Point Blowup Resolution on Arbitrary Regular Surfaces — Examples

The point $p=(0,0)$ is a singular point of the cusp: at $p$ the local ring $R={\mathcal{O}}_{Z,p}$ has dimension one and embedding dimension two, hence is not regular by [F1]. Therefore $Z$ is not regular.

The morphism $\nu :{\mathbf{A}}_k^1\to Z$, $t\mapsto (t^2,t^3)$, is finite: the image $k[t^2,t^3]\subseteq k[t]$ is a $k$-subalgebra over which $k[t]$ is generated as a module by $1$ and $t$ (because $t^2$ and $t^3$ lie in the subalgebra). It is birational: the induced map of fraction fields is $k(t^2,t^3)\hookrightarrow k(t)$, an equality since $t=t^3\mathrm{/}{t}^2$, and $\nu$ is an isomorphism away from the origin with inverse $(x,y)\mapsto y\mathrm{/}x$. It is bijective on scheme points: it is an isomorphism on the complement of the origin by the displayed inverse, and its origin fibre has coordinate ring $k[t]\mathrm{/}(t^2,t^3)$, supported at the single point $t=0$. Normality in [F3] follows directly from unique factorization: a reduced fraction $a\mathrm{/}b$ satisfying a monic integral equation has $b\mid a^n$ after denominators are cleared, so coprimality makes $b$ a unit. Since ${\mathbf{A}}_k^1$ is therefore normal, $\nu$ is a finite normalization of $Z$; by the uniqueness in [F2] the normalization of $Z$ is finite.

On the blowup chart $y=xt$, the strict transform has equation $t^2-x=0$, hence coordinate ring $k[t]$. On the other chart $x=ys$, its equation is $1-y{s}^3=0$, which makes $s$ and $y$ invertible; this portion lies in the overlap with the first chart. Thus the whole strict transform is the regular affine line and its map to $Z$ is $t\mapsto (t^2,t^3)$, the normalization of step 1.2 (Affine blowup standard charts and overlaps).

The curve $Z$ therefore has finite normalization by step 1.2 and is not regular by step 1.1, so the false claim is refuted by the explicit witness $(Z,\nu )$. Moreover $\nu$ is a finite birational morphism which is not an isomorphism over the singular point: if it were an isomorphism at the origin, then $Z$ would be regular at the origin, contradicting step 1.1. Thus finite normalization is not a substitute for the blowup procedure: the regularization theorem requires point blowups, and step 2.1 shows explicitly that one point blowup makes its strict transform regular (Regularization of a one-dimensional integral curve with finite normalization by point blowups, Blowing up a non-regular point strictly increases the finite normalization subalgebra).

The parametrization identifies $R=k[x,y]\mathrm{/}(y^2-x^3)$ with $k[t^2,t^3]$: modulo the monic relation every polynomial has the unique form $f(x)+yg(x)$, and its image $f(t^2)+t^{3}g(t^2)$ is zero only when both polynomials vanish, since their monomials have disjoint even and odd exponents. This ring is a domain, finite integral over $k[x]$ by its monic equation in $y$, and therefore has dimension one (Injective integral extensions preserve Krull dimension). At the origin its maximal ideal has the independent images of $x,y$ as a cotangent basis, because the relation has no linear term, so its embedding dimension is two and the point is not regular. Away from the origin, $x$ is invertible and $t=y\mathrm{/}x$ gives $R[1\mathrm{/}x]=k[t,t^{-1}]$, whose local rings are regular (localisation and polynomial extension of regular rings). Thus the origin is the unique singular point.

In the chart $y=xt$, the strict transform is $Z_1=V(t^2-x)$ and the exceptional curve is $E=V(x)$. The equation is linear in $x$, so this strict transform is regular. In the other chart $x=ys$, the strict-transform equation is $1-y{s}^3=0$, which forces $y$ and $s$ to be invertible; consequently that entire chart portion lies in the overlap with the first chart and adds no point above the origin. Thus the first chart describes the whole strict transform of the affine cusp. Its intersection with $E$ has coordinate ring $k[t]\mathrm{/}(t^2)$, supported at $q=(0,0)$ and of length two, so $m_q(Z_1\cap E)=2$. The curve $Z_1$ is regular and has curve multiplicity one at $q$, whereas the original cusp has multiplicity two because its lowest-degree equation is $y^2$.

