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Let $n$ be a positive integer, which is $4$ or a prime number $l$. Let $E$ be an elliptic curve defined over a number field $K$. Assume that all the $n$-torsion points of $E$ are defined over $K$, i.e. $E$ admits a level-$n$ structure. I have the following questions:

1. Let $v$ be a non-archimedean valuation of $K$, with residue characteristic $p>0$. It is well known that if $n=l$ is a prime number $\neq p$, then $E$ has good or multiplicative reduction at $v$. I am wondering when $n=4, p\neq 2$, does $E$ still have good or multiplicative reduction at $v$? I am especially interested in the case when $p=3$ and $E$ has a model over $\mathbb{Q}$.

2. Let $E'$ be another elliptic curve over $K$, such that $E \times_K \bar{K}$ is $\bar{K}$-isomorphic to $E' \times_K \bar{K}$. Then for $E$ with the level-$n$ structure, do we have a $K$-isomorphism between $E$ and $E'$? If not, can some more properties of $K$ result in the $K$-isomorphism between $E$ and $E'$?

Thank you for the reading.

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1 Answer 1

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  1. Let $K$ be a finite extension of $\mathbb{Q}_p$ with $p\neq 2$. Suppose that $E/K$ is an elliptic curve with additive reduction and such that $E$ has full $4$-torsion over $K$. By the Kodaira classification, the group of components of the Néron model $\mathcal{E}/\mathcal{O}_K$ cannot contain $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$. Therefore, we would have a point of order $2$ in the identity component $\mathcal{E}^0(\mathcal{O}_K)$. The reduction is additive, so the group of non-singular points is a $p$-group, which means that our $2$-torsion point must belong to the kernel of reduction, i.e., the formal group. However the torsion subgroup of a formal group over $\mathcal{O}_K$ is also a $p$-group. This is a contradiction and answers 1) with "yes".

  2. Take any elliptic curve $E/K$ with full $n$-torsion over $K$ and take for $E'$ a quadratic twist by a non-square $D\in K^\times$. Then $E$ and $E'$ are not $K$-isomorphic. This is "no" for the first question in 2. If $n=\ell\neq 2$ is prime and both $E$ and $E'$ have full $n$-torsion, then this is enough to conclude that $E'$ cannot be a non-trivial quadratic twist of $E$, since $E(L)[n] \cong E(K)[n] \oplus E'(K)[n]$ for $L/K$ the corresponding quadratic extension would not work. Along these lines you can get a property that forces $E'$ to be $K$-isomorphic to $E$.

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