Let $E=\mathbb{Q}(\sqrt[7]{5})$ and suppose that you have another field $F$ of degree 7. Then $[F(\sqrt[7]{5}):F][F:\mathbb{Q}]\leq 14$ so we must have $[F(\sqrt[7]{5}):F]=2$. It follows that $\alpha=\sqrt[7]{5}$ satisfies an irreducible polynomial of degree 2 over $F$.
There is now a general fact that states that if a polynomial $x^p-a$ is reducible over some field F for some $p$ prime, then it has a root in F.
More precisely, this polynomial is of the form $(x-\zeta^i \alpha)(x-\zeta^j \alpha)=x^2 -(\zeta^i+\zeta^j)\alpha +\zeta^{i+j}\alpha^2$ where $\zeta$ is a primitive 7 root of unity and $0\leq i<j<7$ - this is because it divides $x^7-5=\prod_{i=1}^7 (x-\zeta^i \alpha)$.
You now have that $\zeta^{i+j}\alpha^2 \in F$ so by taking the 4th power you also have that $\zeta^{4i+4j}\alpha^8=5\zeta^{4i+4j}\alpha \in F$ and therefore $\zeta^{4i+4j}\alpha \in F$. We now have two options - either $\zeta^{4i+4j}=1$ but then $E=F$, or it is some primitive 7 root of unity, which then must be in $F(\alpha)$. But this means that $6\mid [F(\alpha):\mathbb{Q}]$ - contradiction.
It looks like this problem is leading up to the fundamental theorem of Galois theory, which states that there is a one-to-one correspondence between the subfields of a Galois extension and subgroups of its Galois group. So our answer will be, of course, that no such Galois extension exists (which is why it's difficult to find an example).
First, note that if $K$ is a Galois extension of degree $4$ over $\mathbb{Q}$, then it must be the splitting field of a quartic polynomial. Either that polynomial is irreducible, or it can break into a product of irreducible quadratics. Of course, you'll want convince yourself of these cases and that the cases are indeed exhaustive.
If the polynomial breaks into a product of two irreducible quadratics, then I claim that it's rather easy to show that $K$ must have a subfield.
On the other hand, if the polynomial is irreducible, then $K = \mathbb{Q}[\alpha]$, where $\alpha$ is one of its roots. Since the degree of the Galois extension is $4$, we also know the Galois group has order $4$. As such, the Galois group must have an element $\phi$ of order $2$ that sends $\alpha \mapsto \beta$, where $\beta$ is another root of the polynomial.
Claim: $\mathbb{Q}[\alpha\beta] \subsetneq \mathbb{Q}[\alpha]$
To show this, suppose for contradiction they were actually the same field. Then both would be of degree $4$ over $\mathbb{Q}$, and the former would have a basis $\{1, \alpha\beta, (\alpha\beta)^2, (\alpha\beta)^3\}$. Because $\alpha$ would be an element of both fields, we could write:
$$\alpha = c_1 + c_2(\alpha\beta) + c_3(\alpha\beta)^2 + c_4(\alpha\beta)^3$$
What happens when the automorphism $\phi$ acts on both sides of this expression? What can we conclude?
Best Answer
I'll give you two different ways to do this as I don't know if you know the main theorem yet but even if you don't you can revisit this once you do learn it.
Method 1: The simplest method is to use the Fundamental Theorem/ Galois Correspondence Theorem which says the intermediate fields are in bijection with subgroups of the Galois group, which is isomorphic to $C_2 \times C_2$ in this case.
This has $5$ subgroups so we have $5$ intermediate fields which are $\mathbb{Q}$, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{5})$, $\mathbb{Q}(i\sqrt{5})$ and $\mathbb{Q}(i, \sqrt{5})$ which are easily spotted.
Method 2: If you haven't got that far yet then we can use the tower law instead. Note $[\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}]=4$ so any nontrivial subfield will have degree $2$.
It's a standard result that any quadratic field has the form $\mathbb{Q}(\sqrt{D})$ for some squarefree integer $D$. Using your basis $a+bi+c\sqrt{5} +di\sqrt{5}$, we can see $i=\sqrt{-1}$, $\sqrt{5}$ and $i\sqrt{5}=\sqrt{-5}$ all lie in $\mathbb{Q}(i,\sqrt{5})$ so all give quadratic subfields.
Now suppose $\mathbb{Q}(\sqrt{D})$ was also a subfield. Then$\sqrt{D} \in \mathbb{Q}(i,\sqrt{5})$. This means $\sqrt{D} = a+bi+c\sqrt{5} +di\sqrt{5}$ for some $a,b,c,d$.
Squaring both sides we get $D= (a^2 - b^2 +5c^2 - 5d^2) + (2ab+10cd)i + (2ac-2bd)\sqrt{5} +(2ad+2bc)i\sqrt{5}$.
We are then left with solving the simulataneous equations:
$\begin{eqnarray*} D &=& a^2 - b^2 +5c^2 - 5d^2, \\ 0 &=& 2ab+10cd, \\ 0 &=& 2ac-2bd, \\ 0 &=& 2ad+2bc, \end{eqnarray*}$
which then gives you solutions only for $D=-1,5,-5$ (remembering that we only consider squarefree $D$), but this is quite tedious.