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Let $X$ be a scheme. Choose $C\subset X$ be a subscheme of $X$ and let $\mathcal{I}\subset \mathcal{O}$ be the corresponding ideal sheaf. Then $\mathcal{B}=\oplus_{d\ge0}\mathcal{I}^d$ is a sheaf of $\mathcal{O}$-algebra The blow-up of $X$ along $C$ is defined as $$ Y=Proj \mathcal{B} \rightarrow X. $$ My question is, how can one understand $Proj \mathcal{B}$ to $see$ geometric description of blow-up? More precisely, when both $X$ and $C$ are smooth complex variety, $Y$ is obtained by replacing $C$ by $\mathbb{P}(N_{C/X})$, but I cann't really see this description from $Proj \mathcal{B}$.

THank you for your help.

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I believe the blow-up is actually $Proj\;\mathcal{B}$. –  Jeff Tolliver Nov 18 '12 at 3:06
    
You are right. I'll fixe the typo. Thanks. –  M. K. Nov 18 '12 at 3:06

1 Answer 1

up vote 1 down vote accepted

The fiber of $Y\to X$ above $C$ is $$ Y\times_X C=\mathrm{Proj}(\mathcal B\otimes_{O_X} O_X/\mathcal I).$$ We have $$ \mathcal B\otimes_{O_X} O_X/\mathcal I = \oplus_{d\ge 0} (\mathcal I^d\otimes_{O_X} O_X/\mathcal I)=\oplus_{d\ge 0} (\mathcal I^d/\mathcal I^{d+1}).$$ As $C$ is locally complete intersection in $X$, $N_{C/X}:=\mathcal I/\mathcal I^2$ is locally free and $$\mathcal I^d/\mathcal I^{d+1} \simeq \mathcal{Sym}^d_{O_X}(N_{C/X})$$ (symetric power). Therefore $$ \mathcal B\otimes_{O_X} O_X/\mathcal I\simeq \mathcal{Sym}_{O_X}(N_{C/X})$$ (symetric algebra). So the fiber of $Y\to X$ above $C$ is the projective bundle $\mathbb P(N_{C/X})$.

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A great answer! This bridges the two definitions! Thank you very much. –  M. K. Nov 19 '12 at 9:04

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