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Mar 31, 2016 at 17:01 comment added assaferan I'm not sure this is what you are looking for, but if $\mathcal{O}_K=\mathbb{Z}+\mathbb{Z}\tau$ you know that the Hilbert class field should be $K[j(\tau)]$. Therefore, you have your $\alpha$. Briefly - If I didn't know class field theory I would look for it in complex multiplication. :-)
Nov 24, 2012 at 21:21 comment added Dror Speiser Hey Franz, sorry for replying almost who years later. Saw this again while going through favourites. My point was that the first sentence of the second paragraph is false, by giving a counter example. In your reply you used an ideal $\mathfrak{a}=(1)$ that doesn't generate the class group of $K$, which you required in the last sentence of the first paragraph.
Feb 21, 2011 at 17:19 comment added Franz Lemmermeyer @Dror: You've got it backwards. $\alpha = -1$ generates the Hilbert class field, and $-1 = \eta\beta$ for $\eta = -1$ and $(\beta) = (1)^2$. Your claim about $e$ when $1-e$ lies in the square of all prime ideals above $2$ is not true in general, e.g. in ${\mathbb Q}(\sqrt{-6})$ with $e = -1$.
Oct 10, 2010 at 5:23 history edited Bjørn Kjos-Hanssen
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Oct 9, 2010 at 8:49 comment added Dror Speiser Let K be the quadratic imaginary field with discriminant -20. The ideal above 2 generates the class group, its square is the ideal generated by 2. Then the above says that the Hilbert class field is constructed from a square root of 2 or -2. But it is well known that the Hilbert class field of K is constructed by taking a square root of -1. I think that in general the above is only true if there is no unit such that 1-e is in the square of any prime ideal above 2. If there is such a unit, its square root gives the class field.
Oct 8, 2010 at 14:01 history asked Franz Lemmermeyer CC BY-SA 2.5