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place of field
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 place of the field $k$, if it satisfies the following conditions.

The restriction $\varphi_{\mathfrak{o}}$ is a ring homomorphism from $\mathfrak{o}$ to $F$.

If $\varphi(a)=\infty$, then $\varphi(a^{{1}})=0$.
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 $k$:
Theorem.
Let $\mathfrak{o}$ be a valuation domain with field of fractions $k$ and $\mathfrak{p}$ the maximal ideal of $\mathfrak{o}$, consisting of the nonunits 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 $k$.
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 residueclass 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 $\varphi(x^{{1}})=0$.
References
 1 Emil Artin: Theory of Algebraic Numbers. Lecture notes. Mathematisches Institut, Göttingen (1959).
Mathematics Subject Classification
13F30 no label found13A18 no label found12E99 no label found Forums
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