Timeline for divisibility of Tamagawa numbers
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14 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2014 at 10:44 | comment | added | And85 | Maybe this argument should work: (I would post it as a comment but I am new here and I cannot) The newform on $\Gamma_0(N)$ associated to $E$ is $p$-isolated since $p$ does not divide the degree of the modular parametrisation. If $p$ divide a Tamagawa number, then you are in the setting of Ribet's lowering the level theorem. So the form should be congruent to form of lower level, contradicting the fact that $f$ is $p$-isolated. | |
| Jan 7, 2014 at 9:28 | comment | added | Xialu | Dear prof. Wuthrich, thanks for you sketch. Actually I am working with an elliptic curve with analytic rank one, so for I can fix a rank. After the "positive" answer to this question I am planning to study the topic and try to get a proof. | |
| Dec 25, 2013 at 14:52 | comment | added | Chris Wuthrich | Anyway, sorry for my sketchy comment. I think this shows that someone should find a proof for this conjecture, directly without using anything on the analytic rank. | |
| Dec 25, 2013 at 14:50 | comment | added | Chris Wuthrich | ... because there may be ramification. For instance if the conductor $N$ is odd, then $X^0$ is the image of the path from $0$ to $\infty$ then to $1/2$. So the full integral over $X^0$ is $(1-N_2)\cdot [0]$ where $N_2$ is the number of points in the reduction at $2$. So $[0]>0$ but the above is negative, i.e. there will be more than $N_2-1$ preimages of $O$. | |
| Dec 25, 2013 at 14:44 | comment | added | Chris Wuthrich | Now, here is one way to attack the problem if the analytic rank is $0$. Yet I have not been able to solve it completely. Assume BSD (not really necessary if $p$ is nice). The assumptions imply that $[0] = L(E,1)/\Omega$ is divisible by $p$. Now the image of the path from $\infty$ to $0$ on $X =X_0(N)$ is part of the connected component $X^0$ of $X(\mathbb{R})$ containing $0$ and $\infty$. Probably the full integral around this "circle" $X^0$ is a multiple of $[0]$ and hence $C \Omega$ for some integer divisible by $p$. However we can't conclude that there are $C$ preimages of $O$ on $X^0$ ... | |
| Dec 25, 2013 at 14:37 | comment | added | Chris Wuthrich | I would even conjecture : If $p$ does not divide the order of the torsion subgroup of $E(\mathbb{Q})$, but divides a Tamagawa number, then the degree of the modular parametrisation is divisible by $p$. That is slightly stronger, but seems to be ok numerically. | |
| Dec 17, 2013 at 22:29 | comment | added | Tim Dokchitser | Actually there are many curves in the Cremona database whose Tamagawa numbers are divisible by $p\ge 11$, e.g. 147b2 ($p=13$), 190a1 ($p=11$), 262a1 ($p=11$) etc. There are $>6000$ of them of conductor up to 36000, and the claim is true for all of them, even without the "good ordinary at $p$" assumption, i.e. $p$ divides the degree of the modular parametrization in all these cases. So computationally the claim seems very feasible. | |
| Dec 17, 2013 at 18:04 | comment | added | Olivier | Do you have any reason to believe that the answer to your question could be affirmative? It seems very unlikely to me that this is the case but constructing a counterexample might be computationally involved | |
| Dec 17, 2013 at 17:40 | comment | added | Joe Silverman | @stankewicz There's no need to go to quadratic fields. Just take $E:y^2+xy=x^3-36/(j-1728)x-1/(j-1728)$ and substitute $j=1/p^N$ for some prime $p\ge5$. Since $j(E)=j$, its easy to check that $E$ has split multiplicative reduction at $p$ and $\text{ord}_p(\mathcal{D}_E)=N$. So that gives examples over $\mathbb{Q}$ with $c_p(E)$ arbitrarily large. | |
| Dec 17, 2013 at 17:15 | comment | added | stankewicz | Thank you, that was very silly of me. Explicitly, Reichert's paper on torsion structures over quadratic fields gives an example with p= 13. | |
| Dec 17, 2013 at 16:41 | comment | added | Joe Silverman | @stankewicz If $E$ has split multiplicative reduction at $\ell$, then you are correct that $c_\ell(E)=\text{ord}_\ell(\mathcal{D}_E)$, where $\mathcal{D}_E$ is the minimal discriminant. But this valuation can be arbitrarily large, it is not $\le12$ as you assert. So the question is not only about $p=11$. | |
| Dec 17, 2013 at 16:32 | comment | added | stankewicz | A number of people commented that there is a counterexample for $p=5$, which is of course not quite what you're asking about. I'll just note that the Tamagawa number of an elliptic curve over a dvf is essentially the valuation of the minimal discriminant (and thus $\le 12$) if it has split multiplicative reduction and something else (not a large prime) otherwise. So we're really talking about $p=11$ on the nose here and having sage search through (some fraction of) the Cremona database didn't find anything with tamagawa number divisible by 11. | |
| Dec 17, 2013 at 11:26 | review | First posts | |||
| Dec 17, 2013 at 11:29 | |||||
| Dec 17, 2013 at 11:07 | history | asked | Xialu | CC BY-SA 3.0 |