Can someone give me an example of elliptic curve with CM by sqrt(-7) with the action. I've found a list of examples in the following link but not the action.



I am sure others (magma, etc) can do the same. With sage the following lines will produce the rational functions $\bigl(f(x,y), g(x,y)\bigr)$ representing the multiplication by $(1\pm\sqrt{-7})/2$ on $E$ in the Weierstrass equation given at the link. Once you have that you can get all endomorphisms.

sage: E = EllipticCurve([1,-1,0,-2,-1])
sage: K.<t> = NumberField(x^2+7)
sage: EK = E.base_extend(K)
sage: psis = EK.isogenies_prime_degree(2)
sage: [psi.codomain().is_isomorphic(EK) for psi in psis]
[True, True, False]
sage: psi = psis[0]
sage: iota = psi.codomain().isomorphism_to(EK)
sage: psi.set_post_isomorphism(iota)
sage: psi.rational_maps()

whose $x$-coordinate $f(x,y)$ is $$\frac{(\sqrt{-7} - 3)\cdot x^2 + (-2\, \sqrt{-7} - 2)}{8\,x + \sqrt{-7} + 5}$$

  • 3
    $\begingroup$ The formula for the CM by $(1+\sqrt{-7})/2$ is much simpler than the formula for CM by $\sqrt{-7}$, and the former is actually more useful, since it generates the full endomorphism ring, while $\mathbb{Z}[\sqrt{-7}]$ is only a subring. (So you statement that from $[\sqrt{-7}]$ you can get all endomorphisms is not accurate.) $\endgroup$ – Joe Silverman Jan 6 '14 at 13:33
  • $\begingroup$ Absolutely right. I added the simpler, but I will now delete the earlier. $\endgroup$ – Chris Wuthrich Jan 6 '14 at 13:38
  • $\begingroup$ Maybe the sage session will render better and highlight if you add four spaces in the beginning of code. $\endgroup$ – joro Jan 6 '14 at 14:59
  • $\begingroup$ Just a note (mostly for myself): the desired endomorphism must have degree 2 by Corollary II.1.5(b) in Silverman's "AAEC" book. (That's also why $\left[ \sqrt{-7} \right]$ has degree 7 and hence is more complicated...). $\endgroup$ – Watson May 7 at 19:22

See J. Silverman, Advanced Topics in the Arithmetic of Elliptic Curves (Springer GTM), Proposition 2.3.1. Note that since the class number of $\mathbb{Q}(\sqrt{-7})$ is 1, this is the only example defined over $\mathbb{Q}$.

  • 4
    $\begingroup$ Ha!! I was in the middle of typing the same reference when your answer appeared. So you win! Hint for OP: The curve has an endomorphism of degree 2 corresponding to $(1+\sqrt{-7})/2$. There's a general formula for degree 2 isogenies $E\to E'$, so one need merely use that formula and make the additional assumption that $j(E)=j(E')$ to find the desired equation. $\endgroup$ – Joe Silverman Jan 6 '14 at 13:26
  • $\begingroup$ Sorry to steal the answer from you! $\endgroup$ – abx Jan 6 '14 at 13:28
  • $\begingroup$ thanks for the help. going through the above mentioned book now. do I need to know any kind of programming language to get the exact action (not particularly for \sqrt(-7) but for any CM)? $\endgroup$ – user45145 Jan 9 '14 at 7:23

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