How can I find all intermediate fields between $\mathbb{Q}\left(\sqrt[d]{a}\right)$ and $\mathbb{Q}$? Let $a$ be a positive rational number, $n$ be a natural number, and $K$ be extension field of $\mathbb{Q}$ with $\sqrt[n]{a}$.
I guess that intermediate field $E$ between $K$ and $\mathbb{Q}$ with  $\left[E:\mathbb{Q}\right]=d$($d\mid n$) is unique as $\mathbb{Q}\left(\sqrt[d]{a}\right)$. But I can't prove this guess. If $\mathbb{Q}$ has a root of unity $\zeta_{n}$, It should be so easy. But It doesn't. How can I prove that without $\zeta_{n}$?
And is there a formula computing $\text{Gal}\left(x^{n}-a\right)$?
 A: This turns out to be a surprisingly difficult question, but it's true that the intermediate fields are precisely those of the form $\mathbb{Q}\bigl(\sqrt[d]{a}\bigr)$ for $d$ a divisor of $n$.
More generally, if $a_1,\ldots,a_t$ are positive rational numbers and $n_1,\ldots,n_t$ are natural numbers, then every field intermediate between $\mathbb{Q}$ and $\mathbb{Q}\bigl(\sqrt[n_1]{a_1},\ldots,\sqrt[n_t]{a_t}\bigr)$ is generated by monomials in the roots $\sqrt[n_i]{a_i}$.
You can find this result proven immediately after Theorem 1.6 in Greither, Cornelius, and David K. Harrison. "A Galois correspondence for radical extensions of fields." Journal of Pure and Applied Algebra 43.3 (1986): 257-270.  In general, proving results like this apparently involves something called "cogalois theory", which involves a "cogalois correspondence" between intermediate fields of a "cogalois extension" and subgroups of the "cogalois group".  The linked paper from 1986 seems to have been the beginning of this subject, but since then entire books have been written about it.
The splitting field of $x^n - a$ is $\mathbb{Q}\bigl(\sqrt[n]{a},\zeta\bigr)$, where $\zeta$ is a primitive $n$th root of unity.  The Galois group of $x^n - a$ is difficult to analyze in general.  It fits into a short exact sequence
$$
\mathrm{Gal}\bigl(\mathbb{Q}(\sqrt[n]{a},\zeta)\,\bigr/\,\mathbb{Q}(\zeta)\bigr) \;\longrightarrow\; \mathrm{Gal}\bigl(\mathbb{Q}(\sqrt[n]{a},\zeta)\,\bigr/\,\mathbb{Q}\bigr) \;\longrightarrow\; \mathrm{Gal}\bigl(\mathbb{Q}(\zeta)\,\bigr/\,\mathbb{Q}\bigr)
$$
where:


*

*$\mathrm{Gal}\bigl(\mathbb{Q}(\zeta)\,\bigr/\,\mathbb{Q}\bigr) \,\cong\, (\mathbb{Z}/n\mathbb{Z})^\times$, the multiplicative group of units modulo $n$.  

*$\mathrm{Gal}\bigl(\mathbb{Q}(\sqrt[n]{a},\zeta)\,\bigr/\,\mathbb{Q}(\zeta)\bigr) \,\cong\, \mathbb{Z}/r\mathbb{Z}$, where $r$ is the smallest positive integer for which $\bigl(\sqrt[n]{a}\bigr)^r$ lies in the field $\mathbb{Q}(\zeta)$.
Note that $r$ in general may be a divisor of $n$.  For example, even though $x^8-2$ is irreducible over the rationals, the cyclotomic field $\mathbb{Q}\bigl(e^{i\pi/4}\bigr)$ contains $\sqrt{2}$; then $x^8-2$ factors into $(x^4-\sqrt{2})(x^4-\sqrt{2})$ over this field, so in this case $r=4$.  According to the above sequence, the Galois group is some extension of $(\mathbb{Z}/8\mathbb{Z})^\times \cong (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$ by $\mathbb{Z}/4\mathbb{Z}$, and it turns out to be the dihedral group of order 16.
In the case where $r=n$, the sequence splits and the Galois group is isomorphic to the semidirect product $(\mathbb{Z}/n\mathbb{Z}) \rtimes (\mathbb{Z}/n\mathbb{Z})^\times$ with the obvious action.  It's easy to check in this case that all the intermediate fields are generated by powers of $\sqrt[n]{a}$.
For $r<n$, we get some extension of $(\mathbb{Z}/n\mathbb{Z})^\times$ by $\mathbb{Z}/r\mathbb{Z}$, but there doesn't seem to be any simple general description of this extension.  Again, I am not an expert in Galois theory, and I know virtually nothing about cogalois theory, so it's possible that a nice description does exist somewhere in the literature.
