Let X be a seminormal variety and S be the singular locus of X. Is Blow$_S X=$ the normalisation of X?
Is the singular locus given by the conductor ideal?
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2$\begingroup$ That fails for a union of two $2$-planes in $\mathbb{A}^5$ whose intersection equals the origin. $\endgroup$– Jason StarrCommented Sep 5, 2017 at 15:26
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1$\begingroup$ You can actually take the $2$-planes in $\mathbb{A}^4$ in that example. $\endgroup$– Jason StarrCommented Sep 5, 2017 at 16:05
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1$\begingroup$ Continuing Jason's counterexample: $\mathrm{Blow}_SX$ contains the blow up of every branch of $X$ through $S$. So $\mathrm{Blow}_SX\to X$ wouldn't even be finite unless $S$ is of codimension $1$. Do you have that in mind? You also might have a better chance with Nash blowing up. $\endgroup$– Friedrich KnopCommented Sep 5, 2017 at 17:19
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$\begingroup$ S is codimension 1 $\endgroup$– user111251Commented Sep 5, 2017 at 17:38
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$\begingroup$ can you give me some reference where i can get good examples of nash blowup $\endgroup$– user111251Commented Sep 5, 2017 at 17:48
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1 Answer
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First, definitely the singular locus is not equal to the conductor. If $X$ is normal, the singular locus of $X$ is definitely not given by the conductor (which is the unit ideal and doesn't vanish anywhere).
For the first question however, Greco-Traverso actually studied blowing up the conductor in a seminormal variety $X$. For example, they showed the following.
Theorem Suppose $X$ is an S2 seminormal scheme and $B$ is the conductor. If the normalization of $X$ is factorial (for example regular), then the normalization is equal to the blowup of the conductor.
See section 7 of Greco-Traverso's On seminormal schemes for this and some slightly more general results. They also have some references to some other papers where blowing up the conductor is considered, for example Wilson, On blowing up conductor ideals
answered Sep 6, 2017 at 4:50