Proposition 14

Let $A$ be a Krull domain and $B$ a Krull domain containing $A$. Suppose that the following condition holds:

(PDE) For every prime ideal $\mathfrak{P}$ of $B$ of height 1, the prime ideal $\mathfrak{P} \cap A$ is zero or of height 1.

For $p \in P(A)$ the $\mathfrak{P} \in P(B)$ such that $\mathfrak{P} \cap A = p$ are finite in number; we write

where $e(\mathfrak{P}/\mathfrak{p})$ denotes the ramification index of $v_{\mathfrak{p}}$ over $v_{\mathfrak{p}}$ (Chapter VI, § 8, no. 1). Then i defines, by linearity, an increasing homomorphism $i$ of $D(A)$ to $D(B)$, which enjoys the following properties:

(a) for every non-zero element x of the field of fractions of A,

(b) for all D, D' in D(A),

Let $\mathfrak{p} \in P(A)$; consider a non-zero element $a$ of $\mathfrak{p}$; the $\mathfrak{P} \in P(B)$ which contain $a$ are finite in number (no. 6, Theorem 4); a fortiori the $\mathfrak{P} \in P(B)$ such that $\mathfrak{P} \cap A = \mathfrak{p}$ are finite in number.

We now show (a). By additivity, it may be assumed that $x \in A^* = A - \{0\}$.

By definition, $\operatorname{div}_B(x) = \sum_{\mathfrak{P} \in P(B)} v_{\mathfrak{p}}(x) \cdot \mathfrak{P}$. For all $\mathfrak{P} \in P(B)$ such that $v_{\mathfrak{p}}(x) > 0$, $\mathfrak{P} \cap A$ is non-zero (for $x \in \mathfrak{P}$) and is therefore of height 1 by (PDE); setting $\mathfrak{p} = \mathfrak{P} \cap A$, by definition of the ramification index, $v_{\mathfrak{p}}(x) = e(\mathfrak{P}/\mathfrak{p}) v_{\mathfrak{p}}(x)$ (since $v_{\mathfrak{p}}$ and $v_{\mathfrak{P}}$ are normed). As $\operatorname{div}_A(x) = \sum_{\mathfrak{p} \in P(A)} v_{\mathfrak{p}}(x) \cdot \mathfrak{p}$, and $i(q) = 0$ for all $q \in P(A)$ which is not of the form $\mathfrak{Q} \cap A$ where $\mathfrak{Q} \in P(B)$, we deduce (a).

To prove (b) we write

the coefficient of $\mathfrak{p}$ in $\sup(D, D')$ is $\sup(n(\mathfrak{p}), n'(\mathfrak{p}))$. Let $\mathfrak{P}$ be an element of $P(B)$. If $\mathfrak{P} \cap A = (0)$, the coefficients of $\mathfrak{P}$ in $i(D)$ and $i(D')$ and hence also in $\sup(i(D), i(D'))$ are zero; therefore the coefficient of $\mathfrak{P}$ in $i(\sup(D, D'))$ is zero. If $\mathfrak{P} \cap A \neq (0)$, it is a prime ideal $\mathfrak{p}$ of height 1 (by (PDE)); writing $e = e(\mathfrak{P}/\mathfrak{p})$, the coefficients of $\mathfrak{P}$ in $i(D)$, $i(D')$ and $i(\sup(D, D'))$ are respectively $en(\mathfrak{p}), en'(\mathfrak{p})$ and $e \cdot \sup(n(\mathfrak{p}), n'(\mathfrak{p}))$; that of $\sup(i(D), i(D'))$ is

This proves (b).

Under the hypotheses of Proposition 14, it follows from (a) that i defines, by taking quotients, a homomorphism $\bar{i}$, called canonical, of $C(A)$ to $C(B)$, which we shall also sometimes write as $i$, by an abuse of notation.

The condition (PDE) is fulfilled in the two following cases:

(1) B is integral over A; in this case, for the prime ideal $\mathfrak{P}$ of B to be of height 1, it is necessary and sufficient that $\mathfrak{p} = \mathfrak{P} \cap A$ be of height 1. For (0) is the only prime ideal of B lying above the ideal (0) of A (Chapter V, § 2, no. 1, Corollary I to Proposition 1); if $\mathfrak{P}$ is of height 1, then $\mathfrak{p} \neq 0$; if $\mathfrak{p}$ were not of height 1, there would exist a prime ideal $\mathfrak{p}'$ of A distinct from (0) and $\mathfrak{p}$ and such that 0 ⊂ p' ⊂ p; but then, as B is an integral domain and A is an integrally closed domain, there would be a prime ideal $\mathfrak{p}'$ of B such that $\mathfrak{p}' \cap A = p'$ and $\mathfrak{p}' \subset \mathfrak{p}$ (Chapter V, § 2, no. 4, Theorem 3), contrary to the hypothesis. Conversely, if p is of height 1, there can exist no prime ideal $\mathfrak{p}'$ of B distinct from 0 and $\mathfrak{p}$ and such that $0 \subset \mathfrak{p}' \subset \mathfrak{p}$, otherwise $0 \subset \mathfrak{p}' \cap A \subset p$ and $\mathfrak{p}' \cap A$ would be distinct from 0 and p by virtue of Chapter V, § 2, no. 1, Corollary 1 to Proposition 1.

(2) B is aflat A-module. More precisely: