Let all Galois extensions below be finite.
- Show that the splitting field of a separable polynomial is Galois.
- Show that every quadratic extension \(K/F\) with \(\operatorname{char}F\ne2\) is Galois.
- Let \(F\subseteq E\subseteq K\) with \(K/F\) Galois. Show that \(K/E\) is Galois and \[ \operatorname{Gal}(K/E)\le\operatorname{Gal}(K/F). \] Show moreover that \[ E/F\text{ is Galois}\iff \operatorname{Gal}(K/E)\normal\operatorname{Gal}(K/F), \] and in that case \[ \operatorname{Gal}(E/F)\cong \operatorname{Gal}(K/F)/\operatorname{Gal}(K/E). \]
- If \(E/k\) and \(F/k\) are Galois and \(E\cap F=k\), show that \(EF/k\) is Galois and \[ \operatorname{Gal}(EF/k)\cong\operatorname{Gal}(E/k)\times\operatorname{Gal}(F/k). \]
Solution
1. A splitting field is normal by definition: every irreducible polynomial over the base having one root in the field has all of its conjugate roots there. Since the original polynomial is separable, its splitting field is generated by separable algebraic elements and is therefore separable. Finite normal plus separable is Galois.
2. Let \([K:F]=2\) and \(\operatorname{char}F\ne2\). Then \(K=F(\alpha)\) for some \(\alpha\) with irreducible quadratic minimal polynomial\[ f(x)=x^2+bx+c. \]If \(\alpha\) is one root, the other root is\[ -b-\alpha\in K. \]Thus \(f\) splits in \(K\). Since its derivative is not identically zero in characteristic not \(2\), \(f\) is separable. Hence \(K/F\) is Galois.
3. Since normality and separability pass from \(K/F\) to \(K/E\), the extension \(K/E\) is Galois. Its automorphisms are precisely those \(F\)-automorphisms of \(K\) fixing \(E\), so\[ \operatorname{Gal}(K/E)\le\operatorname{Gal}(K/F). \]By the fundamental theorem of Galois theory, the intermediate field \(E\) is Galois over \(F\) exactly when the corresponding subgroup\[ H=\operatorname{Gal}(K/E) \]is normal in\[ G=\operatorname{Gal}(K/F). \]When this holds, restriction induces a surjective homomorphism\[ G\twoheadrightarrow\operatorname{Gal}(E/F) \]with kernel \(H\). Therefore\[ \operatorname{Gal}(E/F)\cong G/H. \]
4. Since \(E/k\) and \(F/k\) are normal and separable, their compositum \(EF/k\) is again normal and separable, hence Galois.
Restriction gives \[ \rho:\operatorname{Gal}(EF/k) \longrightarrow \operatorname{Gal}(E/k)\times\operatorname{Gal}(F/k). \] This map is injective because an automorphism of \(EF\) fixing both \(E\) and \(F\) fixes the field they generate.
Because \(E/k\) and \(F/k\) are finite Galois and \(E\cap F=k\), they are linearly disjoint over \(k\), so \[ [EF:k]=[E:k][F:k]. \] Hence \[ |\operatorname{Gal}(EF/k)| =|\operatorname{Gal}(E/k)|\,|\operatorname{Gal}(F/k)|. \] The injective map \(\rho\) is therefore between finite groups of the same order, so it is an isomorphism.