Understanding the conductor ideal of a ring. Consider the inclusion of a ring $A$ into its integral closure $B$. The conductor ideal $I$ is defined as $I:=\{a\in A~|~aB\subseteq A\}$. This is supposed to describe the locus where the normalization map $\textrm{Spec}(B)\rightarrow \textrm{Spec}(A)$ fails to be an isomorphism.
Can anyone explain to me why this is the case?
Thanks!
 A: Consider the extension as a short exact sequence of $A$-modules.
$$ 0 \rightarrow A\rightarrow \overline{A}\rightarrow \overline{A}/A\rightarrow 0$$
This is telling us that, to get an integrally closed ring, we must extend $A$ by $\overline{A}/A$. We can think of $\overline{A}/A$ as the obstruction to $A$ being integrally closed.
Localization commutes with taking integral closures, so for $p$ any prime ideal in $A$, $\overline{(A_p)}=\overline{A}_{\overline{A}p}$.  Since localization is flat, we see that
$$ \overline{(A_p)}/A_p = \overline{A}_{\overline{A}p}/A_p = (\overline{A}/A)_p$$
So $(\overline{A}/A)_p$ is simultaneously measuring...


*

*the local contribution at $p$ to the global obstruction $\overline{A}/A$, and

*the obstruction to $A_p$ being integrally closed.


In particular, $A_p$ is integrally closed (and $Spec(\overline{A}_p)\rightarrow Spec(A_p)$ is an isomorphism) at those primes where $(\overline{A}/A)_p=0$.  This is the complement of the support of $\overline{A}/A$ (thought of as a coherent sheaf, if you prefer).
An equivalent definition of the conductor $I$ is the annihilator of the $A$-module $\overline{A}/A$.  Thus, $Supp(I)=Supp(\overline{A}/A)$ is the complement of the set of primes where the normalization map is an isomorphism.
