Intersection Products on Smooth Projective Surfaces — Examples

The Euler characteristic of twists. By [F2] only the degrees $q=0$ and $q=2$ contribute to $\chi (X,\mathcal{O}(m))={\sum }_q(-1{)}^q{\dim }_k{H}^q(X,\mathcal{O}(m))$; for $m\ge 0$ one gets $\left(\genfrac{}{}{0px}{}{m+2}{2}\right)$, for $-2\le m\le -1$ all groups vanish and hence $\chi =0$, and for $m\le -3$ the term $(-1{)}^2\left(\genfrac{}{}{0px}{}{-m-1}{2}\right)$ equals $\left(\genfrac{}{}{0px}{}{m+2}{2}\right)$ by the identity $\left(\genfrac{}{}{0px}{}{-m-1}{2}\right)=\left(\genfrac{}{}{0px}{}{m+2}{2}\right)$ for those $m$. In all cases $\chi (X,\mathcal{O}(m))=\left(\genfrac{}{}{0px}{}{m+2}{2}\right)$.

Forms cut out effective divisors. The forms $f$ and $g$ are nonzero global sections of the invertible sheaves $\mathcal{O}(d)$ and $\mathcal{O}(e)$; since $X$ is integral, they are regular sections, so their zero schemes $C=Z(f)$ and $D=Z(g)$ are effective Cartier divisors with $\mathcal{O}(C)\cong \mathcal{O}(d)$ and $\mathcal{O}(D)\cong \mathcal{O}(e)$.

The pairing of twists. By definition and [F1], $\mathcal{O}(d)\cdot \mathcal{O}(e)=\chi (X,{\mathcal{O}}_X)-\chi (X,\mathcal{O}(-d))-\chi (X,\mathcal{O}(-e))+\chi (X,\mathcal{O}(-d-e))\mathrm{.}$ Substituting $\chi (X,\mathcal{O}(m))=\left(\genfrac{}{}{0px}{}{m+2}{2}\right)$ from step 1.1, this is $\frac{(2)(1)}{2}-\frac{(2-d)(1-d)}{2}-\frac{(2-e)(1-e)}{2}+\frac{(2-d-e)(1-d-e)}{2}$, and expanding, the numerator is $[2-(d^2-3d+2)]-[(e^2-3e+2)-((d+e{)}^2-3(d+e)+2)]=d(3-d)-d(3-d-2e)=2de$, so $\mathcal{O}(d)\cdot \mathcal{O}(e)=de$. In particular $l\cdot l=\mathcal{O}(1)\cdot \mathcal{O}(1)=1$.

The divisor computation. By the definition of the pairing, $C\cdot D=\mathcal{O}(C)\cdot \mathcal{O}(D)$ for the effective Cartier divisors of step 1.2, and by [F1] the classes of $\mathcal{O}(C)$ and $\mathcal{O}(D)$ are those of $\mathcal{O}(d)$ and $\mathcal{O}(e)$; hence $C\cdot D=de$ by step 2.1.

Specialisations and the independent cross-check. Taking $d=e=1$ and $f$ linear gives $l\cdot l=1$; taking $d=e=2$ and $g$ a nonzero quadratic form gives $C\cdot C=4$, in particular for a smooth conic. For the line $l=Z(x_0)$ and a conic $C=Z(g)$ of degree two, step 3.1 gives $l\cdot C=2$, and independently the restriction-degree theorem [F4] applied to the effective divisors $l$ and $C$ gives $l\cdot C={\mathrm{deg}}_l(\mathcal{O}(2){\mathrm{\mid }}_l)=2$, since the twisting sequences of [F4] give $\chi (l,\mathcal{O}(2){\mathrm{\mid }}_l)=\chi (X,\mathcal{O}(2))-\chi (X,\mathcal{O}(1))=6-3=3$ and $\chi (l,{\mathcal{O}}_l)=\chi (X,\mathcal{O})-\chi (X,\mathcal{O}(-1))=1-0=1$. Both computations agree, as asserted.

Conclusion and choice accounting. Steps 2.1 and 3.1 give $\mathcal{O}(d)\cdot \mathcal{O}(e)=de$ and $C\cdot D=de$, and step 4.1 records the line, conic and restriction-degree specialisations. The Axiom of Choice enters only through the cohomology and Euler-characteristic suppliers recorded in [F6]; the form $f$, the form $g$ and the line are given data, and steps 1.1–4.1 make no selection.

