How to show this is the minimal polynomial

Question

I'm trying to the following problem. But I can't show some irreducibility of the polynomials.

Put $L=\mathbb{C}(X,Y,Z)$, $\omega=\frac{-1+\sqrt{-3}}{2}$.

Define two automorphism $\sigma, \tau$ of $L$ over $\mathbb{C}$ by

$\sigma(X)=Y, \sigma(Y)=Z, \sigma(Z)=X, \tau(X) = X,\tau(Y)=\omega Y,\tau(Z)=\omega^2 Z$.

$K :=\{x\in L|\sigma(x)=x, \tau(x)=x\}$. Then,

(1) Find the minimal polonominal of $X$ over $K$. (2) Calculate $[K(X):K]$. (3) Calculate $[L:K]$.

First of all, by symmetry of $\sigma$, I suppose the minimal polynomial is something symmetric. So I calculate

$(T^3-X^3)(T^3-Y^3)(T^3-Z^3)=T^9-(X^3+Y^3+Z^3)T^6+(X^3Y^3+Y^3Z^3+Z^3X^3)T^3-X^3Y^3Z^3 =:p(T)$

Since $X^3+Y^3+Z^3,X^3Y^3+Y^3Z^3+Z^3X^3,X^3Y^3Z^3\in K$, this polynomial $p(T)\in K[T]$ and satisfy $p(X)=0$. So I think this is the minimal polynomial. But I can't show this is irreducible. This is the first question.

Next, to calculate $[L:K]$, I remarked $K(X,Y)=L$ because $XYZ\in K$. Then, I think the minimal polynomial of $Y$ over $K(X)$ might be also $p(T)$. But I can't show that this polynomial is irreducible. This is the second question.

Answer 1

You are on a good start. I would argue as follows. I write $X=u_1$, $Y=u_2$, $Z=u_3$ to make a few things notationally simpler.

1. Show that the group $G$ generated by $\sigma$ and $\tau$ consists of the 27 automorphisms of the form $u_i\mapsto \omega^{a_i}u_{\alpha(i)}, i=1,2,3$, where $\alpha$ is a permutation in the subgroup $\langle(123)\rangle\le S_3$ (three choices), and where the vector of exponents $(a_1,a_2,a_3)$ is in the zero sum subgroup $P\le\Bbb{Z}_3^3$ (nine choices).
2. In light of item 1 we see that $X=u_1$ has exactly nine conjugates (its orbit under $G$), namely $\omega^ju_i$, $i,j\in\{1,2,3\}$. Therefore its minimal polynomial must have degree nine. As we are in characteristic zero all those conjugates are simple zeros of the minimal polynomial. This leads to your minimal polynomial.
3. A known result of field theory is that if $\alpha$ is algebraic over a field $K$ and its minimal polynomial has degree $n$, then $[K(\alpha):K]=n$.
4. A known result of Galois theory is that if $L$ is a field, and $G$ is a finite group of automorphisms of $L$, then $[L:K]=|G|$, where $K$ is the fixed field of $G$.