Defn: A field extension $K/F$ is obtained by adjoining $n$th roots if there is a tower of fields $F=K_1\subset\cdots\subset K_n=K$ such that for each $i$, $K_{i+1}=K_i(\alpha_i)$ and there exists some $n_i>1$ such that $\alpha_i^{n_i}\in K_i$.

I'm interested in whether subextensions $K/F$ of an extension $E/F$ obtained by adjoining $n$th roots must also be obtained by adjoining roots.

When $K/F$ is Galois, we can make use of the fact that Galois extensions are obtained by adjoining roots iff their Galois groups are solvable. Extend $E/F$ to its Galois closure, which is also obtained by adjoining roots, so $\mathrm{Gal}(E^{\mathrm{Gal}}/F)$ is solvable. Since $\mathrm{Gal}(K/F)$ is a quotient of $\mathrm{Gal}(E^{\mathrm{Gal}}/F)$ it is solvable as well, thus $K/F$ is obtained by adjoining roots.

This reduces the problem to the case $E=K^{\mathrm{Gal}}$. However, I have no idea how to usefully characterize non-Galois extensions which are obtained by adjoining roots, so I don't know how to approach this case.

I'm mostly interested in the case where $K$ is separable, but I'd also be interested in a counterexample where this is not the case.

  • $\begingroup$ Is $K^G$ your notation for the Galois closure of $K$ ? Pardon me but this might be one of the worst notations I've seen in my life... (EDIT: sorry for my comment about the title -- I got confused by this notation.) $\endgroup$ – darij grinberg Apr 5 '14 at 0:07
  • $\begingroup$ @darijgrinberg Is this not standard? Huh. Would you use $\overline{K}$ instead? $\endgroup$ – Alex Becker Apr 5 '14 at 0:08
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    $\begingroup$ Maybe $K^{\mathrm{Gal}}$ ? The problem with $K^G$ is that it stands for the invariants of $K$ under the action of a group $G$, a situation rather common in Galois theory... $\endgroup$ – darij grinberg Apr 5 '14 at 0:09
  • $\begingroup$ @darijgrinberg That looks reasonable to me. I'll change the notation. $\endgroup$ – Alex Becker Apr 5 '14 at 0:10
  • $\begingroup$ @darijgrinberg Hmm... Now that I think about it, I'm only interested in the case where $K$ is separable, so I've only been considering those cases. Is it alright if I change the question? $\endgroup$ – Alex Becker Apr 5 '14 at 0:19

Let $f(x)=x^3-x-1$, let $\alpha$ be a zero of $f$ in an extension of the rationals, let $E={\bf Q}(\alpha)$, let $K$ be a splitting field for $f$ over $E$. Then $K$ is obtained from the rationals by adjoining $n$th roots, but $E$ isn't.

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    $\begingroup$ Thanks, I see now that this question is trivial. So the Galois assumption is really quite necessary. $\endgroup$ – Alex Becker May 4 '14 at 20:56
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    $\begingroup$ I wouldn't say it's trivial. I remember that the first time I taught Galois Theory out of Stewart's book, and saw the question raised in an exercise, I had some difficulty getting my head around it. $\endgroup$ – Gerry Myerson May 5 '14 at 4:09

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