Using real/complex numbers, it's easy to see that e.g. Is the extension $\mathbb{Q}(\sqrt[m]{n})/\mathbb{Q}$ not Normal for $m > 1$ and $n$ square free?. Another example is Why is $\mathbb{Q}(\sqrt[4]{2}) $ is not normal over $\mathbb{Q}$?. Even in the case of $\mathbb Q(\sqrt[3]2)$ I have seen the proof that this is not normal because $(x^3-2) = (x-\sqrt[3]2) (x^2+x\sqrt[3]2 + \sqrt[3]4)$, and the quadratic has no real roots (by looking at the discriminant in the quadratic formula). But these proofs all rely on properties of $\mathbb R$, chief among them being the fact that $x\in \mathbb R \implies x^2 \geq 0$ (so order properties of $\mathbb R$). Using this strategy, given algebraic $\alpha$ over $\mathbb Q$ s.t. the minimal polynomial $m_\alpha$ has a real root, I can take $\alpha'$ to be that real root, construct the field extension $\mathbb Q(\alpha') \subseteq \mathbb R$ (which obviously does not contain $i\in \mathbb C$), and hence $\mathbb Q(\alpha)$ (the abstract algebraic object) can not contain $i$ (i.e. any root of $x^2+1$) because $\mathbb Q(\alpha) \simeq \mathbb Q(\alpha')$ as fields and $\mathbb Q(\alpha')$ has no roots of $x^2+1 \in \mathbb Q[x]$.
Of course, in the case of $\mathbb Q(\sqrt[3]2)$ one can use the result that if $[F(\alpha):F]$ and $[F(\beta):F]$ are relatively prime, then $[F(\alpha,\beta):F]=[F(\alpha):F]\cdot[F(\beta):F]$ to show that $[\mathbb Q(\sqrt[3]2,i):\mathbb Q] = 6$ and hence $i \not\in \mathbb Q(\sqrt[3]2)$, because $[\mathbb Q(\sqrt[3]2):\mathbb Q] = 3$, the degree of the irreducible (by Eisenstein) polynomial $x^3-2$ with $\sqrt[3]2$ as a root. But one can not even use this strategy to show $i \not\in \mathbb Q(\sqrt 2)$.
My question: there probably isn't a general way to prove the titular result without referring to $\mathbb R$, but e.g. for the "simple rooty field extensions" in the first paragraph, is it possible? The way I phrased the question in the title is a admitedlly bit too broad, since in the very statement of the question I'm forced to use $\mathbb R$. But for "simple rooty extensions", the presence of the real line should not be as crucial, right?
Best Answer
Note that the currently accepted answer has an error: it is claimed that the automorphism sending $\sqrt[m]{n}$ to $\zeta_m \sqrt[m]{n}$ is easily seen to have order $m$ without further argument, whereas that is only clear if that automorphism fixes $\zeta_m$. But it need not fix $\zeta_m$ without further argument. For example, as noted in another answer, the field $\mathbf{Q}(\sqrt[6]{-3})$ is Galois of degree six with Galois group $S_3$ which has no elements of order $6$, and the automorphism sending $\sqrt[6]{-3}$ to $\sqrt[6]{-3} \cdot \zeta_6$ actually has order $2$.
We prove the following stronger statement. Let $m > 0$ and let $n$ be an integer so that $x^m - n$ is irreducible. Suppose that $\mathbf{Q}(\alpha)$ with $\alpha = \sqrt[m]{n}$ is the splitting field of $x^m - n$. Then either:
Conversely, in these cases, $\mathbf{Q}(\alpha)$ is the splitting field.
Note that $\mathbf{Q}(\alpha)$ being the splitting field of $x^m -n$ is equivalent to $\mathbf{Q}(\alpha)$ being Galois over $\mathbf{Q}$.
First some reductions. If $m > 1$ is odd, then $L = \mathbf{Q}(\alpha)$ can never be Galois. This is because $L$ must contain $K = \mathbf{Q}(\zeta_m)$, which has degree $\varphi(m)$ which is even for $m > 2$. (You seem to be happy to accept that the degree of $K$ is $\varphi(m)$, although this is not so easy. The easiest proof to me that the degree is even when $m > 2$ is to use complex conjugation! The only facts about the degree of $\mathbf{Q}(\zeta_m)$ we use is that it is even for $m > 2$ and bigger than $2$ unless $m = 1,2,3,4,6$)
Second, for any $m$ which is not a power of $2$, the extension $L = \mathbf{Q}(\sqrt[m]{n})$ can never be Galois and abelian. If it is abelian, then any subfield is also Galois. But now taking $p > 2$ to be the largest power of $p$ dividing $m$, the extension $L$ contains $\mathbf{Q}(\sqrt[p]{n})$, which we previously have seen is never Galois. (Subfields of abelian Galois fields are always Galois.)
The result is trivially true for $m=1$ and $m = 2$. If $m = 6$, then $n$ is neither a cube nor a square since otherwise $x^6 - n$ is not irreducible. But if $n$ is not a square then the splitting field of $x^6 - n$ contains $(\sqrt[6]{n})^3 = \sqrt{n}$, but also contains $\zeta_6$ and hence $\sqrt{-3}$. If these two fields are the same, then $n = - 3 a^2$ for some integer $a$. If $n$ is not of this form, however, then the splitting field contains two distinct quadratic fields, and hence contains their compositum which has degree $4$. But then the degree $6$ extension given by adjoining a single root cannot contain the degree $4$ extension and hence is not Galois.
