Is $E_{\operatorname{Tor}}(\overline{\mathbb{Q}})$ Zariski-closed?

Question

Let $K$ be a number field of finite degree over $\mathbb{Q}$. Let $E$ be an elliptic curve with coefficients in $K$. Then $E_{\operatorname{Tor}}(L)$ is finite for every algebraic extension $L/K$ of finite degree, hence Zariski-closed.

Is $E_{\operatorname{Tor}}(\overline{\mathbb{Q}})$ Zariski-closed as well?

It's infinite proper subset, so no, since $E$ has dimension $1$. — Wojowu, Mar 21 '20 at 18:36
@Wojowu that looks like an answer to me! Would you care to record it below? — KReiser, Mar 22 '20 at 08:05

score 1 · Accepted Answer · answered Mar 22 '20 at 10:32

$E$ is an algebraic curve, so it has Krull dimension 1. This means that any Zariski closed subset of $E(K)$ (for any field $K$) is either finite or all of $E(K)$. Now, if we let $K=\overline{\mathbb Q}$ and consider the subset of torsion points, then on one hand this subset is infinite (e.g. because it contains points of arbitrary order), but is not equal to all of $E(K)$ - that every elliptic curve over a number field has an algebraic point of infinite point is true, but somewhat tricky, see here for some references.

In fact the set of torsion points of an abelian variety over an algebraically closed field is always Zariski dense. — Pol van Hoften, Mar 22 '20 at 18:03

Is $E_{\operatorname{Tor}}(\overline{\mathbb{Q}})$ Zariski-closed?

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