Let \(K/F\) be a finite cyclic Galois extension of degree \(n\). Prove that for every divisor \(d\mid n\), there is a unique intermediate field \(E\) with \[ [E:F]=d. \]
Solution
Let \[ G=\operatorname{Gal}(K/F)=\langle\sigma\rangle, \qquad |G|=n. \]
1. For every divisor \(d\mid n\), the cyclic group \(G\) has a unique subgroup of index \(d\).
The subgroup \[ H_d=\langle\sigma^d\rangle \] has order \(n/d\) and hence index \(d\). A finite cyclic group has exactly one subgroup of each order dividing its order: if \(H\le G\) has order \(n/d\), then \(H\) is generated by an element of order \(n/d\), and every such element generates \(\langle\sigma^d\rangle\). Thus \(H_d\) is unique.
2. Let\[ E_d=K^{H_d}. \]Then \([E_d:F]=d\).
By the fundamental theorem of Galois theory, \[ [E_d:F]=[G:H_d]=d. \]
3. The field \(E_d\) is the unique intermediate field of degree \(d\) over \(F\).
If \(E\) is another intermediate field with \([E:F]=d\), then its corresponding subgroup \[ H=\operatorname{Gal}(K/E) \] has index \[ [G:H]=[E:F]=d. \] By 1, \(H=H_d\). Galois correspondence is injective, so \[ E=K^H=K^{H_d}=E_d. \]