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Let us work over the complex projective space: consider a smooth variety $X$ and a subvariety $Y$. I learnt that, if we do the blow-up of $X$ with center $Y$, we obtain a new variety $\tilde{X}$, along with a map $\pi: \tilde{X}\to X$, which is an isomorphism outside the exceptional locus, that is $Y$.

Altough I don't have a precise reference, it's been told me that the exceptional divisor of $Y$, that is the inverse image $\pi^{-1}(Y)$, coincides with the projective bundle of the normal bundle, that is,

$$\tilde{Y}=\pi^{-1}(Y)\simeq \mathbb{P}(\mathcal{N}_{Y\mid X}^\vee)=(\mathcal{N}_{Y\mid X}\setminus Y)/\sim,$$

where $\sim$ is the standard action of $\mathbb{C}$.

Questions:

  • What is a good reference of this construction? I know it is the content of Theorem II.8.24 of Hartshorne's Algebraic geometry, but without a knowledge of scheme theory (and proj construction, and coherent sheaves) it is a bit difficult, so maybe there's a more accessible text;
  • In pp. 86-87 of these notes ( https://www.math.ens.fr/~debarre/M2.pdf ), we start with a rational curve $\Gamma^+$ in $X^+$ with normal bundle $\mathcal{O}(-1)\oplus\mathcal{O}(-2)$: then the authors does the blow-up along $\Gamma^+$, and it claims that the exceptional divisor is $$S^+_1=\mathbb{P}(\mathcal{O}\oplus \mathcal{O}(1))$$ but using the above formulas should be $\mathbb{P}(\mathcal{O}(1)\oplus \mathcal{O}(2))$: what am I missing?
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Your statement is missing a hypothesis: we also need $Y$ to be nonsingular. Anyway, an alternate reference is section 22.3 of Vakil's "The Rising Sea: Foundations of Algebraic Geometry", which is generally considered a more accessible reference than Hartshorne. Vakil doesn't give a proof of the statement you want—it's Exercise 22.3.D—but the leadup to the exercise may give some useful guidance on how to prove it.

Unfortunately, I'm not aware of a reference for this statement that doesn't use the language of schemes, the Proj construction, and quasicoherent sheaves at least to some extent.

To answer your second question, the two projective bundles are isomorphic, because projective bundles don't change if you tensor by a line bundle. This is exercise 17.2.G in Vakil's notes, and Lemma II.7.9 in Hartshorne. Another reference for the general statement is Stacks Project, tag 02NB. In particular, for all $c \in \mathbb{Z}$, we have $$\mathbb{P}(\mathcal{O}(a) \oplus \mathcal{O}(b)) \cong \mathbb{P}(\mathcal{O}(a + c) \oplus \mathcal{O}(b + c)).$$

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