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Let $\nu:\tilde{X}\rightarrow X$ be the normalization of an integral scheme $X$. Let $Y$ be the closed subset of $X$ where $\nu$ fails to be an isomorphism, endowed with its reduced subscheme structure.

I'm trying to better understand $\nu^{-1}(Y)$. In particular, I have the following question:

Is it possible for there to be singular points of $\nu^{-1}(Y)$ that map, by $\nu$, to regular points of $Y$? If so, what is an example?

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Consider $X$ the affine variety defined by $y^2=x^2(z^3+x)$ over any of your favorite field $K$. The normalization is obtained by adding $u=y/x$ to $O(X)$. Actually the algebra of the new variety is $K[x, u, z]$ with the relation $u^2=z^3+x$ (note that $y$ disappears because of the relation $y=xu$), so this algebra is equal to $K[u,z]$ which defines the affine plane.

Now the non-normal locus $Y$ on $X$ is $x=y=0$ and is regular (it is an affine line) and its preimage in the normalization is defined by $y^2=z^3$ which contains a singular point $y=z=0$.

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    Great example. Easy a$n$d instructive. Thanks. (I'd upvote, but I don't have enough reputation on this site.)2011-10-24
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If $y \in Y$ is a regular point of $Y$, then you can find an affine open $V$ in $Y$ containing $y$ with $V$ smooth, since the singular locus is closed. But then the normalization map is an isomorphism on $V$ (and hence the entire smooth locus) because a regular local ring is integrally closed. This means that there's only one point $x \in X$ with $\nu(x)=y$ and that $x$ is a regular point of $X$.

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    I see now what you're asking. Intuitively, one shouldn't be able to create new singularities by normalizing.2011-10-22