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:

If $f = j \circ h_{\mathbf{A}}$, the condition $f(x) \neq \infty$ (resp. $f(x) = 0$) is equivalent to the condition $h_{\mathbf{A}}(x) \neq \infty$ (resp. $h_{\mathbf{A}}(x) = 0$) and hence to the condition $x \in \mathbf{A}$ (resp. $x \in m(\mathbf{A})$). Hence $\mathbf{A}$ is determined uniquely and, as $h_{\mathbf{A}}$ is surjective, $j$ is also unique.

Now let $f$ be any place of K with values in L; let $\mathbf{A}$ denote the set of $x \in K$ such that $f(x) \neq \infty$. If $x \in \mathbf{A}$ and $y \in \mathbf{A}$, the compositions $f(x) - f(y)$ and $f(x)f(y)$ are defined and $\neq \infty$, which shows that $x - y \in \mathbf{A}$ and $xy \in \mathbf{A}$; hence $\mathbf{A}$ is a subring of K. If $x \notin \mathbf{A}$, then $f(x) = \infty$, hence $f(x^{-1}) = 0$ and $x^{-1}$ belongs to the kernel $m$ of the homomorphism $f'$ obtained by restricting $f$ to $\mathbf{A}$. Conversely if $y \in m$, then $y^{-1} \notin \mathbf{A}$. This shows that $\mathbf{A}$ is a valuation ring of K and that $m$ is its maximal ideal. Let $j$ be the injective homomorphism from $\kappa(\mathbf{A})$ to L derived from f' by passing to the quotient. Then $f(x) = j(h_A(x))$ for all $x \in A$ and this equation remains true for $x \notin A$, the two sides being then equal to $\infty$.

The decomposition $f = j \circ h_A$ is called the canonical decomposition of the placef. $A$ is called the ring off, $m(A)$ the ideal of $f$ and $\kappa(A)$ the residue field off. For two places $f : \mathbf{K} \to \mathbf{L}$ and $f' : \mathbf{K} \to \mathbf{L}'$ to have the same ring, it is necessary and sufficient that there exist an isomorphisms of the value field off onto that off such that $f' = s \circ f$; then $f$ and $f'$ are called equivalent. It is seen that every result on valuation rings can be translated into a result on places and conversely; this is what we shall do in the following nos.

Examples Of places (1) Let $K$ be a field. The identity mapping on $K$ is a trivial place with ring $K$ and ideal $(0)$. (2) Let $k$ be a field. For all $u \in k((T))^\sim$, let us write $f(u) = \infty$ if $u \notin k[[T]]$ and define $f(u)$ to be the constant term of $u$ if $u \in k[[T]]$. Then $f$ is a place of $k((T))$, with residue field $k$ and ring $k[[T]]$. For $k[[T]]$ is a valuation ring of $k((T))$ (§ 1, no. 4, Example 3 ) and the restriction of $f$ to $k[[T]]$ is identified with the canonical homomorphism of $k[[T]]$ onto its residue field. (3) Let $k$ be a field, $a$ an element of $k$ and $A$ the set of $u \in k(X)$ such that $a$ is substitutable in $u$ (Algebra, Chapter IV, § 3, no. 2). If $p$ denotes the prime ideal $(X - a)$ of $k[X]$, then $A = k[X]_p$, so that $A$ is a valuation ring of $k(X)$ (§ 1, no. 4, Proposition 2). For all $u \in k(X)^\sim$, let us write $f(u) = \infty$ if $u \notin A$ and $f(u) = u(a)$ if $u \in A$. Then $f$ is a place of $k(X)$ with residue field $k$ and ring $A$; for the restriction off to $A$ is a homomorphism of $A$ onto $k$ (Algebra, Chapter IV, § 3, Proposition 2) of kernel $pA = m(A)$. The element $f(u) \in k$ is said to be obtained by putting $X = a$ in $u$. \* (4) Let $S$ be a connected complex analytic variety of dimension 1 and $K$ the field of meromorphic functions on $S$. For all $z_0 \in S$, the mapping $f \mapsto f(z_0)$ from $K$ to $\mathbf{C}$ is a place of $K$ whose ring is the set off $\in K$ which are holomorphic at $z_0$ and whose ideal is the set off $\in K$ which are zero at $z_0$. It is this example and other analogues which are the origin of the term "place".*

Taking account of Proposition 1, this is a translation of Theorem 2 of § 1, no. 2.

Let $V$ be the ring off, $g$ the restriction off to $V$ and $g'$ the extension of $g$ to $V[x_1, \ldots, x_n]$ such that $g'(x_i) = a,$ for $1 \leq i \leq n.$ It is sufficient to take $L'$ to be an algebraic closure of $L$ and apply Proposition 3 to $g'$ and $L'$: we obtain a placef': $K' \to \tilde{L}'$ which extends $g'$; if $x \in K - V,$ then $x^{-1} \in m(V),$ whence $f(x^{-1}) = g(x^{-1}) = 0$ and $f'(x) = \infty = f(x);$ hencef' extendsf.

If $f$ is a place of $K$ extending $h, f$ is finite on $S_p$ and zero on $pS_p$ and hence the ring of the place $f$ dominates $S_p.$ Conversely, if $V$ is a valuation ring of $K$ which dominates $S_p,$ $V$ is the ring of a placef whose restriction to $S$ is a homomorphism with the same kernel as $h;$ replacing $f$ by an equivalent place, it is seen that $V$ is the ring of a place of $K$ which extends $h.$ To say that every place of $K$ extending $h$ is finite at $x$ is equivalent to saying that $x$ belongs to all the valuation rings of $K$ which dominate $S_p.$ The proposition then follows from Theorem 3 of § 1, no. 3.

This is also a consequence of Theorem 3 of § 1, no. 3.