The blowup data. Since $p$ is $k$-rational, $r=1$; the surface $X^{\mathrm{\prime }}$ is integral regular projective and $E$ is an effective Cartier divisor isomorphic to ${\mathbb{P}}_k^1$ with ${\mathcal{O}}_E(E)\cong \mathcal{O}(-1)$, so the intersection product is defined on $X^{\mathrm{\prime }}$ and $E^2=-r=-1$; moreover $E\cdot {\pi }^{\ast }D=0$ and ${\pi }^{\ast }D\cdot {\pi }^{\ast }{D}^{\mathrm{\prime }}=D\cdot D^{\mathrm{\prime }}$ for all Cartier divisors $D,D^{\mathrm{\prime }}$ on $X$.

$l\cdot l=1$. Choose a line $L^{\mathrm{\prime }}$ on $X$; its class satisfies $[{\mathcal{O}}_X(L^{\mathrm{\prime }})]=[{\mathcal{O}}_X(1)]$, so $l\cdot l={\pi }^{\ast }{\mathcal{O}}_X(L^{\mathrm{\prime }})\cdot {\pi }^{\ast }{\mathcal{O}}_X(L^{\mathrm{\prime }})={\mathcal{O}}_X(L^{\mathrm{\prime }})\cdot {\mathcal{O}}_X(L^{\mathrm{\prime }})=1$ by the pullback identity and the plane computation.

$l\cdot E=0$. With $D$ the Cartier divisor of any line on $X$, so that ${\pi }^{\ast }D$ has class $l$, the orthogonality clause gives $E\cdot {\pi }^{\ast }D=0$, that is, $l\cdot E=E\cdot l=0$ by symmetry.

$E\cdot E=-1$. This is the self-intersection formula of step 1.1 with $r=1$.

The matrix. Steps 1.2, 1.3 and 2.1 give $l\cdot l=1$, $l\cdot E=0=E\cdot l$ and $E\cdot E=-1$; if $al+bE=0$ in $\mathrm{Pic}(X\mathrm{\prime })$, pairing with $l$ gives $a=0$ and pairing with $E$ gives $-b=0$. Thus these classes freely generate the stated sublattice; by bilinearity its Gram matrix of the sublattice $\mathbb{Z}l+\mathbb{Z}E$ in the basis $(l,E)$ is $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

The strict transform of a line through $p$. Let $L$ be a line through $p$ with strict transform $m$. The line is reduced and its multiplicity at $p$ is $1$, so ${\pi }^{\ast }L=m+E$ and $m\cdot E=1$ and $m\cdot m=L\cdot L-1=0$; equivalently, taking classes and using ${\mathcal{O}}_{X^{\mathrm{\prime }}}({\pi }^{\ast }L)\cong {\pi }^{\ast }{\mathcal{O}}_X(1)$, the class of $m$ is $l-E$, and bilinearity and steps 1.2–2.1 give $m\cdot E=l\cdot E-E\cdot E=0+1=1$ and $m\cdot m=l\cdot l-2l\cdot E+E\cdot E=1-0-1=0$, in agreement.

Conclusion and choice accounting. Steps 1.2, 1.3 and 2.1 give the matrix of the intersection form on $\mathbb{Z}l+\mathbb{Z}E$, and step 3.2 gives $m=l-E$, $m\cdot E=1$, $m\cdot m=0$ for the strict transform of a line through $p$. The Axiom of Choice enters through the suppliers recorded in [F4]; the point, the lines and the blowup are given data and no selection is made in the computations.

The four charts of $X$. By [F1] the charts of ${\mathbb{P}}_k^3$ give $X\cap D_{+}(w)=\mathrm{Spec}k[x,y,z]\mathrm{/}(xy-z^2)=:\mathrm{Spec}R,X\cap D_{+}(x)=\mathrm{Spec}k[y\mathrm{/}x,z\mathrm{/}x,w\mathrm{/}x]\mathrm{/}(y\mathrm{/}x-(z\mathrm{/}x{)}^2)\cong \mathrm{Spec}k[Z,W],$ $X\cap D_{+}(y)\cong \mathrm{Spec}k[Z,W],X\cap D_{+}(z)=\mathrm{Spec}k[x\mathrm{/}z,y\mathrm{/}z,w\mathrm{/}z]\mathrm{/}((x\mathrm{/}z)(y\mathrm{/}z)-1)\cong \mathrm{Spec}k[X,Y,W]\mathrm{/}(XY-1),$ writing $Z=z\mathrm{/}x$ and so on; each chart is a nonempty affine scheme whose coordinate ring is a domain, and the point $[1:1:1:1]$ lies on $X$ (since $1\cdot 1-1^2=0$) and in all four charts.

