Galois extenstion of rational function field $\mathbb{C}(X,Y,Z)$ Let $F= \mathbb{C}(X,Y,Z)$ be rational function field.
And Let $K$ be the splitting field of $F(T)=T^6 - X T^4 +Y T^2 -Z$.
Please Find all galois extensions $L/F$ such that $L \subset K$ and $[L:F]=6$.
I'm stuck on this problem.  Any idea ,please.
 A: First, we find the degree of the splitting field of $ P(T) = T^6 - XT^4 + YT^2 - Z $. We note that the discriminant $ D $ of $ G(T) = T^3 - XT^2 + YT - Z $ is not a perfect square: indeed, the discriminant is a polynomial of degree $ 3 $ in $ X $, thus can't be a perfect square. Let the roots of $ G $ be $ X_1, X_2, X_3 $, then the splitting field of $ G $ over $ F $ is $ M = F(X_1, X_2, X_3) = F(X_1, \sqrt{D}) $, and has degree $ 6 $ over $ F $. Then, we see that
$$ K = M(\sqrt{X_1}, \sqrt{X_2}, \sqrt{X_3}) $$
Fortunately for us, we actually don't need to know the exact degree of this extension - it suffices to know that it divides $ 8 $, so that the degree $ [K :F] $ divides $ 48 $. (The exact degree is, in fact, $ 48 $.) Why is this helpful? Well, if $ L/F $ is Galois, then any irreducible polynomial in $ F[X] $ splits into factors of equal degree over $ L $. In particular, $ G(T) $ either splits completely or remains irreducible. If the former happens, then by degree considerations we have $ L = M $. Otherwise, $ G $ remains irreducible over $ L $; which implies that the degree $ [K : L] $ is divisible by $ 3 $. But the degree $ [L : F] $ is already divisible by $ 3 $; so we obtain the absurd result that $ [K : F] $, and thus $ 48 $, is divisible by $ 9 $. Thus, $ M $ is the unique intermediate Galois extension of degree $ 6 $ over $ F $.
A: Edit: Starfall's solution is much nicer than mine and I encourage everyone to read that one. But since my attempt is already posted, let me fill in the holes in my attempt using some ideas that Starfall very kindly and patiently explained to me in chat. 
First, let $E$ be the splitting field of the polynomial $U^3 - XU^2 + YU - Z$ over $F$. If $U^3 - XU^2 + YU - Z$ splits as $ (U - \alpha)(U - \beta)(U-\gamma)$ in $E$, then $E = F(\alpha, \beta, \gamma) = \mathbb C(\alpha, \beta, \gamma)$. Note that  $\mathbb C(\alpha, \beta, \gamma)$ has transcendence degree $3$ over $\mathbb C$: the transcendence degree is clearly at most $3$; morevoer it is exactly 3 since it contains $F = \mathbb C (X,Y,Z)$. So  $\mathbb C(\alpha, \beta, \gamma)$ can be viewed as a field of rational functions in variables $\alpha, \beta, \gamma$.
Since the discriminant of $U^3 - XU^2 + YU - Z$ is not a square, $Gal(E:F) = S_3$, the group of permutations on $\{ \alpha, \beta, \gamma \}$.
Next, let's think about $K$, the splitting field for $T^6 - XT^4 + YT^2 - Z$ over $F$.  Clearly $K = E(\sqrt{\alpha}, \sqrt{\beta}, \sqrt{\gamma}) = \mathbb C(\sqrt\alpha, \sqrt\beta, \sqrt\gamma)$. So $[K:E]$ is either 2, 4 or 8. In fact, $[K:E]=8$. To see this, observe that $E$ is the fraction field of $\mathbb C[\alpha, \beta, \gamma]$, which is a UFD. By unique factorization, $\alpha / \beta$ can't be a square in $E$, so certainly $\sqrt{\alpha / \beta} \notin K$. Hence $K(\alpha) \neq K(\beta)$, and$[K:E]$ is at least $4$. Similarly, $\sqrt{\gamma / \alpha}, \sqrt{\gamma / \beta}, \sqrt{\gamma / \alpha \beta} \notin K$ so $[K:E] = 8$.
Therefore, $Gal(K:E) = \mathbb Z_2^3$. The eight elements in this $\mathbb Z_2^3$ are the eight possible sign flips on $\{\pm \sqrt\alpha, \pm \sqrt\beta, \pm\sqrt\gamma\}$.
Now we'll determine  $Gal(K:F)$. This should be a subgroup of the permutation group on the roots $\{+ \sqrt\alpha, - \sqrt\alpha , + \sqrt\beta, - \sqrt\beta, +\sqrt\gamma, -\sqrt\gamma\}$. Since $Gal(K:F)/Gal(K:E) = Gal(E:F) = S_3$ and $Gal(K:E) = \mathbb Z_2$, acting as described above, $Gal(K:F)$ must be the semidirect product of $\mathbb Z_2^3$ with $S_3$ generated by sign flips and $\{\alpha, \beta,\gamma\}$-permutations on $\{\pm \sqrt\alpha, \pm \sqrt\beta, \pm\sqrt\gamma\}$.
To solve the original problem, we now need to find all normal subgroups of index $6$ in $Gal(K:F)$. By the Galois correspondnece, taking the fixed subfield for each such normal subgroup gives a normal extension $L$ of degree $6$ over $F$.
In fact, there is only one normal subgroup of index $6$ in this semidirect product of $\mathbb Z_2^3$ with $S_3$, and this is simply the $\mathbb Z_2^3$ generated by the sign flips. (To see this, consider the image of such a normal subgroup under the quotient map from the semidirect product to its quotient by $\mathbb Z_2^3$, and observe that the only normal subgroup of $S^3$ of order divisible by $8$ is the trivial one.) The fixed subfield associated to this $\mathbb Z_2^3$ is $E$. Hence the only possible $L$ is $E$.
