Showing that a certain field extension is Galois

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!

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.

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.

• Thanks! One question, what does the notation $N_{E/F}(a)$ mean? – User7819 Mar 7 '17 at 4:18
• It's the field norm from $E$ to $F$. – carmichael561 Mar 7 '17 at 4:18
• On second thought, I know I already accepted the answer, but is it at all possible for you to modify your answer so as not to rely on the field norm? Because in the class I'm taking we have not covered that concept. – User7819 Mar 8 '17 at 2:57
• I'm afraid that I don't know of another way to solve the problem. One way to define the field norm is as follows: each element $\beta\in E$ defines an $F$-linear map on $E$ given by left multiplication by $\beta$, and the $N_{E/F}(\beta)$ is the determinant of this linear transformation. – carmichael561 Mar 8 '17 at 5:36
• If you use this definition, then it's not too hard to show that $N_{E/F}(a)=a^{[E:F]}$ if $a\in F$, and that $N_{E/F}(\alpha\beta)=N_{E/F}(\alpha)N_{E/F}(\beta)$. These are the only two properties of $N_{E/F}$ used in my answer. – carmichael561 Mar 8 '17 at 5:38