$X$ is projective over $k$ and proper. Since $X=V_{+}(F)$ is a closed subscheme of ${\mathbb{P}}_k^3$ and ${\mathbb{P}}_k^3\to \mathrm{Spec}k$ is projective in the H-projective convention, the structure morphism $X\to \mathrm{Spec}k$ is projective in that convention, hence proper by [F1].

The plane divisors on ${\mathbb{P}}^3$. Let $k$ be algebraically closed and let $D=V(x_0)$ and $E=V(x_1)$ be the hyperplanes in ${\mathbb{P}}_k^3$; by [F5] both are effective Cartier divisors with $\mathcal{O}(D)\cong \mathcal{O}(E)\cong \mathcal{O}(1)$, and the scheme-theoretic intersection is $D\cap E=V(x_0,x_1)\cong \mathrm{Proj}k[x_2,x_3]={\mathbb{P}}_k^1$, a one-dimensional scheme, hence not finite; replacing $E$ by $D$ leaves $D\cap D=D$ of dimension two.

The cone ring and its invariant model. Every class in $R=k[x,y,z]\mathrm{/}(xy-z^2)$ has a representative $A(x,y)+zB(x,y)$ with $A,B\in k[x,y]$, because the relation $z^2=xy$ reduces every exponent of $z$ modulo two; the substitution $\phi (x)=u^2$, $\phi (y)=v^2$, $\phi (z)=uv$ kills $xy-z^2$ and so induces a $k$-algebra map $\bar{\phi }:R\to k[u,v]$. If $\bar{\phi }(A+zB)=A(u^2,v^2)+uvB(u^2,v^2)$ vanishes, the first summand is supported on the monomials $u^{2i}{v}^{2j}$ and the second on the monomials $u^{2i+1}{v}^{2j+1}$; these supports are disjoint and monomials form a $k$-basis of $k[u,v]$, so $A=B=0$. Hence $\bar{\phi }$ is injective with image the subring $C:=k[u^2,uv,v^2]$ generated by $u^2,uv,v^2$, and $R\cong C$ is a domain. With $G=\{\pm 1\}$ acting on $k[u,v]$ by $(u,v)\mapsto (-u,-v)$, a polynomial is $G$-invariant exactly when the coefficient of every monomial of odd total degree vanishes, since $\mathrm{char}k\mathrm{\ne }2$; the monomials of even total degree are precisely the products of $u^2$, $uv$ and $v^2$, so $C=k[u,v{]}^G$.

$R$ is a Noetherian normal domain of dimension two. By [F2] the quotient $R$ of $S=k[x,y,z]$ is Noetherian, and by step 1.4 it is a domain. Every element of $\mathrm{Frac}(R)=\mathrm{Frac}(C)$ is a quotient of $G$-invariant polynomials, hence $G$-invariant, so $\mathrm{Frac}(R)\subseteq k(u,v{)}^G$; conversely if $f,g\in k[u,v]$ are nonzero and $f\mathrm{/}g$ is $G$-invariant, then $\sigma (f)g=f\sigma (g)$ for the generator $\sigma$ of $G$, so $f\mathrm{/}g=f\sigma (g)\mathrm{/}(g\sigma (g))$ is a quotient of $G$-invariant polynomials and lies in $\mathrm{Frac}(C)$; hence $\mathrm{Frac}(R)=k(u,v{)}^G$. If $\alpha \in \mathrm{Frac}(R)$ is integral over $R$, then $\alpha$ is integral over the integrally closed ring $k[u,v]\supseteq R$, so $\alpha \in k[u,v]$ (Finite-variable polynomial algebras over fields are integrally closed, [F2]); being $G$-invariant and polynomial, $\alpha \in k[u,v{]}^G=C=R$ by step 1.4. Thus $R$ is integrally closed, its localisations are integrally closed domains, and $\mathrm{Spec}R$ is a normal Noetherian scheme ([F2]). Finally $R$ is a finite-type $k$-domain, so $\dim R={\mathrm{trdeg}}_k\mathrm{Frac}(R)$ by [F10]; the field $k(u,v)$ is algebraic over $\mathrm{Frac}(R)=k(u,v{)}^G$ because $u$ and $v$ satisfy the integral equations $T^2-u^2=0$ and $T^2-v^2=0$ over it, so tower additivity [F10] gives $\dim R={\mathrm{trdeg}}_{k}k(u,v)=2$.

