Proof. There is clearly a morphism $C_{+}\amalg _{E}C_{-}\rightarrow C$ . Showing that this map has the desired properties is a local question on $X$ . So it suffices to verify these properties at the level of rings.

If we set $X=\operatorname{Spec}(A)$ , $S(X,D)=\operatorname{Spec}(R)$ and let $I$ be the defining ideal for $D$ , then we have an exact sequence of $R$ -modules

Since $\unicode[STIX]{x1D708}$ is affine, we can write $C=\operatorname{Spec}(B)$ . Let $J\subset B$ be the ideal defining the closed subscheme $E$ . Tensoring (2.6) with $B$ , we get an exact sequence

and this shows that $C_{+}\amalg _{E}C_{-}{\hookrightarrow}C$ is a closed immersion of schemes. It is also clear that this inclusion is an isomorphism in the complement of $E$ . Furthermore, the surjectivity of the map $C_{+}\amalg C_{-}\rightarrow C$ shows that the above inclusion is also surjective on points. We conclude from this that the closed immersion $C_{+}\amalg _{E}C_{-}{\hookrightarrow}C$ induces identity on the underlying reduced schemes. In particular, it is an isomorphism if $C$ is reduced. Equivalently, $\unicode[STIX]{x1D719}_{B}$ is injective.◻

Proof. To prove the proposition, we can assume that $X=\operatorname{Spec}(A)$ is affine. Let $p_{1},p_{2}:A\times A\rightarrow A$ denote the projections and let $q:A\rightarrow A/I$ be the quotient map. Set $q_{i}=q\circ p_{i}$ . Set $\unicode[STIX]{x1D713}_{i}=p_{i}\circ \unicode[STIX]{x1D719}$ for $i=1,2$ . Let $\unicode[STIX]{x1D6FF}:A{\hookrightarrow}R{\hookrightarrow}A\times A$ denote the diagonal map. We then have $\unicode[STIX]{x1D713}_{i}\circ \unicode[STIX]{x1D6FF}=\text{id}_{A}$ for $i=1,2$ and this yields

Since $A\times A$ is a finite free $A$ -module and $R$ is an $A$ -submodule, it follows that $R$ is a finite $A$ -module. This proves (1). The item (2) follows immediately from (2.6).

The ideal of $D$ inside $S(X,D)$ is $\text{Ker}(R\xrightarrow[{}]{q\circ \unicode[STIX]{x1D713}_{i}}A/I)$ , which is $I\times I$ . Since $\unicode[STIX]{x1D6FF}^{\ast }(I)\subseteq I\times I$ , we see that $D\subseteq \unicode[STIX]{x1D6E5}^{\ast }(D)$ and the composite $D{\hookrightarrow}\unicode[STIX]{x1D6E5}^{\ast }(D)\xrightarrow[{}]{\unicode[STIX]{x1D6E5}}D$ is clearly identity. Furthermore, it is clear that $R[(a,b)^{-1}]=A[a^{-1}]\times A[b^{-1}]$ and $\unicode[STIX]{x1D6FF}^{\ast }(I)[(a,b)^{-1}]=R[(a,b)^{-1}]$ , whenever $a,b\in I\setminus \{0\}$ . Hence, we have $|\unicode[STIX]{x1D6E5}^{\ast }(D)|=|D|$ . This proves (3).

To prove (4), we only need to consider the closed part. Let $A^{\prime }=A/J$ , where $J$ is the ideal defining $Y$ and let $R^{\prime }=\{(a^{\prime },b^{\prime })\in A^{\prime }\times A^{\prime }\mid a^{\prime }-b^{\prime }\in (I+J)/J\}$ . Let $\overline{a}$ denote the residue class of $a\in A$ modulo $J$ . Suppose there exist $a,b\in A$ such that $\overline{a}-\overline{b}\in (I+J)/J$ . This means $a-b\in I+J$ and so we can write $a-b=\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}$ , where $\unicode[STIX]{x1D6FC}\in I$ and $\unicode[STIX]{x1D6FD}\in J$ . We set $a^{\prime }=a-\unicode[STIX]{x1D6FD}$ and $b^{\prime }=b$ . This yields $a^{\prime }-b^{\prime }=a-b-\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D6FC}\in I$ and $a^{\prime }-a=\unicode[STIX]{x1D6FD}\in J,~b^{\prime }-b=0\in J$ . We conclude that $(a^{\prime },b^{\prime })\in R$ and it maps to $(\overline{a},\overline{b})\in R^{\prime }$ . Hence $R{\twoheadrightarrow}R^{\prime }$ .

An element of $\unicode[STIX]{x1D6FF}^{\ast }(J)$ is of the form $(a\unicode[STIX]{x1D6FC},b\unicode[STIX]{x1D6FC})$ , where $a,b,\in A,\unicode[STIX]{x1D6FC}\in J$ and $a-b\in I$ . This element clearly dies in $R^{\prime }$ . Hence $S(Y,Y\cap D)\subseteq \unicode[STIX]{x1D6E5}^{\ast }(Y)$ .

To prove (5), let $S_{Y}=S(Y,Y\cap D)$ . Then

where the first equality follows from (3). On the other hand, we have

where the second equality follows from (3) with $X$ replaced by $Y$ . The item (5) now follows. The item (6) follows from (1) and the fact that more than one components of $S(X,D)$ meet along $D$ .

To prove (7), let $Y=\operatorname{Spec}(B)$ and tensor (2.6) with $B$ . The flatness of $B$ over $A$ yields the short exact sequence

and this proves the first part of (7). The second part follows because a base change of a flat (respectively smooth) map is flat (respectively smooth).

The item (8) follows because $\unicode[STIX]{x1D70B}$ is finite and birational and the ideal of $\unicode[STIX]{x1D70B}^{\ast }(D)$ in $X\amalg X$ is $I_{D}\times I_{D}$ which is actually contained in ${\mathcal{O}}_{S(X,D)}$ . So $D$ is a conducting subscheme.◻

Proof. We can again assume that $X=\operatorname{Spec}(A)$ is affine such that $I=(a)$ is a principal ideal such that $a\in A$ is not a zero-divisor. It follows then that $I$ is a free $A$ -module of rank one. We can now apply (2.8) to conclude (1) as the finiteness of $R$ over $A$ is already shown in Proposition 2.3.

To prove (2), let $\mathfrak{m}\subsetneq R$ be a maximal ideal and let $\mathfrak{n}=\unicode[STIX]{x1D6FF}^{-1}(\mathfrak{m})$ . Then $A_{\mathfrak{n}}\rightarrow R_{\mathfrak{n}}$ is a finite and flat map and hence $A_{\mathfrak{n}}\rightarrow R_{\mathfrak{m}}$ is a faithfully flat local homomorphism of noetherian local rings of same dimension. Since $A_{\mathfrak{n}}$ is Cohen–Macaulay and since this local homomorphism preserves regular sequences, it follows that $\text{depth}(R_{\mathfrak{m}})\geqslant \dim (A_{\mathfrak{n}})=\dim (R_{\mathfrak{m}})$ . Hence, $R_{\mathfrak{m}}$ is Cohen–Macaulay.

