By Dirichlet theorem on primes in arithmetic progressions, choose a prime withThe cyclotomic field is Galois over , andwhere the last group is cyclic because the multiplicative group of a finite field is cyclic.
The cyclic group has a unique subgroup of order . By the Galois correspondence, its fixed field satisfiesNormality follows because lies in an abelian group. Thus every occurs as the Galois group of a finite extension of . This is the cyclic Galois extension of the rational numbers of every finite degree.
The products with and span the field compositum over , soThe tower law givesThus the degree is divisible by both and . Since they are coprime, it is divisible by . Combining divisibility with the upper bound givesEquivalently, finite extensions of coprime degrees are linearly disjoint field extensions.
Takeand let . Takethe affine group over a finite field, and let be its translation subgroup. Both normal subgroups have order , andThe groups are not isomorphic: is abelian, while is nonabelian.
Letthe splitting field of . Its Galois group is . The normal subgroup fixes the quadratic discriminant field, so
Now let . Part (a), with , supplies a cyclic Galois extension with group . For , take the splitting fieldof . The prime is unramified in , so at every prime ideal above its valuation is one. It cannot therefore be a th power in the cyclotomic field. HenceThe automorphismsand, for ,satisfyThey generate all automorphisms, provingThis is the affine Galois group of the splitting field of x to the p minus two.
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