place of field

Let $F$ be a field and $\mathrm{\infty }$ an element not belonging to $F$.  The mapping

$$\phi :k\to F\cup \{\mathrm{\infty }\},$$

where $k$ is a field, is called a place of the field $k$, if it satisfies the following conditions.

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

• The restriction  ${\phi |}_{𝔬}$  is a ring homomorphism from $𝔬$ to $F$.

• If  $\phi (a)=\mathrm{\infty }$,  then  $\phi (a^{-1})=0$.

It is easy to see that the subring $𝔬$ 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 $𝔬$ with field of fractions $k$ determines a place of $k$:

Theorem.

Let $𝔬$ be a valuation domain with field of fractions $k$ and $𝔭$ the maximal ideal of $𝔬$, consisting of the non-units of $𝔬$.  Then the mapping

$$\phi :k\to 𝔬/𝔭\cup \{\mathrm{\infty }\}$$

defined by

$$\phi (x):=\{\begin{array}{cc}x+𝔭\hspace{1em}\mathrm{when}x\in 𝔬,\hfill & \\ \mathrm{\infty }\hspace{1em}\mathrm{when}x\in k\setminus 𝔬,\hfill & \end{array}$$

is a place of the field $k$.

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