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.
Best Answer
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 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.