Show that if \(E/k\) and \(F/k\) are finite Galois extensions inside a common field with \(E \cap F = k\), then the compositum \(EF/k\) is Galois and: \[\operatorname{Gal}(EF/k) \cong \operatorname{Gal}(E/k) \times \operatorname{Gal}(F/k).\]
Solution
Since \(E/k\) and \(F/k\) are finite Galois, choose separable polynomials over \(k\) whose splitting fields are \(E\) and \(F\). Their product has splitting field \(EF\), so \[ EF/k \] is finite Galois.
Restriction gives an injective homomorphism \[ \Phi:\operatorname{Gal}(EF/k) \longrightarrow \operatorname{Gal}(E/k)\times\operatorname{Gal}(F/k), \qquad \sigma\mapsto(\sigma|_E,\sigma|_F). \] It is injective because \(E\) and \(F\) generate \(EF\).
Because \(F/k\) is Galois, restriction induces \[ \operatorname{Gal}(EF/E) \cong \operatorname{Gal}(F/E\cap F). \] The hypothesis \(E\cap F=k\) therefore gives \[ [EF:E]=[F:k]. \] Hence \[ [EF:k]=[E:k][F:k]. \] Since all three extensions are Galois, \[ |\operatorname{Gal}(EF/k)| =|\operatorname{Gal}(E/k)|\,|\operatorname{Gal}(F/k)|. \] Thus the injective map \(\Phi\) is between finite groups of the same order, so it is an isomorphism: \[ \boxed{\operatorname{Gal}(EF/k) \cong\operatorname{Gal}(E/k)\times\operatorname{Gal}(F/k)}. \]