Show that if \(E/k, F/k\) are Galois with \(E\intersect F = k\), then \(EF/k\) is Galois and \(G(EF/k) \cong G(E/k)\times G(F/k)\).
Solution
Regard \(E\) and \(F\) as subfields of a common algebraic closure of \(k\).
1. The extensions \(E/k\) and \(F/k\) are linearly disjoint over \(k\).
Let \(E_0\subseteq E\) and \(F_0\subseteq F\) be finite Galois subextensions of \(k\). Then \[ E_0\cap F_0\subseteq E\cap F=k, \] so \(E_0\cap F_0=k\). Since \(E_0/k\) is finite Galois, restriction gives \(\Gal(E_0F_0/F_0)\cong\Gal(E_0/E_0\cap F_0)=\Gal(E_0/k)\), so \([E_0F_0:F_0]=[E_0:k]\) and \[ [E_0F_0:k]=[E_0:k][F_0:k]. \] Equivalently, the multiplication map \[ E_0\otimes_k F_0\longrightarrow E_0F_0 \] is injective. Every finite subset of \(E\) and of \(F\) is contained in such finite Galois subextensions. Hence, passing to the filtered union, the multiplication map \[ E\otimes_k F\longrightarrow EF \] is injective. Thus \(E\) and \(F\) are linearly disjoint over \(k\).
2. The compositum \(EF/k\) is Galois.
Both \(E/k\) and \(F/k\) are algebraic and separable, so every element of the field generated by \(E\cup F\) is separable over \(k\); hence \(EF/k\) is separable. If \(\iota:EF\hookrightarrow\overline{k}\) is a \(k\)-embedding, normality of \(E/k\) and \(F/k\) gives \[ \iota(E)=E, \qquad \iota(F)=F. \] Therefore \[ \iota(EF)=\iota(E)\iota(F)=EF, \] so \(EF/k\) is normal. Hence it is Galois.
3. Restriction defines an injective homomorphism\[ \rho:\Gal(EF/k)\longrightarrow\Gal(E/k)\times\Gal(F/k), \qquad \rho(\gamma)=(\gamma|_E,\gamma|_F). \]
Normality of \(E/k\) and \(F/k\) makes both restrictions well-defined. If \(\rho(\gamma)\) is trivial, then \(\gamma\) fixes both \(E\) and \(F\), hence fixes the field they generate, namely \(EF\). Thus \(\gamma=1\), so \(\rho\) is injective.
4. The restriction map \(\rho\) is surjective.
Let \[ (\sigma,\tau)\in\Gal(E/k)\times\Gal(F/k). \] The map \[ \sigma\otimes\tau:E\otimes_kF\longrightarrow E\otimes_kF, \qquad x\otimes y\longmapsto\sigma(x)\otimes\tau(y), \] is a \(k\)-algebra automorphism. By 1, \(E\otimes_kF\) embeds in \(EF\) by multiplication. Its image is a domain whose fraction field is \(EF\), because it contains the images of both \(E\) and \(F\). Therefore \(\sigma\otimes\tau\) extends uniquely to a \(k\)-automorphism \(\gamma\) of \(EF\). By construction, \[ \gamma|_E=\sigma, \qquad \gamma|_F=\tau. \] Thus \(\rho(\gamma)=(\sigma,\tau)\).
5. Therefore\[ \Gal(EF/k)\cong\Gal(E/k)\times\Gal(F/k). \]
By 3 and 4, the restriction homomorphism \(\rho\) is bijective.