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Let's we have an elliptic curve (EC). Is it possible to construct group $G$ acting on the points of an EC with this property: if $P$ is a rational point on EC then $G(P)$ is also is a rational point? Of course, I means that $G$ is not reduced to addition of $P$ with itself. In other words, can we construct symmetry group?

Michael Galuza
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  • Do you mean, given the curve, can we determine the group of symmetries? And what sort of symmetries? Do they need to be maps of varieties, or just any maps? – Tobias Kildetoft Jun 01 '15 at 19:57
  • @tobias, I meant that group constructed over EC like Galois group over polynom – Michael Galuza Jun 01 '15 at 20:05
  • We just want generate new rational points on EC by group action, not by addition. Nothing more of the group $G$ is not required, so you are right, you are damn right) – Michael Galuza Jun 01 '15 at 20:18

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If $E/K$ is an elliptic curve over a field $K$, then you can construct the automorphism group of $E$, but it is not terribly exciting... $\text{Aut}(E)$ is a finite group of order dividing $24$, and if $j\neq 0,1728$, then $\text{Aut}(E) = \{\pm 1\}$. See Chapter III.10 of Silverman's "The Arithmetic of Elliptic Curves".

If you restrict yourself to torsion subgroups, however, then there are more exciting automorphism groups. Let $E/K$ be an elliptic curve over a field of characteristic $0$, and let $n\geq 2$. Let $E[n]$ be the $n$-torsion subgroup of $E(\overline{K})$. Then, one can construct $\text{Aut}(E[n])\cong \text{Gal}(K(E[n])/K)$ and this group is a subgroup of $\text{GL}(2,\mathbb{Z}/n\mathbb{Z})$. If $n=p$ is a prime number, $K$ is a number field, and $E/K$ is fixed, then we know that $\text{Gal}(K(E[p])/K)\cong \text{GL}(2,\mathbb{Z}/p\mathbb{Z})$ for all but finitely many primes $p$ (a consequence of the so-called Serre's open image theorem).

Álvaro Lozano-Robledo
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  • Very thanks. Automorphism group is cool, but idea was not in it. If we have point $P$ and group $G$ (which determines by coefficients of EC), we can generate new points. For example, $G(P) = P + Q$, and $Q$ is a some point. We are not interested in all points on EC. This idea was inspired Markov diophantine equation: if we have solution $(x, y, z)$, then $(x, y, 3xy-z)$ is also solution. Can we generalize this on elliptic curve case? – Michael Galuza Jun 01 '15 at 21:14
  • I don't understand your notation. If $G$ is a group, what is $G(P)$? Usually if a group $G$ acts on a set $X$, then this means that $g\cdot x$ is in $X$, for any $g\in G$ and $x\in X$. – Álvaro Lozano-Robledo Jun 01 '15 at 21:29
  • Sorry for unclear notation and awful English) At first I wanted to demand that $G$ was arbitrary transformation, but then I decided it would be better if these transformations will be a group. Because if we look at all the points on the curve, it is just a permutation of points, and the question is only in the presentation. Therefore, we do not demand that it was necessarily connected with the operation of addition. – Michael Galuza Jun 02 '15 at 06:36