# Determining if a morphism is a blowup along a given subvariety

Let $$X,\tilde{X}$$ be two smooth projective varieties over $$\mathbb{C}$$, and let $$\pi:\tilde{X}\rightarrow X$$ be a projective morphism. Let us moreover assume that there exists a smooth closed subvariety $$Y\subset X$$, such that $$\pi$$ is isomorphism outside $$Y$$, and $$\pi^{-1}(Y)\rightarrow Y$$ is a projective bundle of rank = codim($$Y,X$$).

I want to show that $$\tilde{X}$$ is indeed the blowup of $$X$$ along $$Y$$. I have seen smoething along this line mentioined in a paper. Of course, there will be a map from $$\tilde{X}$$ to the blowup by the universal perty of blowups, but I can't show it to be an isomorphism. I know that any projective birational morphism is a blowup along some closed subscheme, but it's not clear from the proof that the closed subscheme is indeed $$Y$$.

Any help would be appreciated.

• The induced morphism $\tilde{X}\to Bl_Y(X)$ is a birational morphism of smooth projective varieties with the same Picard number (equal to the Picard number of $X$ plus one). This implies that it is an isomorphism (the equality of Picard numbers implies that there are no exceptional divisors). May 23, 2020 at 9:20
• @OlivierBenoist : Thank you for your prompt response. Could you please explain how you found the picard number of $\tilde{X}$? Also, why does same picard number imply isomorphism? Is there a reference to this? May 23, 2020 at 9:49
• The isomorphism statement follows from this post. May 23, 2020 at 15:51
• @R.vanDobbendeBruyn : In our case, isn't the exceptional subvariety for the map $\tilde{X}\rightarrow Bl_Y(X)$of codimension 1? Could you please explain how you are concluding from the post you have mentioned? May 24, 2020 at 11:54
• If the Picard ranks are the same there cannot be a codimension $1$ exceptional locus, because the exceptional divisor would necessary be linearly independent in Néron–Severi. May 24, 2020 at 14:35