Higher-Dimensional Resolution of Singularities — Examples

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.

By [L1], $\tilde{C}$ is a smooth surface and $f:\tilde{C}\to C$ is proper and birational, isomorphic over $C\setminus \{0\}$, with $f^{-1}(0)=Q=E{\mathrm{\mid }}_{\tilde{C}}$ a smooth conic. Thus one ambient blowup resolves the cone. Its source charts are $k[x,v]$, $k[y,t]$, and $k[p,p^{-1},z]$, so no subsequent normalization is required. The conic is a smooth divisor and the strict transform meets the ambient exceptional divisor transversally.

The regular and smooth loci of $C$ are both $C\setminus \{0\}$: $D(x)$ and $D(y)$ cover this complement with rings $k[x,x^{-1},z]$ and $k[y,y^{-1},z]$, while the chain $(0)⊊(x,z)⊊(x,y,z)$ and $\dim C=2$ give vertex local dimension two, whereas its embedding dimension is three because $xy-z^2$ has no linear term. Since $C$ is normal, integral, affine (hence separated), finite type, and two-dimensional, [L3] applies. Characteristic zero makes $k$ perfect: an irreducible polynomial of positive degree has nonzero derivative of smaller degree, hence is relatively prime to its derivative and is separable. The surface theorem therefore supplies a smooth proper birational resolution isomorphic off the vertex. The explicit map in step 1.1 realizes these existence properties with a single blowup and the displayed conic; those extra descriptions come from the chart computation.

The affine integral finite-type scheme $C$ also satisfies [L4], which supplies a canonical smooth proper birational resolution isomorphic over $C\setminus \{0\}$. Thus both general theorems give the existence properties exhibited in step 1.1. The chart calculation identifies the explicit map $f$, and no identification with the canonical resolution is needed for this comparison.

To exhibit the freedom to use a nonminimal resolution, take the explicit $k$-rational point $q=(1:0:0)$ of the exceptional conic $Q$, and blow it up: $b:S={\mathrm{Bl}}_q\tilde{C}\to \tilde{C}$. By [L5], $S$ is smooth, $b$ is proper, has exceptional curve ${\mathbb{P}}_k^1$, and is an isomorphism away from $q$. The source is integral: $q$ is the origin in the chart $\mathrm{Spec}k[x,v]$ of $\tilde{C}$, so its two point-blowup charts are affine planes with dense complement of the exceptional curve, and off $q$ the source agrees with the integral surface $\tilde{C}$. Consequently $f\circ b$ is proper, isomorphic over $C\setminus \{0\}$, and birational, since this same dense open identifies its function field with that of the integral target. This resolution is nonminimal because the nontrivial morphism $b$ contracts its new exceptional curve back to the smooth surface $\tilde{C}$, which already resolves $C$. The smooth positive-dimensional exceptional fibre asserted in the example is the conic of the original single blowup in step 1.1.

The char-2 computation. In $k[x,y]$ with $\mathrm{char}k=2$ one has $\mathrm{\partial }(x^2)\mathrm{/}\mathrm{\partial }x=2x=0$, $\mathrm{\partial }(x^2)\mathrm{/}\mathrm{\partial }y=0$, $\mathrm{\partial }(y^2)\mathrm{/}\mathrm{\partial }x=0$ and $\mathrm{\partial }(y^2)\mathrm{/}\mathrm{\partial }y=2y=0$. Hence $\mathcal{D}(\mathcal{I})=\mathcal{I}$ by [F1], and iterating ${\mathcal{D}}^i(\mathcal{I})=\mathcal{I}$ for every $i$.

No tangent direction exists and the criterion fails. By [F2, F3] the marked ideal $(\mathcal{I},2)$ is of maximal order with support $\{0\}\mathrm{\ne }\mathrm{\varnothing }$. But ${\mathcal{D}}^1(\mathcal{I})=\mathcal{I}=(x^2,y^2)$, and every element of $(x^2,y^2)$ has order at least $2$ at the origin; hence no section of $\mathcal{D}(\mathcal{I})$ has multiplicity one, so $(\mathcal{I},2)$ admits no tangent direction at the origin, and in particular no hypersurface of maximal contact in the sense of [F4]. Equivalently, the defining criterion of maximal order fails: ${\mathcal{D}}^2(\mathcal{I})=\mathcal{I}\mathrm{\ne }{\mathcal{O}}_{{\mathbb{A}}^2}$, so the characteristic-zero equivalence between the order condition and ${\mathcal{D}}^{\mu }(\mathcal{I})={\mathcal{O}}_X$ breaks down. This refutes the first conjunct of the displayed statement.

For the second conjunct, take an algebraically closed field of characteristic two and $f=x^2+y{z}^3+z{w}^3+y^{7}w$. At a closed point its order is at most two, because its translated polynomial has coefficient $1$ on the square of the $x$-increment. It has order two exactly where $f$ and its first derivatives $z^3+y^{6}w$, $y{z}^2+w^3$, and $z{w}^2+y^7$ vanish. Substituting $(x,y,z,w)=(t^{32},t^7,t^{19},t^{15})$ makes $f$ a sum of four copies of $t^{64}$ and the three derivatives sums of two identical monomials, hence all vanish. The image curve therefore lies in the top locus; equality with the full top locus is unnecessary. Suppose a regular hypersurface germ at the origin contained this curve. Its completed equation would have a nonzero linear part $a_{x}x+a_{y}y+a_{z}z+a_{w}w$. Under substitution the linear terms have exponents $32,7,19,15$, respectively, each distinct and none a sum of at least two of these four positive weights. Thus no nonlinear monomial can cancel any nonzero linear coefficient, so the substituted series cannot vanish. A hypersurface regular at the origin must have a nonzero linear part, and this contradiction proves that no such hypersurface contains the top locus. This verifies the Narasimhan obstruction used here directly, following the polynomial and parametrization in Hauser, §14, Example 1, pp. 387–388.

Conclusion. Neither example disproves resolution of singularities in positive characteristic; both show that the characteristic-zero maximal-contact mechanism and its hypersurface criterion do not extend as stated. In particular this page's characteristic-zero theorem and its invariant cannot be quoted in positive characteristic, which is why no positive-characteristic resolution claim is made.