Proposition 1

Let K be a field, $\mathbf{A}$ a valuation ring of K and $\kappa(\mathbf{A})$ the residue field of $\mathbf{A}$. We extend the canonical mapping of $\mathbf{A}$ onto $\kappa(\mathbf{A})$ to a mapping $h_{\mathbf{A}} : \tilde{K} \to (\kappa(\mathbf{A}))$ by the equation $h_{\mathbf{A}}(x) = \infty$ if $x \notin \mathbf{A}$. The mapping $h_{\mathbf{A}}$ thus defined is a place of K whose valuefield is $\kappa(\mathbf{A})$.

Clearly $h_{\mathbf{A}}(1) = 1$.

We show that $h_{\mathbf{A}}$ is a morphism for addition. Let $x, y$ be two elements of $\tilde{K}$ such that $h_{\mathbf{A}}(x) + h_{\mathbf{A}}(y)$ is defined. One of the two elements $x, y$ belongs then to $\mathbf{A}$ and hence $x + y$ is defined. If $x \in \mathbf{A}$ and $y \in \mathbf{A}$, clearly

holds. If $x \in \mathbf{A}$ and $y \notin \mathbf{A}$, then $x + y \notin \mathbf{A}$ and the two sides of the above formula equal $\infty$.

We show finally that $h_{\mathbf{A}}$ is a morphism for multiplication. Let $x \in K, y \in K$ be such that $h_{\mathbf{A}}(x)h_{\mathbf{A}}(y)$ is defined. If $x \in \mathbf{A}$ and $y \in \mathbf{A}$, clearly $xy$ is defined and $h_{\mathbf{A}}(x)h_{\mathbf{A}}(y) = h_{\mathbf{A}}(xy)$. Suppose now that one of the elements $x, y$, for example $y$, does not belong to $\mathbf{A}$; as $h_{\mathbf{A}}(y) = \infty$, $h_{\mathbf{A}}(x) \neq 0$, that is $x \notin m(\mathbf{A})$, whence $x^{-1} \in \mathbf{A}$; it follows that $xy$ is defined and that $xy \notin \mathbf{A}$, whence

This proves Proposition 1.

If $j$ is an isomorphism of $\kappa(\mathbf{A})$ onto a subfield of a field L, $j \circ h_{\mathbf{A}} : \tilde{K} \to \tilde{L}$ is a place of K with values in L. The above process in fact provides all the places on K. To be precise: