By Dirichlet theorem on primes in arithmetic progressions, choose a prime with
The cyclotomic field is Galois over , and
where 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 satisfies
Normality 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 , so
The tower law gives
Thus the degree is divisible by both and . Since they are coprime, it is divisible by . Combining divisibility with the upper bound gives
Equivalently, finite extensions of coprime degrees are linearly disjoint field extensions.
Take
and let . Take
the affine group over a finite field, and let be its translation subgroup. Both normal subgroups have order , and
The groups are not isomorphic: is abelian, while is nonabelian.
For , let
Then . The fixed field of its subgroup of order three is its unique quadratic subfield,
Let
the 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 field
of . 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. Hence
The automorphisms
and, for ,
satisfy
They generate all automorphisms, proving
This is the affine Galois group of the splitting field of x to the p minus two.

Articles by others on the same topic (0)

There are currently no matching articles.