$ \newcommand{\End}{\mathrm{End}} \newcommand{\Gal}{\mathrm{Gal}} \newcommand{\kb}{\overline{k}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathbb{F}} \newcommand{\Q}{\mathbb{Q}} $

I would like to have an algorithm (possibly very inefficient) that computes the endomorphism ring of a given elliptic curve $E$ over a finite field $k$.

For simplicity, we shall assume that $E$ is ordinary (to avoid maximal orders in quaternion algebras...), so it is enough to compute the conductor of $\End(E)$ in the imaginary quadratic field $K := \Q(\pi)$, where $q = |k|$ and $\pi = \sqrt{a_q^2 - 4q}$. We know that this conductor divides $[O_K : \Z[\pi]]$, the latter being quite easy to compute in SAGE I suppose.

But now, is there a way to check whether, for a given $f \mid [O_K : \Z[\pi]]$, we have $\Z + f O_K = \End(E)$ ? This is where I don't know how to proceed.

I am aware of Kohel's thesis, which involves isogeny graphs, but I'm not sure if one can implement this on SAGE easily.

Ideally, I want to reproduce the table on page 303 here, which lists $\End_{\Bbb F_7}(E)$ for all (isomorphism classes of) elliptic curves over $\Bbb F_7$.

  • 1
    $\begingroup$ You can get some nontrivial information from the "isogenies_prime_degree" method. For instance, this allows you to determine that the endomorphism ring of $y^2 = x^3 + 3x + 2$ over $\mathbb{F}_7$ doesn't contain an element of norm 3. Since we know it's an order in $\mathbb{Q}(\zeta_3)$, and $\mathbb{Z}[\pi] = \mathbb{Z}[3\zeta_3]$, the only possibility is that $End(E) = \mathbb{Z}[\pi]$ in this case. But this falls a bit short of a general algorithm. $\endgroup$ Commented May 23, 2021 at 9:41
  • $\begingroup$ Unrelated to your math question, but possibly of general interest: \operatorname is always better than \mathrm, because "\operatorname{End} E" does the right thing, whereas "\mathrm{End} E" has incorrect spacing. $\endgroup$
    – djao
    Commented May 23, 2021 at 14:05
  • 1
    $\begingroup$ I know you said you don't care about efficiency, but if you change your mind about that, or if someone else reading this does care about efficiency, the most efficient known approach is given in Bisson and Sutherland, arxiv.org/abs/0902.4670, and the idea is to look for products of ideals which would be principal if $f$ is one value and not principal if $f$ is another value, and then do an isogeny computation to see if the product is actually principal. One can tell the difference because a principal ideal induces an isogeny to an isomorphic curve, and a non-principal ideal does not. $\endgroup$
    – djao
    Commented May 23, 2021 at 14:14
  • $\begingroup$ @SamuelLelièvre: yes, thanks, but the link is: ask.sagemath.org/question/57238 $\endgroup$
    – Watson
    Commented May 23, 2021 at 16:32
  • $\begingroup$ Thanks for adding the Ask Sage link. I removed my comment with an incorrect link. If you want to rephrase yours to not be an answer to my now gone comment, I'll also remove this comment. Can you edit your question to spell Sage rather than SAGE (once in the question title, twice in the question body)? I have insufficient reputation to do that on math-stack-exchange, and the edit I submitted was not taken seriously. $\endgroup$ Commented May 25, 2021 at 12:24

1 Answer 1


I couldn't resist trying to code up the algorithm hinted at in my comment. Sage code at:


This is v2 of the code; version 1 used only prime-degree endomorphisms, and thus failed to distinguish between the orders $\mathbf{Z}[\sqrt{-15}]$ and $\mathbf{Z}[\tfrac{1 + \sqrt{-15}}{2}]$ (since the sets of primes arising as norms of elements in these two rings are the same). So it failed on $y^2 = x^3 + x + 8$ over $\mathbf{F}_{19}$. The new implementation uses endomorphisms of all degrees (piecing them together from isogenies of prime degree).

For instance, here we recover the values of $c$ in the table from p.303 in your linked article, using the functions defined in the script linked above:

sage: listofcurves = []
sage: K = GF(7)
sage: for a in K:
....:     for b in K:
....:         if 4*a^3 + 27*b^2 == 0:
....:             continue
....:         E = EllipticCurve([0, 0, 0, a, b])
....:         if not any(E.is_isomorphic(F) for F in listofcurves):
....:             print(E.ainvs(), endomorphism_conductor(E))
....:             listofcurves.append(E)
(0, 0, 0, 0, 1) 1
(0, 0, 0, 0, 2) 1
(0, 0, 0, 0, 3) 1
(0, 0, 0, 0, 4) 1
(0, 0, 0, 0, 5) 1
(0, 0, 0, 0, 6) 1
(0, 0, 0, 1, 0) 2
(0, 0, 0, 1, 1) 1
(0, 0, 0, 1, 3) 1
(0, 0, 0, 1, 4) 1
(0, 0, 0, 1, 6) 1
(0, 0, 0, 3, 0) 1
(0, 0, 0, 3, 1) 2
(0, 0, 0, 3, 2) 3
(0, 0, 0, 3, 3) 1
(0, 0, 0, 3, 4) 1
(0, 0, 0, 3, 5) 3
(0, 0, 0, 3, 6) 2

PS: Important caveat: this computes the endomorphisms of $E$ over the given field $k$ (not over $\overline{k}$). If $E$ is ordinary this makes no difference (all endomorphisms are defined over $k$). In the supersingular case, it makes a difference: if $E$ is supersingular, but $\operatorname{Frob}_E$ isn't in $\mathbf{Z}$, we don't get a quaternion order, but an imaginary quadratic order (the centraliser of the Frobenius in $\operatorname{End}_{\overline{k}}(E)$). If $E$ is supersingular and the Frobenius is an integer, which can only happen if $k$ has even degree over its prime field, then the algorithm fails.

  • $\begingroup$ Dear David Loeffler, thank you very much for your complete and useful answer! I am grateful to the time you spent! $\endgroup$
    – Watson
    Commented May 23, 2021 at 16:44
  • $\begingroup$ (Just a small remark, mostly for myself: The isogenies_prime_degree returns all separable $\ell$-isogenies with domain E, so typically your has_endo(E, p) returns False (but it probably doesn't affect the rest -- I haven't checked carefully yet). $\endgroup$
    – Watson
    Commented May 23, 2021 at 20:32
  • $\begingroup$ Fair point. In practice this code seems to always halt long before it gets to n = p so this bug is unlikely to affect the output, but it would probably be considerably harder to prove that than it would be to fix the bug. :-) $\endgroup$ Commented May 24, 2021 at 12:07

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