Suppose \(F\subset E\subset K\), with \(K/F\) Galois, and both \(E/F\) and \(K/E\) have abelian Galois groups. Must \(\operatorname{Gal}(K/F)\) be abelian? Give counterexamples for number fields and function fields.
Solution
No.
1. Number-field counterexample.
Let \(K\) be the splitting field of \[ x^3-2 \] over \(\QQ\). Then \[ \operatorname{Gal}(K/\QQ)\cong S_3, \] which is nonabelian.
Let \[ E=K^{A_3}=\QQ(\sqrt{-3}). \] Since \(A_3\trianglelefteq S_3\), \[ \operatorname{Gal}(E/\QQ) \cong S_3/A_3 \cong C_2, \] and \[ \operatorname{Gal}(K/E)\cong A_3\cong C_3. \] Both successive Galois groups are abelian, but the total group is \(S_3\).
2. Function-field counterexample.
Let \(\omega\) be a primitive cube root of unity and let \[ K=\CC(x). \] Define automorphisms \[ r(x)=\omega x, \qquad s(x)=x^{-1}. \] They generate a group \[ \langle r,s\rangle\cong S_3. \] Set \[ E=K^{\langle r\rangle}=\CC(x^3), \] and \[ F=K^{S_3}=\CC(x^3+x^{-3}). \] Then \[ \operatorname{Gal}(K/E)\cong C_3, \qquad \operatorname{Gal}(E/F)\cong C_2, \] but \[ \operatorname{Gal}(K/F)\cong S_3 \] is nonabelian.
Thus abelianness of successive Galois groups is not transitive.