If I have a finite field extensions $E:\mathbb{Q}$ and whose Galois closure has Galois group $G$ which is a soluble group then can I use this to show that $E$ is a radical extension?
Specifically if $E=\mathbb{Q}( 2^{1/2} + 2^{1/3})$, then I can show that the Galois closure is a degree 12 extension of $\mathbb{Q}$ and so its Galois group is soluble, But is $E$ a radical extension?
• Your $E$ is not normal. What is your def' of a radical extension? Do you know the main Thm of galois theory? – Martin Brandenburg Dec 8 '13 at 18:18
• I know $E$ is not normal, which is why I said Galois closure. My def of radical extension is that there is a chain of subfields $F<F_1 \cdots < F_n$ such that $F_i(a_i)=F_{i+1}$ with $a_i^{r_i} \in F_i$ for some $r_i$ an integer. And yes I do know the fundamental theorem. The thing is I know that for example all Galois extensions of degree 3 of $\mathbb{Q}$ are not radical extensions. So knowing the Galois group is soluble doesn't seem enough to say the extension is radical, only that its contained in a radical extension. – Chris Birkbeck Dec 8 '13 at 18:26