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.
  • $\begingroup$ 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 $V$. $\endgroup$ – Jyrki Lahtonen Jun 28 at 6:40
  • $\begingroup$ Thank you very much, this is so great!!! I show that the answer of (3) is $5$, is this right? $\endgroup$ – agababibu Jun 29 at 9:15
  • $\begingroup$ List the groups, please @agababibu. $\endgroup$ – Jyrki Lahtonen Jun 29 at 11:15
  • $\begingroup$ 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>$$ $\endgroup$ – agababibu Jun 29 at 11:48
  • 1
    $\begingroup$ Oh, sure! That seems correct to me! $\endgroup$ – Jyrki Lahtonen Jun 29 at 12:08

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