To prove (3), we let $Y=\operatorname{Spec}(B)$ and tensor (2.6) (over $A$ ) with $B$ to get an exact sequence

and $S(B,IB)$ is (by definition) the kernel of the map $B\times B\rightarrow B/IB$ . In particular, we get a surjective map of rings $R\otimes _{A}B{\twoheadrightarrow}S(B,IB)$ . This proves the first part of (3). The transversality of $B$ with $A/I$ means precisely that $\text{Tor}_{A}^{1}(A/I,B)=0$ and we get that $R\otimes _{A}B\xrightarrow[{}]{\simeq }S(B,IB)$ . This proves (3).

Suppose next that $Y$ is Cohen–Macaulay and no irreducible component of $Y$ is contained in $f^{\ast }(D)$ . It suffices to show in this case that $\text{Tor}_{A}^{1}(A/I,B)=0$ . Let $f:A\rightarrow B$ be the map on the coordinate rings. That no component of $Y$ is contained in $f^{\ast }(D)$ means that $f(a)$ does not belong to any minimal prime of $B$ . The Cohen–Macaulay property of $B$ implies that it has no embedded associated prime. In particular, $f(a)$ does not belong to any associated prime and hence is not a zero-divisor in $B$ .

We have a short exact sequence

which says that $\text{Tor}_{A}^{1}(A/I,B)=\text{Ker}(B\xrightarrow[{}]{f(a)}B)$ and we have just shown that the latter group is zero. We have also shown above that $f(a)$ is not a zero-divisor on $B$ and this implies that $f^{\ast }(D)$ is an effective Cartier divisor on $Y$ . This proves (4). The item (5) follows from (1) and an elementary fact that l.c.i. morphisms are preserved under a flat base change.◻

Proof. It follows from the definition of a perfect morphism because if $x\in X$ is such that $s=f(x)$ is a regular point of $S$ , then ${\mathcal{O}}_{X,x}$ has finite Tor-dimension over ${\mathcal{O}}_{S,s}$ . This property for the points over the singular locus of $S$ follows from the hypothesis of the lemma.◻

Proof. Let $U_{1}\subseteq C$ be an open subset of $C$ containing $S_{1}=\unicode[STIX]{x1D708}^{-1}(X_{\text{sing}})$ such that $(C_{\text{sing}}\setminus S_{1})\cap U_{1}=\emptyset$ . This is possible because $S_{1}$ is a finite set. Let $\unicode[STIX]{x1D70B}:(C\setminus S_{1})^{N}\rightarrow C\setminus S_{1}$ denote the normalization map. It follows that $\unicode[STIX]{x1D70B}:\unicode[STIX]{x1D70B}^{-1}(U_{1}\setminus S_{1})\rightarrow U_{1}\setminus S_{1}$ is an isomorphism. Setting $U_{2}=(C\setminus S_{1})^{N}$ , we see that $U_{1}$ and $U_{2}$ glue along $\unicode[STIX]{x1D70B}^{-1}(U_{1}\setminus S_{1})$ to give a unique scheme $C^{\prime }$ and a unique map $p:C^{\prime }\rightarrow C$ . This scheme has the property that $p$ is finite, $p^{-1}(U_{1})\rightarrow U_{1}$ is an isomorphism and $p^{-1}(C\setminus S_{1})=(C\setminus S_{1})^{N}$ .

Setting $Z^{\prime }=p^{-1}(Z)$ and $f^{\prime }=p^{\ast }(f)\in k(C^{\prime })^{\times }$ , we see that $f^{\prime }\in {\mathcal{O}}_{C^{\prime },Z^{\prime }}^{\times }$ and $\text{div}(f)=p_{\ast }(\text{div}(f^{\prime }))$ . If we let $\unicode[STIX]{x1D708}^{\prime }=\unicode[STIX]{x1D708}\circ p$ , we get $\unicode[STIX]{x1D708}_{\ast }(\text{div}(f))=\unicode[STIX]{x1D708}_{\ast }^{\prime }(\text{div}(f^{\prime }))$ . Furthermore, $\unicode[STIX]{x1D708}^{\prime -1}(X_{\text{reg}})\rightarrow X_{\text{reg}}$ is a finite-type morphism of regular schemes and hence is an l.c.i. morphism. Since $\unicode[STIX]{x1D708}$ is l.c.i. over $X_{\text{sing}}$ and $p$ is an isomorphism in a neighborhood of $\unicode[STIX]{x1D708}^{-1}(X_{\text{sing}})$ , we conclude that $\unicode[STIX]{x1D708}^{\prime }$ is an l.c.i. morphism.◻

Proof. We write $S(X,D)$ and $S(C,\unicode[STIX]{x1D708}^{\ast }(D))$ as $S_{X}$ and $S_{C}$ , respectively, in this proof. Since the inclusion $\unicode[STIX]{x1D708}:C{\hookrightarrow}X$ is l.c.i. along $D$ , it follows Proposition 2.4 that the square

is Cartesian. It also follows from Proposition 2.4(5) that $S_{\unicode[STIX]{x1D708}}:S_{C}{\hookrightarrow}S_{X}$ is l.c.i. along $D$ . Moreover, a combination of Proposition 2.4(4) and Proposition 2.3(2) tells us that $S_{C}$ is reduced with two components, both isomorphic to $C$ . We conclude that $S_{C}{\hookrightarrow}S_{X}$ is a (reduced) Cartier curve relative to $D$ .◻

Proof. The map (3.7) is induced by the identity on the set of generators, so we just have to show that it is well defined. Since $X$ is reduced, by Proposition 3.9, we can assume that the Cartier curves defining the rational equivalence on the Levine–Weibel Chow group are reduced. Now, we just note that a reduced Cartier curve is a good curve relative to $(X,Y)$ .◻

Proof. Since $\unicode[STIX]{x1D708}^{-1}(Y)\cup X_{\text{sing}}^{\prime }\subseteq Y^{\prime }$ , we have a push-forward map $\unicode[STIX]{x1D708}_{\ast }:{\mathcal{Z}}_{0}(C,Z)\rightarrow {\mathcal{Z}}_{0}(X,Y)$ , given by $\unicode[STIX]{x1D708}_{\ast }([x])=[k(x):k(\unicode[STIX]{x1D708}(x))]\cdot [\unicode[STIX]{x1D708}(x)]$ . Since $\unicode[STIX]{x1D708}$ is l.c.i. along $X_{\text{sing}}$ , it follows from Lemma 3.3 that the map $\unicode[STIX]{x1D708}:C\rightarrow X$ is perfect. Hence, there is a push-forward map on $K_{0}$ -groups $\unicode[STIX]{x1D708}_{\ast }:K_{0}(C)\rightarrow K_{0}(X)$ by Proposition 3.1.

