# Elliptic curve with only one point

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.

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 $$E(K) = \{0\}$$.
• $y^2=x^3-5$ seems to work as well – byk7 Nov 10 '18 at 15:08
• 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. – KCd Nov 10 '18 at 16:09