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 preimage $\varphi^{-1}(F) = \mathfrak{o}$ is a subring of $k$ .
• 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 1   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

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 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 $\varphi(x^{-1}) = 0$ .

Bibliography

1

Emil Artin: Theory of Algebraic Numbers. Lecture notes. Mathematisches Institut, Göttingen (1959).