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Let $\pi : \tilde{\mathbb{P}^{3}} \longrightarrow \mathbb{P}^{3}$ be the blowup of $\mathbb{P}^{3}$ along a smooth algebraic curve $C$ with exceptional divisor $E$ We know that $\text{Pic}(\mathbb{P}^{3}) \simeq \mathbb{Z}$. How do I determine the Picard Group of $\tilde{\mathbb{P}^{3}}$?

I thank you for your help.

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1 Answer 1

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Whenever $\pi: Y \rightarrow X$ is a blowup of a smooth variety along a smooth subvariety, we have

$$\operatorname{Pic}(Y) = \pi^* \operatorname{Pic}(X) \oplus \mathbf Z[E] $$

where $E$ is the class of the exceptional divisor.

So in your case, $\operatorname{Pic} \tilde{\mathbf P}^3 = \mathbf Z^2$.

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  • $\begingroup$ Thank you very much, Asal Beag. $\endgroup$ Commented Mar 6, 2019 at 12:43
  • $\begingroup$ what is a reference for this statement? $\endgroup$ Commented Nov 21, 2019 at 10:15
  • $\begingroup$ Asal Beag, do you have any reference for the proof of that? $\endgroup$
    – rmdmc89
    Commented Feb 12, 2020 at 19:04

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