To construct the cycle class maps and show that the square commutes, let $x\in C\setminus Z$ be a closed point and set $y=\unicode[STIX]{x1D708}(x)$ . Let $\unicode[STIX]{x1D704}_{x}:\operatorname{Spec}(k(x))\rightarrow C$ and $\unicode[STIX]{x1D704}_{y}:\operatorname{Spec}(k(y))\rightarrow X$ be the closed immersions. Since these maps as well as $\unicode[STIX]{x1D708}$ are perfect (see Lemma 3.3), we have the induced push-forward maps on Grothendieck groups of vector bundles and a commutative diagram by Proposition 3.1.

Setting $\text{cyc}_{C}([x])$ to be ${\unicode[STIX]{x1D704}_{x}}_{\ast }(1)$ , we get the cycle class maps $\text{cyc}_{C}:{\mathcal{Z}}_{0}(C)\rightarrow K_{0}(C)$ and $\text{cyc}_{X}:{\mathcal{Z}}_{0}(X)\rightarrow K_{0}(X)$ such that (3.8) commutes.◻

Proof. Let $\unicode[STIX]{x1D708}:C\rightarrow X$ be a finite map from a reduced curve and let $Z\subsetneq C$ be a closed subset such that $(C,Z)$ is good relative to $(X,Y)$ . By Lemma 3.11, there is a commutative diagram.

Let $f\in {\mathcal{O}}_{C,Z}^{\times }$ . It follows from [Reference Levine and WeibelLW85, Proposition 2.1] that $\text{cyc}_{C}(\text{div}(f))=0$ . In particular, we get $\text{cyc}_{X}\circ \unicode[STIX]{x1D708}_{\ast }(\text{div}(f))=\unicode[STIX]{x1D708}_{\ast }\circ \text{cyc}_{C}(\text{div}(f))=0$ . It follows again from [Reference Levine and WeibelLW85, Proposition 1.4] that $\unicode[STIX]{x1D708}_{\ast }(\text{div}(f))=0$ in $\text{CH}_{0}^{\text{LW}}(X,Y)\simeq \operatorname{Pic}(X){\hookrightarrow}K_{0}(X)$ . We have thus shown that the surjective map $\text{CH}_{0}^{\text{LW}}(X){\twoheadrightarrow}\text{CH}_{0}(X)$ is also injective, hence an isomorphism.◻

Proof. The fact that $\text{cyc}_{X}$ yields a cycle class map $\text{cyc}_{X}^{\text{LW}}:\text{CH}_{0}^{\text{LW}}(X,Y)\rightarrow K_{0}(X)$ is proved in [Reference Levine and WeibelLW85, Proposition 2.1]. To show that $\text{cyc}_{X}$ descends to a map on our modified version of the Chow group, let $\unicode[STIX]{x1D708}:(C,Z)\rightarrow (X,Y)$ be a good curve relative to $(X,Y)$ and let $f\in {\mathcal{O}}_{C,Z}^{\times }$ . We then have $\text{cyc}_{X}\circ \unicode[STIX]{x1D708}_{\ast }(\text{div}(f))=\unicode[STIX]{x1D708}_{\ast }\circ \text{cyc}_{C}(\text{div}(f))$ by Lemma 3.11. On the other hand, it follows from Lemma 3.12 that $\text{cyc}_{C}(\text{div}(f))=0$ . This shows that $\text{cyc}_{X}$ is defined on the Chow groups. The commutativity of (3.11) is clear from the definitions.◻

Proof. Let $\unicode[STIX]{x1D6FC}\in \text{CH}_{0}^{\text{LW}}(X)$ be such that $\text{cyc}_{X}^{\text{LW}}(\unicode[STIX]{x1D6FC})=0$ . By Levine’s theorem recalled above, we know that $\unicode[STIX]{x1D6FC}$ is a torsion class in $\text{CH}_{0}^{\text{LW}}(X)$ .

To show that $\unicode[STIX]{x1D6FC}=0$ , we can use the Lefschetz principle argument and rigidity of the Chow group of zero-cycles over algebraically closed fields and assume that $k=\mathbb{C}$ . Let $\mathbb{H}_{{\mathcal{D}}^{\ast }}^{2d}(X,\mathbb{Z}(d))$ denote the modified Deligne–Beilinson cohomology of $X$ defined in [Reference LevineLev92, §2] (see also §8 below). There is then a short exact sequence

and it was shown in [Reference LevineLev92, § 2] that there is a Chern class map $c_{{\mathcal{D}}^{\ast },X}^{d}:K_{0}(X)\,\rightarrow \,\mathbb{H}_{{\mathcal{D}}^{\ast }}^{2d}(X,\mathbb{Z}(d))$ which induces an Abel–Jacobi map $\text{AK}_{X}^{d}:\text{CH}_{0}^{\text{LW}}(X)_{\text{deg}\,0}\rightarrow A^{d}(X)$ given by $\text{AK}_{X}^{d}=\,c_{{\mathcal{D}}^{\ast },X}^{d}\circ \text{cyc}_{X}^{\text{LW}}$ , where $\text{CH}_{0}^{\text{LW}}(X)_{\text{deg}\,0}:=\text{Ker}(\text{CH}_{0}^{\text{LW}}(X)\rightarrow H^{2d}(X_{\text{an}},\mathbb{Z}(d)))$ .

Since $H^{2d}(X_{\text{an}},\mathbb{Z}(d))$ is torsion-free and $\unicode[STIX]{x1D6FC}$ is torsion, it follows that $\unicode[STIX]{x1D6FC}\in \text{CH}_{0}^{\text{LW}}(X)_{\text{deg}\,0}$ . In particular, it is a torsion class in $\text{CH}_{0}^{\text{LW}}(X)_{\text{deg}\,0}$ . A cycle class map

is also constructed in [Reference Esnault, Srinivas and ViehwegESV99, § 2] and it is shown in [Reference KrishnaKri09, Lemma 2.2] that $\widetilde{\text{AK}}_{X}^{d}=\text{AK}_{X}^{d}$ up to a sign. Now, $\text{cyc}_{X}^{\text{LW}}(\unicode[STIX]{x1D6FC})=0$ implies that $c_{{\mathcal{D}}^{\ast },X}^{d}\circ \text{cyc}_{X}^{\text{LW}}(\unicode[STIX]{x1D6FC})=0$ in $A^{d}(X)$ . We conclude that $\unicode[STIX]{x1D6FC}$ is a torsion class in $\text{CH}_{0}^{\text{LW}}(X)_{\text{deg}\,0}$ such that $\widetilde{\text{AK}}_{X}^{d}(\unicode[STIX]{x1D6FC})=0$ . We now apply [Reference Biswas and SrinivasBS99, Theorem 1.1] to conclude that $\unicode[STIX]{x1D6FC}=0$ . This finishes the proof.◻

