Understanding the conductor ideal of a ring.

Solution 1:

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.