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

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.

$$F$$ 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 $$Gal(E/F) \to \underbrace{\Bbb{Z/nZ} \times \ldots \times \Bbb{Z/nZ}}_R$$.