Proof. In each case, it suffices to show using Lemma 3.13 that $\text{cyc}_{X}^{\text{LW}}$ is injective. The case $d\leqslant 1$ follows from Lemma 3.12. The $d=2$ case follows from [Reference LevineLev85a, Theorem 7], where it is shown that the map $\text{CH}_{0}^{\text{LW}}(X)\rightarrow F^{2}K_{0}(X)$ is an isomorphism. If $X$ is affine, this follows from [Reference KrishnaKri15b, Corollary 7.3] and [Reference LevineLev87, Corollary 2.7]. The last case follows from Proposition 3.15.◻

Proof. It follows from our assumption and Lemma 3.3 that $p$ is perfect and hence there is a push-forward map $p_{\ast }:K_{0}(X^{\prime })\rightarrow K_{0}(X)$ . We have seen before that there is also a push-forward map $p_{\ast }:{\mathcal{Z}}_{0}(X^{\prime },Y^{\prime })\rightarrow {\mathcal{Z}}_{0}(X,Y)$ .

Let us now consider a good curve $\unicode[STIX]{x1D708}^{\prime }:(C,Z)\rightarrow (X^{\prime },Y^{\prime })$ relative to $(X^{\prime },Y^{\prime })$ . It follows from our assumption that $\unicode[STIX]{x1D708}=p\circ \unicode[STIX]{x1D708}^{\prime }:(C,Z)\rightarrow (X,Y)$ is a good curve relative to $(X,Y)$ . We have the push-forward maps ${\mathcal{Z}}_{0}(C,Z)\xrightarrow[{}]{\unicode[STIX]{x1D708}_{\ast }^{\prime }}{\mathcal{Z}}_{0}(X^{\prime },Y^{\prime })\xrightarrow[{}]{p_{\ast }}{\mathcal{Z}}_{0}(X,Y)$ such that $\unicode[STIX]{x1D708}_{\ast }=p_{\ast }\circ \unicode[STIX]{x1D708}_{\ast }^{\prime }$ . This shows that $p_{\ast }(\unicode[STIX]{x1D708}_{\ast }^{\prime }(\text{div}(f)))=\unicode[STIX]{x1D708}_{\ast }(\text{div}(f))$ for any $f\in {\mathcal{O}}_{C,Z}^{\times }$ . This implies that $p_{\ast }$ descends to a push-forward map on the Chow groups. The commutativity of (3.12) is shown exactly as in the proof of Lemma 3.11.◻

Proof. Proposition 3.18 says that there are maps

and Theorem 3.17 says that the last map is an isomorphism. The result follows. ◻

Proof. We can assume that $X$ is connected. Let ${\mathcal{Z}}_{1}(X)_{W}$ denote the free abelian group on integral curves in $X\times \mathbb{P}_{k}^{1}$ which have the following properties:

1. (1) $C\cap (X\times \{0,\infty \})$ is finite;

2. (2) $C\cap (W\times \mathbb{P}^{1})$ is finite;

3. (3) $C\cap (W\times \{0,\infty \})=\emptyset$ ;

4. (4) $C\neq \{x\}\times \mathbb{P}^{1}$ for any $x\in X$ .

Then, the moving lemma of Bloch [Reference BlochBlo86] says that the inclusion of chain complexes

induces isomorphism on $H^{0}$ . In particular, we get exact sequence

On the other hand, there is an isomorphism $(\unicode[STIX]{x2202}_{\infty }-\unicode[STIX]{x2202}_{0})({\mathcal{Z}}_{1}(X)_{W})\rightarrow {\mathcal{R}}_{0}(X)_{W}$ given by $\unicode[STIX]{x2202}([C])\mapsto \text{div}(N(f))$ , where $f$ is the projection map $C\rightarrow \mathbb{P}_{k}^{1}$ , $C^{\prime }$ is its image in $X$ and $N(f)$ is the norm of $f$ under the finite map $k(C^{\prime }){\hookrightarrow}k(C)$ . One checks easily that $N(f)\in {\mathcal{O}}_{C^{\prime },C^{\prime }\cap Y}^{\times }$ .◻

Proof. In view of Lemma 3.20, we need to show that $\unicode[STIX]{x1D6E5}^{\ast }((f)_{C})\in {\mathcal{R}}_{0}(S_{X},D)$ for $C{\hookrightarrow}X$ an integral curve not contained in $D$ and $f\in {\mathcal{O}}_{C,C\cap D}^{\times }$ . Let $\unicode[STIX]{x1D708}:C^{N}\rightarrow X$ denote the induced map from the normalization of $C$ and let $E=\unicode[STIX]{x1D708}^{\ast }(D)$ . We then have $f\in {\mathcal{O}}_{C^{N},E}^{\times }$ and $(f)_{C}=\unicode[STIX]{x1D708}_{\ast }(\operatorname{div}(f))$ . We can thus assume that $C$ is normal and allow the possibility that $\unicode[STIX]{x1D708}$ need not be a closed immersion. With this reduction, we now need to show that $\unicode[STIX]{x1D6E5}^{\ast }\circ \unicode[STIX]{x1D708}_{\ast }(\operatorname{div}(f))\in {\mathcal{R}}_{0}(S_{X},D)$ .

Since $\unicode[STIX]{x1D6E5}$ is flat, it follows from (3.6) that there is a commutative square (see [Reference FultonFul98, Proposition 1.7])

where the bottom horizontal arrow is well defined since $S_{\unicode[STIX]{x1D708}}^{-1}(D)\subseteq E$ . Note that by Proposition 2.4, the map $S_{\unicode[STIX]{x1D708}}:S_{C}\rightarrow S_{X}$ is l.c.i. along $D$ . We also have a commutative square of monomorphisms.

Setting $g=\unicode[STIX]{x1D6E5}_{C}^{\ast }(f)\in {\mathcal{O}}_{S_{C},E}^{\times }$ , it is then clear that $\unicode[STIX]{x1D704}_{S_{C}}(g)=(f,f)\in k(C)^{\times }\times k(C)^{\times }=k(S_{C})^{\times }.$ This yields

Combining this with (4.1), we get

Since $(S_{C},E)$ is clearly a good curve relative to $(S_{X},D)$ , the last term lies in ${\mathcal{R}}_{0}(S_{X},D)$ . This finishes the proof.◻

Proof. For the rational equivalence, we argue as follows. Suppose at first that $\unicode[STIX]{x1D708}:C{\hookrightarrow}S_{X}$ is an l.c.i. curve relative to $D$ and contained in $S_{X}$ (so, it is in particular a Cartier curve on the double). Set $E=\unicode[STIX]{x1D708}^{-1}(D)\cup C_{\text{sing}}$ . Let $C_{\pm }^{\prime }$ denote the unique reduced closed subscheme of $X_{\pm }$ such that $C_{\pm }^{\prime }\setminus D=(C\setminus D)\cap X_{\pm }$ . Then $C_{+}^{\prime }$ is a closed subscheme of $C$ with $\dim (C_{+}^{\prime })\leqslant 1$ . It is easy to check that $\dim (C_{+}^{\prime })=0$ would violate the condition of $C$ being locally defined by a regular sequence at every point of intersection $C\cap D\subset S_{X}$ , so that we can assume that $C_{+}^{\prime }$ (and similarly $C_{-}^{\prime }$ ) is a union of some irreducible components of $C$ .

