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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!

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    $\begingroup$ I believe you can find the answer in this paper: arxiv.org/abs/1402.0031 $\endgroup$
    – Stanley Yao Xiao
    Commented Sep 20, 2015 at 2:19
  • $\begingroup$ 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? $\endgroup$
    – Vesselin Dimitrov
    Commented Sep 20, 2015 at 2:27
  • $\begingroup$ 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 $n=2$. $\endgroup$
    – Joe Silverman
    Commented Sep 20, 2015 at 2:30
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    $\begingroup$ 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). $\endgroup$
    – Jeremy Rouse
    Commented Sep 20, 2015 at 3:08
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    $\begingroup$ 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. $\endgroup$
    – Vesselin Dimitrov
    Commented Sep 20, 2015 at 14:04

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