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.


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.

  • $\begingroup$ I think, on sober reflection, that there’s a much more elementary way of doing this, using the fact that $F$ is Galois over $\Bbb Q$. I may make an addition to the above. Keep posted. $\endgroup$ – Lubin Aug 4 '16 at 2:53
  • $\begingroup$ Really beautiful - I enjoyed reading both of these proofs. I was hunting for the second proof for a while this afternoon, but it never quite materialized for me. $\endgroup$ – Alex Wertheim Aug 4 '16 at 3:31

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