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Is there an elliptic curve $E$ over an infinite field $K$ such that $E(K)=\{\infty\}$?

My original task was to find an elliptic curve over some field $K$ with only one point, which I did for $K=\mathbb{F}_2.$ Now, I'm curious about the case of infinite cardinality, which I am not able to handle.

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According to this database, the elliptic curve $$E : y^2z=x^3−108z^3$$ has only $[0:1:0]$ as rational point, i.e. $E(\Bbb Q)$ is the trivial group. Further examples are given here.

(Notice that if $K$ is an algebraically closed field (hence infinite), then $E(K)$ is always infinite, since it contains $(\Bbb Z/n \Bbb Z)^2$ for every $n \geq 1$, coprime to $\mathrm{char}(K)$ if $K$ has positive characteristic.)


I may add the following related and interesting result, by Mazur and Rubin (theorem 1.1 here): if $K$ is a number field, then there is an elliptic curve $E$ over $K$ with $E(K) = \{0\}$.

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    $\begingroup$ $y^2=x^3-5$ seems to work as well $\endgroup$ – byk7 Nov 10 '18 at 15:08
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    $\begingroup$ If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(\mathbf Z/p\mathbf Z)^2$ even when $K$ is algebraically closed. $\endgroup$ – KCd Nov 10 '18 at 16:09

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