Let \(F\) be a field of characteristic zero and \(\overline{F}\) be an algebraic closure of \(F\). Suppose that \(K\) and \(L\) are fields, with \(F \subseteq K \subseteq L \subseteq \overline{F}\), such that \([K : F] = 2\) and \(L/K\) is a finite Galois extension. Prove that there are at most two fields \(M \subseteq \overline{F}\) conjugate to \(L\) over \(F\) (remember that \(M \subseteq \overline{F}\) is conjugate to \(L\) over \(F\) just in case there is an isomorphism of \(L\) onto \(M\) which is the identity on \(F\)).
Solution
1. The \(F\)-conjugates of \(L\) are the images \(\sigma(L)\) for \(\sigma \in \operatorname{Aut}_F(\overline F)\).
an \(F\)-isomorphism \(L \to M\) extends to an automorphism of \(\overline F\) fixing \(F\), so \(M = \sigma(L)\) for some \(\sigma\).
2. \(K/F\) is Galois (degree \(2\) in characteristic \(0\)).
any degree-\(2\) extension in characteristic \(0\) is Galois (it is the splitting field of a quadratic).
3. Hence \(\sigma(K) = K\) for all \(\sigma \in \operatorname{Aut}_F(\overline F)\).
\(K/F\) is Galois, so it is stable under every \(F\)-automorphism of \(\overline F\).
4. \(L/K\) is Galois, so \(\sigma(L)\) depends only on \(\sigma|_K\).
4.1. For \(\sigma \in \operatorname{Aut}_F(\overline F)\), \(\sigma(L)\) is a \(K\)-conjugate of \(L\) (since \(\sigma(K) = K\)).
\(\sigma|_K\) is an automorphism of \(K\), and \(\sigma(L)\) is a \(K\)-conjugate of \(L\).
4.2. Since \(L/K\) is Galois, \(\sigma(L) = L\) for all \(\sigma\) fixing \(K\).
a Galois extension is stable under all \(K\)-automorphisms of \(\overline F\).
4.3. Hence \(\sigma(L)\) depends only on \(\sigma|_K\).
if \(\sigma|_K = \tau|_K\), then \(\sigma^{-1}\tau\) fixes \(K\), so \(\sigma^{-1}\tau(L) = L\), i.e. \(\sigma(L) = \tau(L)\).
5. There are at most two choices for \(\sigma|_K\).
5.1. \(\operatorname{Aut}_F(K)\) has order \(2\) (since \([K:F] = 2\) and \(K/F\) is Galois).
\(|\operatorname{Aut}_F(K)| = [K:F] = 2\).
5.2. Hence \(\sigma|_K\) is one of at most two automorphisms.
\(\sigma|_K \in \operatorname{Aut}_F(K)\), which has two elements.
6. Hence there are at most two \(F\)-conjugates of \(L\).
By step 4, \(\sigma(L)\) is determined by \(\sigma|_K\), and by step 5 there are at most two such restrictions.
7. Q.E.D.
Step 6 is the claim.