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Let $E/\mathbb{Q}$ be an elliptic curve over $\mathbb{Q}$ and $\Delta_E$ denote the discriminant of $E$. We say an elliptic curve has entanglement fields if the intersection of the $m_1$ and $m_2$ division fields $\mathbb{Q}(E[m_1]) \, \cap \, \mathbb{Q}(E[m_2])$ is non-trivial where $\gcd(m_1,m_2) = 1$.

One can show that if an elliptic curve $E$ has non-square discriminant,then $E$ will always have entanglement fields. Indeed, since $\mathbb{Q}(\sqrt{\Delta_E}) \subseteq \mathbb{Q}(E[2])$ and $\mathbb{Q}(\sqrt{\Delta_E})$ is an abelian extension, we have that $\mathbb{Q}(\sqrt{\Delta_E})\subseteq \mathbb{Q}(\zeta_n)$ for some $n$ by Kronecker--Weber. The Weil-pairing tells us that $\mathbb{Q}(\zeta_n) \subseteq \mathbb{Q}(E[n])$, and so $\mathbb{Q}(\sqrt{\Delta_E}) \subseteq \mathbb{Q}(E[2]) \cap \mathbb{Q}(E[n])$.

Edit As mentioned below, the above $n$ from Kronecker--Weber could in fact be even. In this case, we do not consider $E$ to have entanglement.

For a more precise statement see Proposition 22 of Serre's work .

I am giving a presentation on this topic and some related work, and I want to make a ``precise as possible" statement about the quantity of elliptic curves that have such entanglement. Hence, my question is as follows:

What percentage/proportion/density of elliptic curves have non-square discriminant?

Thank you in advance for your time!

Jackson Morrow
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    I believe you can find the answer in this paper: http://arxiv.org/abs/1402.0031 – Stanley Yao Xiao Sep 20 '15 at 02:19
  • 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? – Vesselin Dimitrov Sep 20 '15 at 02:27
  • 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 $n=2$. – Joe Silverman Sep 20 '15 at 02:30
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    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). – Jeremy Rouse Sep 20 '15 at 03:08
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    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. – Vesselin Dimitrov Sep 20 '15 at 14:04
  • @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. – Jackson Morrow Sep 20 '15 at 14:09
  • @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. – Jackson Morrow Sep 20 '15 at 14:13
  • @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. – Jackson Morrow Sep 20 '15 at 14:15
  • @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 $\gcd(m_1,m_2) = d$. – Jackson Morrow Sep 20 '15 at 14:22

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