Blowing up vector bundles in the zero section Assume we are given a scheme $X$ (feel free to add all the needed hypotheses, at this point I’m working with smooth schemes, but the fewer is needed, the better) and a vector bundle $E$ over $X$. I will denote $|E|$ the total space of this vector bundle. I know that if we blow up $|E|$ in the zero section, we get that $Bl_B|E| = |\mathcal{O}_{\mathbb{P}(E)}(-1)|$ (if possible, I would like a proof or reference for this fact). The question is whether there is any relation between $\mathcal{O}_g(-1)$ and $p^{*}(\mathcal{O}_{\mathbb{P}(E)}(-1))$, where $g$ is the map of the blow up, and $p$ is the map from the total space of $\mathcal{O}_{\mathbb{P}(E)}(-1)$ to $\mathbb{P}(E)$. I would hope them to be isomorphic, but I don’t have any clue on how to prove it.
 A: You may have a look at remark 11.10 in Huybrechts' Fourier Mukai transform (or SGA6 or Fulton's Intersection Theory : appendix B5 and B6). I think your expectation is correct. In a nutshell a "complicated" proof would be the following:
i) $\mathrm{Pic}(|\mathcal{O}_{\mathbb{P}(E)}(-1)|) = \mathrm{Pic}(\mathbb{P}(E))$
ii) $\mathrm{Pic}(\mathbb{P}(E)) = \mathrm{Pic}(X) \oplus \mathbb{Z}. \mathcal{O}_{\mathbb{P}(E)}(-1)$
iii) $\mathrm{Pic}(\mathrm{Bl}_{0}(|E|)) = \mathrm{Pic}(|E|) \oplus \mathbb{Z}.\mathcal{O}_g(1)$
iv) $\mathrm{Pic}(|E|) = \mathrm{Pic}(X)$.
The identification $\mathrm{Bl}_0(|E|) = |\mathcal{O}_{\mathbb{P}(E)}(-1)|$ identifies the zero section of $\mathcal{O}_{\mathbb{P}(E)}(-1)$ with the exceptional divisor or $\mathrm{Bl}_0(|E|)$. Since the restriction of both $\mathcal{O}_g(-1)$ and $p^*\mathcal{O}_{\mathbb{P}(E)}(-1)$ to $\mathbb{P}(E)$ (seen as the zero section) is the relative tautological bundle, these two line bundles coincide up to the pull back of a line bundle on $X$. The formula in ii), ii) and iv) show that this line bundle is trivial
