# Finding an elliptic curve with CM by $\mathbb{Z}[\sqrt{-17}]$

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!

• This should help: wstein.org/books/bsd/bsd/node38.html – Ricardo Buring Sep 6 '17 at 13:35
• I don't see how that helps me to get an exact elliptic curve. – anama Sep 6 '17 at 14:01
• 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/… – Lord Shark the Unknown Sep 6 '17 at 14:53
• 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 $\mathbb{Z}[\sqrt{-17}]$ – anama Sep 6 '17 at 15:14
• For the first look I don't see anything in Pari, do you knoe the command? – anama Sep 6 '17 at 15:15

## 2 Answers

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.

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);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

• 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). – Ricardo Buring Aug 24 '18 at 6:54
• Yeah you're right, I multiplied 17 by 4 wrong. It's the last one that has the right CM order. Corrected now. – Prometheus Aug 24 '18 at 8:10
• Your answer's perfectly fine. I just wanted to add a different method that doesn't rely on approximation. – Prometheus Aug 24 '18 at 8:12
• 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). – Ricardo Buring Aug 24 '18 at 9:31