Prove that $i\not\in \mathbb Q(\alpha)$ for algebraic $\alpha \in \mathbb R$, WITHOUT using that $i \not\in \mathbb R$ (or referencing $\mathbb R$)? 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?
 A: 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:

*

*$m = 1$ or $m = 2$,

*$m = 6$ and $n = - 3 a^2$ for some integer $a$ where $n$ is not a cube.

*$m = 4$ and $n = - a^2$ for some integer $a$.

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.
A: This is based on my comments to the question.
There is an algebraic way to get something similar to $\mathbb {R} $ and Artin and Schreier developed this whole concept of formally real and real closed fields. Let us first define an ordered field as a pair $F, P$ where $F$ is a field and $ P\subseteq F$ such that

*

*$0\notin P$

*if $a\in F$ then either $a\in P$ or $a=0$ or $-a\in P$

*$P$ is closed under addition and multiplication operations of $F$.

One can observe that any non-zero square must lie in $P$ and thus $1\in P$ and thus an ordered field is necessarily of characteristic $0$. Further if $a_1,a_2,\dots,a_n\in F$ then $$\sum_{i=1}^n a_i^2=0\implies a_i=0 \text{ for each }i $$ This property is used to develop the idea of a formally real field.
A field $F$ is said to be formally real if the only relations of the form $\sum_{i=1}^n a_i^2=0$ in $F$ are those for which $a_i=0$. Note again that a formally real field must be of characteristic $0$. And then we can say that every ordered field is formally real.
It is a bit surprising that the converse is also true that every formally real field can be ordered. The order may not necessarily be unique (as it is for $\mathbb{Q} $). For example $\mathbb{ Q} (\sqrt{2})$ is formally real and it has two distinct orderings.
It can also be noted that definition of a formally real field is equivalent to the condition that $-1$ can not be expressed as a sum of squares in such a field. And thus we can't have $i$ (square root of $-1$) in any formally real field.
The theory of formally real fields becomes interesting when we try to find out a maximal formally real field in the following sense. A formally real field $F$ is said to be real closed if $F$ has no proper algebraic extension which is formally real.
Next we can we can start with any formally real field $F$ and generate its algebraic closure $\overline{F} $ via algebraic processes. And then we can consider the collection $\mathcal{C} $ of fields $K$ given by $$\mathcal{C} =\{K\mid F\subseteq K\subseteq \overline {F}, K \text{ is formally real} \} $$ and using Zorn's lemma find a maximal element $F_{rc} $ of $\mathcal{C} $. It can be proved that $F_{rc} $ is a real closed field and is essentially unique and has a unique ordering. The field $F_{rc} $ is called the real closure of $F$ (I have used the subscript "rc" to mean real closure). The same process can be applied for an ordered field $F$ (as it is also formally real).
Applying the above process to $\mathbb{Q} $ gives us the field $\mathbb {Q} _{rc} $ and its elements are called real algebraic numbers. Real closed fields have many properties which are similar to that of $\mathbb {R} $.
If $F$ is a real closed field then

*

*$F$ is ordered with the squares of non-zero elements being considered positive and thus every positive member of $F$ has a square root in $F$.

*if $f(x) \in F[x] $ is of odd degree then $f(x) $ has a root in $F$.

*because of the above two properties $F(i) $ is an algebraically closed field.

*polynomials in $F[x] $ satisfy intermediate value theorem, Rolle's theorem, mean value theorem and extreme value theorem.

The proofs of above properties are fully algebraic and really very interesting. For details one may have a look at last chapter on Artin-Schreier Theory in Jacobson's Lectures in Abstract Algebra, Vol. 3.
The development of these ideas allows us to handle the algebraic numbers (both real and complex) without actually using any analysis. Dealing with polynomials of the form $x^n-a\in \mathbb {Q} [x] $ is then straightforward. If $n$ is odd then this polynomial has only one root in $\mathbb {Q} _{rc} $ and rest of the roots lie in $\overline {\mathbb{Q}} $. If $n$ is even then we have either $0$ or $2$ roots in $\mathbb {Q} _{rc} $ depending on $a<0$ or $a\geq 0$ and rest of the roots lie in $\overline {\mathbb {Q}} $.
A: The question is a little vague because it's not entirely clear what the rules are when you say you can "not use $\mathbf{R}$". One pretty strict way to formulate this is to say that you cannot use any inequalities. But one remark is that if you assume $x^m - n$ is irreducible (say by assuming that $n$ has at least one prime divisor and is squarefree) and you want to prove that this is not normal when $m > 2$ you will have to assume that $n > 0$ rather than $n < 0$ since it is not true otherwise. For example, $x^6 + 3$ is irreducible and yet the splitting field is given by taking any root of this polynomial. So in this strict sense you have to invoke the fact that $n > 0$ and implicitly use $\mathbf{R}$.
A: Ring extension equals field extension: $\Bbb Q[\alpha] = \Bbb Q(\alpha)$.
So if this extension has degree $n$, then each element of $\Bbb Q(\alpha)$ has the form
$a_{n-1} \alpha^{n-1} + \ldots a_1\alpha + a_0$ with all $a_i\in\Bbb Q$.
This expression lies in $\Bbb R$ by hypothesis
