p$-isogeny, which is an {\em a priori} unrelated set of primes. E.g. if $E = X_0(11)$, it admits a $5$-isogeny (and no $p$-isogeny for any other $p$). | |
| Oct 12, 2010 at 23:47 | comment | added | Emerton | ... not even be abstractly isomorphic as $G_K$-modules, at least in the non-CM case.) | |
| Oct 12, 2010 at 23:46 | comment | added | Emerton | Dear Basic, Your assertion that "The Tate modules ... are isomorphic ... if [they are so] for just one prime" is not true. (I don't have Serre's book in front of me, but I presume he states this after tensoring with $\mathbb Q$.) E.g. if $E_1 \to E_2$ via a cyclic 5 isogeny (think about $X_1(11) \to X_0(11)$) then this will induce an isomorphism of $\ell$-adic Tate modules for every $\ell \neq 5$, but will not induce an isomorphism of $5$-adic Tate modules. (It will induce an embedding of one into the other, and the image will be of index 5. Furthermore, the $5$-adic Tate modules will ... | |
| Oct 12, 2010 at 22:24 | comment | added | natura | @Ekedahl. Thank you for the answer, though I need some time to think about it. But also, for my original question, I was thinking if it is possible to characterize the elliptic curves just by its Tate modules and some other invariants ( As we already see, Tate modules fully characterize elliptic curves up to isogeny). What you described seems to be some kind of description of the isogenous class of elliptic curves, right? | |
| Oct 12, 2010 at 22:14 | comment | added | natura | The Tate modules of two elliptic curves are isomorphic G_K modules for all primes if and only if for just one prime. (See Serre's Abelian l-adic Galois representations, Page IV-15). | |
| Oct 12, 2010 at 21:42 | comment | added | Dror Speiser | Oops. Of course. The latter hints that there is actually a finite list of primes that need to be checked. I'll throw a guess once more: primes that divide conductor are enough? | |
| Oct 12, 2010 at 21:33 | comment | added | Torsten Ekedahl | The kernel of the isogeny is finite and hence can not be an ideal in $\mathrm{End}(E_1)$. If $E_1$ and $E_2$ are just isogeneous, then their Tate modules are isomorphic for all but a finite number of $\ell$'s so density $1$ is not sufficient. | |
| Oct 12, 2010 at 21:30 | history | edited | Torsten Ekedahl | CC BY-SA 2.5 |
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| Oct 12, 2010 at 21:29 | comment | added | Dror Speiser | Also, in the spirit of the question, do we need all Tate modules to be isomorphic? Or is it enough that for a positive density of $\ell$'s they are isomorphic? | |
| Oct 12, 2010 at 21:27 | comment | added | Dror Speiser | If I understand it correctly, in simpler terms, the above means: The two are isomorphic if and only if the kernel of the isogeny is a principal ideal in $\text{End} (E1,E2)$ (which is independent of which isogeny we take). Is this correct? | |
| Oct 12, 2010 at 20:00 | history | edited | Cam McLeman | CC BY-SA 2.5 |
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| Oct 12, 2010 at 19:58 | history | edited | Torsten Ekedahl | CC BY-SA 2.5 |
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| Oct 12, 2010 at 19:33 | history | answered | Torsten Ekedahl | CC BY-SA 2.5 |