# Picard group of resolution

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 $$E_1,\dots,E_k$$?

• 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 $E$. – Olivier Benoist May 17 at 8:55
• 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? – AmMo May 17 at 10:47
• 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. – Sasha May 17 at 15:10
• A similar question is discussed extensively here: mathoverflow.net/questions/122227/… – AG learner May 17 at 16:19