Let \(K/F\) be a finite Galois extension and suppose \(\Gal(K/F)\) is cyclic. Let \(F\subseteq E\subseteq K\).
- Show that both \(K/E\) and \(E/F\) are cyclic Galois extensions.
- If \([K:F]=n\), show that for every divisor \(d\mid n\) there is a unique intermediate field \(E_d\) with \[ [E_d:F]=d. \]
Solution
Let \[ G=\Gal(K/F), \qquad H=\Gal(K/E). \] Since \(G\) is cyclic, every subgroup of \(G\) is cyclic and normal.
1. Both intermediate extensions are cyclic Galois.Because \(H\le G\) and \(G\) is cyclic, \(H\) is cyclic. Hence\[ \Gal(K/E)=H \]is cyclic, so \(K/E\) is cyclic Galois.
Because every subgroup of a cyclic group is normal, \[ H\trianglelefteq G. \] By the fundamental theorem of Galois theory, this is equivalent to \(E/F\) being Galois, and restriction gives \[ \Gal(E/F)\cong G/H. \] A quotient of a cyclic group is cyclic, so \(E/F\) is cyclic Galois as well.
2. There is a unique intermediate field of every divisor degree.Write\[ G=\langle g\rangle, \qquad |G|=n. \]For each divisor \(d\mid n\), a cyclic group has a unique subgroup of index \(d\), namely\[ H_d=\langle g^d\rangle, \qquad |H_d|=n/d. \]Let\[ E_d=K^{H_d} \]be its fixed field. By Galois correspondence,\[ [E_d:F]=[G:H_d]=d. \]Uniqueness of \(H_d\) gives uniqueness of \(E_d\).
Thus the intermediate-field lattice of a finite cyclic Galois extension is exactly the divisor lattice of \([K:F]\).