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Recently I was considering cubic extensions $K/Q$ that have discriminant negative of a perfect square. Classifying these curves reduces to solving a Diophatine equation of the form $4a^3+27b^2=c^2$ which is not that difficult. However, another approach is to note that the unique quadratic subfield of the Galois closure of $K$ is $Q[i]$, hence we are looking for cubic cyclic extensions of $Q[i]$. We should be able to write these down by CFT, and in particular, they should all be subfields of $Q[i][E_{tors}]$ where $E:y^2=x^3+x$ (if I'm not mistaking). However, I'm having a hard time finding the correct torsion points that generate a field containing some of my K.

As a particular example, consider the field generated by the cubic polynomial $x^3-11x+194/3$. This field has discriminant $-324=-18^2$. However, as far as I can tell $Q[i][E[n]]$ does not contain this field for $n<30$. How can I find this value of $n$, or where am I going wrong?

Thanks, Soroosh

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If your calculations are correct, the field is ramified only at $2$ and $3$. Every element in $\mathbb Z_2[-1]^{\times}$ is a a third power, so the ramification can't come from there. Every element in $\mathbb Z_3[-1]^{\times}$ that's $1$ mod $3$ has a log, so every element that's $1$ mod $9$ is a third power. Thus the conductor should be $9$, so $E[9]$ should be sufficient. –  Will Sawin Nov 9 '12 at 18:29
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Hello Soroosh. I checked, and ${\mathbb Q}[i](E[9])$ does contain your cubic field, as Will explains it should. To find such cubics efficiently, there are perhaps better ways than CM theory: there are families of cubics with a given quadratic field, e.g. there is one in Jensen-Ledet-Yui's book "Generic polynomials", p.111, example under Prop. 5.5.2. (I can post it, if you want.) Or you can use the RayClassField machinery, e.g. in Magma. –  Tim Dokchitser Nov 10 '12 at 9:57
    
Ahh... I actually looked at Q[i](E[9]), but somehow I missed the non-cyclotomic degree three extension. Thanks Tom and Will. As far as "Generic polynomials" book, thanks for the reference. I found it. –  Soroosh Nov 10 '12 at 19:12

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