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May 25, 2012 at 11:07 comment added Dror Speiser This is an elementary exercise. Remember that you're not calculating the norm of arbitrary elements, but of generators. My favorite solution involves the terms quadratic forms and fundamental domain. I will point out that it is crucial the discriminant is prime. If after trying a bit more you still haven't solved this problem, I suggest you post it on math.stackexchange.
May 25, 2012 at 9:46 comment added MPI Thank you for the reponse. I agree with what you said, but indeed the point that is blocking me is precisely how to determine the degree of the elements of this fractional ideal. I know that this degree is going to be N$(x)$/N$(\mathfrak{a})$ for every $x\in \mathfrak{a}$ ($\mathfrak{a}$ being the Hom space.) but since I don't explicitely know $\mathfrak{a}$, I don't know how to calculate the degree of its elemtents.
May 25, 2012 at 8:59 comment added Dror Speiser * When $l=3$ the class number is one and the unique j-invariant is 0, so that $P_3(X)=X$, and indeed $X=X-12^3\pmod{3}$.
May 25, 2012 at 8:47 comment added Dror Speiser Finally, elements of $\mathbb{Z}[\frac{1+F}{2},I]$ which aren't in $\mathbb{Z}[I]$, quite clearly have degree at least $\frac{l+1}{4}$, the smallest element being $\frac{1+F}{2}$ with that degree.
May 25, 2012 at 8:44 comment added Dror Speiser By choosing an embedding of $K_l$ into $\mathbb{C}$, then looking at the parameterising tori and their lattices, and after taking a suitable homothety, the lattices are ideals in $\mathcal{O}_l$, which cannot lie in the same class (in the class group) since they have different j-invariants. The Hom space is now a fractional ideal, that cannot be principal. That the ideal is spanned by two elements of reduced norm less than $\frac{l+1}{4}$, for $l\ge 7$ is an exercise. For $l=3$, there should be a sperate computation.
May 24, 2012 at 23:18 history edited MPI
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May 24, 2012 at 20:32 history asked MPI CC BY-SA 3.0