Let $f\in {\mathcal{O}}_{C,E}^{\times }$ . Let $C=C_{+}^{\prime }\cup C_{-}^{\prime }=(\bigcup _{i=1}^{r_{+}}C_{+}^{i})\cup (\bigcup _{j=1}^{r_{-}}C_{-}^{j})$ . We can clearly assume that we have $E_{\text{red}}=C_{+}^{\prime }\cap C_{-}^{\prime }$ and a commutative square.

We can write $\unicode[STIX]{x1D703}_{C}(f)=(f_{+},f_{-})$ with $f_{\pm }\in k(C_{\pm }^{\prime })^{\times }$ and $\text{div}_{C}(f)~(=(f)_{C})=(f_{+})_{C_{+}}+(f_{-})_{C_{-}}$ , by definition. We now consider the following diagram.

This diagram is clearly commutative and yields

On the other hand, ${\unicode[STIX]{x1D708}_{+}}_{\ast }((f_{+})_{C_{+}^{\prime }})=\sum _{i=1}^{r_{+}}\operatorname{div}(f_{+}^{i})$ by definition, where

Since each $\operatorname{div}(f_{+}^{i})\in {\mathcal{R}}_{0}(X)$ , we conclude that $\unicode[STIX]{x1D704}_{+}^{\ast }\circ \unicode[STIX]{x1D708}_{\ast }((f)_{C})\in {\mathcal{R}}_{0}(X)$ . In particular, we obtain that the maps $\unicode[STIX]{x1D704}_{\pm }^{\ast }$ descend to group homomorphisms $\unicode[STIX]{x1D704}_{\pm }^{\ast }:\text{CH}_{0}^{\text{LW}}(S_{X})\rightarrow \text{CH}_{0}(X)$ from the Levine–Weibel Chow group.

For the general case of a good curve $\unicode[STIX]{x1D708}:C\rightarrow S_{X}$ which is not necessarily an embedding, we can assume by Lemma 3.5 that $C\rightarrow S_{X}$ is a finite l.c.i. morphism. We now factor $\unicode[STIX]{x1D708}$ as $C{\hookrightarrow}\mathbb{P}_{S_{X}}^{N}\xrightarrow[{}]{\unicode[STIX]{x1D70B}}S_{X}$ , where $\unicode[STIX]{x1D70B}$ is the projection and $\unicode[STIX]{x1D707}:C{\hookrightarrow}\mathbb{P}_{S_{X}}^{N}$ is a regular embedding. By Proposition 2.3, we can identify $\mathbb{P}_{S_{X}}^{N}=\mathbb{P}_{S(X,D)}^{N}$ with $S(\mathbb{P}_{X}^{N},\mathbb{P}_{D}^{N})$ . We have then a commutative diagram

by the definition of $\unicode[STIX]{x1D704}_{\pm }^{\ast }$ . For $f\in {\mathcal{O}}_{C,\unicode[STIX]{x1D708}^{\ast }D}^{\times }$ , we have $\unicode[STIX]{x1D704}_{\pm }^{\ast }(\unicode[STIX]{x1D707}_{\ast }((f)_{C}))\in {\mathcal{R}}_{0}(\mathbb{P}_{X}^{N})$ by the embedded case. In particular, we get $\unicode[STIX]{x1D704}_{\pm }^{\ast }(\unicode[STIX]{x1D708}_{\ast }((f)_{C}))=\unicode[STIX]{x1D704}_{\pm }^{\ast }\unicode[STIX]{x1D70B}_{\ast }\unicode[STIX]{x1D707}_{\ast }((f)_{C})=\unicode[STIX]{x1D70B}_{\ast }\unicode[STIX]{x1D704}_{\pm }^{\ast }(\unicode[STIX]{x1D707}_{\ast }((f)_{C}))\in {\mathcal{R}}_{0}(X)$ , completing the proof.◻

Proof. Let $\unicode[STIX]{x1D6FF}:{\mathcal{Z}}_{0}(S_{X},D)\rightarrow \text{CH}_{0}(X|D)$ be the composition

We have to show that $\unicode[STIX]{x1D6FF}$ factors through $\text{CH}_{0}(S_{X})$ , defined using good l.c.i. curves as in Definition 3.4. Using again Lemma 3.5, we have to show more precisely that $\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D708}_{\ast }(\text{div}_{C}(f)))=0$ for every $\unicode[STIX]{x1D708}:C\rightarrow S_{X}$ finite l.c.i. morphism from a reduced curve $C$ that is good relative to $(S_{X},D)$ and for every rational function $f$ on $C$ that is regular and invertible along $E=(\unicode[STIX]{x1D708}^{-1}(D)\cup C_{\text{sing}})$ . We factor $\unicode[STIX]{x1D708}$ as composition $\unicode[STIX]{x1D708}=\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D707}$ , where $\unicode[STIX]{x1D707}:C{\hookrightarrow}\mathbb{P}_{S_{X}}^{N}=S(\mathbb{P}_{X}^{N},P_{D}^{N})$ (using Proposition 2.3) is a regular embedding and $\unicode[STIX]{x1D70B}:\mathbb{P}_{S_{X}}^{N}\rightarrow S_{X}$ is the projection. In particular, $\unicode[STIX]{x1D707}(C)=C$ is a Cartier curve on the double $S(\mathbb{P}_{X}^{N},P_{D}^{N})$ relative to $\mathbb{P}_{D}^{N}$ .

It follows from (4.8) and the formula $\unicode[STIX]{x1D6FF}=\unicode[STIX]{x1D704}_{+}^{\ast }-\unicode[STIX]{x1D704}_{-}^{\ast }$ that the square

commutes, where $\unicode[STIX]{x1D6FF}_{\mathbb{P}_{S_{X}}^{N}}$ is the composition ${\mathcal{Z}}_{0}(\mathbb{P}_{S_{X}}^{N},\mathbb{P}_{D}^{N})\rightarrow \text{CH}_{0}^{\text{LW}}(S(\mathbb{P}_{X}^{N},\mathbb{P}_{D}^{N}))\rightarrow \text{CH}_{0}(\mathbb{P}_{X}^{N}|\mathbb{P}_{D}^{N})$ . That is, $\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D708}_{\ast }((f)_{C}))=\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D70B}_{\ast }(\unicode[STIX]{x1D707}_{\ast }(f)_{C}))=\unicode[STIX]{x1D70B}_{\ast }(\unicode[STIX]{x1D6FF}_{\mathbb{P}_{S_{X}}^{N}}(\unicode[STIX]{x1D707}_{\ast }((f)_{C})))$ .

