Proposition 15

Let A and B be Krull domains such that B contains A and is aflat A-module. Then: (a) condition (PDE) of Proposition 14 is fulfilled; (b) for every divisorial ideal $a \in A$, Ba is the divisorial ideal of B which corresponds to the divisor $i(\operatorname{div}_A(a))$.

To show (a), suppose that there exists a prime ideal $\mathfrak{p}$ of B of height 1 such that $\mathfrak{p} \cap A$ is neither 0 nor of height 1. Take an element $x \neq 0$ in $\mathfrak{p} \cap A$. The ideals $p_i$ of A of height 1 which contain x are finite in number and none contains $\mathfrak{p} \cap A$; there therefore exists an element y of $\mathfrak{p} \cap A$ such that $y \notin p_i$ for all $i$ (Chapter II, § 2, no. 1, Proposition 2). Thus $\operatorname{div}_A(x)$ and $\operatorname{div}_A(y)$ are relatively prime elements of the ordered group P(A), so that $\sup(\operatorname{div}_A(x), \operatorname{div}_A(y)) = \operatorname{div}_A(x) + \operatorname{div}_A(y) = \operatorname{div}_A(xy)$; as the ideals $Ax \cap Ay$ and $Axy$ are divisorial, we deduce that $Ax \cap Ay = Axy$. Since B is a flat A-module, $Bx \cap By = Bxy$ (Chapter I, § 2, no. 6, Proposition 6). This implies that $\sup(v_{\mathfrak{p}}(x), v_{\mathfrak{p}}(y)) = v_{\mathfrak{p}}(xy) = v_{\mathfrak{p}}(x) + v_{\mathfrak{p}}(y)$, which contradicts the inequalities $v_{\mathfrak{p}}(x) > 0, v_{\mathfrak{p}}(y) > 0$ (which hold since x and y are in $\mathfrak{p}$). Thus (a) has been proved by reductio ad absurdum.

We now show (b). If $a$ is a divisorial ideal of A, it is the intersection of two fractional principal ideals (no. 5, Corollary 2 to Proposition 9), say

where $a, b,$ dare in $A^*$; as B is flat over A, $Ba = d^{-1}(Ba \cap Bb)$ (Chapter I, § 2, no. 6, Proposition 6), which shows that Ba is divisorial. This shows also that $\operatorname{div}_B(Ba) = \sup(\operatorname{div}_B(a), \operatorname{div}_B(b)) - \operatorname{div}_B(d)$; using Proposition 14 (a) and (b), it is seen that