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Let $Y$ be a nonsingular subvariety of a normal, Cohen-Macaulay variety $X$. Further, let $\pi:X'\to X$ be the blowup of $X$ along $Y$.

Question: Is there a formula for the Picard group of $X'$ involving the Picard group of $X$ and the exceptional locus of the blowup?

For example when $X$ is smooth it is well known that $\textbf{Pic}\, X'\cong \textbf{Pic}\, X \oplus \mathbb{Z}$ and the isomorphism is induced by the pullback $\pi ^*: \textbf{Pic}\, X \to \textbf{Pic}\, X' $ and the class of the exceptional divisor.

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    $\begingroup$ A major problem to do this is that the exceptional locus maybe reducible and the irreducible components may not be Cartier divisor. Or some are and some are not. $\mathrm{Pic} X'$ will be generated by $\mathrm{Pic} X$ and the collection of Cartier divisors supported on the exceptional locus, but this will likely not be free in general (in other words, you should not expect a direct sum here always). $\endgroup$
    – Sándor Kovács
    Commented Nov 5, 2019 at 0:55
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    $\begingroup$ Your assumptions of $X$ being CM and $Y$ being non-singular are not really the best for this purpose. What you need to know is how $Y$ is embedded in $X$. For instance, $Y$ could be singular, but if it is a local complete intersection, then you have much better control. In fact, if $Y$ is also irreducible, then you can expect a formula similar to the smooth case. $\endgroup$
    – Sándor Kovács
    Commented Nov 5, 2019 at 0:58
  • $\begingroup$ If $Y$ is a locally-complete intersection in $X$, then the usual formulas for Chow groups of the blow up will take place, with the same proof as in the smooth case. The point is that there is a well-defined pull-back from $X$ to $Y$ and from $X$ to $X'$ in this case. This applies in particular to the Picard group. $\endgroup$
    – Evgeny Shinder
    Commented Nov 5, 2019 at 21:43
  • $\begingroup$ To elaborate on Sandor's comment on the non-lci case: if $X$ is a nodal threefold, and $Y = Sing(X)$, then the blow up $X'$ is the resolution of singularities of $X$, and the exact relation between $Pic(X)$ and $Pic(X')$ depends on the positions of the nodes (this phenomenon goes under the name of a defect of a nodal threefold), so there will be no simple formula in general. $\endgroup$
    – Evgeny Shinder
    Commented Nov 5, 2019 at 21:53
  • $\begingroup$ The answer here gives an interesting example : mathoverflow.net/questions/164214/… $\endgroup$
    – Watson
    Commented Sep 23, 2020 at 15:51

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