Suppose $L|\mathbb{Q}$ is an abelian extension of number fields. Then, all the roots of unity are certainly contained in the maximal abelian extension $L^{ab}$ of $L$. Why is it obvious that if $L \ne \mathbb{Q}$ then $L^{ab} \ne \mathbb{Q}^{ab}$.



Non-cyclotomic abelian extensions
Pick some $\gamma\_1\in L\setminus\mathbb Q$ which is not a square. Pick some $\gamma\in L^\times/(L^\times)^2$ which is not fixed by $\operatorname{Gal}(L/\mathbb Q)$ and fix a lift $\gamma\_1\in L$. Let $\gamma\_1,\ldots,\gamma\_n$ be the orbit of $\gamma\_1$ under $\operatorname{Gal}(L/\mathbb Q)$. Then it is an easy exercise in Galois theory to show that:


1. $L(\sqrt{\gamma\_1},\ldots,\sqrt{\gamma\_n})/L$ is Galois with abelian Galois group $\subseteq(\mathbb Z/2\mathbb Z)^n$.
2. $L(\sqrt{\gamma\_1},\ldots,\sqrt{\gamma\_n})/\mathbb Q$ is Galois with nonabelian Galois group $\subseteq\operatorname{Gal}(L/\mathbb Q)\ltimes(\mathbb Z/2\mathbb Z)^n$.


