Exceptional divisor of the blowup of A^3 at the origin is P^2

The three charts of the blowup are $\mathrm{Spec}A[I\mathrm{/}x]$, $\mathrm{Spec}A[I\mathrm{/}y]$, $\mathrm{Spec}A[I\mathrm{/}z]$ by [F1]; by [F2] the first is $A$-generated by $y\mathrm{/}x$ and $z\mathrm{/}x$, with $IA[I\mathrm{/}x]=xA[I\mathrm{/}x]$, the map from $k[x,s,t]$ sending $s$ to $y\mathrm{/}x$ and $t$ to $z\mathrm{/}x$ is injective, since after inverting $x$ it is the coordinate change $y=xs$, $z=xt$ in $k[x,x^{-1},y,z]$. Thus it is the polynomial ring $k[x,y\mathrm{/}x,z\mathrm{/}x]\cong k[x,s,t]$, whose spectrum is ${\mathbb{A}}_k^3$; symmetrically the other two charts are $k[x\mathrm{/}y,y,z\mathrm{/}y]$ and $k[x\mathrm{/}z,y\mathrm{/}z,z]$, each isomorphic to ${\mathbb{A}}_k^3$, glued by the transition ratios of [F1] and [F3].

The same blowup is presented inside ${\mathbb{A}}^3\times {\mathbb{P}}^2$ by the $2\times 2$ minors of $\left(\begin{array}{ccc}x& y& z\\ T_0& T_1& T_2\end{array}\right)$: on the chart $D_{+}(T_0)$ set $T_0=1$. The minor equations become $x{T}_1=y$, $x{T}_2=z$, the third minor being a consequence of these two. Eliminating $y,z$ gives exactly the first polynomial chart of step 1.1. The other charts are symmetric; their ratio transitions coincide with those of the blowup, so the chart isomorphisms glue to the claimed closed incidence subscheme, without needing a separate assertion about the kernel of the entire Rees presentation.

The exceptional divisor is cut by $x$ in the first chart, by $y$ in the second and by $z$ in the third: on each chart $I\mathcal{O}$ is generated by the corresponding variable, by [F2] and [F4]. By [F5] and [F6] it is ${\mathrm{Proj}}_k({\mathrm{gr}}_{(x,y,z)}k[x,y,z{]}_0)={\mathrm{Proj}}_{k}k[X,Y,Z]={\mathbb{P}}_k^2$, the projectivized cotangent space in the quotient convention, since the conormal space $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ is free of rank three over $k$.

Finally [F7] gives ${\mathcal{O}}_{\mathrm{Bl}}(E)=\mathcal{O}(-1)$ and ${\mathcal{O}}_{\mathrm{Bl}}(-E)=I{\mathcal{O}}_{\mathrm{Bl}}=\mathcal{O}(1)$; restricting the first identity to $E$ and using the identification $E={\mathrm{Proj}}_k{\mathrm{gr}}_{\mathfrak{m}}k[x,y,z{]}_0={\mathbb{P}}_k^2$ of step 3.1, the restricted twist is the standard ${\mathcal{O}}_{{\mathbb{P}}_k^2}(-1)$, so ${\mathcal{O}}_E(E)={\mathcal{O}}_{{\mathbb{P}}_k^2}(-1)$ in the quotient convention.