- If \(F\) is some (not necessarily Galois) field extension over \(K\) such that \([F: K]=6\) and \(\operatorname{Aut}_K F \simeq S_{3}\), then \(F\) is the splitting field of an irreducible cubic over \(K[x]\).
- Recall the definition of the join \(H_1\vee H_2=\langle H_1,H_2\rangle\) of two subgroups. For \(F\) a finite dimensional Galois extension over \(K\) and let \(A\) and \(B\) be intermediate extensions. Prove that
- \(\operatorname{Aut}_{A B} F=\mathrm{Aut}_{A} F \cap \mathrm{Aut}_{B} F\)
- \(\operatorname{Aut}_{A\cap B}F=\mathrm{Aut}_{A} F \vee \mathrm{Aut}_{B} F\)
Solution
1. In part 7, the extension \(F/K\) is Galois with Galois group \(S_3\).
For every finite extension, \[ |\operatorname{Aut}_K(F)|\le [F:K], \] with equality exactly for a Galois extension. Here \[ |\operatorname{Aut}_K(F)|=|S_3|=6=[F:K]. \] Hence \(F/K\) is Galois and \(\operatorname{Gal}(F/K)\cong S_3\).
2. Choose a subgroup \(H\le\operatorname{Gal}(F/K)\) of order \(2\) and let \(E=F^H\). Then \([E:K]=3\).
By Galois correspondence, \[ [E:K]=[\operatorname{Gal}(F/K):H]=6/2=3. \] Because \(F/K\) is finite Galois, \(E/K\) is finite separable.
3. There is \(\alpha\in E\) such that \(E=K(\alpha)\) and the minimal polynomial \(f\) of \(\alpha\) over \(K\) is irreducible of degree \(3\).
The primitive element theorem applies to the finite separable extension \(E/K\). Thus \(E=K(\alpha)\) for some \(\alpha\), and \[ \deg f=[K(\alpha):K]=3. \]
4. The splitting field of \(f\) is all of \(F\).
Since \(F/K\) is normal, all \(K\)-conjugates of \(\alpha\) lie in \(F\), so \(f\) splits in \(F\). Let \(L\) be its splitting field inside \(F\). The pointwise stabilizer of all conjugates of \(\alpha\) is the core \[ \bigcap_{g\in G}gHg^{-1} \] of \(H\) in \(G=S_3\). A subgroup of order \(2\) in \(S_3\) is generated by a transposition; its three conjugates intersect trivially. Hence \[ \operatorname{Gal}(F/L)=1, \] so \(L=F\). Therefore \(F\) is the splitting field of the irreducible cubic \(f\).
Now let \(F/K\) be finite Galois for part 8, set \[ G=\operatorname{Gal}(F/K), \qquad H_A=\operatorname{Gal}(F/A), \qquad H_B=\operatorname{Gal}(F/B). \]
5. One has\[ \operatorname{Gal}(F/AB)=H_A\cap H_B. \]
An automorphism \(\sigma\in G\) fixes the compositum \(AB\) pointwise if and only if it fixes both \(A\) and \(B\) pointwise. Thus \[ \sigma\in\operatorname{Gal}(F/AB) \Longleftrightarrow \sigma\in H_A\cap H_B. \]
6. The fixed field of the join \(H_A\vee H_B=\langle H_A,H_B\rangle\) is \(A\cap B\).
An element \(x\in F\) is fixed by every element of the subgroup generated by \(H_A\) and \(H_B\) exactly when it is fixed by every element of \(H_A\) and every element of \(H_B\). Therefore \[ F^{H_A\vee H_B}=F^{H_A}\cap F^{H_B}=A\cap B, \] using Galois correspondence for \(F^{H_A}=A\) and \(F^{H_B}=B\).
7. Hence\[ \operatorname{Gal}(F/A\cap B)=H_A\vee H_B. \]
By 6, the subgroup \(H_A\vee H_B\) has fixed field \(A\cap B\). The finite Galois correspondence is a bijection between subgroups of \(G\) and intermediate fields, so the unique subgroup with this fixed field is \[ \operatorname{Gal}(F/A\cap B). \] Thus the two groups are equal.