By assumption, we have $\unicode[STIX]{x1D6FF}_{\mathbb{P}_{S_{X}}^{N}}(\unicode[STIX]{x1D707}_{\ast }((f)_{C}))=0\in \text{CH}_{0}(\mathbb{P}_{X}^{N}|\mathbb{P}_{D}^{N})$ . Since the push-forward map $\unicode[STIX]{x1D70B}_{\ast }:\text{CH}_{0}(\mathbb{P}_{X}^{N}|\mathbb{P}_{D}^{N})\rightarrow \text{CH}_{0}(X|D)$ is well defined, we can finish the proof.◻

Notation 5.3. Let $S$ be a quasi-projective $k$ -scheme and let ${\mathcal{L}}$ be a line bundle on $S$ . For a global section $t\in H^{0}(S,{\mathcal{L}})$ , we write $(t)$ for the divisor of zeros of $s$ , that we consider as a closed subscheme of $S$ .

Proof. Let $C{\hookrightarrow}S_{X}$ be a reduced Cartier curve relative to $D$ and let $f\in {\mathcal{O}}_{C,C\cap D}^{\times }$ . Since $C$ is Cartier along $D=(S_{X})_{\text{sing}}$ , it follows that it is Cartier in $S_{X}$ along each of its generic points.

We will replace $C$ by the curves of desired type using a combination of [Reference LevineLev87, Lemma 1.3] and [Reference LevineLev87, Lemma 1.4] as follows. We also refer to [Reference JouanolouJou83, ch. I, § 6] for the Bertini theorem for geometrically reduced schemes and [Reference SeidenbergSei50, Theorem 12] for irreducible schemes.

Since $S_{X}$ is reduced quasi-projective, $X$ is integral and $C{\hookrightarrow}S_{X}$ is Cartier along $D$ , we can apply [Reference LevineLev87, Lemma 1.3] to find a locally closed embedding $S_{X}{\hookrightarrow}\mathbb{P}_{k}^{N}$ and distinct hypersurfaces $H_{1},\ldots ,H_{d-2}{\hookrightarrow}\mathbb{P}_{k}^{N}$ such that the following hold:

1. (1) $Y=S_{X}\cap H_{1}\cap \cdots \cap H_{d-2}$ is a complete intersection which is (geometrically) reduced (note that since $k$ is perfect, $Y$ is reduced if and only if is geometrically reduced);

2. (2) $X_{\pm }\cap Y=X_{\pm }\cap H_{1}\cap \cdots \cap H_{d-2}:=Y_{\pm }$ are integral;

3. (3) no component of $Y$ is contained in $D$ ;

4. (4) $C\subset Y$ ;

5. (5) $C$ is locally principal in $Y$ in a neighborhood of the finite set $C\cap D$ and at each generic point of $C$ .

Since $X_{\pm }=X$ are smooth, we can apply [Reference Kleiman and AltmanKL79, Theorem 1] to further assume that $Y_{\pm }$ are smooth away from $C_{\pm }$ . Here $C_{\pm }$ denotes $C\cap X_{\pm }=C\times _{S_{X}}X_{\pm }$ .

Since $Y{\hookrightarrow}S_{X}$ is constructed as a complete intersection of hypersurfaces of arbitrarily large degrees (see [Reference LevineLev87, Lemma 1.4]), we can furthermore find a locally principal closed subscheme $C_{1}$ of $Y$ such that:

1. (a) $C\subset C_{1}$ ;

2. (b) $C_{1}$ equals $C$ at every generic point of $C$ ;

3. (c) $\overline{(C_{1}\setminus C)}\cap C\cap D=\emptyset$ .

Since $Y$ is geometrically reduced, it follows from Lemma 2.2 that $Y$ is the join of $Y_{\pm }$ along $D$ and there is a short exact sequence

Since $C_{1}$ is locally principal in $Y$ , it follows that $C_{1}\cap X_{\pm }=C_{1}\cap Y_{\pm }$ are locally principal in $Y_{\pm }$ . Since $Y_{\pm }$ are integral surfaces, it follows that $C_{1}\cap X_{\pm }$ are Cartier divisors in $Y_{\pm }$ . It follows from (5.2) that $C_{1}$ is a Cartier divisor in $Y$ . Since $S_{X}$ is Cohen–Macaulay and $Y\subset S_{X}$ is a complete intersection, it follows that $Y$ is also Cohen–Macaulay. Since $C_{1}\subset Y$ is a Cartier divisor, we conclude that $C_{1}$ is Cohen–Macaulay.

It follows from (c) that $f$ extends to a function $g\in {\mathcal{O}}_{C_{1},C_{1}\cap D}^{\times }$ by setting $g=f$ on $C$ and $g=1$ on $C_{1}\setminus C$ . In particular, we have $\operatorname{div}(f)=\operatorname{div}(g)$ . This finishes the proof.◻

Proof. In this proof, we shall assume that $Y=S_{X}$ if $d=2$ . In this case, a Cartier curve on $S_{X}$ along $D=(S_{X})_{\operatorname{sing}}$ must be an effective Cartier divisor on $S_{X}$ . Hence we can assume that $C$ is an effective Cartier divisor on $Y$ for any $d\geqslant 2$ . Notice that $Y=Y_{+}\amalg _{D}Y_{-}$ , again by Lemma 2.2.

Arguing as in Lemma 5.4, since $S_{X}$ (and hence $Y$ ) is quasi-projective over $k$ , we can find an effective Cartier divisor $\tilde{C}$ on $Y$ such that:

1. (1) $C\subset \tilde{C}$ ;

2. (2) $\overline{\tilde{C}\setminus C}\cap (C\cap D)=\emptyset$ ;

3. (3) ${\mathcal{O}}_{Y}(\tilde{C})$ is a very ample line bundle on $Y$ .

We can extend the function $f$ on $C$ to a function $\tilde{f}$ on $\tilde{C}$ by setting $\tilde{f}=f$ on $C$ and $\tilde{f}=1$ on $\tilde{C}\setminus C$ . Condition (2) guarantees that $\tilde{f}$ is regular and invertible at each point of $\tilde{C}\cap D$ , and it is clear by construction that $\operatorname{div}(\tilde{f})=\operatorname{div}(f)$ . Replacing $C$ with $\tilde{C}$ (and changing the notation for simplicity), we can thus assume that $C$ is a Cartier divisor on $Y$ such that the associated line bundle ${\mathcal{O}}_{Y}(C)$ is very ample. Choose $t_{0}\in H^{0}(Y,{\mathcal{O}}_{Y}(C))$ such that $C=(t_{0})$ . Since $Y$ is geometrically reduced, standard Bertini (see the proof of [Reference LevineLev87, Lemma 1.3]) allows us to choose another divisor $C_{\infty }$ in the linear system $H^{0}(Y,{\mathcal{O}}_{Y}(C))$ such that:

1. (1) $C_{\infty }$ is reduced;

2. (2) $C_{\infty }\cap C\cap D=\emptyset$ ;

3. (3) $C_{\infty }$ contains no component of $C$ ;

4. (4) $D$ contains no component of $C_{\infty }$ .

