# Blow-up along an ideal sheaf

Let $k^2=\operatorname{Spec} \; k[x,y]$ where $k$ is an algebraically closed field. Let $\mathcal{I}$ be the ideal sheaf defined by $(x,y)$. Then

$$Bl_{\mathcal{I}}k^2$$ is covered by two open charts $\operatorname{Spec} \; k[x, y/x] \cup \operatorname{Spec}\; k[y,x/y]$.

Q1: Why can each chart be described by $$\operatorname{Spec} \; k[x,y][t]/(tx-y) \mbox{ and } \operatorname{Spec} \; k[x,y][t]/(ty-x)?$$

Q2: Isn't $Bl_{\mathcal{I}}k^2=\operatorname{Proj}(\oplus_{i\geq 0} (Rx\oplus Ry)^i t^i)$?

Q3.a: Now let $k^3 =\operatorname{Spec} \; k[x,y,z]$ with $\mathcal{I}$ being defined by $(x,y,z).$ Then isn't $$Bl_{\mathcal{I}}k^3 = \operatorname{Spec} \; k[x,y/x,z/x] \cup \operatorname{Spec}\; k[y,x/y,z/y] \cup \operatorname{Spec}\; k[z,x/z,y/z]?$$

Q3.b: How can one see that the charts $$\operatorname{Spec}\; k[x,y,z][t_1, t_2]/(t_1 x - y, t_2 x-z)$$

$$\operatorname{Spec}\; k[x,y,z][t_1, t_2]/(t_1 y - x, t_2 y-z)$$

$$\operatorname{Spec}\; k[x,y,z][t_1, t_2]/(t_1 z - x, t_2 z-y)$$ also cover $Bl_{\mathcal{I}}k^3$?



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1. Let $A=k[x,y]$, then $$k[x,y/x]=k[x,y,y/x]=A[y/x]=A[t]/(y-tx).$$ For the last equality, see the comments below. Similar equalities apply to the other $k[y,x/y]$.
2. By definition the blowing-up of $k^2$ along the origin should be $$\operatorname{Proj}(\oplus_{i\geq0}I^i)\longrightarrow k^2,$$ where $I=(x,y)$.
3. The blowing-up of the affine space $k^n$ along the origin is (see Liu,8.1.13) $$\operatorname{Proj}(k[t_1,...,t_n][T_1,...,T_n]/(t_iT_j-t_jT_i))\longrightarrow k^n.$$ For your $n=3$ case, we have $$D_+(T_1)=\operatorname{Spec}(k[t_1,t_2,t_3][\frac{T_2}{T_1},\frac{T_3}{T_1}]/(t_2-t_1\frac{T_2}{T_1},t_3-t_1\frac{T_3}{T_1})),$$ and similar for $D_+(T_2),D_+(T_3)$. By re-writting the symbols carefully these are the three open charts you mentioned exactly.
Thanks Zhe! A question about the equality: $A[y/x]=A[t]/(y-tx)$. It's clear that in $A[y/x]$, $x$ is invertible. But in $A[t]/(y-tx)$, how do you know that $x$ is invertible? – math-visitor Jun 9 '12 at 23:14
@math-visitor, it doesn't matter $x$ is invertible or not: consider the homomorphism from $A[t]$ to A[y/x]: $t\mapsto y/x$. It is surjective with kernel contains $(y-tx)$. Try to show $A[t]/(y-tx)\rightarrow A[y/x]$ is injective. – Ch Zh Jun 9 '12 at 23:19
Thanks again! $\;\;\;$ – math-visitor Jun 9 '12 at 23:21