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Sep 19, 2015 at 5:37 vote accept Will Chen
Sep 16, 2015 at 2:20 answer added Will Sawin timeline score: 4
Sep 16, 2015 at 0:46 history edited Will Chen CC BY-SA 3.0
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Sep 16, 2015 at 0:31 history edited Will Chen CC BY-SA 3.0
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Sep 16, 2015 at 0:28 comment added Will Chen @JasonStarr I'm not necessarily asking to reconstruct the normalization. I suppose I shouldn't have used the word "reconstruct" since that assumes there is a unique/canonical/best candidate for a finite map $p$ extending $p^\circ$. My basic two questions are - is there a functor that is at least as good as normalization at "extending $p^\circ$?", and failing that, I was hoping for someone to explain the intuition behind how normalization behaves above the possibly ramified locus.
Sep 15, 2015 at 11:54 comment added Jason Starr The question is, in what category are you looking for a "functorial" way of reconstructing the normalization? Since normalization is not compatible with arbitrary base change, you should not expect a functor that extends, for instance, to the scheme of an Artin ring. So that obstructs Question 3. You mention smooth morphisms to $X$. If you work with the larger category of normal schemes together with a dominant morphism to $X$, it is straightforward to give a functor that is represented by $X'$: the functor of rational transformations to $Y^o$ over $X$.
Sep 14, 2015 at 17:19 comment added Will Chen @JasonStarr Thanks for pointing that out. I guess it only commutes with smooth base change.
Sep 14, 2015 at 17:14 history edited Will Chen CC BY-SA 3.0
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Sep 14, 2015 at 17:10 comment added Jason Starr I see that you want the complement of the $X^o$ to be a union of sections. So base change by $\mathbb{C}[t] \mapsto \mathbb{C}[s]$, $t\mapsto - s^2$. Then the complement of $X^o$ is the union of the sections $g_+$ and $g_{- }$ with $g_+^*s = +x$ and $g_{- }^*s = -x$.
Sep 14, 2015 at 17:06 comment added Jason Starr One thing to bear in mind: the formation of $X'$ is not compatible with base change. If $S$ is $\text{Spec}\ \mathbb{C}[t]$, if $X$ is $\text{Spec}\ \mathbb{C}[t,x]$, if $X'$ is $\text{Spec}\ \mathbb{C}[t,x,y]/\langle t - (y-x)(y+x) \rangle$, and if $X^o$, $Y^o$ are the corresponding etale loci, then the fiber over $t=0$ of the normalization $X'$ of $X$ in $Y^o$ is different from the normalization of the fiber over $t=0$ of $X$ in the fiber over $t=0$ of $Y^o$.
Sep 14, 2015 at 16:57 history asked Will Chen CC BY-SA 3.0