## Moving the origin of an elliptic curve

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Suppose one has an elliptic curve $E = (C,O)$ over a field $k$ where $C$ is a non-singular genus one curve over $k$ and $O$ is a $k$-rational point on $C$. By moving $O$ on $C$ one gets a family of elliptic curves. Is this family trivial in the sense that all the curves are mutually isomorphic? If not (as I suspect), what are the isomorphism classes of elliptic curves one gets this way and how are they related to the original curve $E$?

Special cases of $k = \mathbb{C}$ and $\mathbb{Q}$ would be specially helpful to know about, for $\mathbb{C}$ in terms of lattices, and for $\mathbb{Q}$ in terms of the associated modular forms after Wiles et al.

(Depending on the answer) should CM curves be distinguished?

Thanks!

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Yes, they're isomorphic, and this is true for any algebraic group, because translation by a given ($k$-rational) group element is an isomorphism of algebraic varieties. This should be particularly easy to visualize in the case of ${\bf C}/L$ (translation by some complex number modulo the lattice $L$). – Noam D. Elkies Sep 27 2011 at 20:01
$C(k)$ is a group, and it acts on itself by multiplication. So if you have $E=(C,O)$ and $E'=(C,O')$ as you suggest on your question, wouldn't it be enough to look at the action of $C' - O$ ? – Reimundo Heluani Sep 27 2011 at 20:05

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Is this a fishing expedition? – András Bátkai Sep 27 2011 at 22:01
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