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Let $X$ be a complex, affine variety. Let $\nu:\tilde X\to X$ be the normalization of $X$ and denote by $D\subseteq X$ the closed set of points where $\nu$ fails to be an isomorphism, i.e. $D$ is the non-normal locus of $X$.

Question 1. Under which conditions is $D\subseteq X$ pure of codimension one? I can remove all codimension one components of $D$ and end up with a variety whose nonormal locus has codimension at least two. So, if $X$ was Cohen-Macaulay then I would already know that $D$ must be pure of codimension one. Can I do with less than the Cohen-Macaulay assumption?

Question 2. Let $Z\subseteq D$ be an irreducible component of $D$. What can be said about the dimension of the irreducible components of $\nu^{-1}(Z)$?
More specifically, what can be said if $D$ is pure of codimension one - is $\nu^{-1}(D)$ also of codimension one in this case?

Thanks a lot in advance!

Jesko Hüttenhain
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  • I don't have an answer at the moment, just a nitpick: your description of the non-normal locus as the set of points with more than one preimage is incorrect, as the cuspidal curve shows. –  Apr 21 '15 at 15:41
  • @AsalBeagDubh: A very good nitpick, I fixed that. – Jesko Hüttenhain Apr 21 '15 at 15:55

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