Moreover, if there exists a single index $k$ such that $v(x_k) = \inf_{1 \leq i \leq n} v(x_i)$, the two sides of (1) are equal. In particular, if $v(x) \neq v(y)$, then $v(x + y) = \inf(v(x), v(y))$.

Relation (1) follows from axiom (VL_{II}) by induction on $n$. If there exists a single index $k$ such that $v(x_k) = \inf_{1 \leq i \leq n} v(x_i)$, then, writing $y = \sum_{i \neq k} x_i$ and $z = \sum_{i=1}^n x_i$, $v(y) > v(x_k)$ and $v(z) \geq v(x_k)$ by (1); if $v(z) > v(x_k)$, the relation $x_k = z - y$ would give $v(x_k) \geq \inf(v(z), v(y)) > v(x_k)$, which is absurd; whence $v(z) = v(x_k)$, which proves the second assertion.

If there were only a single index $k$ such that $v(x_k) = \inf_{1 \leq i \leq n} v(x_i)$, Proposition 1 would show that $v(x_k) = v(0) = +\infty$, whence $v(x_i) = +\infty$ for all $i$, contrary to the relation $n \geq 2$ and the hypothesis made on $k$.

Remarks (1) If $v : C \to \Gamma_\infty$ is a valuation on $C$ and $u : B \to C$ a homomorphism of a ring $B$ to $C$, it is immediate that the composite mapping $B \xrightarrow{u} C \xrightarrow{v} \Gamma_\infty$ is a valuation on $B$ with values in $\Gamma$.

(2) Conditions (VL$_I$) and (VL$_II$) show immediately that the set $\bar{v}^{-1}(+\infty)$ is a two-sided ideal $\mathfrak{p}$ in $C$ distinct from $C$ by virtue of (VL$_III$); moreover, if $x, y$ are two elements of $C$ such that $v(xy) = +\infty$, it follows from (VL,) that necessarily $v(x) = +\infty$ or $v(y) = +\infty$; in other words, the quotient ring $C/\mathfrak{p}$ has no divisor of 0 other than 0; it is immediately verified that the mapping $\bar{v} : C/\mathfrak{p} \to \Gamma_\infty$ derived from $v$ by passing to the quotient is a valuation on $C/\mathfrak{p}$, the inverse image of $+\infty$ under this valuation reducing to 0.

Assertion (i) follows from the fact that $\bar{v}^{-1}(+\infty)$ is an ideal of $K$ not equal to $K$. The verification of the fact that $A$ is a ring and the $V$, and $V'_a$ two-sided ideals is trivial by virtue of axioms (VL$_I$), (VL$_II$) and (VL$_III$). If $a$ is a (left, for example) ideal of $A$ and $x \neq 0$ belongs to $A$, every $y \in A$ such that $v(y) \geq v(x)$ can be written $y = zx$ where $z = yx^{-1}$, hence $v(z) = v(y) - v(x) \geq O$ and therefore $z \in A$; in other words the left ideal $Ax$ contains the $V'_a$ for $a > v(x)$. The set $U(A) = A - m(A)$ is the set of $x \in K$ such that $v(x) = 0$; if $x \in U(A)$, then

whence $x^{-1} \in U(A)$; conversely, if $y \in A$ is invertible in $A$, then $v(y) \geq 0$, $v(y^{-1}) \geq 0$ and $v(y) + v(y^{-1}) = 0$, whence $v(y) = 0$ and $y \in U(A)$; this proves (iv) and (v) follows immediately from the definitions.

$A$ (resp. $m(A)$, $\kappa(A)$) is called the ring (resp. ideal, residue field) of the valuation $v$ on $K$.

Clearly U(A) is the kernel of the homomorphism $v : K^* \to \Gamma$ and the image $v(K^*)$ under $v$ of the multiplicative group $K^*$ is a subgroup of the additive group $\Gamma$, called the order group or value group of $v$, which is therefore isomorphic to $K^*/U(A)$; for $x \in K$, the element $v(x)$ of $\Gamma_\infty$ is sometimes called the valuation or order of $x$ for $v$. Two valuations $v, v'$ on $K$ are called equivalent if they have the same ring.

Suppose $v$ and $v'$ are equivalent; by hypothesis, the ring $A$ of the valuation $v$ being the same as that of $v'$, $v$ and $v'$ (restricted to $K^*$) factor into homomorphisms $K^* \to K^*/U(A) \xrightarrow{\mu} v(K^*)$, $K^* \to K^*/U(A) \xrightarrow{\nu} v'(K^*)$, where $\mu$ and $\nu$ are isomorphisms; moreover, the set of positive elements of $v(K^*)$ (resp. $v'(K^*)$) is the image under $\mu$ (resp. $\nu$) of the set of classes mod. $U(A)$ of elements $\neq 0$ of $m(A)$; we conclude that $A = v \circ \mu^{-1}$ solves the problem, the converse being obvious.

