Blowing up a point on a singular surface need not be smooth

The original ring is a domain: substitution $x=t^3,y=t^5$ identifies $k[x,y]\mathrm{/}(y^3-x^5)$ with $k[t^3,t^5]$. Division by the monic polynomial in $y$ reduces to $y$-degree at most two; the exponents $3i+5j$, $0\le j<3$, are distinct, proving injectivity. Similarly $k[x,s]\mathrm{/}(s^3-x^2)$ embeds as $k[t^3,t^2]$ by the distinct exponents $3i+2j$, $0\le j<3$. The first chart is $A[I\mathrm{/}x]=A[y\mathrm{/}x]$: putting $s=y\mathrm{/}x$, the relation $y^3-x^5=0$ becomes $x^3(s^3-x^2)=0$ in $A_x$, and since $A[I\mathrm{/}x]$ is the $A$-subalgebra of $A_x$ generated by $s$ (by [F2] applied to $a=x$), it is the domain $k[x,s]\mathrm{/}(s^3-x^2)$; its local ring at the origin $(x,s)=(0,0)$ is $R=k[x,s{]}_{(x,s)}\mathrm{/}(s^3-x^2)$. The second chart is computed with $t=x\mathrm{/}y$ and $u=y{t}^3$: from $y^3=x^5$ one gets $t^5{y}^2=1$, hence $u^2=y^2{t}^6=t$ and $y=u\mathrm{/}{t}^3=u^{-5}$, so $A[I\mathrm{/}y]=k[t,y]=k[u,u^{-1}]$, a localization of the polynomial ring $k[u]$, all of whose local rings are fields or discrete valuation rings and hence regular.

The local ring $R$ is one-dimensional: $(s^3-x^2)$ is a principal ideal generated by the nonzerodivisor $s^3-x^2$ in the two-dimensional local ring $k[x,s{]}_{(x,s)}$, so its minimal primes have height one by [F6], and $\dim R=1$ since $k[x,s{]}_{(x,s)}$ has dimension two; alternatively [F7]. Its associated graded ring is ${\mathrm{gr}}_{\mathfrak{m}}R\cong k[X,S]\mathrm{/}(X^2)$ by [F3], since the lowest homogeneous part of $s^3-x^2$ is $x^2$; this graded ring is not a domain, because $X\mathrm{\ne }0$ while $X^2=0$. Were $R$ regular local, [F4] would make its associated graded ring a polynomial ring, in particular a domain; hence $R$ is not regular, and the total space $X^{\mathrm{\prime }}$ is not regular.

The exceptional divisor is ${\mathrm{Proj}}_{\kappa (p)}({\mathrm{gr}}_{\mathfrak{m}}{\mathcal{O}}_{X,p})$ by [F5], where $\mathfrak{m}=(x,y)\subseteq A$; by [F3] its associated graded ring is $k[X,Y]\mathrm{/}(Y^3)$, the lowest homogeneous part of $y^3-x^5$ being $y^3$. Hence $E=\mathrm{Proj}k[X,Y]\mathrm{/}(Y^3)$: a single closed point, corresponding to the prime $(Y)$ of the graded ring $k[X,Y]\mathrm{/}(Y^3)$, whose local ring has length three and nonzero nilpotents. Thus $E$ is a nonreduced subscheme of ${\mathbb{P}}_k^1$ and is not smooth over $k$, by [F8].

To refute the stated surface claim, take the cylinder $S=\mathrm{Spec}(A[z])$ and its origin ideal $(x,y,z)$. On the $x$-chart the incidence substitution $y=xs$, $z=xv$ gives $k[x,s,v]\mathrm{/}(x^3(s^3-x^2))$; removing its $x$-power torsion gives $D=k[x,s,v]\mathrm{/}(s^3-x^2)$, a domain by step 1.1. This is the chart algebra. It has transcendence degree two, hence dimension two. The origin local ring has dimension two by the maximal-height formula [F7] applied to $D$, but cotangent dimension three because its relation lies in $(x,s,v{)}^2$. It is not regular. The tangent cone of $S$ at its origin is $k[X,Y,Z]\mathrm{/}(Y^3)$, so [F5] identifies its exceptional subscheme with its Proj. On $D_{+}(X)$ this has coordinate ring $k[Y\mathrm{/}X,Z\mathrm{/}X]\mathrm{/}((Y\mathrm{/}X{)}^3)$, with a nonzero nilpotent; hence the exceptional divisor is not smooth. The center is the regular point $\mathrm{Spec}k$. If its immersion into $S$ were regular, its rank-three conormal space and [F9] would give the reduced projective plane as exceptional divisor, contradicting the computed triple line. Thus the immersion is not regular. Thus the surface example refutes the universal claim, while steps 1.1–3.1 retain the curve calculation.