Proposition 3

For two valuations $v, v'$ over a (not necessarily commutative) field $K$ to be equivalent, it is necessary and sufficient that there exist an isomorphism $\lambda$ of the ordered group $v(K^*)$ onto the ordered group $v'(K^*)$ such that $v' = A \circ v$.

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 $v$.