The alternating sum on ${\mathbb{P}}^3$. By [F6] applied with $n=3$ one has $\chi ({\mathbb{P}}_k^3,\mathcal{O}(d))=\left(\genfrac{}{}{0px}{}{d+3}{3}\right)$ for every $d$; hence $\chi (\mathcal{O})=1$, $\chi (\mathcal{O}(-1))=\left(\genfrac{}{}{0px}{}{2}{3}\right)=0$ and $\chi (\mathcal{O}(-2))=\left(\genfrac{}{}{0px}{}{1}{3}\right)=0$, so the alternating sum of statement (b) for the pair $D,E$ of step 1.3 equals $\chi (\mathcal{O})-\chi (\mathcal{O}(-1))-\chi (\mathcal{O}(-1))+\chi (\mathcal{O}(-2))=1-0-0+0=1.$

The ruling is a prime divisor. In the grading of [F8] the maximal ideal $\mathfrak{m}=(x,y,z)$ is the set $R_{\ge 1}$ of positive-degree elements and $P=(x,z)$ is homogeneous. The quotient $R\mathrm{/}P\cong k[y]$ is a domain, so $P$ is prime; the quotient $R\mathrm{/}(x)\cong k[y,z]\mathrm{/}(z^2)$ has nilradical $(z)$ because $z^2=0$ and every element of $k[y,z]\mathrm{/}(z^2)$ is $a(y)+zb(y)$ with $(a+zb{)}^n=a^n+n{a}^{n-1}zb$, so $(z)$ is the unique minimal prime of $R\mathrm{/}(x)$ and $P$ is the unique minimal prime of $R$ over the principal ideal $(x)$. By Krull's principal ideal theorem [F9] $\mathrm{ht}P\le 1$, and $\mathrm{ht}P=1$ because $R$ is a domain by step 2.1 and $0\mathrm{\ne }x\in P$, so $\mathrm{ht}P\ge 1$. Hence $V(P)$ is a prime divisor of the normal Noetherian scheme $\mathrm{Spec}R$ of step 2.1 (Weil divisor normal noetherian scheme).

$X$ is integral of pure dimension two. By step 1.1 the charts of $X$ are spectra of domains, hence reduced and irreducible, and they pairwise meet in the common point $[1:1:1:1]$; a space covered by pairwise intersecting irreducible open subspaces is irreducible, and reducedness is local, so $X$ is nonempty, reduced and irreducible, that is, integral (Integral schemes). The charts have dimension two — the cone chart by step 2.1, the polynomial charts by [F1] — so $X$ has pure dimension two (Chain dimension and the empty-space convention).

$X$ is normal. The local rings of the charts $X\cap D_{+}(x)$, $X\cap D_{+}(y)$ and $X\cap D_{+}(z)$ of step 1.1 are localisations of polynomial rings over $k$, hence regular local rings, hence integrally closed domains by [F4]; the local rings of the cone chart $X\cap D_{+}(w)=\mathrm{Spec}R$ are integrally closed domains because $R$ is a normal domain of step 2.1. Normality of a Noetherian scheme is the local condition that all local rings are integrally closed domains, so $X$ is normal.

Statement (b) fails. Take $X={\mathbb{P}}_k^3$, $D=V(x_0)$ and $E=V(x_1)$; by step 1.3 the scheme $X$ is a smooth projective $k$-scheme of dimension three and the divisors are effective Cartier, while $D\cap E\cong {\mathbb{P}}_k^1$ is not finite, so already the finiteness assertion of (b) is false; moreover by step 2.2 the alternating sum equals $1$, whereas $D\cap E$ itself has no finite length over $k$, and the value does not change when $E$ is replaced by $D$, although $D\cap D$ has dimension two.

The vertex chart is the affine quadric cone. The open subscheme $X\cap D_{+}(w)$ of $X$ is isomorphic to $\mathrm{Spec}R$ with $R=k[x,y,z]\mathrm{/}(xy-z^2)$, the vertex $v$ corresponds to the maximal ideal $\mathfrak{m}=(x,y,z)$, and under this isomorphism $Z\cap D_{+}(w)$ corresponds to $V(P)$ with $P=(x,z)$, the ruling of the cone. By steps 2.1 and 3.1 the ring $R$ is a normal domain and $P$ is a prime of height one, so $V(P)$ is a prime divisor of the normal Noetherian scheme $\mathrm{Spec}R$.