The strict transform is ${\mathbf{A}}_k^1$, parametrized by $t$ with $x=t^2$ and $y=t^3$. The map $k[t^2,t^3]\hookrightarrow k[t]$ is finite, because $1,t$ generate the latter as a module, and is birational, since $t=y\mathrm{/}x$ in the fraction field. Its source is normal: by [F5] it is a UFD, and if a reduced fraction $a\mathrm{/}b$ in its fraction field is integral, multiplying a monic integral equation by $b^n$ shows that $b$ divides $a^n$; coprimality forces $b$ to be a unit. Hence the finite birational parametrization is the normalization by [F3]. It is bijective on scheme points: it is an isomorphism where $x\mathrm{\ne }0$, and the fibre over the origin has support only $t=0$; there are no other points with $x=0$ on the cusp. This proves the claimed affine normalization and curve-multiplicity drop.

The contact of $Z_1$ with $E$ has multiplicity $2$, so the support is not yet SNC by [F2]. Blow up the point $q$ in the chart of step 1.2, writing $x=tu$: the strict transform of $Z_1=V(t^2-x)$ becomes $V(t^2-tu)=V(t(t-u))$, with strict transform $V(t-u)$; the strict transform of $E=V(x)$ becomes $V(u)$; and the new exceptional curve is $V(t)$. These are three distinct lines through the origin, meeting pairwise only there with multiplicity $1$ (their three tangent directions are distinct): a transverse triple point.

Blow up that triple point. Each of the three lines is regular there, and pairwise they meet with multiplicity $1$, so the strict transform of each meets the new exceptional curve in its own point with multiplicity $1$ and the three strict transforms become pairwise disjoint over the blown-up point. The resulting support has regular components, and at every closed point at most two components pass, meeting transversally; hence it is a strict normal crossings divisor by [F2], reached after three point blowups in total.

The projective cubic $\bar{Z}=V(Y^{2}Z-X^3)$ is regular away from its cusp. On the affine chart $Z=1$ outside the origin, $x$ is invertible and $t=y\mathrm{/}x$ identifies the curve with $\mathrm{Spec}k[t,t^{-1}]$. The only point at infinity is $[0:1:0]$; on the chart $Y=1$ its equation is $v-u^3=0$, whose coordinate ring is $k[u]$. Thus the defect is supported only at the cusp and is the module $k[t]\mathrm{/}k[t^2,t^3]$, with basis the class of $t$. By [F3], its global section space has dimension one, so ${\delta }_k(\bar{Z})=1$. The projective strict transform after the first blowup is regular at the cusp by step 1.2 and is unchanged elsewhere. Its map to $\bar{Z}$ is the intrinsic point blowup (Strict transforms of closed subschemes are blowups of the subscheme), hence finite by The blowup of a one-dimensional integral Noetherian scheme at a closed point is finite; it is birational and normal, so it is the normalization of $\bar{Z}$ and has defect zero. Formula [F4] applies on ${\mathbb{P}}_k^2$, a regular proper surface with ample $\mathcal{O}(1)$, at the $k$-rational cusp with residue degree $r=1$ and multiplicity $m=2$. It gives ${\delta }_k({\bar{Z}}^{\mathrm{\prime }})=1-1\left(\genfrac{}{}{0px}{}{2}{2}\right)=0$, agreeing with the direct calculation.

Collecting: one point blowup makes the strict transform the regular normalization of the cusp and drops the multiplicity at the point over the origin from $2$ to $1$; three point blowups make the total-transform support SNC. The example therefore realizes the regularization theorem after one step and the embedded SNC conclusion after finitely many, with the defect drop ${\delta }_k=1\to 0$ confirming the formula $rm(m-1)\mathrm{/}2=1$.

