Let $$E = \mathbb{C}(x, y, z)$$, $$F = \mathbb{C}(x^2y, y^2z, z^2x)$$, $$L$$ the subfield of $$E$$ fixed by $$S_3$$, and $$K = F \cap L$$. Then,
(1) Is $$E/F$$ Galois? And what is its Galois group $$G$$?
(2) What is $$[E:K]$$?
(3) Calculate the number of intermediate fields of $$L/K$$.

Here is what I have tried: Since $$E = F(x)$$ and $$x^9 \in F$$, for $$\sigma_i : x \mapsto \zeta^i x$$, $$G = \{ \sigma_i \}_{i=0, \dots, 8}$$ (where $$\zeta$$ is a primitive 9th root of unity). And $$\operatorname{Gal}(E/K) = H := \left< S_3, G\right>$$ (the group generated by these two groups). But I don't understand what is this group, and its cardinality.

You have made very good headway essentially answering part (1). I recap the argument. Partly for my own benefit, partly to frame my solution to (2).

• Let $$\zeta=e^{2\pi i/9}$$ be a primitive complex ninth root of unity. As $$x,y,z$$ freely generate the field $$E$$ over $$\Bbb{C}$$, for all $$j\in\{0,1,\ldots,8\}$$ the rules $$x\mapsto \zeta^jx$$, $$y\mapsto\zeta^{-2j}y$$, $$z\mapsto \zeta^{4j}z$$ describe an (obviously bijective) automorphism $$\sigma_j$$ of $$E$$. Clearly $$\sigma_j\circ\sigma_\ell=\sigma_{j+\ell}$$ with addition done modulo nine, so the automorphisms $$\sigma_j$$ form a group $$G$$ isomorphic to $$\Bbb{Z}_9$$.
• By general Galois theory the fixed field of $$G$$, call it temporarily $$\tilde{F}$$, is such that $$E/\tilde{F}$$ is Galois with Galois group $$G$$. In particular, $$[E:\tilde{F}]=9$$.
• Clealy the monomials $$u=x^2y, v=y^2z, w=z^2x$$ are invariant under $$G$$, so the field $$F$$ they generate over $$\Bbb{C}$$ is contained in $$\tilde{F}$$.
• You observed that $$E=F(x)$$ and that $$x^9=u^4v^{-2}w\in F$$, so $$[E:F]\le9$$. Together with the preceding observations this tells us that we must have $$F=\tilde{F}$$, implying that $$E/F$$ is Galois with Galois group $$G$$ (and also that you have found the minimal polynomial of $$x$$ over $$F$$).

The field $$E$$ also has permutations of $$\{x,y,z\}$$ as automorphisms. Let's call this group $$S=Sym(\{x,y,z\})$$, obviously isomorphic to $$S_3$$. Let $$\Omega=\langle G,S\rangle$$ be the group of automorphisms of $$E$$ these two groups jointly generate.

• Both groups, $$G$$ and $$S$$, act faithfully and linearly on the (linear) complex span $$V=\langle x,y,z\rangle$$. As $$V$$ also generates $$E$$ as a field, the same holds for all the automorphisms in $$\Omega$$. Therefore we can do the calculations involving $$\Omega$$ inside the group of $$3\times3$$ complex matrices.
• The elements of $$G$$ then become identified with diagonal matrices, $$\sigma_j=diag(\zeta^j,\zeta^{7j},\zeta^{4j})$$, and the elements of $$S$$ are similarly identified with permutation matrices with a single $$1$$ on each row and column and six zeros. All the matrices in $$\Omega$$ are thus contained in the group $$\Gamma$$ of monomial matrices with the single non-zero entry of each row/column being a ninth root of unity $$\in \mu_9$$ (so $$\Gamma=\mu_9\wr S_3$$, but we don't really need the wreath product here).
• It is more helpful to view $$\Gamma$$ as a semi-direct product $$\Gamma\cong \Bbb{Z}_9^3\rtimes S_3$$ with $$\Bbb{Z}_9^3$$ identified with the diagonal matrices, and the conjugation action by $$S_3$$ permuting the diagonal entries. Doing it this way we see that $$\Delta=\Omega\cap \Bbb{Z}_9^3$$ is generated by all the componentwise permutations of the vector $$z=(1,7,4)\in \Bbb{Z}_9^3$$.
• A key observation is that all the components of $$z$$ are congruent to each other modulo $$3$$. A bit of experimenting reveals (leaving the verification to you) that the abelian group $$\Delta\le\Bbb{Z}_9^3$$ is generated by $$z=(1,7,4)$$ and the vector $$u=(0,3,6)$$. For example $$(4,7,1)=4z+u$$. It follows that $$\Delta\cong \Bbb{Z}_9\oplus \Bbb{Z}_3$$. In particular, $$\Delta$$ has order $$27$$. It may be worth observing that in $$\Bbb{Z}_9^3$$ there are $$81$$ vectors with the property that all the components are congruent to each other modulo $$3$$. But, for example $$(0,0,3)\notin\Delta$$, for all the "deviations" from multiples of $$z$$ must also be in the zero sum subgroup of $$\Bbb{Z}_9^3$$ as that is stable under all the permutations of components.
• At this point we can conclude that $$\Omega=\Delta\rtimes S_3$$ has order $$|\Omega|=27\cdot6=162$$. Galois theory then tells us that $$E$$ is a degree $$162$$ extension of $$K=Inv(\Omega)$$. Question (2) is thus settled.
• By Galois correspondence Question (3) can be answered by carrying out a census of the subgroups of $$\Omega$$ containing $$S$$. Those are in bijective correspondence with subgroups $$K\le\Delta$$ such that $$K$$ is stable under the action of $$S$$ (so the actual subgroups are $$K\rtimes S$$). Leaving the details to you. I'm sure you can manage.
• In retrospect, mentioning the $3\times3$-matrices was not necessary. If I were a great mathematician I would remove that as a form of scaffolding. However, that is the route my feeble mind followed. So I leave it there thinking that it may have some pedagogical benefit. After all, not all automorphisms of $E$ act on $V$. – Jyrki Lahtonen Jun 28 at 6:40
• Thank you very much, this is so great!!! I show that the answer of (3) is $5$, is this right? – agababibu Jun 29 at 9:15
• List the groups, please @agababibu. – Jyrki Lahtonen Jun 29 at 11:15
• I list the subgroups of $\Delta$. $0$, $\Delta$, $\left< (3,3,3) \right>$, $\left< (1,1,1) \right>$, $\left< (1,7,4) \right> \times \left< (0,3,6) \right>$\$ – agababibu Jun 29 at 11:48
• Oh, sure! That seems correct to me! – Jyrki Lahtonen Jun 29 at 12:08