Let $G$ be a finite abelian group.
Is there a field $K$, and an elliptic curve $E$ over $K$ such that $E(K)_{tor} \cong G$?
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4$\begingroup$ I assume that you do realize that a necessary condition is that $G\cong \mathbb Z/n\mathbb Z \times \mathbb Z/mn\mathbb Z$ for positive integers $n$ and $m$. Also, will you be satisfied with any field $K$, or do you want a field $K$ of a particular type, e.g., $K$ a finite field or $K$ a number field? $\endgroup$– Joe SilvermanSep 16, 2017 at 17:09
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1$\begingroup$ @JoeSilverman Yes, there are obvious counterexamples. I should have been clear about that. And preferably $K$ is a number field, but I'm open other suggestions. $\endgroup$– Eva_MWSep 16, 2017 at 21:35
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2 Answers
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If $n \geq 1$ and $2|mn$ then the Tate curve $E = L^*/2^{\mathbb Z}$ with $L={\mathbb Q}_2(\zeta_{mn}, 2^{1/n})$ has $E(L)_{tor} = {\mathbb Z}/n{\mathbb Z} \times {\mathbb Z}/mn{\mathbb Z}$. A 2-adically close elliptic curve over a suitable (i.e. containing the curve's corresponding torsion) number field $K \subset$ algebraic closure of ${\mathbb Q}$ in $L$ also has $E(K)_{tor} = {\mathbb Z}/n{\mathbb Z} \times {\mathbb Z}/mn{\mathbb Z}$.
answered Sep 17, 2017 at 1:59
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Even better, there exists an elliptic curve $E$ over a number field $K$, such that for any group of the form $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/mn\mathbb{Z}$, there is a finite extension $K'/K$ such that $$E(K')_{\text{tors}}\simeq\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/mn\mathbb{Z}.$$ Indeed, take any $(E,K)$ such that the natural Galois representation on the total Tate module of $E$, $$\rho: \text{Gal}(\bar K/K)\to GL_2(\hat{\mathbb{Z}})$$ is surjective (such curves exist by e.g. this paper of Zywina). Now choose a subgroup $$\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/mn\mathbb{Z}\subset E(\bar K)_{\text{tors}}$$ and let $K'$ be the field fixed by its stabilizer under $\rho$. By the surjectivity of $\rho$, the desired subgroup is all that is fixed, and we're done.
answered Sep 17, 2017 at 10:57