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Assume that $X$ is a projective variety. Let $Pic(X)$ be its Picard group. Let $E$ be a vector bundle over $X$ say of rank $r$ (for example TX).

What is the picard group of the total space of $E$? can we say something about it?

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Let $f:E\to X$ be the natural map (where I have used the same letter to denote the total space of the vector bundle). Then the natural map $f^*:\mathrm{Pic}\, X\to\mathrm{Pic}\, E$ is an isomorphism. I would suggest that you look up a proof (which can be found in many places, not difficult) say in Fulton's book on Intersection Theory.

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    $\begingroup$ Just a little remark : the question is a bit more interesting if one want to compare $\rm{Pic}(\mathbb P(E))$ and $\rm{Pic}(X)$ and also probably in Fulton's book. $\endgroup$ Jan 14, 2018 at 16:53
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    $\begingroup$ Another remark : the Picard group is in general not homotopy invariant. If $X$ is not normal, it may very well happen that $\operatorname{Pic}(X)\rightarrow\operatorname{Pic}(X\times\mathbb{A}^1)$ is not an isomorphism. $\endgroup$
    – Roland
    Jan 14, 2018 at 17:18
  • $\begingroup$ @Roland You are absolutely right. I was tacitly assuming it, but should have mentioned it. $\endgroup$
    – Mohan
    Jan 14, 2018 at 18:32

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