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?

• If we have $x=\sqrt{1+\sqrt{2}}$, we have $(x^2-1)^2=2$ , so $\mathbb Q(x)$ should be a radical extension. There seems to be a problem with your definition, but I am not an expert on this topic. – Peter Dec 5 '16 at 16:27
• I think the working you show is to show that $x$ is algebraic – yoyostein Dec 5 '16 at 16:29
• As I already said, I am not an expert on this topic, but as far as I can remember we do not need that some power of $x$ is in $\mathbb Q$ to have a radical extension – Peter Dec 5 '16 at 16:31

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.