1
$\begingroup$

I have the imaginary quadratic field $K= \mathbb{Q}(\sqrt{-17})$ with $\mathcal{O}_K = \mathbb{Z}[\sqrt{-17}]$. Now I want to have the $j$-Invariant of an elliptc curve $E$ with complex multiplication by $\mathcal{O}_K$. For a given elliptic curve I know how to compute the $j$-invariant with PARI. Therefore just a hint how to get this curve would be perfect!

$\endgroup$
  • $\begingroup$ This should help: wstein.org/books/bsd/bsd/node38.html $\endgroup$ – Ricardo Buring Sep 6 '17 at 13:35
  • $\begingroup$ I don't see how that helps me to get an exact elliptic curve. $\endgroup$ – anama Sep 6 '17 at 14:01
  • $\begingroup$ All you have to do is to find an elliptic curve with the $j$-invariant you have already found. See math.stackexchange.com/questions/362128/… $\endgroup$ – Lord Shark the Unknown Sep 6 '17 at 14:53
  • $\begingroup$ Sorry I was not clear. I just know how to compute the $j$-invariant from an elliptic curve. Maybe I can look for a solution to get the invariant to the lattice $\mathbb{Z}[\sqrt{-17}]$ $\endgroup$ – anama Sep 6 '17 at 15:14
  • $\begingroup$ For the first look I don't see anything in Pari, do you knoe the command? $\endgroup$ – anama Sep 6 '17 at 15:15
4
$\begingroup$

The class group $\operatorname{Cl}(\mathcal{O}_K)$ has order $4$ and consists of $ \langle 1 \rangle, \langle 2, 1+\sqrt{-17} \rangle, \langle 3, 1 + \sqrt{-17} \rangle$ and $\langle 3, 2 + \sqrt{-17} \rangle$.

One elliptic curve with CM by $\mathbb{Z}[\sqrt{-17}]$ is just $\mathbb{C}/\mathbb{Z}[\sqrt{-17}]$. From the action of $\operatorname{Cl}(\mathcal{O}_K)$ on elliptic curves with CM by $\mathcal{O}_K$ we get three more elliptic curves, given by lattices homothetic to $\mathbb{Z}\big[\frac{1+\sqrt{-17}}{2}\big]$, $\mathbb{Z}\big[\frac{1+\sqrt{-17}}{3}\big]$, and $\mathbb{Z}\big[\frac{2+\sqrt{-17}}{3}\big]$.

The $j$-invariant of $\mathbb{C}/\mathbb{Z}[\sqrt{-17}]$ is $j(\sqrt{-17})$ which is an algebraic integer of degree $\#\operatorname{Cl}(\mathcal{O}_K) = 4$ whose conjugates are exactly [notes, Theorem 11] the $j$-invariants of the elliptic curves above, i.e. $j\big(\frac{1+\sqrt{-17}}{2}\big), j\big(\frac{1+\sqrt{-17}}{3}\big)$ and $j\big(\frac{2+\sqrt{-17}}{3}\big)$.

The approximate values of these numbers can be found using PARI/GP's ellj:

? ellj(sqrt(-17))
%1 = 178211465582.57236317285989152242246715
? ellj((1+sqrt(-17))/2)
%2 = -421407.46393796828425027276471142244105
? ellj((1+sqrt(-17))/3)
%3 = -2087.5542126022878206248288778733953807 - 4843.8060029534518331212466414598790536*I
? ellj((2+sqrt(-17))/3)
%4 = -2087.5542126022878206248288778733953807 + 4843.8060029534518331212466414598790536*I

Now $j(\sqrt{-17})$ is a root of $f = \prod (x-j_k) \in \mathbb{Z}[x]$ where the $j_k$ are all the conjugates. We can approximate $f$ by replacing the $j_k$ with their numerical approximations and expanding the product. In this way we find that $j(\sqrt{-17})$ is very probably the unique positive real root of $$x^4 - 178211040000x^3 - 75843692160000000x^2 - 318507038720000000000x - 2089297506304000000000000,$$ namely $$8000 (5569095 + 1350704 \sqrt{17} + 4 \sqrt{3876889241278 + 940283755330 \sqrt{17}}),$$ which can be verified numerically to very high precision.

As mentioned in the comments we can also find an explicit Weierstrass form for an elliptic curve with given $j$-invariant.

$\endgroup$
1
$\begingroup$

To compute this sort of thing in Sage, you can use cm_j_invariants(). For example:

K.<a>=NumberField(x^2+17) H0.<b>=NumberField(K.hilbert_class_polynomial()) cm_j_invariants(H0)

The output is:

[-12288000,
 54000,
 0,
 287496,
 1728,
 16581375,
 -3375,
 8000,
 -32768,
 -884736,
 -884736000,
 -147197952000,
 -262537412640768000,
 2911923/16469087501000000000*b^3 - 1621681416684/51465898440625*b^2 - 6605556732936/50210632625*b - 850332053302272/968769853,
 -2911923/16469087501000000000*b^3 + 1621681416684/51465898440625*b^2 + 6605556732936/50210632625*b - 5367201819436127232/968769853,
 1124064022653/7750158824000000*b^3 - 626003413105165524/24219246325*b^2 - 2549884963677490296/23628533*b - 696625107656760443520000/968769853,
 -1124064022653/7750158824000000*b^3 + 626003413105165524/24219246325*b^2 + 2549884963677490296/23628533*b - 4402688233312810879896960000/968769853,
 b,
 -5041/886477376000000*b^3 + 701846857/692560450*b^2 + 58591882490/13851209*b + 22967913600000/814777]

The last number in the list is the right one:

j=-5041/886477376000000*b^3 + 701846857/692560450*b^2 + 58591882490/13851209*b + 22967913600000/814777

Now

j.minpoly()

identifies it as the root of:

x^4 - 178211040000*x^3 - 75843692160000000*x^2 - 318507038720000000000*x - 2089297506304000000000000

To test it, you can define a number field with that minimal polynomial, or alternatively

L.<b0>=H0.subfield(j)[0];L

which gives:

Number Field in b0 with defining polynomial x^4 - 178211040000*x^3 - 75843692160000000*x^2 - 318507038720000000000*x - 2089297506304000000000000

Then

E=EllipticCurve_from_j(b0);E

shows

Elliptic Curve defined by y^2 = x^3 + (-3*b0^2+5184*b0)*x + (-2*b0^3+6912*b0^2-5971968*b0) over Number Field in b0 with defining polynomial x^4 - 178211040000*x^3 - 75843692160000000*x^2 - 318507038720000000000*x - 2089297506304000000000000

Checking that it has the right CM order:

E.cm_discriminant()

outputs

-68
$\endgroup$
  • $\begingroup$ Isn't the desired CM discriminant $-4\cdot 17 = -68$? From e.g. cm_j_invariants_and_orders(H0) one sees that b itself is such a $j$-invariant (and its minimum polynomial is the quartic in my answer). $\endgroup$ – Ricardo Buring Aug 24 '18 at 6:54
  • $\begingroup$ Yeah you're right, I multiplied 17 by 4 wrong. It's the last one that has the right CM order. Corrected now. $\endgroup$ – Prometheus Aug 24 '18 at 8:10
  • $\begingroup$ Your answer's perfectly fine. I just wanted to add a different method that doesn't rely on approximation. $\endgroup$ – Prometheus Aug 24 '18 at 8:12
  • $\begingroup$ Sage's K.hilbert_class_polynomial() uses the same approximation (with a proven bound on the error), at least when algorithm is arb (the default) or sage (not sure about magma). $\endgroup$ – Ricardo Buring Aug 24 '18 at 9:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.