Is it true that for any proper dominant morphism $\pi:X \rightarrow Y$, there is an open set $U\subset Y$ such that $\pi|_U$ is a composition of smooth morphisms and radicial morphisms (assuming $X$ is regular, also can assume that $X$ and $Y$ are of finite type over a finite field)?
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2$\begingroup$ I'm confused, do you mean a dense nonempty open set $U \subseteq X$ instead of a subset of $Y$? $\endgroup$– Karl SchwedeJul 1, 2012 at 0:45
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$\begingroup$ Yes, I do mean dense, nonempty open set. (Nonempty is the hard one I guess). But I today learnt that this is not possible. (Need to look at example myself, before I say more). However, in that case I will be happy with a "radicial modification" that is a morphism $X'\rightarrow Y'$ satisfying the property in question such that we have maps $X\rightarrow X'$ and $Y\rightarrow Y'$ radicial, making the diagram commute. (Or even if it is $X'\rightarrow X$ and $Y'\rightarrow Y$, with the same property). $\endgroup$– vavaJul 1, 2012 at 8:23
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$\begingroup$ Ok. So the example seems to be incorrect, the original question stands. $\endgroup$– vavaJul 1, 2012 at 11:39
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2$\begingroup$ Valbhav -- You still need to make $U$ an open subset of $X$ rather than $Y$, otherwise just take $X$ to be any singular variety and take $Y$ to be a point. $\endgroup$– Jason StarrJul 1, 2012 at 15:06
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$\begingroup$ In your comment above, you seem to be allowing radicial maps to and from things. What if the maps go the other way, i.e. can we assume that $X$ and $Y$ are reduced? $\endgroup$– Karl SchwedeJul 1, 2012 at 19:43
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