Proposition 6

Let $A$ be a Dedekind domain, $K$ its field of fractions, $L$ a finite extension of $K$ and $B$ the integral closure of $A$ in $L$. Let $p$ be a non-zero prime ideal of $A$, $v$ the corresponding essential valuation of $K$ and

the decomposition of the ideal $Bp$ as a product of prime ideals. Then: (a) the prime ideals of $B$ lying above $p$ are the $p_i$ such that $e(i) > 0$; (b) the valuations $v_i$ on $L$ corresponding to these ideals $p_i$ are, up to equivalence, the valuations on $L$ extending $v$; (c) $[B/p_i : A/p] = f(v_i/v);$ (d) $e_i = e(v_i/v)$ (cf. Chapter VI, § 8, no. 1, Definitions 1 and 2).

(a) To say that a prime ideal $q$ of $B$ lies above $p$ amounts to saying that $q \supseteq p$, hence that $q \supseteq Bp$ and that $q$ contains one of the $p_i$ such that $e(i) > 0$ (Chapter 11, § 1, no. 1, Proposition 1). (b) This follows, taking account of (a), from § 1, no. 8, Corollary to Proposition 12. (c) The residue field of $v$ is identified with $A/p$ and that of $v_i$ with $B/p_i$ (§ 1, no. 4, Corollary 1 to Proposition 6).

(d) Let $a$ (resp. $a_i$) be a uniformizer for $v$ (resp. $v_i$). Then

since $p_j B_{p_i} = B_j$ for $j \neq i$; whence (d), since $e(v_i/v) = v_i(a)$.