Determine all subfields of $\mathbb{Q}(\alpha)$, where $\alpha$ is real fourth root of $2$ Determine all subfields of $\mathbb{Q}(\alpha)$, where $\alpha$ is real fourth root of $2$ containing $\mathbb{Q}$
I proceed as follows. It is known that $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$ and hence if $E$ is a non-trivial subfield of $\mathbb{Q}(\alpha)$ containing $\mathbb{Q}$, then $ [\mathbb{Q}(\alpha):E]=[E:\mathbb{Q}]=2$
I know that there exists such a field, that is $\mathbb{Q}(\alpha^2)$ but I am not able to prove that this is the unique such field.
What I tried here is since $[\mathbb{Q}(\alpha):E]=2$, then $\alpha \in \mathbb{Q}(\alpha\setminus E)$ satisfies a degree $2$ polynomial over $E$. Say $f(x)=x^2+ax+b$ where $a,b \in \mathbb{Q}(\alpha) \cap E$. Now I assumed $a=a_0+a_1 \alpha+a_2 \alpha^2 \dots a_n \alpha^m $ and $b=b_0+b_1 \alpha+b_2 \alpha^2 \dots b_n \alpha^m $ and tried putting in the equation $x^2+ax+b=0$
Then I was able to conclude that $$b_0+(a_0+b_1)\alpha+(a_1+b_2+1)\alpha^2+(a_2+b_3)\alpha \dots (a_n)\alpha^n=0$$
Now I think this is getting too complicated.
I also think that using Galois Correspondence, I can do it but it will be overkill. I want to take an elementary approach. And I havent used the fact that $\alpha$ is a real fourth root of $2$ above. Hence I would appreciate if someone can help.
 A: The field has the obvious intermediate field $\Bbb{Q}(\sqrt2)$. As can be seen by Galois theory, it is the only intermediate field.
A possibly more elementary argument leading to this conclusion is the following.

The compositum of two quadratic extensions would be normal (or Galois, but I try to avoid Galois theory, so...). This quartic extension is not. Therefore there cannot be more than one intermediate field.

The details are based on the following two relatively simple facts. Ask, if you have not seen these facts and cannot prove them. The latter needs manipulations with numbers of the form $a+b\sqrt n$. My argument also needs the first properties of normal extensions.
Fact 1. If $K$ is a quadratic extension field of $\Bbb{Q}$ then $K=\Bbb{Q}(\sqrt n)$ for some square-free integer (that is, $n\neq 0,1$ and $n$ is not divisible by the square of any prime number).
Facst 2. Let $m,n$ be two distinct square-free integers. Then the fields $K_1=\Bbb{Q}(\sqrt n)$ and $K_2\Bbb{Q}(\sqrt m)$ (both subfields of $\Bbb{C}$) intersect trivially. That is, $K_1\cap K_2=\Bbb{Q}$.
On with the proof of the main claim. Assume contrariwise that the field $L=\Bbb{Q}(\alpha)$ has at least two subfields other than $\Bbb{Q}$. Both of them are then necessarily quadratic, so in view of the above facts we can find two distinct square-free integers $m$ and $n$ such that $\sqrt m$ and $\sqrt n$ are both elements of $L$. But, by Fact 2, $\sqrt m\notin\Bbb{Q}(\sqrt n)$, so we see that
$$[\Bbb{Q}(\sqrt m,\sqrt n):\Bbb{Q}]=4.$$
Undoubtedly you have proven this as an exercise (or as an example in class) when $m=2, n=3$. The general case is similar but I use the facts instead.
Because $\Bbb{Q}(\sqrt m,\sqrt n)\subseteq L$, and we also have $[L:\Bbb{Q}]=4$, our contrapositive assumption has lead to the conclusion that $L=\Bbb{Q}(\sqrt m,\sqrt n)$.
Drums, please.
But the field $\Bbb{Q}(\sqrt m,\sqrt n)$ is the splitting field (over $\Bbb{Q}$) of the polynomial $(x^2-m)(x^2-n)$. Hence it is a normal extension of $\Bbb{Q}$. However, the irreducible polynomial $x^4-2$ (over $\Bbb{Q}$) has only two of its zeros in $L$. Hence $L/\Bbb{Q}$ cannot be normal. Contradiction.
A: If you know any Galois theory, this is a nice introductory exercise.
The minimal polynomial of $\alpha$ is $x^4-2$. The splitting field of this polynomial is $\Bbb Q(\alpha, i)$. The Galois group of this field over $\Bbb Q$ is the dihedral group $D_4$, generated by the "reflection" that is complex conjugation and the "$90^\circ$ rotation" that is $\alpha^ai^b\mapsto \alpha^ai^{a+b}$.
This dihedral group has exactly two subgroups of index 2, which means there are exactly two subextensions of degree 2:

