# $K$-rational points on an elliptic curve

Let $$E$$ be an elliptic curve defined over a number field $$K$$. Denote with $$E(K)$$ the set of $$K$$-rational points. Is $$E(K)$$ always a cyclic group?

My attempt:

I think this is not true and I am showing it taking $$K=\mathbb{Q}$$ and $$E : y^2=x^3+3x$$. Now $$P=\Big(\frac{7}{8},\frac{1}{4}\Big) \in E(\mathbb{Q})$$ and $$P$$ can't be a torsion point by Nagell-Lutz, hence $$E(\mathbb{Q})$$ has rank at least $$1$$. Moreover we have $$E(\mathbb{Q})_{torsion} \neq \emptyset$$, since $$(0,0)$$ is a rational $$2$$-torsion point. We get that $$E(\mathbb{Q})$$ is not cyclic in this case.

Am I right?

• It's quite easy to write down an elliptic curve over $\Bbb Q$ with three rational $2$-torsion points. – Lord Shark the Unknown Apr 4 at 21:04
• @LordSharktheUnknown and what happens if I find such elliptic curve? That I can just write as $y^2=(x-e_1)(x-e_2)(x-e_3)$ with $e_1 < e_2 < e_3$? – user289143 Apr 4 at 21:06

There is no need to use points of infinite order. Take $$E/K : y^2=x(x-1)(x-2)$$. Then, the torsion subgroup over $$K$$ contains $$E[2] = \langle (0,0),(1,0)\rangle \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$, and therefore $$E(K)$$ cannot be cyclic.
You are correct. Taking $$K=\mathbb{Q}$$ there are known to exist curves of rank more than $$1$$ and in fact it is conjectured that there exist curves of arbitrarily large rank. Your method is correct: a point with non integer rational coordinates will always by outside the torsion subgroup in general and thus to show an elliptic curve has positive rank it is sufficient to show the existence of one such point.
• Indeed! Thank you for pointing out! The fact the average rank is proven to be bounded above by $1.5$ (to my best knowledge) and the rarity of examples with rank larger then $20$ certainly doesn’t help – Μάρκος Καραμέρης Apr 4 at 21:21