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10
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'ideal norm'
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| Title of object: |
ideal norm |
| Canonical Name: |
IdealNorm |
| Type: |
Definition |
| Created on: |
2006-03-04 03:55:52 |
| Modified on: |
2006-10-27 10:42:13 |
| Classification: |
msc:11R04 |
| Keywords: |
residue class, congruence |
| Defines: |
congruent modulo the ideal, residue classes modulo the ideal, absolute norm of ideal |
| Synonyms: |
ideal norm=norm of an ideal
ideal norm=norm of ideal
ideal norm=absolute norm |
Revision comment (for changes between this and next version):
Preamble:
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\usepackage{amssymb}
\usepackage{amsmath}
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%\usepackage{psfrag}
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\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
\def\N{\mathrm{N}} |
Content:
Let $\alpha$ and $\beta$ be algebraic integers in an algebraic number field $K$ and $\mathfrak{m}$ a non-zero ideal in the ring of integers of $K$.\, We say that $\alpha$ and $\beta$ are {\em congruent modulo the ideal} $\mathfrak{m}$ in the case that\, $\alpha\!-\!\beta \in \mathfrak{m}$.\, This is denoted by
$$\alpha \equiv \beta \pmod{\mathfrak{m}}.$$
This congruence relation \PMlinkescapetext{divides} the ring of integers of $K$ into equivalence classes, which are called the {\em residue classes modulo the ideal} $\mathfrak{m}$.
\textbf{Definition.}\, Let $K$ be an algebraic number field and\, $\mathfrak{a}$\, a non-zero ideal in $K$.\, The {\em absolute norm of ideal} $\mathfrak{a}$ means the number of all residue classes modulo $\mathfrak{a}$.
\textbf{Remark.}\, The \PMlinkescapetext{norm} of any ideal $\mathfrak{a}$ of $K$ is finite --- it has the expression
$$\N(\mathfrak{a}) = \sqrt{\frac{\Delta(\mathfrak{a})}{d}}$$
where $\Delta(\mathfrak{a})$ is the discriminant of the ideal and $d$ the fundamental number of the field.
\PMlinkescapetext{\textbf{Some properties}}
\begin{itemize}
\item $\N(\mathfrak{ab})
= \N(\mathfrak{a})\!\cdot\!\N(\mathfrak{b})$
\item $\N(\mathfrak{a}) = 1 \quad\Leftrightarrow\quad \mathfrak{a} = (1)$
\item $\N((\alpha)) = |\N(\alpha)|$
\item $\N(\mathfrak{a}) \in \mathfrak{a}$
\item If $\N(\mathfrak{p})$ is a rational prime, then $\mathfrak{p}$ is a prime ideal.
\end{itemize} |
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