$Z$ is a prime divisor. On the charts of step 1.1 the subscheme $Z=V_{+}(x,z)\cap X$ is cut out by the images of $x$ and $z$: in the $w$-chart it is $V(P)\subseteq \mathrm{Spec}R$ with $R\mathrm{/}(x,z)\cong k[y]$; in the $y$-chart it is $V(Z)\subseteq \mathrm{Spec}k[Z,W]$, with quotient $k[W]\cong k[Z,W]\mathrm{/}(Z)$; in the $x$-chart and the $z$-chart it is empty, because $x=0$ or $z=0$ is incompatible with the chart. The two nonempty charts are spectra of domains, meet at the point $[0:1:0:1]$ of $Z$, and have dimension one, so $Z$ is a nonempty reduced irreducible closed subscheme of $X$ of dimension one, i.e. an integral closed subscheme of codimension one: a prime divisor of the normal Noetherian scheme $X$ (Weil divisor normal noetherian scheme, Integral schemes).

The ideal $P{R}_{\mathfrak{m}}$ is not principal. Since $P=Rx+Rz$ we have $P{R}_{\mathfrak{m}}=R_{\mathfrak{m}}x+R_{\mathfrak{m}}z$, and $\mathfrak{m}P$ is generated by $x^2,xy,xz,yz,z^2$, all of degree two in the grading of step 3.1; hence $\mathfrak{m}P\subseteq R_{\ge 2}$ and $\mathfrak{m}P\cap R_1=0$, and $\mathfrak{m}P$ is generated by homogeneous elements, so the degree-one component of every element of $\mathfrak{m}P$ vanishes. If $cx+dz\in \mathfrak{m}P$ with $c,d\in R$, write $c=c_0+c_{+}$, $d=d_0+d_{+}$ with $c_0,d_0\in k=R_0$ and $c_{+},d_{+}\in \mathfrak{m}$; then the degree-one component $c_{0}x+d_{0}z$ of $cx+dz$ lies in $\mathfrak{m}P$, hence is zero, so $c_0=d_0=0$ and $c,d\in \mathfrak{m}$. Now let $a,b\in R$ and $s,t\mathrm{\notin }\mathfrak{m}$ with $(a\mathrm{/}s)x+(b\mathrm{/}t)z\in \mathfrak{m}{R}_{\mathfrak{m}}P{R}_{\mathfrak{m}}=(\mathfrak{m}P)R_{\mathfrak{m}}$; multiplying by $st$ and clearing denominators gives $u\mathrm{\notin }\mathfrak{m}$ and $q\in \mathfrak{m}P$ with $u(atx+bsz)=q$, so $uat,ubs\in \mathfrak{m}$ by the previous paragraph; since $u,s,t\mathrm{\notin }\mathfrak{m}$ and $\mathfrak{m}$ is prime, $a,b\in \mathfrak{m}$, that is, $a\mathrm{/}s,b\mathrm{/}t\in \mathfrak{m}{R}_{\mathfrak{m}}$. Therefore the images of $x$ and $z$ span $P{R}_{\mathfrak{m}}\mathrm{/}\mathfrak{m}{R}_{\mathfrak{m}}P{R}_{\mathfrak{m}}$ and are linearly independent over $R_{\mathfrak{m}}\mathrm{/}\mathfrak{m}{R}_{\mathfrak{m}}=k$: this $k$-vector space has a basis of two elements. If $P{R}_{\mathfrak{m}}$ were a principal ideal, this space would be spanned by the image of one element and have dimension at most one; contradiction.

