2
$\begingroup$

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.

| cite | improve this question | | | | |
$\endgroup$
  • 3
    $\begingroup$ 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). $\endgroup$ – Olivier Benoist yesterday
  • $\begingroup$ @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? $\endgroup$ – yojusmath yesterday
  • 1
    $\begingroup$ The isomorphism statement follows from this post. $\endgroup$ – R. van Dobben de Bruyn yesterday
  • $\begingroup$ @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? $\endgroup$ – yojusmath 14 hours ago
  • $\begingroup$ 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. $\endgroup$ – R. van Dobben de Bruyn 12 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.