Timeline for On discriminants of elliptic curves
Current License: CC BY-SA 3.0
12 events
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| Sep 20, 2015 at 14:22 | comment | added | Jackson Morrow | @JoeSilverman My apologies for the poor phrasing of my question. I have made the appropriate changes over. Yes, you and Vesselin Dimitrov are both correct that I do want to consider m1 and m2 to be co-prime. Also, thank you for your comments about what happens when $\gcd(m_1,m_2) = d$. | |
| Sep 20, 2015 at 14:18 | history | edited | Jackson Morrow | CC BY-SA 3.0 |
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| Sep 20, 2015 at 14:15 | comment | added | Jackson Morrow | @JeremyRouse Yes I did mean to ask ``Did you mean to ask what proportion have non-square discriminant? " In which case, your answer does make total sense. I believe that using your answer and @VesselinDimitrov 's I know have a much better understanding of when this type of entanglement can occur. | |
| Sep 20, 2015 at 14:13 | comment | added | Jackson Morrow | @VesselinDimitrov Once again, you are completely correct! I did not think about what the possible value of $n$ is for which $\mathbb{Q}(\sqrt{\Delta_E}) \subset \mathbb{Q}(\zeta_n)$. As you mention, if $n$ is even, then I don't count $\mathbb{Q}(E[2])$ and $\mathbb{Q}(E[n])$ as entangled. Once again, thank you for your comments. | |
| Sep 20, 2015 at 14:09 | comment | added | Jackson Morrow | @StanleyYaoXiao Thank you for the reference. The article did answer my poorly phrased question. As it turns out, I did in fact mean non-square discriminant, however, it is nice to know about such a result. | |
| Sep 20, 2015 at 14:04 | comment | added | Vesselin Dimitrov | One other typo: It is not true that $\mathbb{Q}(E[2]) \subset \mathbb{Q}(\zeta_n)$ (it is only solvable, not abelian over $\mathbb{Q}$). Only $\mathbb{Q}(\sqrt{\Delta}_E)$ is contained in a cyclotomic field. And then, the minimum $n$ can be odd or even with frequency $1/2$ each, and in the latter case you don't count $\mathbb{Q}(E[2])$ and $\mathbb{Q}(E[n])$ as entangled, so the statement in your second paragraph is true only half of the time. | |
| Sep 20, 2015 at 13:59 | history | edited | Jackson Morrow | CC BY-SA 3.0 |
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| Sep 20, 2015 at 3:08 | comment | added | Jeremy Rouse | Did you mean to ask what proportion have non-square discriminant? If so, the answer is "100%", as those curves with square discriminant are those with j-invariant in the image of $1728+t^2$ (and is hence a thin set). | |
| Sep 20, 2015 at 2:30 | comment | added | Joe Silverman | Did you mean to put some conditions on $m_1$ and $m_2$. Certainly you don't want $m_1=m_2$. But actually, if $d=\gcd(m_1,m_2)$, then your intersection always trivially contains $\mathbb Q(E[d])$. So the interesting case is probably when $\gcd(m_1,m_2)=1$. Also your proof of your example needs to note that if the 2-torsion is rational, then you want to take $n=1$, not $n=2$. | |
| Sep 20, 2015 at 2:27 | comment | added | Vesselin Dimitrov | By "non-trivial" in the first paragraph, do you mean strictly larger than $\mathbb{Q}(E[\mathrm{gcd}(m_1,m_2)])$? (Or do you consider $m_1,n_1$ co-prime in the definition of entanglement?) But then it seems to me some care is needed in the second paragraph. It could happen that the conductor $n$ of the quadratic field is even; in this case we simply have $E[2] \subset E[n]$, and the division fields are not entangled. Do I miss something? | |
| Sep 20, 2015 at 2:19 | comment | added | Stanley Yao Xiao | I believe you can find the answer in this paper: arxiv.org/abs/1402.0031 | |
| Sep 20, 2015 at 1:44 | history | asked | Jackson Morrow | CC BY-SA 3.0 |