If $n = -3 a^2$, then $\mathbf{Q}(\sqrt[6]{n})$ contains $(\sqrt[6]{n})^3 = \sqrt{-3 a^2}$ and thus contains $\sqrt{-3}$ and $\zeta_6$, and hence these extensions are Galois.
If $m = 4$, then $n$ is not a square since otherwise $x^4 - n$ is not irreducible. But if $n$ is not a square then the splitting field of $x^4 - n$ contains $(\sqrt[4]{n})^2 = \sqrt{n}$, but also contains $\zeta_4$ and hence $\sqrt{-1}$. If these two fields are the same, then $n = - a^2$ for some integer $a$. If $n = a^2$ then, as above, the field $\mathbf{Q}(\sqrt[4]{n})$ contains $\sqrt{-a^2}$ and thus $\sqrt{-1}$ and is hence Galois. If not, then $K$ must contain the compositum of $\sqrt{n}$ and $\sqrt{-1}$. But unlike the case $m = 6$ where $4$ did not divide $6$, in this case we can only deduce that there is an equality
$$\mathbf{Q}(\sqrt[4]{n}) = \mathbf{Q}(\sqrt{-1},\sqrt{n}).$$
Since the Galois group has order $4$ and acts transitively on the roots of $x^4 - n$, any non-trivial element must act non-trivially on every root. (Otherwise there is not enough elements to sent that root to every other root.) By looking at the Galois group of the right hand side, we see that there must be an order $2$ element $\sigma$ which fixes $\sqrt{n}$ but not $i = \sqrt{-1}$. If $\sigma \sqrt[4]{n} = \pm i \sqrt[4]{n}$, then $\sigma \sqrt{n} = - \sqrt{n}$ contradicting the assumption that $\sigma$ fixes $\sqrt{n}$. Hence $\sigma \sqrt[4]{n} = - \sqrt[4]{n}$ (since it cannot fix any root) and $\sigma i \sqrt[4]{n} = - i \sqrt[4]{n}$. But then, taking the ratio, we find that
$$\sigma i = \frac{\sigma i \sqrt[4]{n}}{\sigma \sqrt[4]{n}} = \frac{- i \sqrt[4]{n}}{-\sqrt[4]{n}} = i,$$
a contradiction. This completes the case of $m = 4$.
We now proceed by induction. Having considered the cases $m=1$, $m = 2$, $m=3$ ($m$ odd), $m = 4$, and $m = 6$, we may assume that $\varphi(m) > 2$.
Suppose that $x^m - n$ is irreducible. Let $\alpha = \sqrt[m]{n}$, and let $\zeta = \zeta_m$ be a primitive $m$th root of unity. Let $L = \mathbf{Q}(\alpha)$. The roots of $x^m - n$ are of the form $\alpha \zeta^i$ for some $i = 0,1,\ldots,m-1$. Assume that $L$ is Galois. Then $L$ contains $\zeta$ and hence the field $K = \mathbf{Q}(\zeta)$. Note that $[K:\mathbf{Q}] = \varphi(m)$, so $m = r s$ for some integer $r = [L:K]$ and $s = \varphi(m)$.
Let $G = \mathrm{Gal}(L/\mathbf{Q})$ and let $H = \mathrm{Gal}(L/K)$ which has order $r$. Any element of $H$ fixes $\zeta$ by definition and sends $\alpha$ to $\alpha \zeta^k$ for some $k$. But since $\sigma$ fixes $\zeta$, it then acts on all the roots of $x^m - n$ as multiplication by $\zeta^k$. It follows that $H$ is cyclic of order $d$ where $d$ is the smallest integer such that $dk \equiv 0 \bmod m$. Since $|H| = r$, we have $r = d$.
We see that $\alpha^r = (\sqrt[m]{n})^r = (\sqrt[rs]{n})^r = \sqrt[s]{n}$ is fixed by $H$ and thus $\mathbf{Q}(\sqrt[s]{n}) \subseteq L^{H} = K$. But we have
$$r \ge [\mathbf{Q}(\alpha):\mathbf{Q}(\alpha^r)] = [L:K] \cdot [K:\mathbf{Q}(\alpha^r)] = r \cdot [K:\mathbf{Q}(\alpha^r)],$$
and hence $\mathbf{Q}(\sqrt[s]{n}) = K$. In particular, $\mathbf{Q}(\sqrt[s]{n})$ is Galois of degree $s = \varphi(m)$ and equal to $K = \mathbf{Q}(\zeta_m)$. Moreover, $\sqrt[s]{n}$ is a root of $x^s - n$, which is irreducible because $\sqrt[s]{n}$ has degree $s$. So $\sqrt[s]{n}$ generates a Galois extension $K$ which must therefore be the splitting field of this polynomial. In addition, this splitting field $K$ is abelian. By our preliminary remarks, this is a contradiction unless $s$ is a power of $2$. But by induction we obtain a contradiction unless $s \le 2$, or unless $\varphi(m) \le 2$. But this implies that $m=1,2,3,4,6$, cases which we have already considered.