# When is possible to factor a field extension into one which adds no roots of unity, followed by one which adds only roots of unity?

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?

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..

• 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... – Alex Jan 11 '16 at 19:06
• 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. – David E Speyer Jan 11 '16 at 20:14
• 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. :) – David E Speyer Jan 11 '16 at 20:15
• 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. – David E Speyer Jan 11 '16 at 20:16