Showing that a certain field extension is Galois

Question

I'm trying to solve the following problem.

Let $E = F(\alpha)$ with $\alpha^n \in F$. Assume that char($F$) does not divide $n$ and that GCD($n$, $[E : F]$) = $1$. Show that $E$ is Galois over $F$ and that $Gal(E/F)$ is abelian.

My first intuition was to show that $E$ is the splitting field over $F$ of a separable polynomial, and so I considered the polynomial $x^n - \alpha^n$ over F. But for $E$ to be the splitting field over $F$ of this polynomial, it would need to contain all $n$th roots of unity, right? And I'm not sure whether that is true. Any hints would be appreciated!

Answer 1

Let $g=\mathrm{Irr}(\alpha,F), \text{ and } d=[E:F]=\mathrm{deg }\,g$. Then

$g$ divides $X^n-\alpha^n$ and so the roots of $g$ are of the form $\zeta^i \alpha$ where $\zeta $ is an $n$-th root of $1$. It follows that $g(0)=\zeta^k\alpha^d $ for some $k$. Hence $\zeta^k\alpha^d \in F$.

Now use the fact that $\mathrm{gcd}(d,n)=1$, as in carmichael561 answer.

Answer 2

Let $a=\alpha^n\in F$, and let $d=[E:F]$. Then $$ a^d=N_{E/F}(a)=N_{E/F}(\alpha^n)=N_{E/F}(\alpha)^n $$ Therefore $a^d$ is the $n$th power of an element of $F$. But since $(d,n)=1$ there are integers $r$ and $s$ such that $rd+sn=1$, hence $$ a=a^{rd}a^{sn}=N_{E/F}(\alpha)^{rn}a^{sn} $$

Therefore $a$ is an $n$th power in $F$, say $a=b^n$ with $b\in F$. This means that $\alpha=\zeta^kb$ where $\zeta$ is a primitive $n$th root of unity and $k$ is an integer.

Since $\mathrm{char}(F)$ does not divide $n$, it follows that $F(\zeta)$ is a Galois extension of $F$ with an abelian Galois group. And $E$ is a subfield of $F(\zeta)$, so $E/F$ is also Galois with an abelian Galois group.