Showing a certain polynomial of odd degree is irreducible over quadratic extension I am studying for qualifying exams and am having trouble with this question.
$m>1$ is a square-free integer, $n\ge1$ odd integer.
Suppose $F/\mathbb{Q}$ is any field extension with $[F:\mathbb{Q}]=2$.
Then $x^n-m$ is irreducible in $F[x]$.

Professor Lubin's answer is definitely superior to what I'm about to write, but I thought I'd offer an approach that uses no Galois theory at all. Since $m > 1$ is square-free, pick your favorite prime $p$ dividing $m$; by Eisenstein at $p$, $f(x) = x^{n}-m$ is irreducible over $\mathbb{Q}$. Let $\alpha$ be a root of $f(x)$, and let $K := \mathbb{Q}(\alpha)$. Let $F = \mathbb{Q}(\beta)$ for some $\beta \in \mathbb{C}$. 
Consider the composite extension $L = \mathbb{Q}(\alpha, \beta)$. Since $L$ contains both $K$ and $F$ as subfields, $[K:\mathbb{Q}]$ and $[F:\mathbb{Q}]$ both divide $[L:\mathbb{Q}]$, so $2$ and $n$ both divide $[L:\mathbb{Q}]$. Since $2$ and $n$ are coprime, however, this means $2n \mid [L:\mathbb{Q}]$, and since $[L:\mathbb{Q}] \leqslant 2n$, we get $[L:\mathbb{Q}] = 2n$. 
Then as $[L:\mathbb{Q}] = [L:F][F:\mathbb{Q}]$, it follows that $[L:F] = n$. Since $L = F(\alpha)$, this means the minimal polynomial of $\alpha$ over $F$ has degree $n$. Obviously, $x^{n}-m \in F[x]$ is still a monic polynomial of degree $n$ for which $\alpha$ is a root, so it must be the minimal polynomial for $\alpha$ over $F$, whence it is irreducible. 
