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Let $F:V\to W$ be a birational transformation of elliptic curves; let $g$ be a generator of $V$. Is necessarily $F(g)$, a generator of $W$?

Surb
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Piquito
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  • What do you mean by a generator of an elliptic curve? Are you looking at the group of $\Bbb{F}_q$-rational points on $V$, and that happens to be a cyclic? Or what? – Jyrki Lahtonen Mar 01 '15 at 22:05
  • I mean as a generator an element of the elliptic curve the number of all of them is the rank of the curve. Yhanks for your answer sir Laktonen. – Piquito Mar 03 '15 at 17:11

1 Answers1

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No, this is not necessarily the case. Here is a silly example. Using the same argument one can come up with more exotic examples.

An elliptic curve is a pair $(E,\mathcal{O})$ consisting of a smooth projective curve $E$ of genus $1$, and a point $\mathcal{O}$ that will serve as identity of the group. Suppose $V=(E,[0:1:0])$, where $E: y^2=x^3-2$, with $\mathcal{O}=[0:1:0]$ the point at infinity. The group $V(\mathbb{Q})\cong \mathbb{Z}$, generated by $P=(3,5)=[3:5:1]$. Now let $W=(E,P)$, and $F:V\to W$ to be the identity map, i.e., $F(x,y)=(x,y)$. Then, $F(P)=P$, but $P\in V$ has infinite order and generates $V(\mathbb{Q})$, while $P\in W$ has order $1$, being the identity element of $W$, so $F(P)$ does not generate $W(\mathbb{Q})$.

Álvaro Lozano-Robledo
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  • Hi Álvaro, Did you know that your E was given by Fermat as a challenge to england mathematicians asking to prove that your point P is the only integer point of E? – Piquito Apr 27 '15 at 17:37
  • Yes, that's well-known. – Álvaro Lozano-Robledo Apr 27 '15 at 22:12
  • Yes, Álvaro, by the way, it was time of “cold war” between Cartesian and Newtonian. Your answer leads me to the following: Change of zeros in an elliptic curve C can become C in quite another one, even not birationally equivalent by the identity? It is maybe true but sounds to me not “euphonic”. Regards. – Piquito Apr 28 '15 at 14:19