*

*The subgroup consisting by all rotations. The fixed subfield is $\Bbb Q(i)$, and thus not valid for our purposes

*The subgroup generated by $180^\circ$ rotation $\alpha\mapsto -\alpha$ and complex conjugation. The fixed subfield is $\Bbb Q(\alpha^2)$, which you already found

Alternately, $\Bbb Q(\alpha)$ is the fixed subfield of the subgroup generated by complex conjugation. Any subfield must correspond to a subgroup of $D_4$ that contains complex conjugation. There are 3, only one of which is of index 2. Again, we find that there is only one suitable intermediate field.
A: Let the basic field $K$ of characteristic $\ne 2$, and
$$K \subset K(\sqrt{d_1}, \sqrt{d_2})=\colon L$$
where $\sqrt{d_1}$, $\sqrt{d_2}$ are linearly independent over $K$. We want to show that the field $L$ does not contain any $4$-th roots. Now, this will happen under some extra conditions. For in the case $K = \mathbb{Q}$, $d_1 = -1$, $d_2 = 2$, we have $L = \mathbb{Q}(\sqrt[4]{-1})$.
Now, we have $[L\colon K]=4$ ( use the great observation of Jyrki for a quick proof). Therefore, every element of $L$ can be written uniquely as
$$\alpha + \beta \sqrt{d_1} + \gamma \sqrt{d_2} + \delta \sqrt{d_1 d_2}$$
Moreover, due to the uniqueness of the writing, the maps $\sqrt{d_i} \mapsto \pm \sqrt{d_i}$ are automorphisms of $L$ over $K$.
Now, let $u \in L$ such that $u^4= k \in K$. Since the number of roots in $L$ of $x^4-k=0$ is at most $4$, and the negative of a root is again a root, we conclude that at most $2$ of the coefficient $\alpha$, $\beta$, $\gamma$, $\delta$ of $u$ are $\ne 0$. We may ( by multpliying by some square root) reduce to the case
$$u = \alpha + \beta\sqrt{d_1}$$
But now for $v = u^2$ we have $v^2\in K$, so the rational or the irrational part of $v$ is $0$. Now, if the irrational part is $0$, we conclude $u$ is in fact a square root. However, if $\alpha^2 + \beta^2 d_1  = 0$, then $d_1 = - p^2$, with $p\in K$. This is the situation that is "not allowed", since then $\sqrt{d_1}$ is "not real". So "reality" of $L$ boils down to this. We left out some details  for the diligent reader.
$\bf{Added:}$  Based on the very generous hint from Jyrki, let's try to generalize this.
Consider $K$ a field, $n$ a natural number not divisible by the characteristic of $K$, and $d\in K$ such that $x^n - d$ is irreducible. Consider the field extension of degree $n$
$$K\subset K(\sqrt[n]{d})$$
We would like to show that all the intermediate subfields are of the form $K(\sqrt[n]{d}^r)$, where $r \mid n$. Now, this is not true always, as our previous example show. Consider $L$ the decomposition field of $X^n - d$. It will be a Galois extension of $K$, and is the composite of $K(\sqrt[n]{d})$, and $F$  the field obtained from $K$ by adjoining all the $n$-th roots of $1$. Let us assume that $K(\sqrt[n]{d})$ and $F$ are linearly disjoint. This is equivalent (since $F/K$ is normal) to $F\cap K(\sqrt[n]{d}) = K$  (use that for $L \supset K_1$, $K_2 \supset K$, $L/K$ Galois and one of the $K_i/K$ again Galois, then $[K_1 K_2\colon K_1] = [K_2\colon K_1 \cap K_2]$, and this is proved using Galois theory and a basic fact from group theory). Let $\omega$ a primitive $n$-th root of $1$ in $F$. We have $F= K(\omega)$. Now $\operatorname{Gal}(K(\omega)/K) = G\subset (\mathbb{Z}/n)^{\times}$. We conclude that
$$\operatorname{Gal}(K(\omega, \sqrt[d]{n})/K) = \mathbb{Z}/n \rtimes G$$ by the correspondence
$$\sqrt[n]{d}\cdot \omega^x \mapsto \sqrt[n]{d} \cdot \omega^{b+ a x}$$
Now, $K(\sqrt[n]{d})$ is the subfield invariated by $0 \rtimes G$. We conclude that the subfields of $K(\sqrt[n]{d})$ are as indicated.
