Let \(f(x)\) be an irreducible polynomial of degree 4 in \(\mathbb{Q}[x]\) whose splitting field \(K\) over \(\mathbb{Q}\) has Galois group \(G = \operatorname{Gal}(K/\mathbb{Q}) \cong S_4\). Let \(\theta\) be a root of \(f(x)\). Prove that \(\mathbb{Q}(\theta)\) is an extension of \(\mathbb{Q}\) of degree 4 and that there are no intermediate fields strictly between \(\mathbb{Q}\) and \(\mathbb{Q}(\theta)\).
Prove that if \(K\) is a Galois extension of \(\mathbb{Q}\) of degree 4, then there is an intermediate subfield strictly between \(K\) and \(\mathbb{Q}\).
Solution
Goal: Prove that \(\mathbb{Q}(\theta)\) for an \(S_4\)-quartic has no non-trivial intermediate subfields using maximal subgroups in (a), and that every degree 4 Galois extension has intermediate subfields via order 2 subgroups in (b).
1. Part (a): Degree \([\mathbb{Q}(\theta) : \mathbb{Q}] = 4\).
1.1. The polynomial \(f(x) \in \mathbb{Q}[x]\) is irreducible and of degree 4.
1.2. Since \(f(\theta) = 0\), \(f(x)\) is a non-zero constant multiple of the minimal polynomial \(\operatorname{irr}(\theta, \mathbb{Q})\).
1.3. Therefore, \([\mathbb{Q}(\theta) : \mathbb{Q}] = \deg \operatorname{irr}(\theta, \mathbb{Q}) = \deg f = 4\).
2. Part (a): No intermediate fields strictly between \(\mathbb{Q}\) and \(\mathbb{Q}(\theta)\).
2.1. Let \(K\) be the splitting field of \(f(x)\) over \(\mathbb{Q}\), so \(K/\mathbb{Q}\) is Galois with Galois group \(G = \operatorname{Gal}(K/\mathbb{Q}) \cong S_4\).
2.2. The roots of \(f(x)\) are \(\theta = \theta_1, \theta_2, \theta_3, \theta_4 \in K\).
2.3. Under the isomorphism \(G \cong S_4\) acting on the 4 roots, the subgroup fixing \(\theta = \theta_1\) is the stabilizer \(H = \operatorname{Stab}_G(\theta_1) \cong S_3\).
2.4. By the Fundamental Theorem of Galois Theory, intermediate fields \(E\) with \(\mathbb{Q} \subseteq E \subseteq \mathbb{Q}(\theta)\) correspond bijectively (via \(E \mapsto \operatorname{Gal}(K/E)\)) to subgroups \(M \le G\) containing \(H\):\[H \le M \le G.\]
2.5. The index \([G : M]\) divides \([G : H] = \frac{|S_4|}{|S_3|} = \frac{24}{6} = 4\). Thus \([G : M] \in \{1, 2, 4\}\).
2.6. Case \([G : M] = 1\): Then \(M = G\), which corresponds to the base field \(K^G = \mathbb{Q}\).
2.7. Case \([G : M] = 4\): Then \([M : H] = 1\), so \(M = H\), which corresponds to the fixed field \(K^H = \mathbb{Q}(\theta)\).
2.8. Case \([G : M] = 2\):- A subgroup of index 2 in \(S_4\) must be normal, and the alternating group \(A_4\) is the unique subgroup of index 2 in \(S_4\).- However, \(H \cong S_3\) is generated by transpositions of \(\{2, 3, 4\}\), which are odd permutations in \(S_4\).- In particular, \((2~3) \in H\) is an odd permutation, so \((2~3) \notin A_4\).- Thus \(H \not\subseteq A_4\), meaning no subgroup \(M\) of index 2 contains \(H\).
2.9. Therefore, the only subgroups containing \(H\) are \(H\) and \(G\) themselves (\(H\) is a maximal subgroup of \(S_4\)).
2.10. Hence, there are no intermediate fields strictly between \(\mathbb{Q}\) and \(\mathbb{Q}(\theta)\).
3. Part (b): Existence of an intermediate field for degree 4 Galois extensions.
3.1. Let \(K/\mathbb{Q}\) be a Galois extension of degree \([K : \mathbb{Q}] = 4\).
3.2. The Galois group \(G = \operatorname{Gal}(K/\mathbb{Q})\) has order \(|G| = [K : \mathbb{Q}] = 4\).
3.3. Every group of order 4 is isomorphic to either the cyclic group \(\mathbb{Z}/4\mathbb{Z}\) or the Klein 4-group \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\).
3.4. In both cases, \(G\) contains a proper non-trivial subgroup \(H\) of order 2:- If \(G = \langle \sigma \rangle \cong \mathbb{Z}/4\mathbb{Z}\), then \(H = \langle \sigma^2 \rangle = \{\operatorname{id}, \sigma^2\}\) has order 2.- If \(G \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\), every non-identity element has order 2, so \(H = \{\operatorname{id}, \tau\}\) has order 2 for any \(\tau \in G \setminus \{\operatorname{id}\}\).
3.5. By the Fundamental Theorem of Galois Theory, the fixed field \(E = K^H\) is an intermediate field with\[[E : \mathbb{Q}] = [G : H] = \frac{4}{2} = 2, \quad [K : E] = |H| = 2.\]
3.6. Thus \(\mathbb{Q} \subsetneq E \subsetneq K\) is an intermediate field strictly between \(\mathbb{Q}\) and \(K\).
4. Conclusion:
$\mathbb{Q}(\theta)$ has no non-trivial intermediate subfields because $S_3$ is maximal in $S_4$, whereas every Galois extension of degree 4 contains an intermediate subfield of degree 2.