Class Field Theory for Imaginary Quadratic Fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:35:47Z http://mathoverflow.net/feeds/question/23092 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields Class Field Theory for Imaginary Quadratic Fields Barinder Banwait 2010-04-30T10:38:26Z 2010-05-04T00:00:40Z <p>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:</p> <p>K(j,E[c]) / K(j,h(E[c])) / K(j) / K,</p> <p>where h here is any Weber function on E. (Note that K(j) is the Hilbert class field of K). </p> <p>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). </p> <p>Questions:</p> <ol> <li>Does the biggest one have to be abelian? Give a proof or counterexample.</li> </ol> <p>My suspicion: No, it doesn't. I've been trying an example with K = Q($\sqrt{-15}$), E = C/O_K, and c = 3; it just requires me to factorise a quartic polynomial over Q-bar, which SAGE apparently can't do.</p> <ol> <li>What about if I replace E[c] in the above by E_tors, the full torsion group? </li> </ol> http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields/23105#23105 Answer by Álvaro Lozano-Robledo for Class Field Theory for Imaginary Quadratic Fields Álvaro Lozano-Robledo 2010-04-30T13:26:54Z 2010-04-30T13:26:54Z <p>Hello,</p> <p>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 $K(j)=K$.</p> <p>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.</p> <p>Alvaro</p> http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields/23106#23106 Answer by Franz Lemmermeyer for Class Field Theory for Imaginary Quadratic Fields Franz Lemmermeyer 2010-04-30T13:29:45Z 2010-04-30T13:29:45Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields/23150#23150 Answer by Junkie for Class Field Theory for Imaginary Quadratic Fields Junkie 2010-05-01T00:53:54Z 2010-05-01T00:53:54Z <p>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.</p> <pre><code>&gt; jrel:=PowerRelation(jInvariant((1+Sqrt(-15))/2),2 : Al:="LLL"); &gt; K:=QuadraticField(-15); &gt; Kj&lt;j&gt;:=ext&lt;K|jrel&gt;; &gt; A:=AbsoluteField(Kj); &gt; C:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]); &gt; b, d := HasComplexMultiplication(C); assert b and d eq -15; &gt; E:=QuadraticTwist(C, 7*11); // conductor at 3, 5 &gt; E:=ChangeRing(WeierstrassModel(ChangeRing(E,A)),Kj); &gt; c4, c6 := Explode(cInvariants(E)); &gt; f:=Polynomial([-c6/864,-c4/48,0,1]); &gt; poly:=DivisionPolynomial(E,3); // Linear x Cubic &gt; r:=Roots(poly)[1][1]; &gt; Kj2:=ext&lt;Kj|Polynomial([-Evaluate(f,r),0,1])&gt;; // quadratic ext for linear &gt; KK:=ext&lt;Kj2|Factorization(poly)[2][1]&gt;; // cubic x-coordinate &gt; assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here &gt; f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1]; &gt; // assert IsIsomorphic(ext&lt;K|f&gt;,KK); // taking too long ? &gt; // SetVerbose("GaloisGroup",2); &gt; GaloisGroup(f); Permutation group acting on a set of cardinality 12 Order = 12 = 2^2 * 3 &gt; IsAbelian($1); true </code></pre> <p>The Magma has as online calculator for this. <a href="http://magma.maths.usyd.edu.au/calc" rel="nofollow">http://magma.maths.usyd.edu.au/calc</a></p> http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields/23390#23390 Answer by Junkie for Class Field Theory for Imaginary Quadratic Fields Junkie 2010-05-04T00:00:40Z 2010-05-04T00:00:40Z <p>Here is a case where it is non-Abelian. I use$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.</p> <pre><code>&gt; K&lt;s&gt;:=QuadraticField(-23); &gt; jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2); &gt; jrel:=PowerRelation(jinv,3 : Al:="LLL"); &gt; Kj&lt;j&gt;:=ext&lt;K|jrel&gt;; &gt; E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]); &gt; HasComplexMultiplication(E); true -23 &gt; c4, c6 := Explode(cInvariants(E)); // random twist with this j &gt; f:=Polynomial([-c6/864,-c4/48,0,1]); &gt; poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic &gt; R:=Roots(poly); &gt; Kj2:=ext&lt;Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])&gt;; &gt; KK:=ext&lt;Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])&gt;; &gt; assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here &gt; f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1]; &gt; GaloisGroup(f); /* not immediate to compute */ Permutation group acting on a set of cardinality 12 Order = 48 = 2^4 * 3 &gt; IsAbelian($1); false </code></pre> <p>This group has $A_4$ and $Z_2^4$ as normal subgroups, but I don't know it's name if any.</p> <p>PS. 5-torsion is too long to compute most often.</p>