Suppose now that $K$ is a commutative field; then, for all valuations $v$ on $K$, the ring $A$ of the valuation $v$ is a valuation ring of $K$ in the sense of § 1, no. 2, Definition 2 (which justifies the terminology); this follows immediately from Proposition 2 (c) and § 1 no. 2, Theorem 1 (c). Conversely, recall that for every integral domain $B$ whose field of fractions is $K$ the relation of divisibility $x|y$ (equivalent to $y \in Bx$) makes $K^*$ into a preordered group, whose associated ordered group $\Gamma_B$ is the quotient $K^*/U(B)$ of $K^*$ by the group $U(B)$ of invertible elements of $B$, the positive elements of this group being those of $B^*/U(B)$ (where $B^* = B - \{0\}$); the mapping $x \mapsto Bx$ defines, by passing to the quotient, an isomorphism of the ordered group $K^*/U(B)$ onto the group (ordered by the relation $\supset$) of non-zero principal fractional ideals of $K$ (Algebra, Chapter VI, § 1, no. 5). The rings $A$ with field of fractions $K$ and for which the group $\Gamma_A = K^*/U(A)$ is totally ordered are precisely the valuation rings of K (§ 1, no. 2, Theorem 1 (d)). If $v_A$ denotes the canonical homomorphism of $K^*$ onto $\Gamma_A$, it is immediate that $v_A$ (extended by $v_A(0) = +\infty$) is a valuation (called canonical) on $K$ whose ring is $A$; every valuation equivalent to $v_A$ may be written $v = \sigma \circ v_A$, where $\sigma$ is an isomorphism of $\Gamma_A$ onto a subgroup of the group where $v$ takes its values (Proposition 3); $\sigma \circ v_A$ is called the canonical factorization of $v$.

By Theorem 2 of Algebra, Chapter I, § 2, no. 7, there exists a unique homomorphism $w$ from $K^*$ to $\Gamma$ which extends $v|C^*$ and $w(K^*)$ is generated by $v(C^*)$. It remains to prove that $w$ satisfies axiom (VL$_{\text{II}}$). Then let $x' \in K^*, y \in K^*$ be such that $x + y \in K^*$; there exists $a \in \mathbf{C}^*$ such that $ax \in \mathbf{C}^*$ and $ay \in \mathbf{C}^*$, whence $a(x + y) \in \mathbf{C}^*$. Since the restriction of $w$ to $\mathbf{C}^*$ satisfies (VL$_{\text{II}}$),

Eliminating $w(a)$ from both sides, we obtain

Let $\Gamma_v$ and $\Gamma_{v'}$ be the order groups of $v$ and $v'$. Since the restriction of $v'$ to $K$ is equivalent to $v$, there exists an isomorphism $A$ of $\Gamma_v$ onto a subgroup of $\Gamma_{v'}$, such that $v' = A \circ v$ on $K$. If $\Gamma_v$ is identified with $\lambda(\Gamma_v)$ by means of $\lambda$, it is seen that $v'$ extends $v$.

Note that $\Gamma_{v'}$ is in general distinct from $\lambda(\Gamma_v)$ and the equivalence class of $v'$ is not necessarily unique. We shall return to this in § 8.

Translating Theorem 3 of § 1, no. 3 (or Proposition 6 of § 2, no. 5), we obtain:

From now on, we shall in general leave to the reader the trouble of performing translations analogous to the above.

Let $b$ be a sub-A-module of $K$. The set of $v(x)$ for $x \in b - (0)$ is a major subset $M(b)$ of $G$. Proposition 7 will be shown if the following equations are proved:

(2) $M(a(N)) = N$ for every major subset $N$ of $G$; (3) $a(M(b)) = b$ for every sub-A-module $b$ of $K$.

Formula (2) is easy, since, for all $m \in N$, there exists $x \in K$ such that $v(x) = m$. Then obviously $b \subset a(M(b))$; conversely, let $x \in a(M(b))$ and suppose $x \neq 0$;

then $v(x) \in M(b)$ and therefore there exists $y \in b$ such that $v(x) = v(y)$; whence $x = uy$ where $v(u) = 0$, which proves that $x \in A$ $y \subset b$ and completes the proof.

As $A = a(G_+)$, $a(M) \subset A$ is equivalent to $M \subset G_+$.

For example the maximal ideal $m(A)$ is equal to $a(S)$, where $S$ denotes the set of strictly positive elements of $G$.

It may be assumed that $v$ is normed, so that $v(u) = 1$. For all $x \in K^*$, there is an integer $n \in \mathbf{Z}$ such that $(v(x) = n = v(u^n)$ and hence we may write

where $z, z'$ are two invertible elements of the ring $A$; whence the proposition.

(d) $A$ is a Noetherian ring and $m(A)$ is principal. (e) $A$ is a Noetherian valuation ring.

Proposition 8 shows that (a) implies (b), (d) and (e). If $A$ is a principal ideal domain, then $m(A) = Au$ and every non-zero ideal of $A$ is of the form $Au^n$ since $A$ is local (Algebra, Chapter VII, § 1, no. 3, Theorem 2); therefore $\bigcap_{n=1}^\infty m(A)^n = 0$; this shows that (b) implies (c). On the other hand (d) implies (c) (Chapter III, § 3, no. 2, Corollary to Proposition 5); by Proposition 2 of § 1, no. 4, (c) implies (a). Thus conditions (a), (b), (c), (d) are equivalent and imply (e). Finally suppose (e) holds and let us show that (b) holds; it will be sufficient to prove the following lemma:

Let $E$ be a finitely generated torsion-free $A$-module and let $x_1, \ldots, x_n$ be generators of $E$ which are minimal in number; we show that they are linearly independent. If $\sum_{i=1}^n a_i x_i = 0$ ($a_i \in A$) is a non-trivial relation between the $x_i$, one of the $a_i$, say $a_1$, divides all the others since the set of principal ideals of $A$ is totally ordered by inclusion (§ 1, no. 2, Theorem 1); then $a_1 \neq 0$ since the relation is non-trivial. As $E$ is torsion-free, we can divide by $a_1$, which amounts to assuming that $a_1 = 1$. But then $x_1$ is a linear combination of $x_2, \ldots, x_n$, contrary to the minimal character of $n$. Hence $E$ is free.

In particular every finitely generated ideal $\mathfrak{a}$ of $A$ is principal, all the elements of a system of generators of $\mathfrak{a}$ being multiples of one of them. Proposition 3 of Chapter I, § 2, no. 4 then shows that every torsion-free $A$-module is flat.