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The answer to whether this is possible for general fields is no. However, the counterexamples used two ingredients:

1) $\Bbb Q_p$, whose extensions $K$ containing $\Bbb Q_p(\sqrt[p^e]{u})$ might not be not Galois even if $\zeta_{p^e}\in K$.

2) Ramified extensions.

IF one or both of these is not allowed, then will this result then hold? For example:

If $k$ is a number field, and $L/k$ is an unramified / Abelian extension, can we decompose $L = K(\zeta_n)$, where $K$ and $k$ have the same number of roots of unity?

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No. Let $k = \mathbb{Q}(\sqrt{-17})$. The class group of $k$ is $\mathbb{Z}/4 \mathbb{Z}$, generated by $\langle 3, 1+\sqrt{-17} \rangle$. (I checked this table for a field of class number $4$ and then checked that $\langle 3, 1+\sqrt{-17} \rangle$ and $\langle 3, 1+\sqrt{-17} \rangle^2$ are not principal. To do this, note that $a^2+17 b^2=3$ is not solvable in integers, and $a^2+17 b^2=9$ only has the solution $(\pm 3, 0)$.)

Let $L$ be the class field of $k$. The only fields between $L$ and $k$ are $L$, $k$ and an intermediate quadratic extension $K$, corresponding to the three subgroups of $\mathbb{Z}/4 \mathbb{Z}$. But $k(\sqrt{-1})$ is abelian and unramified, so $K$ must be $k(\sqrt{-1})$. Thus, all nontrivial extensions of $k$ within $L$ contain a root of unity.


The field $L$ is $k(j(\sqrt{-17}))$, where $j$ is the $j$-function. The minimal polynomial of $j(\sqrt{-17})$ is

x^4 - 178211040000 x^3 -75843692160000000 x^2 -318507038720000000000 x -2089297506304000000000000,

and an explicit formula in radicals is $$8000 \left(\ 5569095 + 1350704 \sqrt{17} + 4 \sqrt{2 (1938444620639 + 470141877665 \sqrt{17})}\ \right).$$ So this is clearly a quadratic extension of $\mathbb{Q}(\sqrt{-17}, i)$. It isn't easy to me to see that it is Galois over $\mathbb{Q}(\sqrt{-17})$, though.

I note that the element $2(1938444620639 + 470141877665 \sqrt{17})$ in $\mathbb{Q}(\sqrt{17})$ has norm $-421496^2 = - 2^6 19^2 47^2 59^2$ in $\mathbb{Q}$.


I think the prettiest way to describe the top field is $\mathbb{Q}(\sqrt{1+4i}, \sqrt{1-4i})$. This is clearly Galois over $\mathbb{Q}$ with Galois group dihedral of order $8$. Writing $\rho$ for the order $4$ rotation in this dihedral group, the fixed fields of $\rho$ and $\rho^2$ are $\mathbb{Q}(\sqrt{-17})$ and $\mathbb{Q}(\sqrt{17}, i)$ respectively, and it isn't too bad to check that $\mathbb{Q}(\sqrt{1+4i}, \sqrt{1-4i})/\mathbb{Q}(\sqrt{-17})$ is unramified..

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  • $\begingroup$ I thought that $L$, as a Hilbert Class Field of $k$, would be automatically Abelian (Galois) over $k$? p.s., what command did you use to get that polynomial? I think such polynomials might help me in research... $\endgroup$ – Alex Jan 11 '16 at 19:06
  • $\begingroup$ The Galois conjugates of $j(\sqrt{-17})$ are itself, $j((1+\sqrt{-17})/2)$, $j((1+\sqrt{-17})/3)$ and $j((-1+\sqrt{-17})/3)$. (Coming from representatives of the ideal classes.) I computed these numerically, and then used them to compute the coefficients of the minimal polynomial numerically. Since I knew those would be integers, I could recognize them from their numerical values. $\endgroup$ – David E Speyer Jan 11 '16 at 20:14
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    $\begingroup$ A warning if you use Mathematica: KleinInvariantJ[] is off by a factor of 1728 from everyone else in the world. For example, Mathematica thinks KleinInvariantJ[Sqrt[-1]] is 1, not 1728. Discovering this was the longest part of the computation. :) $\endgroup$ – David E Speyer Jan 11 '16 at 20:15
  • $\begingroup$ As regards "not obvious" I just mean that, if you told me to consider the extension $k(\sqrt{2(1938444620639+47014187766517\sqrt{17})})$ of $k = \mathbb{Q}(\sqrt{-17})$, I wouldn't recognize that it was Galois. $\endgroup$ – David E Speyer Jan 11 '16 at 20:16

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