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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?

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  • $\begingroup$ 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. $\endgroup$ – Peter Dec 5 '16 at 16:27
  • $\begingroup$ I think the working you show is to show that $x$ is algebraic $\endgroup$ – yoyostein Dec 5 '16 at 16:29
  • $\begingroup$ 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 $\endgroup$ – Peter Dec 5 '16 at 16:31
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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.

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  • $\begingroup$ +1. How about the main question? I know it is false now, is there a simple counter Example? $\endgroup$ – yoyostein Dec 5 '16 at 17:54
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    $\begingroup$ @yoyostein I think such a counterexample must involve fields of prime characteristic. Right now, I don't have any counterexamples, but I'll think about it. $\endgroup$ – Xam Dec 5 '16 at 18:14

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