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'$p$-extension'
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| Title of object: |
$p$-extension |
| Canonical Name: |
PExtension |
| Type: |
Definition |
| Created on: |
2005-02-17 16:46:08 |
| Modified on: |
2005-02-17 16:46:08 |
| Classification: |
msc:12F05 |
| Keywords: |
field extension |
| Synonyms: |
$p$-extension=p-extension |
Preamble:
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\usepackage{amssymb}
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%\usepackage{psfrag}
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\newcommand{\Nats}{\mathbb{N}}
\newcommand{\Ints}{\mathbb{Z}}
\newcommand{\Reals}{\mathbb{R}}
\newcommand{\Complex}{\mathbb{C}}
\newcommand{\Rats}{\mathbb{Q}} |
Content:
\begin{defn}
Let $p$ be a prime number. A Galois extension of fields $E/F$, with $G=\operatorname{Gal}(E/F)$, is said to be a $p$-extension if $G$ is a $p$-group.
\end{defn}
\begin{exa}
Let $d$ be a square-free integer. Then the field extension $\Rats(\sqrt{d})/\Rats$ is a $2$-extension.
\end{exa}
\begin{exa}
Let $p>2$ be a prime and, for any $n$, let $\zeta_{p^n}$ be a primitive $p^n$th root of unity. The extension:
$$\Rats(\zeta_{p^n})/\Rats(\zeta_p)$$
is a $p$-extension. Indeed:
$$G_n=\operatorname{Gal}(\Rats(\zeta_{p^n})/\Rats)\cong (\Ints/p^n\Ints)^\times$$
Thus, $|G_n|=\varphi(p^n)=p^{(n-1)}(p-1)$ and $|G_1|=\varphi(p)=p-1$, where $\varphi$ is the Euler phi function. Therefore the extension above is of degree $p^{(n-1)}$.
\end{exa} |
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