Denote by $t_{\infty }$ the section of ${\mathcal{O}}_{Y}(C)$ with $C_{\infty }=(t_{\infty })$ . As before, we extend the function $f$ on $C$ to a function $h$ on $C_{\infty }\cup C$ by setting $h=f$ on $C$ and $h=1$ on $C_{\infty }$ . Notice that $h$ is meromorphic on $C_{\infty }\cup C$ and regular invertible in a neighborhood of $(C_{\infty }\cup C)\cap D$ by (2). Let $S$ denote the finite set of points $S=\{\text{poles of }f\}\cup (C\cap C^{\infty })$ . Note that $S\cap D=\emptyset$ .

Now choose a very ample line bundle ${\mathcal{L}}=j^{\ast }({\mathcal{O}}_{\mathbb{P}^{r}}(1))$ on $S_{X}$ , corresponding to an embedding $j:S_{X}\rightarrow \mathbb{P}_{k}^{r}$ for $r\gg 0$ of $S_{X}$ as locally closed subscheme. We can assume that $r$ is sufficiently large so that ${\mathcal{L}}_{|Y}\otimes _{{\mathcal{O}}_{Y}}{\mathcal{O}}_{Y}(C)$ is also very ample on $Y$ . Since $k$ is infinite, there is a dense open subset $V_{X}$ of the dual projective space $(\mathbb{P}_{k}^{r})^{\vee }$ such that for $L\in V_{X}$ , the scheme theoretic intersections $L\cdot S_{X}=L\times _{\mathbb{P}_{k}^{r}}S_{X}$ and $L\cdot Y=L\times _{\mathbb{P}_{k}^{r}}Y$ satisfy the following list of properties:

1. (a) $L\cdot S_{X}$ and $L\cdot Y$ are geometrically reduced (since both $S_{X}$ and $Y$ are);

2. (b) $Y\not \subset L\cdot S_{X}$ ;

3. (c) $L\cdot X_{\pm }=L\times _{\mathbb{P}_{k}^{r}}X_{\pm }$ and $L\cdot Y_{\pm }=L\times _{\mathbb{P}_{k}^{r}}Y_{\pm }$ are integral (since $X_{\pm }$ and $Y_{\pm }$ are);

4. (d) $L\cdot Y\cap (C\cup C_{\infty })\cap D=\emptyset$ ;

5. (e) $L\cdot Y\supset S$ ;

6. (f) $L\cdot Y$ intersects $C_{\infty }\cup C$ in a finite set of points.

The hyperplane $L$ corresponds to a section $l$ of the linear system $H^{0}(\mathbb{P}_{k}^{r},{\mathcal{O}}_{\mathbb{P}^{r}}(1))$ . Write $s_{\infty }$ for the global section of ${\mathcal{L}}$ that is the restriction of $l$ to $S_{X}$ . Then $L\cdot S_{X}=(s_{\infty })$ . Note that we can assume that $s_{\infty }$ does not have poles on $(C\cup C_{\infty })\cap D$ . Let $\overline{S_{X}}$ be the closure of $S_{X}$ in $\mathbb{P}_{k}^{r}$ and let ${\mathcal{I}}$ be the ideal sheaf of $\overline{S_{X}}\setminus S_{X}$ in $\mathbb{P}_{k}^{r}$ . We can find a section $s_{\infty }^{\prime }$ of the sheaf ${\mathcal{I}}\otimes {\mathcal{O}}_{\mathbb{P}_{k}^{N}}(m)$ for some $m\gg 0$ , which restricts to a section $s_{\infty }$ on $S_{X}$ satisfying the properties (a)–(f) on $S_{X}=\overline{S_{X}}\setminus V({\mathcal{I}})$ . This implies in particular that $S_{X}\setminus (s_{\infty })=\overline{S_{X}}\setminus (s_{\infty }^{\prime })$ is affine. Thus, up to taking a further Veronese embedding of $\mathbb{P}_{k}^{r}$ (replacing $s_{\infty }$ with $s_{\infty }^{\prime }$ ) we can assume that (a)–(f) as above as well as the following hold:

1. (g) $S_{X}\setminus (s_{\infty })=S_{X}\setminus L\cdot S_{X}$ is affine.

We now know that $Y_{\pm }$ is smooth away from $C$ and (d) tells us that $(s_{\infty })$ intersects $Y_{\pm }$ along $D$ at only those points which are away from $C$ . It follows that $(s_{\infty })\cap Y_{\pm }\cap D\subset (s_{\infty })\cap (Y_{+})_{\text{reg}}$ . We can then use the Bertini theorem of Altman and Kleiman [Reference Kleiman and AltmanKL79, Theorem 1] to ensure that $(s_{\infty })\cap (Y_{+})_{\text{reg}}$ is smooth away from the subscheme $(C\cup C_{\infty })\cap X_{+}$ where we ask the containment condition. In particular, we can assume that $(s_{\infty })\cap Y_{+}$ is smooth along $D$ . The same holds for $Y_{-}$ as well. We conclude that we can moreover achieve the following property:

1. (h) $(s_{\infty })\cap Y_{\pm }$ are smooth along $D$ .

Consider again the function $h$ on $C\cup C_{\infty }$ . By our choice of $S$ , $h$ is regular on $(C\cup C_{\infty })\setminus S$ . By (g) above, $h$ extends to a regular function $H$ on the affine open $U=S_{X}\setminus (s_{\infty })$ . Since $H$ is a meromorphic function on $S_{X}$ which has poles only along $(s_{\infty })$ , it follows that for $N\gg 0$ the section $Hs_{\infty }^{N}$ is an element of $H^{0}(U,{\mathcal{L}}^{\otimes N})$ which extends to a section $s_{0}$ of ${\mathcal{L}}^{\otimes N}$ on all $S_{X}$ . Since $h$ is regular and invertible at each point of $C\cup C_{\infty }\cap D$ and since $s_{\infty }$ does not have zeros or poles on $C\cup C_{\infty }\cap D$ , it follows that $(s_{0})\cap (C\cup C_{\infty })\cap D=\emptyset$ and (using (f) above) that $(s_{0})$ does not contain any component of $C\cup C_{\infty }$ . Note that up to replacing $s_{0}$ by $s_{0}s_{\infty }^{i}$ , we are free to choose $N$ as large as needed.

