Let $K$ be a quadratic imaginary field, and $E$ an elliptic curve whose endomorphism ring is isomorphic to the full ring of integers of $K$. Let $j$ be its $j$-invariant, and $c$ an integral ideal of $K$. Consider the following tower:

$$K(j,E[c])\ \ /\ \ K(j,h(E[c]))\ \ /\ \ K(j)\ \ /\ \ K,$$

where $h$ here is any Weber function on $E$. (Note that $K(j)$ is the Hilbert class field of $K$).

We know that all these extensions are Galois, and any field has ABELIAN galois group over any smaller field, EXCEPT POSSIBLY THE BIGGEST ONE (namely, $K(j,E[c]) / K$).


  1. Does the biggest one have to be abelian? Give a proof or counterexample.

My suspicion: No, it doesn't. I've been trying an example with $K = \mathbf Q(\sqrt{-15})$, $E = C/O_K$, and $c = 3$; it just requires me to factorise a quartic polynomial over $\bar{ \mathbf Q}$, which SAGE apparently can't do.

  1. What about if I replace $E[c]$ in the above by $E_{tors}$, the full torsion group?

4 Answers 4


Here is a case where it is non-Abelian. I use $K$ of class number 3. If I use the Gross curve, it is Abelian. If I twist in $Q(\sqrt{-15})$, it is Abelian for every one I tried, maybe because it is one class per genus. My comments are not from an expert.

> K<s>:=QuadraticField(-23);                                                    
> jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2);                             
> jrel:=PowerRelation(jinv,3 : Al:="LLL");                                      
> Kj<j>:=ext<K|jrel>;                                                           
> E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]);
> HasComplexMultiplication(E);
true -23                 
> c4, c6 := Explode(cInvariants(E)); // random twist with this j                
> f:=Polynomial([-c6/864,-c4/48,0,1]);                                          
> poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic                 
> R:=Roots(poly);                                                               
> Kj2:=ext<Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])>;                          
> KK:=ext<Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])>;                          
> assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here         
> f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1];  
> GaloisGroup(f); /* not immediate to compute */                                
Permutation group acting on a set of cardinality 12
Order = 48 = 2^4 * 3
> IsAbelian($1);                                                                

This group has $A_4$ and $Z_2^4$ as normal subgroups, but I don't know it's name if any.

PS. 5-torsion is too long to compute most often.

  • $\begingroup$ Thank you Junkie; this example shows that there is an error in my original question. When I assumed that K(j,E[c]) is always galois over K, I was wrong! In this example, [K(j,E[3]) : K] = 12 (in your notation this is [KK : K] = 12). However, you show that the Galois closure of KK has degree 48 over K. I may start thinking about precisely when this extension is Galois and when it isn't. Thanks again! $\endgroup$ May 7, 2010 at 11:07
  • $\begingroup$ The Galois group is the direct product $Z_2^2 \times A_4$ (I simply asked the magma command GroupName). $\endgroup$
    – Aurel
    Jan 3 at 12:56


In general $K(j,E[c])$ will not be abelian over $K$ (the reason being that $K(j,h(E[c]))$ is the ray class field of $K$ of conductor $c$, therefore maximal for abelian extensions of conductor $c$). However, $K(j,E[c])$ is always abelian over $K(j)$. In particular, if the class number of $K$ is $1$, the answer is yes to both your questions, because $K(j)=K$.

For more on this, see Silverman's "Advanced topics in the AEC". In particular, see pages 135-138, and Example 5.8 discusses briefly this question.


  • $\begingroup$ Is it possible for some example that $K(j,E[c])$ is always nontrivial over $K(j)$ when $c\geq 2$? $\endgroup$ Jul 5, 2013 at 22:52

The extension K(j,E_{tors}) is abelian over the Hilbert class field of K, hence over K if K has class number 1. Silverman (Advanced topics, p. 138) says that, in general, the extension is not abelian. For getting a counterexample, looking at ${\mathbb Q}(\sqrt{-15})$ is the right idea. Instead of factoring the quartic you might simply want to compute its Galois group, which you probably can read off the discriminant and the cubic resolvent.

The field generated by $E_{tors}$ is the union of the fields generated by the $E[c]$, so for $c$ large enough you should see the nonabelian group already there.

  • $\begingroup$ I don't think the Galois group is enough. The quartic I referred to was the 3-division polynomial, defined over K(j), NOT K. Its splitting field would give me K(j,h(E[3])) (since h just picks out the x-coordinates). But since I'm trying to get K(j,E[3]), I need the y-coordinates as well. Note that, using the universal elliptic curve, I can write down a Weierstrass equation for E := C/O_K (well, this requires an explicit computation of j, which is possible, and is explicitly stated on page 143 of Silverman's advanced topics). $\endgroup$ Apr 30, 2010 at 14:35

Magma is not facile here but works, but maybe SAGE can do the same. You get $K(j,E[3])/K$ to be a degree 12 and cyclic Galois group, for the $E$ I think you want.

> jrel:=PowerRelation(jInvariant((1+Sqrt(-15))/2),2 : Al:="LLL");
> K:=QuadraticField(-15);
> Kj<j>:=ext<K|jrel>;
> A:=AbsoluteField(Kj);
> C:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]);
> b, d := HasComplexMultiplication(C); assert b and d eq -15;
> E:=QuadraticTwist(C, 7*11); // conductor at 3, 5
> E:=ChangeRing(WeierstrassModel(ChangeRing(E,A)),Kj);
> c4, c6 := Explode(cInvariants(E));
> f:=Polynomial([-c6/864,-c4/48,0,1]);
> poly:=DivisionPolynomial(E,3); // Linear x Cubic
> r:=Roots(poly)[1][1];
> Kj2:=ext<Kj|Polynomial([-Evaluate(f,r),0,1])>; // quadratic ext for linear
> KK:=ext<Kj2|Factorization(poly)[2][1]>; // cubic x-coordinate
> assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here
> f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1];
> // assert IsIsomorphic(ext<K|f>,KK); // taking too long ?
> // SetVerbose("GaloisGroup",2);
> GaloisGroup(f);
Permutation group acting on a set of cardinality 12
Order = 12 = 2^2 * 3
> IsAbelian($1);

The Magma has as online calculator for this. http://magma.maths.usyd.edu.au/calc

  • $\begingroup$ Darn, that means my example was wrong. Presumably if I replace 3 by something bigger, and try the above calculation, the last line should return "false" and I'm happy. I will find a friend (who has MAGMA) to do this. I'll keep you posted. $\endgroup$ May 3, 2010 at 20:42
  • $\begingroup$ I do not know, but my impression is that for 3-torsion with the Gross curve the tendency is to be Abelian. I tested the same for $d=-23$ (the Gross curve) and got a cyclic group of order 12. But my comments are not from an expert. Computing with 5-torsion is time-consuming. If you look at twists (preserving $j$-invariant), they have larger Galois group. See below. $\endgroup$
    – Junkie
    May 3, 2010 at 23:51

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