Proof. For part (a), we follow the approach of [Reference Hartshorne and Polini24, Theorem 4.5]. Let $\hat X$ be the formal completion of X along Y. We recall the proof of [Reference Hartshorne20, IV, Theorem 3.1], which shows that if Y meets every effective divisor of X and U is any open neighborhood of Y, then the restriction $\textrm{Pic}\ X\to \textrm{Pic}\ U$ is an isomorphism and, moreover, if the effective Lefschetz condition $\textrm{Leff}(X,Y)$ holds (see [Reference Hartshorne20, IV, §1]), then the natural map $\textrm{Pic}\ U\to \textrm{Pic}\ \hat X$ is an isomorphism.

In the current situation, since Y is ample on X, clearly Y must meet every effective divisor on X, and $\textrm{Leff}(X,Y)$ holds by [Reference Grothendieck16, X, Example 2.2]. Therefore, both isomorphisms in the preceding paragraph hold and the natural map $\textrm{Pic}\ X \to \textrm{Pic}\ \hat X$ is an isomorphism.

Next, we use the isomorphism $\displaystyle {\textrm{Pic}\ \hat X \cong \lim _{\longleftarrow }\textrm{Pic}\ Y_n}$ [Reference Hartshorne19, Lemma 8.2]. Kodaira vanishing implies that $H^1({\mathcal I}_Y^n/{\mathcal I}_Y^{n+1}) =0$ for all $n\ge 1$ and hence the cohomology fragment

$$\begin{align*}H^1({\mathcal I}_Y^n/{\mathcal I}_Y^{n+1}) \to \textrm{Pic}\ Y_{n+1} \to \textrm{Pic}\ Y_{n} \to H^2({\mathcal I}_Y^n/{\mathcal I}_Y^{n+1}) \end{align*}$$

shows that $\textrm{Pic}\ Y_{n+1}$ injects into $\textrm{Pic}\ Y_n$ for all $n\ge 1$ , so that we can interpret the inverse limit $\displaystyle {\lim _{\longleftarrow } \textrm{Pic}\ Y_n}$ as the intersection $\bigcap \textrm{Pic}\ Y_n$ . Moreover, each quotient is contained in a ${\mathbb C}$ -vector space and thus torsion-free, so the torsion subgroup of $\textrm{Pic}\ Y_{n+1}$ maps isomorphically to the torsion subgroup of $\textrm{Pic}\ Y_n$ . Therefore, $\displaystyle {\textrm{Pic}\ X \cong \textrm{Pic}\ \hat X \cong \lim _{\longleftarrow } \textrm{Pic}\ Y_n} = \cap \textrm{Pic}\ Y_n$ injects into $\textrm{Pic}\ Y$ and the torsion subgroup of $\textrm{Pic}\ Y$ lifts to $\textrm{Pic}\ X$ , so this injection is an isomorphism on torsion.

For part (b), the Kodaira vanishings $h^1 ({\mathcal O}_X (-Y))=h^2 ({\mathcal O}_X (-Y))=0$ combined with the exact sequence $0 \to {\mathcal O}_X (-Y) \to {\mathcal O}_X \to {\mathcal O}_Y \to 0$ yield $H^1 ({\mathcal O}_Y)=0$ . The exponential sequence on Y gives the long exact cohomology sequence

$$\begin{align*}0 \to H^1(Y_h,{\mathbb Z}) \to H^1 ({\mathcal O}_Y) \to \textrm{Pic}\ Y \to H^2 (Y_h, {\mathbb Z}) \to H^2 ({\mathcal O}_Y) \to \cdots, \end{align*}$$

where $Y_h$ is the complex analytic space associated with Y. This shows that $\textrm{Pic}\ Y$ is a finitely generated abelian group, since $H^2 (Y_h, {\mathbb Z})$ is.

Let $L \subset Y$ be the conductor with inverse image $\Gamma = \pi ^{-1} (L) \subset S$ . These are projective curves of pure dimension $1$ by the theory of generalized divisors [Reference Hartshorne22], as L is the inverse to ${\mathcal O}_S$ , thought of as an ${\mathcal O}_Y$ -module. In this setting, there is a commutative diagram of short exact sequences

(2.1)

in which the third vertical map is an isomorphism [Reference Hartshorne and Polini24, proof of Theorem 4.1]. Since Y and S are integral projective surfaces, the map $H^0 ({\mathcal O}_Y^*) \to H^0 ({\mathcal O}_S^*)$ is an isomorphism, both being equal to the unit group $\mathbb C^*$ . The long exact cohomology sequence associated with the top row thus shows that $H^0 (\mathcal N)$ is the kernel of the map $\textrm{Pic}\ Y \to \textrm{Pic}\ S$ , so $H^0 (\mathcal N)$ is a finitely generated abelian group.

