Let \(E\) be a splitting field of \(g(x) \in \mathbb{Q}[x]\) over \(\mathbb{Q}\). Suppose its intermediate-field lattice consists of \(\mathbb Q\), \(E\), and exactly four distinct proper intermediate fields \(F_1,F_2,F_3,F_4\). Each satisfies \(\mathbb Q\subsetneq F_j\subsetneq E\), and no \(F_j\) contains another \(F_k\) for \(j\ne k\). No degrees are specified. Which of the following groups could be \(\text{Gal}(E/\mathbb{Q})\)? Justify your answers.
\(\mathbb{Z}_2 \times \mathbb{Z}_2\)
\(\mathbb{Z}_6\)
The dihedral group with 6 elements
\(S_3\)
\(\mathbb{Z}_3 \times \mathbb{Z}_3\)
Solution
The possible answers are (c), (d), and (e). Parts (c) and (d) describe the same isomorphism type. The unlabeled lattice does not distinguish that type from \(C_3\times C_3\).
1. The required subgroup lattice has exactly four nontrivialproper subgroups, all mutually incomparable.
A splitting field over \(\mathbb Q\) is a finite Galois extension, since characteristic zero gives separability. The Galois correspondence is an inclusion-reversing bijection between its intermediate fields and the subgroups of its Galois group [@DF04]. It exchanges the two endpoints and sends the four incomparable proper fields to four incomparable nontrivial proper subgroups. Conversely, a Galois extension with this subgroup lattice has the specified intermediate-field lattice.
2. Neither (a) nor (b) has the required subgroup lattice.
In \(C_2\times C_2\), each of the three nonidentity elements generates a different subgroup of order \(2\). By Lagrange’s theorem these are all the nontrivial proper subgroups. There are three, not four, so (a) is impossible.
A cyclic group has exactly one subgroup of each order dividing its order [@DF04]. Thus \(C_6\) has just two nontrivial proper subgroups, of orders \(2\) and \(3\). This rules out (b).
3. Both (c) and (d) have the required subgroup lattice andare realized over \(\mathbb Q\).
The symmetry group of an equilateral triangle acts faithfully on its three vertices. It has six elements, so this action identifies the dihedral group of order \(6\) with \(S_3\).
The nontrivial proper subgroups of \(S_3\) can only have orders \(2\) and \(3\). Its three transpositions generate three distinct order-two subgroups. Its two \(3\)-cycles are inverse and generate one order-three subgroup. These four subgroups are exhaustive and mutually incomparable: distinct order-two subgroups cannot contain one another, and neither \(2\) nor \(3\) divides the other.
For realization, let \(a=\sqrt[3]{2}>0\) and let \(\zeta\) be a nonreal cube root of unity. The splitting field of \(x^3-2\) is \(\mathbb Q(a,\zeta)\), because its roots are \(a,\zeta a,\zeta^2a\) and their ratios recover \(\zeta\). Eisenstein’s criterion at \(2\) gives \([\mathbb Q(a):\mathbb Q]=3\), and adjoining the nonreal root \(\zeta\) of \(x^2+x+1\) to the real field \(\mathbb Q(a)\) has degree \(2\). Its Galois group therefore has order \(6\) and acts faithfully on the three roots, so is \(S_3\). Step 1 now gives the required field lattice. This realizes both descriptions.
4. The group in (e) also has the required subgroup latticeand is realized over \(\mathbb Q\).
View \(C_3\times C_3\) as the additive group of \(\mathbb F_3^2\). Each nonzero vector spans a subgroup of order \(3\) containing exactly two nonzero vectors. The eight nonzero vectors are thus partitioned among \(8/2=4\) such subgroups. Lagrange’s theorem rules out any other nontrivial proper subgroup. Distinct subgroups of order \(3\) are incomparable, giving exactly the required lattice.
For realization over \(\mathbb Q\), let \(L\) be the cyclotomic field \(\mathbb Q(\zeta_{91})\). The cyclotomic Galois theorem and the Chinese remainder theorem give \[ \operatorname{Gal}(L/\mathbb Q) \cong(\mathbb Z/91\mathbb Z)^\times \cong(\mathbb Z/7\mathbb Z)^\times\times (\mathbb Z/13\mathbb Z)^\times \cong C_6\times C_{12} \] [@DF04]. For the last identification, \(3\) modulo \(7\) has successive powers \(1,3,2,6,4,5\), and \(2\) modulo \(13\) has successive powers \(1,2,4,8,3,6,12,11,9,5,10,7\); these lists contain every unit in the respective fields.
Take the subgroup \(H=C_2\times C_4\) in these cyclic factors, and put \(E=L^H\). It is normal because the whole Galois group is abelian. Therefore the Galois correspondence yields \[ \operatorname{Gal}(E/\mathbb Q) \cong(C_6\times C_{12})/(C_2\times C_4) \cong C_3\times C_3. \] The field \(E\) is the splitting field of a polynomial over \(\mathbb Q\). Indeed, choose a finite basis \(e_1,\ldots,e_d\) of \(E\) over \(\mathbb Q\) and let \(g\) be the product of their minimal polynomials. Normality of \(E/\mathbb Q\) implies that all those polynomials split in \(E\), and their roots include the basis elements, which generate \(E\) as a field. Thus the splitting field of \(g\) inside \(E\) is exactly \(E\). Step 1 then gives the prescribed six-element lattice.