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 $F[x]$.

Call $R$ the ring of algebraic integers of $F$. Let $p$ be a prime number dividing $m$.
If $p$ is inert in $F$, then $X^n-m$ is Eisenstein with respect to the prime $pR$ of $R$.
If $p$ splits in $F$, i.e. $pR=\mathfrak p\mathfrak q$ with $\mathfrak p\ne\mathfrak q$, $\text{ord}_{\mathfrak p}(m)=1$ and again $X^n-m$ is Eisenstein with respect to $\mathfrak p$.
Notice that I haven’t yet used oddness of $n$.
The interesting case is where $p$ ramifies in $F$, i.e. $pR=\mathfrak p^2$. Now, $\text{ord}_{\mathfrak p}(m)=2$, and you’d think that Eisenstein didn’t apply. But look at the Newton polygon of $X^n-m$: it has only two vertices, at $(0,2)$ and $(n,0)$. The segment between these two contains no other integral points because $n$ is odd. If you’re Newton-savvy, this is enough to show irreducibility, but let me fill in some details. If you had an $F$-factorization $X^n-m=f(X)g(X)$, say $f$ was of degree $n'<n$. Over the $\mathfrak p$-completion, all roots $\rho$ of $X^n-m$ have $\text{ord}_{\mathfrak p}(\rho)=2/n$ and so the constant term of $f$ would have order $1=2n'/n$, showing $n$ to be even, a contradiction.
EDIT — Addition
I ought to be ashamed of myself that I offered such an overcomplicated proof. Here is what’s really happening:
Proposition. Let $k$ be a field, and $f(X)\in k[X]$ be irreducible of odd degree. Let $K$ be a quadratic separable extension of $k$. Then $f$ is irreducible in $K[X]$.
Proof. Let $g$ be any monic factor of $f$, irreducible in $K[X]$. If $\sigma$ is the nonidentity $k$-automorphism of $K$, there are two cases. If $g=g^\sigma$, then $g\in k[X]$ and so $g=f$, and $f$ is thus $K$-irreducible. If $g\ne g^\sigma$, then $gg^\sigma\in k[X]$, and since $g|f$, we also have $g^\sigma|f$. This allows us to say $gg^\sigma|f$, which by $k$-irreducibility of $f$ implies that $f=gg^\sigma$, making $f$ of even degree. So this case doesn’t occur.
