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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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I'm confused, do you mean a *dense nonempty open set $U \subseteq X$* instead of a subset of $Y$? –  Karl Schwede Jul 1 '12 at 0:45
    
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). –  vava Jul 1 '12 at 8:23
    
Ok. So the example seems to be incorrect, the original question stands. –  vava Jul 1 '12 at 11:39
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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. –  Jason Starr Jul 1 '12 at 15:06
    
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? –  Karl Schwede Jul 1 '12 at 19:43
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