$Z$ is not Cartier at $v$. Suppose for contradiction that $Z$ were Cartier at $v$: there are an open neighbourhood $U\subseteq X$ of $v$ and a Cartier divisor $D_0$ on $U$ whose associated Weil divisor is ${\mathrm{cyc}}_U(D_0)=[Z\cap U]$. The chart $D_{+}(w)\cap X\cong \mathrm{Spec}R$ is an open neighbourhood of $v$ corresponding to an open neighbourhood of $\mathfrak{m}$ in $\mathrm{Spec}R$ (step 4.1), so replacing $U$ by $U\cap D_{+}(w)$ and restricting $D_0$ gives a Cartier divisor $D$ on an open neighbourhood $U^{\mathrm{\prime }}\subseteq \mathrm{Spec}R$ of $\mathfrak{m}$ with ${\mathrm{cyc}}_{U^{\mathrm{\prime }}}(D)=[V(P)\cap U^{\mathrm{\prime }}]$, by the restriction compatibility of [F3] and the identification of $Z\cap D_{+}(w)$ with $V(P)$ in step 4.1. The open subscheme $U^{\mathrm{\prime }}$ of $\mathrm{Spec}R$ is integral, normal and Noetherian: it is integral as an open subscheme of an integral scheme (Integral schemes), its local rings are localisations of the integrally closed local rings of $R$ (step 2.1), and it is Noetherian by [F13]. So the Cartier divisor $D$ is represented on an open cover $\{U_i\}$ of $U^{\mathrm{\prime }}$ by local equations $f_i\in {\mathcal{K}}_{U^{\mathrm{\prime }}}(U_i{)}^{\times }$ (Cartier divisor), and ${\mathcal{K}}_{U^{\mathrm{\prime }}}$ is the constant sheaf with value $\mathrm{Frac}(R)$ by [F13]; fix an index $i_0$ with $\mathfrak{m}\in U_{i_0}$ and put $f:=f_{i_0}\in \mathrm{Frac}(R{)}^{\times }$. Every height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$ of $R$ lies in $U_{i_0}$: its closure $V(\mathfrak{q})$ in $\mathrm{Spec}R$ contains $\mathfrak{m}$ because $\mathfrak{q}\subseteq \mathfrak{m}$, so if $\mathfrak{q}\mathrm{\notin }{U}_{i_0}$ then the closed set $\mathrm{Spec}R\setminus U_{i_0}$ would contain $\mathfrak{q}$ and hence its closure and the point $\mathfrak{m}$, contradicting $\mathfrak{m}\in U_{i_0}$. Hence, by the coefficient formula of Cartier divisors on a normal Noetherian scheme give Weil divisors applied to the normal Noetherian scheme $U^{\mathrm{\prime }}$ and the local equation $f_{i_0}$, the coefficient of the prime divisor $V(\mathfrak{q})$ in ${\mathrm{cyc}}_{U^{\mathrm{\prime }}}(D)=[V(P)\cap U^{\mathrm{\prime }}]$ equals $v_{\mathfrak{q}}(f)$ (Order codimension one rational function, [F12]); therefore $v_P(f)=1$ and $v_{\mathfrak{q}}(f)=0$ for every height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$ with $\mathfrak{q}\mathrm{\ne }P$. The local ring $R_{\mathfrak{m}}$ is a Noetherian integrally closed domain by [F2] and step 2.1, hence satisfies $(S_2)$ by [F11], and its height-one primes are the primes $\mathfrak{q}\subseteq \mathfrak{m}$ of height one, with $(R_{\mathfrak{m}}{)}_{\mathfrak{q}{R}_{\mathfrak{m}}}=R_{\mathfrak{q}}$ by [F13]; the intersection theorem [F11] gives $R_{\mathfrak{m}}={\bigcap }_{\mathrm{ht}\mathfrak{q}=1,\mathfrak{q}\subseteq \mathfrak{m}}{R}_{\mathfrak{q}}$ inside $\mathrm{Frac}(R_{\mathfrak{m}})=\mathrm{Frac}(R)$. Since $v_{\mathfrak{q}}(f)\ge 0$ for each such $\mathfrak{q}$, [F12] gives $f\in R_{\mathfrak{q}}$ for all of them, hence $f\in R_{\mathfrak{m}}$; and $v_P(f)=1\ge 1$ gives $f\in P{R}_P$, so $f\in P{R}_P\cap R_{\mathfrak{m}}=P{R}_{\mathfrak{m}}$ (Localisation at a prime ideal: $R_{\mathfrak{p}}=(R\setminus \mathfrak{p}{)}^{-1}R$). If $g\in P{R}_{\mathfrak{m}}$, then $v_P(g)\ge 1=v_P(f)$ and $v_{\mathfrak{q}}(g)\ge 0=v_{\mathfrak{q}}(f)$ for every other height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$, so $v_{\mathfrak{q}}(g\mathrm{/}f)\ge 0$ for all of them and [F11] gives $g\mathrm{/}f\in R_{\mathfrak{m}}$, that is, $g\in f{R}_{\mathfrak{m}}$; hence $P{R}_{\mathfrak{m}}=f{R}_{\mathfrak{m}}$ is principal, contradicting step 4.3. Therefore no such $D_0$ exists: the prime divisor $Z$ is not Cartier at $v$, ${\mathcal{O}}_X(Z)$ is not invertible near $v$, and $Z\cdot Z$ is not defined by the alternating-sum formula.