Blowup charts of the quadric cone at its vertex

Integrality and dimension. Put $R=k[x,y,z]\mathrm{/}(xy-z^2)$ and $B=k[a,b]$. The map $R\to B$ given by $(x,y,z)\mapsto (a^2,b^2,ab)$ is injective: reducing monomials with $xy=z^2$ leaves monomials $x^i{z}^j$ for $i,j\ge 0$ and $y^i{z}^j$ for $i\ge 1,j\ge 0$, whose images are distinct monomials in $a,b$. Its image is $k[a^2,b^2,ab]$, a domain, and $B$ is finite integral over it because $a^2=x$ and $b^2=y$. Thus $\dim R=\dim B=2$, and $C$ is integral.

The three charts of the blowup are $U_x=\mathrm{Spec}k[x,u,v]$, $U_y=\mathrm{Spec}k[s,y,t]$, and $U_z=\mathrm{Spec}k[p,q,z]$ with the substitutions in [F1]. Their exceptional equations are $x=0$, $y=0$, and $z=0$, and globally $E\cong {\mathbb{P}}_k^2$.

Normality and the singular vertex. Under the involution $\sigma (a,b)=(-a,-b)$, the invariant polynomials in $B$ are precisely the even-total-degree monomials, hence $B^{\sigma }=R$. If $h\in \mathrm{Frac}(R)$ is integral over $R$, its monic equation also makes it integral over $B$; by [F6] it belongs to $B$, and, being fixed by $\sigma$, it belongs to $R$. Therefore $R$ and its localizations are integrally closed, so $C$ is normal. On $D(x)$ and $D(y)$ the coordinate rings are $k[x,x^{-1},z]$ and $k[y,y^{-1},z]$, respectively, so $C\setminus \{0\}$ is smooth. At $\mathfrak{m}=(x,y,z)$ the chain $(0)⊊(x,z)⊊\mathfrak{m}$ and step 1.1 give local dimension two, whereas $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ has basis $x,y,z$ because the defining relation is quadratic. The vertex is therefore singular by the regular-local-ring definition.

The total transforms of $xy-z^2$ on the three charts are $x^2(u-v^2)$, $y^2(s-t^2)$, and $z^2(pq-1)$. Modulo $u-v^2$, the first chart ring is $k[x,v]$, where multiplication by $x$ is injective; thus saturation by $x$ removes precisely the factor $x^2$. The same argument gives saturation $(s-t^2)$ in the second chart and $(pq-1)$ in the third, whose quotient is $k[p,p^{-1},z]$ and has no $z$-torsion. Hence these are exactly the strict-transform equations in (2).

The strict-transform chart rings are $k[x,v]$, $k[y,t]$, and $k[p,p^{-1},z]$, so they are smooth surfaces over $k$. Each is a domain and its open complement of the exceptional parameter is nonempty and dense. Those complements glue to the integral scheme $C\setminus \{0\}$ by [F4]; consequently their common dense open makes $\tilde{C}$ integral. This proves smoothness and assertion (2).

Intersecting the three equations with $E$ gives $u=v^2$, $s=t^2$, and $pq=1$ on its projective charts; these are the charts of the conic $\{xy=z^2\}\subseteq {\mathbb{P}}_k^2$. The first two cover this conic, since $x=y=0$ forces $z=0$, and each is an affine line. Thus the exceptional intersection is a smooth conic.

On $U_x$, the polynomial coordinate change $(x,u,v)\mapsto (x,u-v^2,v)$ identifies $E$ and $\tilde{C}$ with two coordinate hyperplanes. On $U_y$ use $(s,y,t)\mapsto (s-t^2,y,t)$ instead. These charts cover their intersection by step 3.2, so the divisors meet transversally everywhere; on $\tilde{C}$ their intersection is cut out by the coordinate $x$ or $y$. It is therefore a reduced effective Cartier divisor, proving (3).

The blowup $X\to {\mathbb{A}}_k^3$ is proper by [F10]. Its base change $X{\times }_{{\mathbb{A}}_k^3}C\to C$ is proper, and the closed inclusion of $\tilde{C}$ into that fibre product is proper, so $\tilde{C}\to C$ is proper. It is an isomorphism over the dense open $C\setminus \{0\}$ by [F4] and the strict-transform construction, hence birational by integrality and [F11]. Its smooth source and transverse smooth exceptional divisor prove (1) and the embedded-resolution assertion (4). The displayed source chart rings are integrally closed by [F6] and localization, so no further normalization is needed. The Axiom of Choice is inherited only from the cited constructions.