Suppose $K\subseteq L\subseteq M$ is a tower of fields.
If $M$ is a radical extension of $K$, is $L$ a radical extension of $K$?
My attempt (resulting in a paradox):
I don't know much about radical extensions other than the definition. I tried cooking up this counter-example:
$K=\mathbb{Q}$
$L=\mathbb{Q}(\sqrt{1+\sqrt{2}})$
$M=\mathbb{Q}(\sqrt{2},\sqrt{1+\sqrt{2}})$
$M/K$ is radical since $\sqrt{2}^2$ lies in $\mathbb{Q}$, while $\sqrt{1+\sqrt 2}^2$ lies in $\mathbb{Q}(\sqrt{2})$.
$L/K$ is not radical since any power of $\sqrt{1+\sqrt 2}$ does not lie in $\mathbb{Q}$.
This seems to result in a paradox since $L=M$?
The definition of radical extension I am using is: An extension field $F$ of a field $K$ is a radical extension of $K$ if $F=K(u_1,\dots,u_n)$, some power of $u_1$ lies in $K$ and for each $i\geq 2$, some power of $u_i$ lies in $K(u_1,\dots,u_{i-1})$.
Thanks for any help.
Main question: Is the statement true or false and why?
Secondary question: Why is there a "paradox"? Radical extension depends on the generators used?
We have $L=\mathbb{Q}(\sqrt{1+\sqrt{2}})$ and $M=L(\sqrt{2})$, so it should be $\sqrt{1+\sqrt{2}}^a\in \mathbb{Q}$ for some $a\in \mathbb{N}$, and $\sqrt{2}^b\in L$, for sombe $b\in \mathbb{N}$. But, there is no such $a$, hence $M/K=L/K$ isn't radical and there is no paradox.