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Viewing Version
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'Galois groups of finite abelian extensions of $\mathbb{Q}$'
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| Title of object: |
Galois groups of finite abelian extensions of $\mathbb{Q}$ |
| Canonical Name: |
GaloisGroupsOfFiniteAbelianExtensionsOfMathbbQ |
| Type: |
Theorem |
| Created on: |
2006-10-09 02:22:37 |
| Modified on: |
2006-10-09 06:37:02 |
| Classification: |
msc:12F10, msc:11R32, msc:11R20, msc:11N13 |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
\newtheorem*{thm*}{Theorem}
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Content:
\PMlinkescapeword{order}
\begin{thm*}
Let $G$ be a finite abelian group with $|G|>1$. Then there exist infinitely many number fields $K$ with $K/\mathbb{Q}$ Galois and $\operatorname{Gal}(K/\mathbb{Q}) \cong G$.
\end{thm*}
\begin{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 $p \equiv 1 \operatorname{mod} n$. Let $\zeta_p$ denote a \PMlinkescapetext{primitive} $p^{\text{th}}$ root of unity. Let $L=\mathbb{Q}(\zeta_p)$. Then $L/\mathbb{Q}$ is Galois with $\operatorname{Gal}(L/\mathbb{Q})$ cyclic of \PMlinkname{order}{OrderGroup} $p-1$. Since $n$ divides $p-1$, there exists a subgroup $H$ of $\operatorname{Gal}(L/\mathbb{Q})$ such that $\displaystyle |H|=\frac{p-1}{n}$. Since $\operatorname{Gal}(L/\mathbb{Q})$ is cyclic, it is abelian, and $H$ is a normal subgroup of $\operatorname{Gal}(L/\mathbb{Q})$. Let $K=L^H$, the subfield of $L$ \PMlinkname{fixed}{FixedField} by $H$. Then $K/\mathbb{Q}$ is Galois with $\operatorname{Gal}(K/\mathbb{Q})$ cyclic of order $n$. Thus, $\operatorname{Gal}(K/\mathbb{Q}) \cong G$.
Let $p$ and $q$ be distinct primes with $p \equiv 1 \operatorname{mod} n$ and $q \equiv 1 \operatorname{mod} n$. Then there exist subfields $K_1$ and $K_2$ of $\mathbb{Q}(\zeta_p)$ and $\mathbb{Q}(\zeta_q)$, respectively, such that $\operatorname{Gal}(K_1/\mathbb{Q}) \cong G$ and $\operatorname{Gal}(K_2/\mathbb{Q}) \cong G$. Since $\mathbb{Q} \subseteq K_1 \cap K_2 \subseteq \mathbb{Q}(\zeta_p) \cap \mathbb{Q}(\zeta_q)=\mathbb{Q}$, then $K_1 \cap K_2=\mathbb{Q}$. Thus, $K_1 \neq K_2$. Therefore, for every prime $p$ with $p \equiv 1 \operatorname{mod} n$, there exists a distinct number field $K$ such that $K/\mathbb{Q}$ is Galois and $\operatorname{Gal}(K/\mathbb{Q}) \cong G$. The theorem in the cyclic case follows from using the full \PMlinkescapetext{force} of Dirichlet's theorem on primes in arithmetic progressions: There exist infinitely many primes $p$ with $p \equiv 1 \operatorname{mod} n$.
The general case follows immediately from the above \PMlinkescapetext{argument} and the \PMlinkname{fundamental theorem of finite abelian groups}{FundamentalTheoremOfFinitelyGeneratedAbelianGroups}.
\end{proof} |
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