How to find all intermediate fields of $\mathbb{Q}(2^{1/4})$ over $\mathbb{Q}$ without the Galois correspondence? As a $\mathbb{Q}$-vector space, $\mathbb{Q}(2^{1/4})$ has a dimension 4. 
So, any intermediate subfields have a dimension 2 over $\mathbb{Q}$.
I'm already know that $\mathbb{Q}(2^{1/2})$ is such a subfield. But, without the Galois correspondence, the uniqueness is not obvious, I think. 
Since such an intermediate field is a simple extension having a primitive element $\beta$, I tried to express $\beta$ as $a+b\alpha+c\alpha^2+d\alpha^3$, where $\alpha=2^{1/4}$, but it's too messy. 
 A: This can be done without much tedious calculation. See Example 1.3 of http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisappn.pdf.
A: As you noted, we must have $\beta=a+b\alpha+c\alpha^2+d\alpha^3$.
As $\mathbb Q[\beta-a]=\mathbb Q[\beta]$ we may assume wlog. that $a=0$, so $$\tag0\beta=b\alpha+c\alpha^2+d\alpha^3.$$
As $\beta\ne0$, at least one of $b,c,d$ is nonzero.
Then 
$$ \beta^2=2c^2+4cd\alpha+(b^2+2d^2)\alpha^2+2bc\alpha^3.$$
As $\beta$ must obey a quadratic equation, we conclude that $\beta^2-2c^2$ is a rational multiple of $\beta$. Thus from the coefficients of $\alpha$ and $\alpha^2$
$$\tag14cd\cdot c=(b^2+2d^2) \cdot b$$
and from those of $\alpha$ and $\alpha^3$
$$\tag24cd\cdot d = 2bc\cdot b. $$
One option for $(2)$ is $c=0$. Then in $(1)$, $b^2+2d^2>0$ and hence $b=0$, so that $d\ne 0$ and $\beta=\alpha^3$. In that case $\mathbb Q[\beta]=\mathbb Q[\alpha^3]=\mathbb Q[\alpha]$ is the full extension field.
Hence we may assume $c\ne 0$. Then from $(2)$, $2d^2=b^2$. If $d\ne 0$ this implies that $\sqrt 2=\frac bd$, which is absurd. Hence $d=0$, $b=0$, and $\mathbb Q[\beta]=\mathbb Q[\alpha^2]$.
