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单词 GaloisGroupsOfFiniteAbelianExtensionsOfmathbbQ
释义

Galois groups of finite abelian extensions of


Theorem.

Let G be a finite abelian group with |G|>1. Then there exist infinitely many number fieldsMathworldPlanetmath K with K/Q Galois and Gal(K/Q)G.

Proof.

This will first be proven for G cyclic.

Let |G|=n. By Dirichlet’s theorem on primes in arithmetic progressions, there exists a prime p with p1modn. Let ζp denote a pth root of unityMathworldPlanetmath. Let L=(ζp). Then L/ is Galois with Gal(L/) cyclic of order (http://planetmath.org/OrderGroup) p-1. Since n divides p-1, there exists a subgroupMathworldPlanetmathPlanetmath H of Gal(L/) such that |H|=p-1n. Since Gal(L/) is cyclic, it is abelianMathworldPlanetmath, and H is a normal subgroupMathworldPlanetmath of Gal(L/). Let K=LH, the subfieldMathworldPlanetmath of L fixed (http://planetmath.org/FixedField) by H. Then K/ is Galois with Gal(K/) cyclic of order n. Thus, Gal(K/)G.

Let p and q be distinct primes with p1modn and q1modn. Then there exist subfields K1 and K2 of (ζp) and (ζq), respectively, such that Gal(K1/)G and Gal(K2/)G. Note that K1K2= since K1K2(ζp)(ζq)=. Thus, K1K2. Therefore, for every prime p with p1modn, there exists a distinct number field K such that K/ is Galois and Gal(K/)G. The theorem in the cyclic case follows from using the full of Dirichlet’s theorem on primes in arithmetic progressions: There exist infinitely many primes p with p1modn.

The general case follows immediately from the above , the fundamental theorem of finite abelian groups (http://planetmath.org/FundamentalTheoremOfFinitelyGeneratedAbelianGroups), and a theorem regarding the Galois groupMathworldPlanetmath of the compositum of two Galois extensionsMathworldPlanetmath.∎

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