Determine the Galois group of $F$ over $E$

Question

Let $\Bbb{Q}\subseteq E\subseteq F$ be fields such that

• $F$ is generated over $\Bbb{Q}$ by an n-th root
• $E$ is generated over $\Bbb{Q}$ by all roots of unity in $F$ Determine. $Gal(F/E)$

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My attempt: Let $\alpha= \sqrt[n]{d}$ for some $d\in\Bbb{Q}$ and $\omega=e^{2\pi i/n}$. Then $F=\Bbb{Q}(\omega^k\alpha)$ for some $k=1,2,...,n$.

The question was asked in a qualifying exam. I think that there is a missing point in this question. Suppose $n=3=k$ and $d=2$ and $F=\Bbb{Q}(\sqrt[3]{2})$. Then the only roots of unity in $F$ is 1 and -1. So we have $E=\Bbb{Q}$. But $F$ is not Galois over $E$.

Can anyone check my attempt? In general case how can we determine $E$? Thanks!

Answer 1

The question didn't say to show $F$ is Galois over $E$ (which i interpret as showing $F$ is a splitting field over $E$), but to determine $Gal(F/E)$, the group of automorphisms of $F$ over $E$. In your example the only cube root of 2 in $F$ is the real one, so only the identity automorphism 'works', hence $Gal(F/E)$ is the trivial group.

I believe the general case when the extracted n-th root in F is real is $Gal(F/E)$ trivial when $n$ is odd, and is $Z_2$ when $n$ is even (because -1 is then also a root of unity in $F$). When the extracted $n$-th root is complex, the possibility of other roots of unity has to be considered because the conjugate of an $n$-th root is again an $n$-th root, but on the same grounds I again suspect the answer is $Z_2$ (or possibly $Z_3$?).

Answer 2

This is a tricksy question. In my usage, you don’t speak of $\text{Gal}(K/k)$ unless $K\supset k$ is Galois. Others use that notation for the group of $k$-automorphisms of $K$.

Let me pretend that the authors of the question meant for $F\supset E$ to be Galois, as I would have it. Then it seems to me that there are two cases, namely whether the “root” is a square root, or is $d^{1/m}$ for $m>2$.

In the case that the root in question is $\sqrt d$, there are two subcases, namely $d$ is a nonunit, or $d=\pm1$. If $d$ is a nonunit, then $E=\Bbb Q$ and the Galois group is either trivial (if $d$ is a square) or cyclic of order $2$. If $d=-1$, then $E=F=\Bbb Q(i)$ and the Galois group is again trivial.

In the case that $m>2$, then ignoring messy cases like $\sqrt[4]4$, we see that the only way for the extension to be Galois is for $d$ to be $\pm1$, and we’re adjoining some root of unity. In that case too, $E=F$, and the Galois group is trivial.