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Let $X$ be a normal variety and $f:Y\rightarrow X$ a birational morphism, contracting exceptional divisors $E_1,\dots,E_k$ onto the singular locus of $X$, with $Y$ smooth.

In this situation is $Pic(Y)$ generated by $f^{*}Pic(X)$ and the exceptional divisors $E_1,\dots,E_k$?

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    $\begingroup$ The answer is negtive. Let $X$ be a cone over an elliptic curve, and let $f:Y\to X$ be the blowup of the vertex with exceptional divisor $E$. Then $f^*Pic(X)$ consists of classes that restrict to $0$ to $E$. Since the image of the restriction map $Pic(Y)\to Pic(E)$ is not cyclic, $Pic(Y)$ cannot be generated by $f^*Pic(X)$ and $E$. $\endgroup$ – Olivier Benoist May 17 at 8:55
  • $\begingroup$ Thank you very much for your example. On the other hand, if we replace the elliptic curve with a rational curve it works. Would it be enough to assume that $Y$ is rational? $\endgroup$ – AmMo May 17 at 10:47
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    $\begingroup$ Rationality of $Y$ doesn't help. For instance, let $X$ be a quadratic cone in $\mathbb{P}^3$ and take $Y$ to be the blowup of its vertex. Then $Pic(X) = \mathbb{Z}$, while $Pic(Y) = \mathbb{Z}^3$, so it cannot be generated by $f^*Pic(X)$ and the class of the exceptional divisor. $\endgroup$ – Sasha May 17 at 15:10
  • $\begingroup$ A similar question is discussed extensively here: mathoverflow.net/questions/122227/… $\endgroup$ – AG learner May 17 at 16:19

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