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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?


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Have you tried any examples with $A$ not integrally closed, such as $A = {\mathbf Z}[3i]$ or $A = k[X^2,X^3]$? – KCd Jun 3 '12 at 23:49
$\TeX$ tip: one never writes ~|~. Use \mid or a good ol' colon there. – Mariano Suárez-Alvarez Jun 4 '12 at 0:02
$\TeX$ tip continued: for proper usage of |, \mid, \vert etc see here. – Bill Dubuque Jun 4 '12 at 0:16
To follow up @KCd's point, if you post an example which you've attempted to work out and indicate where you are stuck, you are much more likely to get a helpful reply. (Plus, just as important, you're likely to figure some things out just in the process of working out and writing up your attempt at the example!) – Michael Joyce Jun 4 '12 at 1:07
up vote 6 down vote accepted

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.

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(PS: Happy Birthday!) – Greg Muller Jun 4 '12 at 18:58
Thanks Greg! Nice answer. :) Alternatively, we can do this directly and localize the conductor $I$ at a prime that doesn't contain it. Then the localized conductor is trivial and the result follows. – jrajchgot Jun 4 '12 at 20:26
Though the direct computation doesn't give the same intuition that your explanation does. So thanks again! – jrajchgot Jun 4 '12 at 20:32
Does the direct computation ever give the same intuition that the explanation does? ;) – Chris Cunningham Jun 4 '12 at 21:13

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