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Assume that you have proved the two inequalities of class field theory, and that you want to show that the Hilbert class field, i.e., the maximal unramified abelian extension, of a number field $K$ actually exists. Assume for simplicity, as Hilbert did, that $K$ is totally complex with class number $2$, and that the ideal class of ${\mathfrak a}$ generates the class group.

We know that the Hilbert class field $L = K(\sqrt{\alpha})$ satisfies $\alpha = \eta \beta$ for some unit $\eta$, where $(\beta) = {\mathfrak a}^2$. The most natural construction would therefore show that among the elements in the "Selmer group" $S$ generated by units and squares of ideals there is an element congruent to a square modulo $4$, but apparently this does not work as directly as we would wish, and Hilbert had to enlarge $S$ by elements which had no chance of generating the class field but without which his proof does not work (he made $S$ so large that two elements had to lie in the same residue class mod $4$, used Dirichlet's box principle, and then showed that the resulting element does not involve any elements outside of $S$).

My question is whether there are other ways of showing the existence of such an $\alpha$ that have at least a chance of working. To put it another way: if you didn't know class field theory, where would you look for a proof of the existence of $\alpha$?

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    $\begingroup$ 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. $\endgroup$
    – Dror Speiser
    Commented Oct 9, 2010 at 8:49
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    $\begingroup$ @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 $e = -1$. $\endgroup$
    – Franz Lemmermeyer
    Commented Feb 21, 2011 at 17:19
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    $\begingroup$ 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. $\endgroup$
    – Dror Speiser
    Commented Nov 24, 2012 at 21:21
  • $\begingroup$ 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. :-) $\endgroup$
    – assaferan
    Commented Mar 31, 2016 at 17:01

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