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Feb 24, 2018 at 15:38 comment added Jason Starr @byu. "Are there obstructions in dimensions $1$ and $2$?" Yes. In dimension $2$, the Grothendieck-Lefschetz Theorem on algebraic fundamental groups implies that $X'$ has trivial algebraic fundamental group. Thus the, degree of the finite part of the Stein factorization of $f$ is an upper bound on order of the algebraic fundamental group of $X$. Thus, if the algebraic fundamental group of $X$ is infinite, then there exists no pair $(X',f:X'\to X)$.
Feb 24, 2018 at 14:33 comment added byu Are there obstructions in dimensions 1 and 2?
Feb 24, 2018 at 14:24 comment added Jason Starr There does not exist such a pair $(X',f:X'\to X)$ if the dimension of $X$ is at least $3$ and the Picard rank of $X$ is $\geq 2$. By the Grothendieck-Lefschetz Theorem on Picard groups, the restriction homomorphism of Picard groups, $\text{Pic}(\mathbb{P}^n)\to \text{Pic}(X')$, is a bijection. In particular, every invertible sheaf on $X'$ is either trivial, ample, or antiample. So if there is such a morphism, it is finite of some degree $d>0$. Now you can use norms to conclude that $\text{Pic}(X)[1/d]\to \text{Pic}(X')[1/d]$ is injective. But $\text{Pic}(X')$ is rank $1$.
Feb 24, 2018 at 13:56 history edited user119470
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Feb 24, 2018 at 13:34 history asked user119470 CC BY-SA 3.0