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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