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Let $F$ be a field which contains $n$ distinct $n$th roots of $1$. Let $E$ be the splitting field over $F$ of a polynomial

$$f(x) = (x^n−a_1)···(x^n−a_r)$$

with $a_i∈F$. Show that $E/F$ is an abelian Galois extension such that $G=\text{Gal}(E/F)$ has exponent $m$ dividing $n$.

I can't find this anywhere on stack exchange. I know the title is really bad but I can't think of anything better other than something extremely vague like "Galois Extension Question".

Attempt: Let us first try the case where $r=1$. That is to say we have $f(x)=x^n-a_1$. Let the nth roots of $1$ which are not $1$ be denoted by $\theta_1,\theta_2,...,\theta_{n-1}$. If there exists $b_1\in F$ such that $b_1^n=a_1$ then $F$ is the splitting field of $f(x)=x^n-a_1\in F[x]$ and we are done, so let us assume there exists no such $b_1\in F$ such that $b_1^n=a_1$. Adjoin an $n$th root of $a_1$, $b_1$ say, to $F$ to obtain $F(b_1)$. Then we have that $F(b_1)$ is the splitting field of $f(x)\in F[x]$ since $F$ contains $n$ distinct $n$th roots of 1.

Thus we have that $F(b_1)/F$ is a Galois extension as $f$ is separable since it has roots $b_1, \theta_1 b_1, \theta_2 b_1,...,\theta_{n-1}b_1$ which must be distinct by hypothesis and $F(b_1)$ is the splitting field of $f(x)\in F$. Then we must have $\vert\text{Gal}(F(b_1)/F)\vert=[F(b_1):F]$. Since $b_1$ is a root of $f(x)=x^n-a_1$ it follows that the minimal polynomial of $b_1$ over $F$ has degree at most $n$. Hence $G_1:=\vert\text{Gal}(F(b_1)/F)\vert=[F(b_1):F]\leq n$. The exponent of $G_1$, denoted by $\text{Exp}(G_1)$ will be at most $[F(b_1):F]$ and hence will be less than $n$, but I see no reason why it should divide $n$. Thus I would like help in order to show that $\text{Exp}(G_1)$ divides $n$. Also, I don't see why $G_1$ should be abelian.

For the general case, assuming we have indeed shown the theorem to be true for $r=1$, assume we have shown the theorem to be true up to $r=k$, we then wish to show it is true for $r=k+1$. To this end, let $g(x)=(x^n-a_1)(x^n-a_2)...(x^n-a_k)$ and thus $f(x)=g(x)(x^n-a_{k+1})$. Then let $E'$ be the splitting field of $g(x)\in F[x]$ and hence $E'/F$ is a Galois extension. Let $G_{k}:=\text{Gal}(E'/F)$, then we know that $\text{Exp}(G_k)|n$ by assumption. Let $E$ be the splitting field of $f(x)\in E'[x]$. Then we know that $\text{Gal}(E/E')$ has exponent which divides $n$. I would like to show that $\text{Gal}(E/F)$ has exponent which dives $n$...but I don't know how to relate the exponent of $\text{Gal}(E/F)$ to the exponent of $\text{Gal}(E/E')$ and $\text{Gal}(E'/F)$.

Any help or hints would be greatly appreciated.

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$F$ is a field containing a primitive $n$-th of unity $\zeta_n$.

For $a \in F$ since $x^n-a = \prod_{m=1}^n (x-\zeta_n^m a^{1/n})$ then $F(a^{1/n})/F$ is normal and separable thus Galois. Let $g \in Gal(F(a^{1/n})/F)$ then $(g(a^{1/n}))^n-a= g((a^{1/n})^n-a)=0$ so $g(a^{1/n}) = \zeta_n^{e_g} a^{1/n}$ and $g(\zeta_n^m a^{1/n}) = \zeta_n^{m+e_g}a^{1/n}$ so the map $g \mapsto e_g$ is an injective homomorphism $G \to \Bbb{Z/nZ}$ and $g^n = Id \in G$.

Let $d = gcd( \{e_g, g \in G\})$ then $G$ will be isomorphic to the subgroup $\Bbb{(dZ)/(nZ)}$ which is cyclic with $n/d$ elements.

For $E=F(a_1^{1/n}, \ldots,a_R^{1/n})$ the idea is the same, with $g(a_r^{1/n}) = \zeta_n^{e_{g,r}}a_r^{1/n}$ the map $g \mapsto (e_{g,1},\ldots,e_{g,R})$ will be an injective homomorphism $Gal(E/F) \to \underbrace{\Bbb{Z/nZ} \times \ldots \times \Bbb{Z/nZ}}_R$.

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