Let $X$ be a smooth projective variety of Picard number one, and let $f:X\dashrightarrow X$ be a birational automorphism which is not an automorphism. 


Must $f$ necessarily contract a divisor?



Birational automorphisms of varieties of Picard number one
In contrast to an earlier answer on this question, I claim the answer here is **yes**.


In the accepted answer to this question:


[Pseudo-automorphisms on Fano varieties](https://mathoverflow.net/questions/179070/pseudo-automorphisms-on-fano-varieties?rq=1)


abx explains that for any smooth variety $X$ of Picard number 1, any pseudo-automorphism of $X$ (i.e. a birational automorphism which is an isomorphism in codimension 1) must in fact be an automorphism.


So it remains to argue that if $f$ does not contract any divisor, then it is a pseudo-automorphism. The only thing to check is that $f^{-1}$ does not contract a divisor either. You can do this by looking at a resolution of $f$ : 


$$ X \leftarrow^p \widetilde{X} \rightarrow^q X $$


where $\widetilde{X}$ is smooth: the numbers of $p$-exceptional prime divisors and $q$-exceptional prime divisors must be equal, but the hypothesis that $f$ doesn't contract a divisor says that every $q$-exceptional divisor is $p$-exceptional. Hence the two sets are the same, and so $f^{-1}$ doesn't contract a divisor either. 


