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In "3264 & All ThatIntersection Theory in Algebraic Geometry" by Eisenbud & Harris, we have the following :

Let $ A^{1}(X) $ denote the Picard group of a variety $ X. $

Let $ S $ be a smooth projective surface and $ \pi: \widetilde{S} \rightarrow S $ the blow-up of $ S $ at a point $ p$; let $ e \in A^{1}(\widetilde{S}) $ be the class of the exceptional divisor

  1. $ A(\widetilde{S}) = A(S) \oplus \mathbb{Z}e $ as abelian groups

  2. $ \pi^{*}\alpha \cdot \pi^{\beta} = \pi^{*}(\alpha\beta) $ for any $ \alpha,\beta \in A^{1}(S). $

  3. $ e \cdot \pi^{*}\alpha = 0 $ for any $ \alpha \in A^{1}(S). $

  4. $ e^{2} = -[q] $ for any point $ q \in E $(in particular, $ \operatorname{deg}(e^{2}) = -1 $).

A generalisation of (4) is the fact that in some sense $ E|_{E} = E^{2} = -H $ in $ E, $ and I would appreciate some insight as to why this is.

AG Novice
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  • Are you familiar with the fact that the exceptional divisor is given by $O(-1)$ on the blowup? (That is, $O(E)\cong O(-1)$.) – KReiser Jul 25 '20 at 21:34
  • @KReiser I have seen this before, in particular in the comments to the answer to this question: https://math.stackexchange.com/questions/2817595/restriction-of-the-exceptional-divisor-to-itself-as-a-line-bundle – AG Novice Jul 26 '20 at 00:47

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