Finding all intermediate fields STATEMENT: Let $\alpha$ be the real positive fourth root of 2. Find all intermediate fields in the extension $\mathbb{Q}(\alpha)$ of $\mathbb{Q}$.
QUESTION: I basically used the tower law to show that $\mathbb{Q}(\alpha),\mathbb{Q}(\alpha^2),\mathbb{Q}$ are all intermediate fields by tower law, and $\mathbb{Q}(\alpha^3)=\mathbb{Q}(\alpha)$.  I am just not sure how to show that these are all of the intermediate fields. I would really appreciate a hint or suggestion.
 A: Here is an alternative argument that does not use Galois theory. Instead one can make use of the linear algebraic structure:
Observation 1: $\operatorname{irr}(\alpha,\Bbb{Q},X)=X^4-2$ is the irreducible polynomial of $\alpha$ over $\Bbb{Q}$, for example by Eisenstein's criterion. Thus if $F$ is an intermediate field of $\Bbb{Q}(\alpha)$ over $\Bbb{Q}$, i.e. if $F$ a field such that
$$\Bbb{Q}< F<\Bbb{Q}(\alpha),$$
then $4=\operatorname{deg}(\operatorname{irr}(\alpha,\Bbb{Q},X))=[\Bbb{Q}(\alpha):\Bbb{Q}]=[\Bbb{Q}(\alpha):F][F:\Bbb{Q}]$, so that
$$[\Bbb{Q}(\alpha):F]=2=[F:\Bbb{Q}].$$
Observation 2: Since $\Bbb{Q}<\Bbb{Q}(\alpha^2)<\Bbb{Q}(\alpha)$ and $\operatorname{irr}(\alpha^2,\Bbb{Q},X)=X^2-2$ is the irreducible polynomial of $\alpha^2$ over $\Bbb{Q}$, we have that $\Bbb{Q}(\alpha^2)$ is an intermediate field.
Observation 3: $\Bbb{Q}(\alpha)$ is a $4$-dimensional $\Bbb{Q}$-vector space with basis $\{1,\alpha,\alpha^2,\alpha^3\}$ and $\Bbb{Q}(\alpha^2)$ is a $2$-dimensional $\Bbb{Q}$-vector space with basis $\{1,\alpha^2\}$. Rearranging the (standard?) basis of $\Bbb{Q}(\alpha)$ so that it is subordinate to the filtration $\Bbb{Q}<\Bbb{Q}(\alpha^2)<\Bbb{Q}(\alpha)$, we have
\begin{align}
\Bbb{Q}(\alpha)=\Bbb{Q}\oplus \alpha\Bbb{Q} \oplus \alpha^2\Bbb{Q} \oplus \alpha^3\Bbb{Q}=\Bbb{Q}\oplus \alpha^2\Bbb{Q} \oplus \alpha\Bbb{Q} \oplus \alpha^3\Bbb{Q} = \Bbb{Q}(\alpha^2) \oplus \alpha\Bbb{Q} \oplus \alpha^3\Bbb{Q}.
\end{align}

Claim: $\Bbb{Q}(\alpha^2)$ is the only intermediate field.
Proof: Suppose there is another intermediate field $F$. Then $F\cap \Bbb{Q}(\alpha^2)=\Bbb{Q}$, and since $F$ is a $2$-dimensional $\Bbb{Q}$-subspace of $\Bbb{Q}(\alpha)$, $\exists a,b,c\in\Bbb{Q}$ not all $0$ such that
$$F=\Bbb{Q}\oplus (a\alpha+b\alpha^2+c\alpha^3)\Bbb{Q}.$$
Regarding the values of $a,b$ and $c$ there are exactly three possibilities:

*

*$b=0$.

*$b\neq0,$ WLOG $a=1$.

*$b\neq0,$ WLOG $c=1$.

In the first case,
\begin{align}
F\ni &(a\alpha+c\alpha^3)^2=(4ac)+(a^2+2c^2)\alpha^2\in \Bbb{Q}\oplus \alpha^2\Bbb{Q}=\Bbb{Q}(\alpha^2)\\
&\implies (4ac)+(a^2+2c^2)\alpha^2\in F\cap\Bbb{Q}(\alpha^2)=\Bbb{Q}\\
&\implies (a^2+2c^2)=0 \stackrel{a,c\in\Bbb{Q}}{\implies} a=0=c,{\bf\large\unicode{x21af}}.
\end{align}
The other two cases follow similarly: In the second case we have
\begin{align}
F\ni &(\alpha+b\alpha^2+c\alpha^3)^2=(4c+2b^2)+(4bc)\alpha+(1+2c^2)\alpha^2
+(2b)\alpha^3\\
&\implies (4bc)\alpha+(1+2c^2)\alpha^2+(2b)\alpha^3 \in (\alpha+b\alpha^2+c\alpha^3)\Bbb{Q}\\
&\implies 2b=4bc^2 \stackrel{b\neq0}{\implies} c^2=\dfrac{1}{2}\stackrel{c\in\Bbb{Q}}{\implies} \sqrt{2}=\alpha^2\in\Bbb{Q},{\bf\large\unicode{x21af}};
\end{align}
and in the third case we have
\begin{align}
F\ni &(a\alpha+b\alpha^2+\alpha^3)^2=(4a+2b^2)+(2b)\alpha+(a^2+2)\alpha^2
+(2ab)\alpha^3\\
&\implies (2b)\alpha+(a^2+2)\alpha^2+(2ab)\alpha^3 \in (a\alpha+b\alpha^2+\alpha^3)\Bbb{Q}\\
&\implies 2b=2a^2b, a^2+2=2ab^2 \stackrel{b\neq0}{\implies} a^2=1,a=\dfrac{3}{2b^2}>0\\
&\implies a=1, b^2=\dfrac{3}{2}\implies \sqrt{\dfrac{3}{2}}=\vert b\vert\in\Bbb{Q},{\bf\large\unicode{x21af}}.
\end{align}
Hence the overlap of $F$ and $\Bbb{Q}(\alpha^2)$ can not be $1$-dimensional, $\checkmark$.

Note: For further reference this is Exercise V.6.4 from Lang's Algebra (p. 253 of 3e). Since this exercise is before the (official) Galois chapter in the book I believe an argument that does not (explicitly) use Galois theory is what was asked.
A: Are you familiar with the Galois Correspondence? Because you could compute the Galois group of the polynomial $x^4-2$ and write all its subgroups, and then by the correspondence you would obtain all the subfields between $\mathbb{Q}(\sqrt[4]{2},i)$ (the splitting field of the previous polynomial) and $\mathbb{Q}$: in particular one of the ramifications would be the subfields of $\mathbb{Q}\sqrt[4]{2}) \subset \mathbb{Q}(\sqrt[4]{2},i)$,
