Timeline for On the conductor of the Groessencharacter of a CM elliptic curve
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 15, 2014 at 3:24 | vote | accept | Hugo Chapdelaine | ||
| Jul 15, 2014 at 3:18 | comment | added | Hugo Chapdelaine | I see, so then may be I should add an hypothesis like semi-stability and then this means the $E/L$ should have good reduction everywhere... | |
| Jul 15, 2014 at 0:01 | history | edited | GH from MO |
edited tags
|
|
| Jul 14, 2014 at 23:24 | answer | added | Will Sawin | timeline score: 3 | |
| Jul 14, 2014 at 23:19 | comment | added | Kestutis Cesnavicius | An isogeny over $\overline{\mathbb{Q}}$ need not be defined over $K$. Another way to see that the answer to both of your questions is `no' is to take a single CM elliptic curve over $L$ and look at the family of its quadratic twists (in which the (norms of the) conductors of its members are unbounded). | |
| Jul 14, 2014 at 19:09 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
added 22 characters in body
|
| Jul 14, 2014 at 19:07 | comment | added | Hugo Chapdelaine | So if we look at the set of CM elliptic curves by $K$defined over $\overline{\mathbf{Q}}$, then since they are all isogeneous, they share the same set of primes of good reduction. So is it possible to determine the set of primes of bad reduction (necessarily additive since CM) stricly in terms of $K$. For example if $K$ has class number one a naive guess would be to say that the conductor is supported on primes dividing $12\cdot disc(K)$.... But this is probably to naive. | |
| Jul 14, 2014 at 16:19 | comment | added | Ari Shnidman | The conductor of $E$ is the square of the conductor of $\psi$. Even if you fix the $j$-invariant, there are infinitely many iso classes of CM elliptic curves over $L$, and only finitely many of bounded conductor (by Shafarevich's theorem, for example). | |
| Jul 14, 2014 at 15:33 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |