Timeline for Ranks of elliptic curves depend only on the field?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 13, 2015 at 6:50 | vote | accept | Pablo | ||
| Aug 6, 2015 at 15:22 | comment | added | Pablo | mathoverflow.net/questions/213033/… mathoverflow.net/questions/212906/… | |
| Aug 6, 2015 at 15:20 | comment | added | Pablo | @NoamD.Elkies Please note that if we confine ourselves to compistums of quadratic extensions, then by a 1974 result of Frey and Jarden it is not possible for all primes to ramify (they prove that an elliptic curve over the compositum of all quadratic extensions of $\mathbb{Q}$ has infinite rank). In light of that I conjecture that if many primes ramify (say with respect to some density) then all elliptic curves should have rank $\infty$. How big can you make the set primes ramifying here? Some related questions are: | |
| Aug 6, 2015 at 15:06 | comment | added | Noam D. Elkies | to Pablo's question: the field will be the compositum of infinitely many quadratic extensions, and thus will have infinitely many ramified prime and a Galois group that's abelian of exponent 2. | |
| Aug 6, 2015 at 14:59 | comment | added | Noam D. Elkies | It seems that this kind of construction should even work unconditionally, replacing Tate-Shafarevich by the Heegner construction (for positive rank) and Kolyvagin etc. or even an explicit 2-descent (for rank zero). | |
| Aug 6, 2015 at 14:31 | comment | added | Pablo | @YonatanHarpaz This sounds amazing! Can you point out some properties of the arising field like how many rational primes ramify there, or what is the Galois group over $\mathbb{Q}.$ | |
| Aug 6, 2015 at 13:17 | comment | added | Lubin | Thanks for this. It’s what I would have expected, but what do I know? | |
| Aug 6, 2015 at 13:15 | history | answered | Yonatan Harpaz | CC BY-SA 3.0 |