Intermediate Galois fields I want to find two different fields $K_1,K_2$ such that $\Bbb Q\subset K_i \subset \Bbb Q(\alpha,\zeta)$ such that $K_i /\Bbb Q$ are Galois.
A few things:


*

*$\alpha$ is the real $6^{th}$ roof of $2$, $\zeta$ is a primitive $6^{th}$ root of unity.

*Let $L= \Bbb Q(\alpha,\zeta)$, I have shown here that $\text{Gal}(L/\Bbb Q)\cong D_6$

*$L$ is the splitting field of $x^6-2$ over $\Bbb Q$



I have attempted to solve this in the answer below based on the above things. 
Is there a faster way at the end there, than seeing what is fixed as an arbitrary element in the basis?
 A: Perhaps a quicker way is by direct computation. You know that the primitive sixth roots of unity are $\frac{1\pm\sqrt{-3}}2$, so that two obvious Galois subextensions are $\Bbb Q(\sqrt{-3}\,)$ and $\Bbb Q(\alpha^3)$, both of them quadratic over the rationals, so Galois.
A: We know that $L=\Bbb Q(\alpha,\zeta)$ has Galois group $\text{Gal}(L/\Bbb Q)\cong D_6$.
We want to find two intermediate fields that are Galois extensions of $\Bbb Q$. Since $\Bbb Q$ has characteristic $0$, it is perfect, and hence any algebraic extension over $\Bbb Q$ is separable. Also, any non-trivial normal subgroup of $D_6$ will correspond to a normal extension of $\Bbb Q$.
Hence we need to find two normal subgroups of $D_6$.
I believe $\langle r^2\rangle$ and $\langle r^3\rangle$ are both normal subgroups of $D_6$.
$K_1=\langle r^2\rangle$ where $r^2:\alpha\mapsto \alpha\zeta^2$ and $r^4:\alpha \mapsto \alpha\zeta^4$, so we want $$L^{K_1}=\{x\in L: r^2(x)=r^4(x)=x\}$$
And $r^3:\alpha \mapsto \alpha\zeta^3$
$$L^{K_2} =\{x\in L : r^3(x)=x\}$$

Now using this, we can find what the fields are probably inefficiently by seeing what is fixed in the basis:
$$\{1,\alpha,\alpha^2,\alpha^3,\alpha^4,\alpha^5,\zeta,\alpha\zeta,\alpha^2\zeta,\alpha^3\zeta,\alpha^4\zeta,\alpha^5\zeta\}$$
(which I found here)
