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We are given a field $F$ that contains all the roots of order $n$ of unity and $K=F(\alpha)$, whenever $\alpha^n \in F$. Show that $G(K,F)$ is a cyclic group.

I believe the Galois group is composed of $\{ \alpha,\alpha \omega,\ldots, \alpha \omega^{n-1}\}$ which is a cyclic group.

Am I correct? if not, what is missing? Thanks!

MathematicalPhysicist
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    It can be that $\alpha$ or $\alpha^2$ is in $F$. Any automorphism is of the form $\alpha\mapsto \omega^j \alpha$ for some $j$, not hard to see that it gives a cyclic group (of order $m|n$) – reuns Jul 22 '22 at 09:26

1 Answers1

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First, as a matter of notation, you cannot just list the elements in the field, since the Galois group is full of automorphisms.

Consider the map from $G(K|F)$ to $F^{\times}$ such that $\sigma\mapsto \frac{\sigma(\alpha)}{\alpha}$ (assume $\alpha\not=0$, otherwise the problem is trivial)

First this is well-defined: $(\frac{\sigma(\alpha)}{\alpha})^n=\frac{\sigma(\alpha^n)}{\alpha^n}=\frac{\alpha^n}{\alpha^n}=1$. Since $F$ contains all the $n$-th roots of unity, we know $\sigma(\alpha)/\alpha\in F$. Secondly, we show it's a group homomorphism. Indeed, since $\sigma_1(\alpha)/\alpha\in F$, we must have $\sigma_2(\frac{\sigma_1(\alpha)}{\alpha})=\frac{\sigma_1(\alpha)}{\alpha}$ and hence $\sigma_2(\sigma_1(\alpha))=\frac{\sigma_1(\alpha)\sigma_2(\alpha)}{\alpha}, \sigma_1\sigma_2\mapsto\frac{\sigma_2(\sigma_1(\alpha))}{\alpha}=\frac{\sigma_2(\alpha)}{\alpha}\frac{\sigma_1(\alpha)}{\alpha}$. Finally if $\frac{\sigma(\alpha)}{\alpha}=1$, then $\sigma(\alpha)=\alpha$, hence $\sigma$ must fix all elements of $F(\alpha)=K$, that is the homomorphism is injective. Therefore $G(K|F)$ is isomorphic to a subgroup of $F^{\times}$.

Now we may cite a theorem that any finite subgroup of the multiplicative group of a field must be cyclic.

Note that we didn't show $K|F$ is a Galois extension. This is not necessarily true as the extension may not be separable (but it's always normal).

Just a user
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