Normalization of a non-normal surface is not a point blowup

The polynomial $x^2-y^{2}z$ is irreducible: over $k(y,z)$ it is a quadratic in $x$ and can factor only if $z$ is a square, which it is not because its valuation at the prime $(z)$ is one; Gauss's lemma then gives irreducibility over $k[y,z]$. Thus $A$ is a domain, and $k[y,z]$ embeds in it while $x$ is algebraic over $k(y,z)$, so its fraction field has transcendence degree two and $S$ is a surface by [F8]. The partial derivatives are $(2x,-2yz,-y^2)$, whose common zero locus on $S$ is exactly $x=y=0$. At the generic point $P=(x,y)$ of this axis, $z$ is invertible and $A_P\cong k(z)[x,y{]}_{(x,y)}\mathrm{/}(x^2-z{y}^2)$. Before localization this is a one-dimensional affine domain by [F8]; the chain $(0)⊊(x,y)$ shows the local ring has dimension one. Its maximal ideal has cotangent-space basis the classes of $x$ and $y$, since the relation is in $(x,y{)}^2$; hence it is not regular by [F7].

Let $\psi :A\to B$ be the displayed homomorphism. After inverting $y$, it is an isomorphism $A_y\cong k[y,y^{-1},u]=B_y$, with inverse $u=x\mathrm{/}y$; since $A$ is a domain and $y\mathrm{\ne }0$, $\psi$ is injective. Thus we identify $A=k[y,yu,u^2]\subseteq B$ and their fraction fields agree, so $\nu$ is birational by [F4]. The ring $B=A[u]$ is generated as an $A$-module by $1,u$, because $u^2=z\in A$, so $\nu$ is finite by [F3]. The element $u$ is integral over $A$ by $u^2-z=0$, but $u\mathrm{\notin }A$: modulo $y$, the image of $A$ in $B\mathrm{/}(y)=k[u]$ is $k[u^2]$, which does not contain $u$. Hence $A$ is not integrally closed and $S$ is not normal. Since $B$ is integrally closed by [F5], every element of the common fraction field integral over $A$ is also integral over $B$ and therefore lies in $B$; conversely every element of $B$ is integral over $A$. Thus $B$ is the integral closure of $A$, and [F1] identifies $\nu$ as the normalization.

The inverse $u=x\mathrm{/}y$ proves that $\nu$ is an isomorphism over $D(y)$. For a geometric point on the $z$-axis with $z=z_0\mathrm{\ne }0$, the fibre is given by $y=0$ and $u^2=z_0$; it has two distinct points because the geometric residue field has characteristic not two. At the origin $m=(x,y,z)$, the local ring $A_m$ is not integrally closed either: if $u=a\mathrm{/}s$ with $a\in A$ and $s\in A\setminus m$, then $su=a$ in $B$. Reducing modulo $y$ gives $q(u^2)u=r(u^2)$ in $k[u]$, where $q(u^2)$ is the image of $s$ and has nonzero constant term, while $r(u^2)$ is the image of $a$. The left side is a nonzero polynomial containing only odd powers of $u$, and the right side contains only even powers, a contradiction. Thus the normalization is not an isomorphism over any neighbourhood of the origin. Together with the two-point fibres this shows that its isomorphism locus is exactly $D(y)$.

By [F6], each point blowup is an isomorphism away from its centre. Therefore a finite composition of point blowups is an isomorphism over the complement of the finite set of images of its centres in $S$. Any proper modification that is an isomorphism outside a finite set has the same property. But [step 3.1] shows that $\nu$ is not an isomorphism over the complement of any finite set, since infinitely many points of the $z$-axis have fibres with two distinct geometric points. Hence it is not any of these point-blowup modifications.