Let \(K\) be a Galois extension of \(\QQ\) with Galois group \(G\), and let \(E_1 , E_2\) be intermediate fields of \(K\) which are the splitting fields of irreducible \(f_i (x) \in \QQ[x]\).
Let \(E = E_1 E_2 \subset K\).
Let \(H_i = \Gal(K/E_i)\) and \(H = \Gal(K/E)\).
Show that \(H = H_1 \cap H_2\).
Show that \(H_1 H_2\) is a subgroup of \(G\).
Show that \[ \Gal(K/(E_1 \cap E_2 )) = H_1 H_2 .\]
- The Galois correspondence:
- \(H_1 \intersect H_2 \mapstofrom E_1 E_2\),
- \(H_1 H_2 \mapstofrom E_1 \intersect E_2\).
Solution
1. \(H=H_1\cap H_2\).
An automorphism \(\sigma\in G\) lies in \(H\) exactly when it fixes the compositum \(E_1E_2\) pointwise. This certainly implies that it fixes the subfields \(E_1\) and \(E_2\), so \[ H\subseteq H_1\cap H_2. \]
Conversely, if \(\sigma\in H_1\cap H_2\), then \(\sigma\) fixes both \(E_1\) and \(E_2\) pointwise. Every element of \(E_1E_2\) is obtained from elements of \(E_1\cup E_2\) by finitely many field operations, and a field automorphism respects these operations. Hence \(\sigma\) fixes \(E_1E_2\) pointwise, so \(\sigma\in H\).
2. Each \(H_i\) is normal in \(G\).
Each \(E_i\) is a splitting field over \(\QQ\). Thus \(E_i/\QQ\) is normal, and since \(\operatorname{char}\QQ=0\), it is separable as well. Hence \(E_i/\QQ\) is Galois.
By the fundamental theorem of Galois theory for \(K/\QQ\), this is equivalent to \[ H_i=\Gal(K/E_i)\triangleleft G. \]
3. \(H_1H_2\) is a subgroup of \(G\).
Let \(h_1,h_1'\in H_1\) and \(h_2,h_2'\in H_2\). Since \(H_1\triangleleft G\) by 2, \[ h_2h_1'h_2^{-1}\in H_1. \] Therefore \[ (h_1h_2)(h_1'h_2') =h_1(h_2h_1'h_2^{-1})(h_2h_2')\in H_1H_2. \] Similarly, \[ (h_1h_2)^{-1} =h_2^{-1}h_1^{-1} =(h_2^{-1}h_1^{-1}h_2)h_2^{-1}\in H_1H_2. \] Thus \(H_1H_2\le G\).
4. The fixed field of \(H_1H_2\) is \(E_1\cap E_2\).
Since \(H_i\subseteq H_1H_2\), \[ K^{H_1H_2}\subseteq K^{H_1}\cap K^{H_2}=E_1\cap E_2. \]
Conversely, if \(x\in E_1\cap E_2\), then every \(h_1\in H_1\) and every \(h_2\in H_2\) fixes \(x\). Hence every product \(h_1h_2\) fixes \(x\), so \[ E_1\cap E_2\subseteq K^{H_1H_2}. \]
5. \(\Gal(K/(E_1\cap E_2))=H_1H_2\).
By 3, \(H_1H_2\) is a subgroup of \(G\), and by 4 its fixed field is \(E_1\cap E_2\). The Galois correspondence gives the claimed identity.