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In the question (Direct Image by a Blow up), follows the following statements

1) $\text{Sym}(A^{r}) \longrightarrow \bigoplus_{m \geq 0}I_{Y}^{m}$ corresponding to the closed immersion $\widetilde{X} \longrightarrow \mathbb{P}(A^{r})$.

2) The exceptional divisor $E$, correponds to the line bundle $\mathcal{O}_{\mathbb{P}}(A^{r})(-1)|_{\widetilde{X}}$

3) There is a canonical isomorphism $\mathcal{O}_{\widetilde{X}}(1) \simeq \mathcal{O}_{\widetilde{X}}(-E)$.

4) Where was Serre vanishing used?

5) Why can we identify $\pi_{*}\mathcal{O}_{\mathbb{P}(A^{r})} = \text{Sym}^{r}(A)$ and $\pi_{*}\mathcal{O}_{\widetilde{X}}(-nE) = I^{n}$?

I would like to understand such statements, so thank you in advance for your suggestions and references.

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1) The blowup of a sheaf of ideals $\mathcal{I}\subset \mathcal{O}_X$ is the relative Proj of the blowup algebra $\bigoplus_{m\geq 0} \mathcal{I}^m$. This is a definition, and we're just applying it to the affine case here where we replace $\mathcal{O}_X$ by $A$ and $\mathcal{I}$ by $I$. Choosing a set of generators $f_1,\cdots,f_r$ for $I$ gives you a surjection from $A^r\to I$, which turns in to a surjection of graded rings $\operatorname{Sym}(A^r)\to \bigoplus_{m\geq 0} I^m$, which corresponds to a closed immersion of their Projs by the general properties of that construction.

2) Every algebraic geometry book which covers blowing up should have a proof. For instance, Griffiths and Harris pg 184, Stacks 02OS, or this MSE question for a low-dimensional example (which generalizes).

3) This is just dualizing the statement of 2). Saying "$E$ corresponds to $\mathcal{O}(-1)$" means that $\mathcal{O}(E)\cong\mathcal{O}(-1)$, and dualizing gives $\mathcal{O}(-E)\cong\mathcal{O}(1)$ as requested.

4) Serre vanishing is not actually used in the proof: it's only meant to point out that the claim $\mathcal{O}_{\widetilde{X}}(-nE)\cong I_Y^n$ holds for any $X,Y$ once $n\gg 0$. For the specific situation of the proof, we are able to use the argument involving the exact sequence from the (now doubly) linked post to show this for $n\geq 1$ directly.

5) You've missed a $(n)$ here: the correct statement should be that global sections of $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n)$ are exactly $\operatorname{Sym}^n(A^r)$.

Both of these claims follow from the property that for nice rings $R$, the sheaf $\mathcal{O}(n)$ on $\operatorname{Proj} R$ has global sections exactly $R_n$, the degree-$n$ portion of $R$. See for instance the discussion in Stacks 01QG.

In the first case, the ring $R$ is $\operatorname{Sym}(A^r)$, with graded components $\operatorname{Sym}^n(A^r)$, and in the second, the ring is $\bigoplus_{m\geq 0} I^m$ with graded component $I^n$, and we use the isomorphism $\mathcal{O}_{\widetilde{X}}(n)\cong\mathcal{O}_{\widetilde{X}}(-nE)$.

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  • $\begingroup$ Hi, @KReiser. Thank you for your answer. See if I'm right : I think it is not necessary for $ $ to be a projective scheme, because I didn't see this hypothesis being used in the prove. $\endgroup$
    – Emanuell
    Jan 18, 2020 at 21:58
  • $\begingroup$ Again, @KReiser. The statement could be like this: Let $X$ be a smooth normal noetherian scheme and $Y \subset X$ a smooth closed subscheme and $$ \pi : \widetilde{X} \longrightarrow X$$ be the blow up along $Y$. Then.... $\endgroup$
    – Emanuell
    Jan 18, 2020 at 22:04
  • $\begingroup$ 1. You don't need to tag folks if you're commenting on their post. 2. You're correct that projectiveness is not a necessary hypothesis. 3. Smooth implies normal, so there's no need to assume something is "smooth and normal". $\endgroup$
    – KReiser
    Jan 18, 2020 at 23:28
  • $\begingroup$ Truth. Smooth $\rightarrow$ regular $\rightarrow$ normal. Another doubt. Is the push forward of line bundles a line bundle? Why the homomorphism of sheaves $$\pi_{*}\mathcal{O}_{\mathbb{P}(A^{r})} \longrightarrow \mathcal{O}_{\widetilde{X}}(-nE)$$ is an isomorphism? Thank you so much. $\endgroup$
    – Emanuell
    Jan 19, 2020 at 2:24
  • $\begingroup$ In general the pushforwards of a line bundle is not a line bundle: indeed, in the context of the original question, if $Y$ is not a divisor, $I_Y$ is not a line bundle, but $\pi_*\mathcal{O}_\widetilde{X}(-nE)=I_Y^n$. You're still missing the $(n)$ on the $\pi_*\mathcal{O}_{\Bbb P(A^r)}$. The claim you're curious is a general instance of the following fact: any surjective morphism of line bundles is an isomorphism. Proof: look on the level of stalks and use Nakayama's lemma. $\endgroup$
    – KReiser
    Jan 19, 2020 at 2:44

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