<?xml version="1.0" encoding="UTF-8"?>

<record version="13" id="6640">
 <title>place of field</title>
 <name>PlaceOfField</name>
 <created>2005-01-13 08:15:08</created>
 <modified>2006-12-12 12:52:44</modified>
 <type>Theorem</type>
<parent id="4506">valuation domain</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <classification>
	<category scheme="msc" code="12E99"/>
	<category scheme="msc" code="13A18"/>
	<category scheme="msc" code="13F30"/>
 </classification>
 <defines>
	<concept>place of field</concept>
 </defines>
 <synonyms>
	<synonym concept="place of field" alias="place"/>
	<synonym concept="place of field" alias="spot of field"/>
 </synonyms>
 <related>
	<object name="KrullValuation"/>
	<object name="ValuationDeterminedByValuationDomain"/>
	<object name="IntegrityCharacterizedByPlaces"/>
	<object name="RamificationOfArchimedeanPlaces"/>
 </related>
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\newtheorem*{thmplain}{Theorem}</preamble>
 <content>Let $F$ be a field and $\infty$ an element not belonging to $F$.\, The mapping
               $$\varphi: \,k\to F\cup\{\infty\},$$
where $k$ is a field, is called a {\em place of the field} $k$, if it satisfies the following conditions.
\begin{itemize}
 \item The preimage\, $\varphi^{-1}(F) = \mathfrak{o}$\, is a subring of $k$.
 \item The restriction\, $\varphi|_\mathfrak{o}$\, is a ring homomorphism from $\mathfrak{o}$ to $F$.
 \item If\, $\varphi(a) = \infty$,\, then\, $\varphi(a^{-1}) = 0$.
\end{itemize}

It is easy to see that the subring $\mathfrak{o}$ of the field $k$ is a valuation domain; so any place of a field determines a unique valuation domain in the field.\, Conversely, every valuation domain $\mathfrak{o}$ with field of fractions $k$ determines a place of $k$:

\begin{thmplain}
\,Let $\mathfrak{o}$ be a valuation domain with field of fractions $k$ and $\mathfrak{p}$ the maximal ideal of $\mathfrak{o}$, consisting of the non-units of $\mathfrak{o}$.\, Then the mapping
   $$\varphi: \,k\to \mathfrak{o/p}\cup\{\infty\}$$
defined by
$$
\varphi(x):=
\begin{cases}
x+\mathfrak{p} \quad \mathrm{when} \,\,\, x \in\mathfrak{o}, \\
\infty \quad \mathrm{when} \,\,\, x \in k\smallsetminus\mathfrak{o},
\end{cases}
$$
is a place of the field $k$.
\end{thmplain}

{\em Proof.}\, Apparently,\, $\varphi^{-1}(\mathfrak{o/p}) = \mathfrak{o}$\, and the restriction\, $\varphi|_\mathfrak{o}$\, is the canonical homomorphism from the ring $\mathfrak{o}$ onto the residue-class ring $\mathfrak{o/p}$.\, Moreover, if\, $\varphi(x) = \infty$,\, then $x$ does not belong to the valuation domain $\mathfrak{o}$ and thus the inverse element $x^{-1}$ must belong to it without being its unit.\, Hence $x^{-1}$ belongs to the ideal $\mathfrak{p}$ which is the kernel of the homomorphism\, 
$\varphi|\mathfrak{o}$.\, So we see that\, $\varphi(x^{-1}) = 0$.

\begin{thebibliography}{7}
\bibitem{artin} Emil Artin: {\em \PMlinkescapetext{Theory of Algebraic Numbers}}.\, Lecture notes.\, Mathematisches Institut, G\"ottingen (1959).
\end{thebibliography}</content>
</record>