The parametrization of [F2] identifies $R=k[x,y]\mathrm{/}(y^2-x^2(x+1))$ with $k[t^2-1,t(t^2-1)]$. Indeed every polynomial reduces uniquely to $a(x)+yb(x)$, and its image $a(t^2-1)+t(t^2-1)b(t^2-1)$ is zero only when both polynomials vanish: the first term has even powers of $t$, the second odd powers, and $k[t]$ is a domain. Thus $R$ is a domain, finite integral over $k[x]$, and has dimension one by Injective integral extensions preserve Krull dimension. At the origin the local maximal ideal has the independent classes of $x,y$ modulo its square, since the defining equation has no linear term; the embedding dimension is two, so this closed point of the integral curve is not regular. Elsewhere $x$ is invertible, because $x=0$ on the curve forces $y=0$; putting $t=y\mathrm{/}x$ gives $R[1\mathrm{/}x]=k[t,1\mathrm{/}(t^2-1)]$, a localization of the regular affine line (localisation and polynomial extension of regular rings). Therefore the origin is the unique singular point over every field of characteristic different from two, including characteristic three.

The completed ambient local ring is the ring of formal power series in $x,y$: compatible residues modulo $(x,y{)}^n$ specify its coefficients, and denominators with nonzero constant term are inverted by formal geometric series. Construct the formal power series $u=1+{\sum }_{n\ge 1}{a}_n{x}^n$ recursively by $u^2=1+x$. The coefficient equation gives $a_1=1\mathrm{/}2$, and for each $n>1$ determines $a_n$ by $2a_n+{\sum }_{1\le i<n}{a}_i{a}_{n-i}=0$, so only powers of two need to be inverted. Thus this construction works in every allowed characteristic. In the ring of formal power series in $x$ and $y$, the equation factors as $(y-xu)(y+xu)$. Each factor has nonzero linear term $y-x$ or $y+x$, and its quotient is the formal power-series ring in $x$, a DVR with uniformizer $x$; thus it defines a regular formal branch; their ideal together is $(x,y)$ because two and $u$ are units. Hence the branches have distinct tangent directions and intersection length one. They are formal branches of the single integral global curve established in step 1.1.

On the $x$-chart $y=xt$, the exceptional curve is $E=V(x)$ and the strict transform is $V(t^2-x-1)$, a regular curve with parameter $t$. Its exceptional intersection is $V(x,t^2-1)$, the two reduced points $t=1,-1$ because two is invertible; each has intersection length one. On the other chart $x=ys$, the strict-transform equation is $1-s^2-y{s}^3=0$. It makes $s$ invertible, since $s^2(1+ys)=1$, so this entire portion belongs to the overlap with the $x$-chart, where $t=1\mathrm{/}s$. Thus the $x$-chart describes the entire strict transform, including all points above the origin, and its two transverse exceptional contacts are precisely the two formal tangent directions of step 2.1.

The strict transform is ${\mathbf{A}}_k^1$ with $x=t^2-1$ and $y=t(t^2-1)$, so its map to $Z$ is the finite birational map of [F2]. The normality invoked there follows directly from the UFD assertion: for an integral reduced fraction $a\mathrm{/}b$, a monic equation implies $b\mid a^n$ after clearing denominators, and coprimality forces $b$ to be a unit. Hence the map is the normalization. The original curve has multiplicity two at the origin, from its lowest-degree term $y^2-x^2$; the strict transform is regular and has curve multiplicity one at each of the two points above the origin. The contact calculation of step 3.1 is direct and does not apply the regular-source multiplicity lemma to the singular original curve or to nonexistent global branch components.

By [F1] the surface $S_1$ is regular and $E$ is a regular curve. The support of the total transform of $Z$ is $Z^{\mathrm{\prime }}\cup E$: two regular curves meeting exactly in the two points of step 3.1, each with multiplicity $1$. At every closed point at most two components pass, and when two pass they meet transversally, so by [F3] the support is a strict normal crossings divisor; therefore the resolution sequence of Embedded strict-normal-crossings resolution of a reduced curve on a regular surface terminates after this single blowup, and no further blowups are needed.