A square-conic blowup has cubic-controlled singular successors

A square tangent conic means that the quadratic $q$ is the square of a linear form, so the exceptional fibre is twice the reduced exceptional line $C\cong {\mathbb{P}}_{\kappa }^1$; in the $x_1$-chart of the standard blowup charts put $y=x_2\mathrm{/}{x}_1$ and $w=z\mathrm{/}{x}_1$, so that dividing the relation by $x_1^2$ gives $w^2=x_{1}G$, where $G$ is the cubic expression in the chart coordinates with coefficients from $A$. Its restriction to $C$ is $h(y)$, obtained by setting $w=0$ and reducing coefficients modulo $\mathfrak{m}$. Homogenizing gives the cubic $H$ on $C$; thus $h$ is the restricted bracket, rather than the entire chart equation.

By the rational-domain definition [F10], $A$ is in the field or complete equicharacteristic finite-type class. The blowup is normal with trivial canonical module by [F4]. Every closed successor local ring $B$ has dimension two by [F11], has the same fraction field and is essentially of finite type over $A$, hence remains in that class and is rational by [F9]. Its canonical module is the stalk of the trivial module supplied by [F4]. Finally [F8] gives normal completion of $B$. These establish all hypotheses needed to apply [F3] at a singular successor.

At the generic point of $C$ the local ring is a discrete valuation ring with maximal ideal $(x_1,w)$; the exceptional divisor has multiplicity two, so $v(x_1)=2$. Since $(x_1,w)$ generates the DVR maximal ideal, $v(w)=1$, and $w^2=x_{1}G$ gives $v(G)=0$, that is, the bracket is a unit and $H$ is not identically zero on the reduced line; the $x_2$-chart gives the corresponding homogeneous cubic on $C$.

At a closed point of $C$ whose local maximal ideal is generated by $x_1$, $w$ and a lift $g$ of the prime polynomial on the affine line, nonvanishing of $H$ lets the relation express $x_1$ modulo the square of that maximal ideal; Nakayama's lemma then leaves at most two generators of the cotangent space, so the normal local surface ring is regular; hence every singular successor is a zero of $H$.

If the prime polynomial of such a point divides $H$ exactly once, its tangent quadratic has the form $w^2-x_1(\alpha x_1+\beta w+ug)$ with $u\mathrm{\ne }0$ in the residue field; this is not a scalar multiple of a square, because the vanishing of the $g^2$ coefficient would force the $g$-coefficient of a proposed linear square to vanish, contradicting the nonzero $x_{1}g$ coefficient; the nonsquare-conic termination lemma therefore resolves that branch, even when its residue field extends $\kappa$.

A remaining square-conic successor must be a multiple closed zero of $H$; a nonzero homogeneous cubic on ${\mathbb{P}}^1$ has total zero-degree three, so it has at most one multiple closed zero, and that zero has degree one; hence there is at most one possible square-conic successor and it is $\kappa$-rational.

Thus the zero scheme of $H$ on the reduced exceptional line contains all singular successors, simple closed zeros have nonsquare successor conic and finite resolution, and at most one $\kappa$-rational square-conic successor can occur; this is a branching statement and does not by itself terminate the continuing square branch, and the Axiom of Choice and the Axiom of Dependent Choice are inherited from the cited suppliers.