Corollary 4

Let $A$ be an integral domain, $K$ its field of fractions and $S$ a multiplicative subset of $A$ such that $0 \notin S$.

(i) Let $B$ be a subring of $K$ which is integral over $A$ and let $\mathfrak{f}$ be the annihilator of the $A$-module $B/A$. Then $S^{-1}\mathfrak{f}$ is contained in the annihilator of the $(S^{-1}A)$-module $S^{-1}B/S^{-1}A$ and is equal to this annihilator if $B$ is a finitely generated $A$-module.

(ii) Let $A'$ be the integral closure of $A$. For $S^{-1}A$ to be integrally closed, it is sufficient that the annihilator $\mathfrak{f}$ of the $A$-module $A'/A$ meet $S$. This condition is also necessary if $A'$ is a finitely generated $A$-module.

(i) As $\mathfrak{f}B \subset A$, $(S^{-1}\mathfrak{f})(S^{-1}B) \subset S^{-1}A$ and hence $S^{-1}\mathfrak{f}$ is contained in $\operatorname{Ann}(S^{-1}B/S^{-1}A)$. If $B$ is a finitely generated $A$-module, the equation $S^{-1}\mathfrak{f} = \operatorname{Ann}(S^{-1}B/S^{-1}A)$ is a special case of formula (9) of Chapter II, § 2, no. 4, $S^{-1}B/S^{-1}A$ being canonically identified with $S^{-1}(B/A)$.

(ii) By Corollary 1 $S^{-1}A'$ is the integral closure of $S^{-1}A$. As the relations $\mathfrak{f} \cap S \neq \varnothing$ and $S^{-1}\mathfrak{f} = S^{-1}A$ are equivalent (Chapter II, § 2, no. 5, Remark) (ii) is an immediate consequence of (i).

If $B$ is a subring of $K$ which is integral over $A$, the annihilator $f$ of $B/A$ (equal by definition to the transporter $A : B$ (Chapter I, § 2, no. 10)) is sometimes called the conductor of $B$ in $A$.