Write ${\mathcal{I}}_{C\cup C_{\infty }}$ for the ideal sheaf of $C\cup C_{\infty }$ in $S_{X}$ . We can then find sections $s_{1},\ldots ,s_{m}$ of $H^{0}(S_{X},{\mathcal{L}}^{\otimes N}\otimes {\mathcal{I}}_{C\cup C_{\infty }})$ such that the rational map $\unicode[STIX]{x1D719}:S_{X}{\dashrightarrow}\mathbb{P}_{k}^{m-1}$ that they define is a locally closed immersion on $S_{X}\setminus (C\cup C_{\infty })$ . In particular, there exists an affine open neighborhood $U_{x}$ of every $x\in S_{X}\setminus (C\cup C_{\infty })$ where at least one of the $s_{i}$ is not identically zero and where the $k$ -algebra $k[s_{1}/s_{i},\ldots ,s_{m}/s_{i}]$ generated by $s_{1}/s_{i},\ldots ,s_{m}/s_{i}$ coincides with the coordinate ring of $U_{x}$ . But then the same must be true for the algebra $k[s_{0}/s_{1},s_{1}/s_{i},\ldots ,s_{m}/s_{i}]$ obtained by adding the element $s_{0}/s_{i}$ . Hence, the rational map $\unicode[STIX]{x1D713}:S_{X}{\dashrightarrow}\mathbb{P}_{k}^{m}$ given by the sections $(s_{0},s_{1},\ldots ,s_{m})$ of $H^{0}(S_{X},{\mathcal{L}}^{\otimes }N)$ is also a locally closed immersion on $S_{X}\setminus (C\cup C_{\infty })$ , and since the base locus of the linear system associated to $(s_{0},s_{1},\ldots ,s_{m})$ is $(s_{0})\cap (C\cup C_{\infty })$ , it is in fact a morphism away from $(s_{0})\cap (C\cup C_{\infty })$ .

In particular, $\unicode[STIX]{x1D713}$ is birational (hence separable) and has image of dimension at least two, so that the linear system $V=(s_{0},s_{1},\ldots ,s_{m})$ is not composite with a pencil. By the classical Theorem of Bertini (see, for example, [Reference ZariskiZar58, Theorem I.6.3]), a general divisor $E$ in $V$ is generically geometrically reduced. Moreover, $E$ is itself a Cohen–Macaulay scheme (since $S_{X}$ is Cohen–Macaulay). But a locally noetherian Cohen–Macaulay scheme that is generically reduced is in fact reduced by [Reference Görtz and WedhornGW10, Proposition 14.124]. Hence the general divisor $E$ in $V$ is indeed geometrically reduced (hence reduced). We can apply the same argument to $E\cdot Y$ , noting that being a complete intersection in $S_{X}$ , the surface $Y$ is Cohen–Macaulay.

Since $Y_{\pm }$ and $X_{\pm }$ are all integral, a general divisor $E$ of $V$ will be moreover irreducible when restricted to $X_{\pm }$ and $Y_{\pm }$ . We can therefore assume that there is a global section $s_{0}^{\prime }$ of $H^{0}(S_{X},{\mathcal{L}}^{\otimes N})$ of the form $s_{0}^{\prime }=s_{0}+\unicode[STIX]{x1D6FC}$ with $\unicode[STIX]{x1D6FC}\in (s_{1},\ldots ,s_{m})$ , that satisfies the following properties:

(a $^{\prime }$ )

$(s_{0}^{\prime })$ and $(s_{0}^{\prime })\cap Y$ are (geometrically) reduced;

(b $^{\prime }$ )

$Y\not \subset (s_{0}^{\prime })$ ;

(c $^{\prime }$ )

$(s_{0}^{\prime })\cap X_{\pm }$ and $(s_{0}^{\prime })\cap Y_{\pm }$ are integral;

(d $^{\prime }$ )

$(s_{0}^{\prime })\cap (C\cup C_{\infty })\cap D=\emptyset$ ;

(e $^{\prime }$ )

$C_{\infty }\cup C$ contains no component of $(s_{0}^{\prime })\cap Y$ ;

( $\text{f}^{\prime }$ )

$(s_{0}^{\prime })\cap Y_{\pm }$ are smooth along $D$ .

The items (a $^{\prime }$ ) and (c $^{\prime }$ ) follow from the previous discussion. Properties (b $^{\prime }$ ), (d $^{\prime }$ ) and (e $^{\prime }$ ) are clearly open conditions on the space of sections $V$ , and are therefore satisfied by the general divisor. As for the last item ( $\text{f}^{\prime }$ ), it follows from the classical Theorem of Bertini on smoothness applied to $Y_{\pm }$ , the assumption on $Y_{\pm }$ and the item (d $^{\prime }$ ).

We then have

Since $\unicode[STIX]{x1D6FC}$ vanishes along $C\cup C_{\infty }$ and $s_{\infty }$ does not vanish identically on $U\cap (C\cup C_{\infty })$ by (f), it follows that $H_{|(C\cup C_{\infty })\cap U}^{\prime }=H_{|(C\cup C_{\infty })\cap U}=h_{|U}$ . In other words, we have $s_{0}^{\prime }/s_{\infty }^{N}=h$ as rational functions on $C\cup C_{\infty }$ . We can now compute

Setting $(s_{\infty }^{Y})=(s_{\infty })\cap Y$ and $(s_{0}^{\prime Y})=(s_{0}^{\prime })\cap Y$ , we get

where $f^{\prime }=(t_{0}/t_{\infty })|_{(s_{0}^{\prime Y})}\in {\mathcal{O}}_{(s_{0}^{\prime Y}),D\cap (s_{0}^{\prime Y})}^{\times }$ (by (d $^{\prime }$ )) and $f^{\prime \prime }=(t_{0}/t_{\infty })|_{(s_{\infty }^{Y})}\in {\mathcal{O}}_{(s_{\infty }^{Y}),D\cap (s_{\infty }^{Y})}^{\times }$ (by (d)).

It follows from (h) and ( $\text{f}^{\prime }$ ) that $(s_{0}^{\prime Y})_{|X_{+}},~(s_{0}^{\prime Y})_{|X_{-}},~(s_{\infty }^{Y})_{|X_{+}}$ and $(s_{\infty }^{Y})_{|X_{-}}$ are all smooth along $D$ . Setting ${\mathcal{L}}^{\prime \prime }=({\mathcal{L}}^{\prime })^{N},~t^{\prime \prime }=(s_{0}^{\prime })$ and $t^{\prime }=(s_{\infty })$ , the curves $C^{\prime }=(t^{\prime })\cap Y$ and $C^{\prime \prime }=(t^{\prime \prime })\cap Y$ together with the functions $f^{\prime }$ and $f^{\prime \prime }$ satisfy the conditions of the lemma.◻