Let $E/\mathbb{Q} = E_{a,b}$ $$\displaystyle y^2 = x^3 + ax + b, a,b \in \mathbb{Z}$$ be an elliptic curve defined over the field of rational numbers, and let $n \geq 3$ be an integer. Let $K_n$ be the $n$-torsion field of $E$, namely the field obtained by adjoining all of the coordinates of $n$-torsion points of $E$ to $\mathbb{Q}$. Let $p$ be a prime divisor of the discriminant $\Delta(E)$ of $E$. Does it follow that $K_n$ is ramified at $p$? If so, what can be said about the splitting behaviour of $p$ in $K_n$?
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asked Aug 22, 2023 at 12:09
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2$\begingroup$ Dumb example: what if $K_n = \mathbf Q$, i.e. all points of $E[n]$ are defined over $\mathbf Q$? $\endgroup$– R. van Dobben de BruynCommented Aug 22, 2023 at 13:47
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$\begingroup$ Another counter example is a prime $n=p$ of split multiplicative reduction such that $p$ divides the Tamagawa number $c_p$. There is a lot of literature on these questions and googling will reveal some references. $\endgroup$– Chris WuthrichCommented Aug 22, 2023 at 14:28
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4$\begingroup$ If $n>2$ then $\mathbf{Q}$ does not contain a primitive $n$th root of unity. In light of the nondegeneracy and Galois equivariance of the Weil pairing, not all points of $E[n]$ are defined over $\mathbf{Q}$. Hence, the ``Dumb example" does not work. $\endgroup$– Yuri ZarhinCommented Aug 22, 2023 at 16:52
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$\begingroup$ Edited in view of @YuriZarhin's comment $\endgroup$– Stanley Yao XiaoCommented Aug 22, 2023 at 21:47
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$\begingroup$ @YuriZarhin ah right, I was indeed thinking about $n=2$. But when there are trivial counterexamples, there are almost certainly nontrivial ones :) $\endgroup$– R. van Dobben de BruynCommented Aug 22, 2023 at 23:50
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2 Answers
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No. The modular curve $X_E(n)$, which paramaterizes pairs $(E',\iota)$, where $E'$ is another elliptic curve and $\iota \colon E'[n] \to E[n]$ is an isomorphism (of group schemes, or equivalently, of Galois modules) is a twist of $X(n)$, and thus has genus 0 for $n = 2,3,4,5$ and genus 1 for $n = 6$. In particular, for $n = 3,4,5$, $X_E(n)$ is isomorphic to $\mathbb{P}^1$ and has infinitely many rational points; i.e., there are infinitely many $E'/\mathbb{Q}$ such that $\mathbb{Q}(E'[n]) = \mathbb{Q}(E[n])$. But $\mathbb{Q}(E[n])$ is unramified outside of the primes dividing the conductor of $E$ or $n$, and most such $E'$ have bad reduction at new primes.
For primes greater than 6, the genus of $X(p)$ is at least 3. Anyway, this is related to the Frey--Mazur conjecture; but even if that conjecture is true, I don't think it implies what you're asking for $n > 6$. I do think what you are asking is true for large $n$ (independent of $E$), but as far as I know this is unknown.
answered Aug 22, 2023 at 23:54
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Going in the other direction, the Néron-Ogg-Shafarevich criterion and Weil pairing imply that the Tate module $T_{\ell}E$ is a Galois representation which is ramified at $p$. So if $n$ is large enough, then $p$ does indeed ramify in $K_n$ (because $n$ is then either divisible by a prime $\ell$ sufficiently large so that no level-lowering can occur or by a power of a small prime $\ell$ large enough so that the image of the inertia group at $p$ is not contained in the kernel of reduction modulo $\ell^{n}$). But I have the feeling you already knew that.
The splitting question seems very hard to me.