Now, we use the bottom row in Diagram (2.1). If $D = \textrm{Supp}\ \Gamma $ with connected components $D_1, \dots , D_r$ , then $H^0 ({\mathcal O}_D^*) \cong (k^*)^r$ and the exact sequence

$$\begin{align*}0 \to H^0 (\mathcal I) \to H^0 ({\mathcal O}_{\Gamma}^*) \to H^0 ({\mathcal O}_D^*) \to 0 \end{align*}$$

from [Reference Hartshorne and Polini24, Proposition 5.9] show that $H^0 ({\mathcal O}^*_{\Gamma })$ is a direct sum of copies of k and $k^*$ , because $H^0 (\mathcal I)$ is divisible and hence injective, so that the sequence splits. Looking at the bottom row of Diagram (2.1), we see that the image of $H^0 ({\mathcal O}_{\Gamma }^*)$ in $H^0 (\mathcal N_0)$ must be divisible; since finitely generated divisible groups are trivial, $H^0 ({\mathcal O}_{\Gamma }^*)\to H^0 (\mathcal N_0)$ is the zero map. Therefore, $H^0 (\mathcal N_0)$ is the kernel of the map $\textrm{Pic}\ L \to \textrm{Pic}\ \Gamma $ . Using the decomposition $\textrm{Pic}\ L \cong {\mathbb Z} \oplus \textrm{Pic} ^0 L$ associated with the degree of a line bundle and the similar decomposition for $\textrm{Pic}\ \Gamma $ shows that the associated map on ${\mathbb Z}$ is injective. Thus, $H^0 (\mathcal N_0)$ is the kernel of the induced map $\textrm{Pic} ^0 L \to \textrm{Pic} ^0 \Gamma $ , but this is a map of group schemes of finite type over ${\mathbb C}$ [Reference Kleiman25, Proposition 5.3] and hence the kernel $H^0 (\mathcal N_0)$ is also a group scheme. Being a finitely generated abelian group, it is discrete; since it is of finite type over ${\mathbb C}$ , it must be finite, proving part (b).

For part c(i), if $\textrm{Pic}\ X$ is torsion-free, then so is $\textrm{Pic}\ Y$ by part (a), so its finite subgroup $H^0 (\mathcal N_0)$ is zero and p is injective. For part c(ii), if $\textrm{Pic}\ L$ is finitely generated, then it is discrete, so the connected component of the identity $\textrm{Pic} ^0 L$ must be zero, so again its subgroup $H^0(\mathcal N_0) = 0$ .

Hypothesis 3.1. Let C be a smooth connected curve on a smooth complex threefold X with very ample line bundle ${\mathcal O}_X (1)$ satisfying $H^1 ({\mathcal O}_X (1))=0$ . Define $Z \subset X$ by ${\mathcal I}_Z = {\mathcal I}_C^m$ for $m>1$ and assume d large enough that ${\mathcal I}_Z (d-1)$ is globally generated.

Observe the following consequences of Hypothesis 3.1. Since ${\mathcal I}_Z (d-1)$ is globally generated, so is ${\mathcal I}_Z (d)$ so that Z is the scheme-theoretic base locus of the linear system ${\mathbb P} H^0 ({\mathcal I}_Z (d))$ , giving a morphism $X - Z \to {\mathbb P} H^0 ({\mathcal I}_Z (d))^*$ , which extends to the blowup $\pi : \widetilde X \to X$ at Z [Reference Hartshorne21, II, Example 7.17.3], defined by the line bundle $dH - F$ , where F is the exceptional divisor and $\widetilde H = \pi ^* H$ is the hyperplane section. Moreover, global generation of ${\mathcal I}_Z (d-1)$ ensures that the extension $\sigma : \widetilde X \to {\mathbb P} H^0 ({\mathcal I}_Z (d))^*$ is a closed immersion [Reference Beltrametti and Sommese1, Proposition 2.1]. An element $f \in H^0 ({\mathcal I}_Z (d))$ modulo nonzero scalar corresponds to a surface $Y \subset X$ containing Z and also a hyperplane $H_f \subset {\mathbb P} H^0 ({\mathcal I}_Z (d))^*$ . The connection between these is that $S = \sigma ^{-1} (H_f)$ is the strict transform of Y under the map $\pi $ so that $\pi ^{-1} (Y) = S \cup E$ , where E is the exceptional divisor for the blowup $\widetilde X \to X$ . Note that the blowup $\widetilde X \to X$ at Z is isomorphic to the blowup at C [Reference Hartshorne21, II, Exercise 7.11(a)], but if E is the exceptional divisor for the blowup at C, then $F=mE$ is the exceptional divisor for the blowup at Z, thus the embedding $\sigma $ is defined by the invertible sheaf $d \widetilde H - mE$ , and we get the following diagram:

(3.1)

Letting $\Gamma = E \cap S$ , we obtain a further diagram

(3.2)

Proof. By [Reference Hartshorne21, II, Exercise 7.11(a)], the blowup $\pi : \widetilde X \to X$ along Z is isomorphic to the blowup along C, hence is smooth [Reference Hartshorne21, II, Theorem 8.24]. We can understand the strict transform S of Y to be the hyperplane section under $\sigma $ of its equation $f \in H^0 ({\mathcal I}_Z (d))$ , because the two are isomorphic away from C and S is smooth and connected by Bertini’s theorem.

Here, $\pi \vert _S:S\to Y$ is quasi-finite, since it is an isomorphism away from $\Gamma $ and $\pi \vert _\Gamma $ is a (finite) morphism of curves. Since $\pi $ is projective, it is finite [Reference Hartshorne21, III, Exercise 11.2], so locally ${\mathcal O}_Y \subset {\mathcal O}_S$ is module-finite and thus an integral ring extension in the fraction field of ${\mathcal O}_Y$ . It follows that ${\mathcal O}_S$ is contained in the integral closure of ${\mathcal O}_Y$ , but since regular rings are integrally closed, ${\mathcal O}_S$ is equal to the integral closure of ${\mathcal O}_Y$ , hence S is the normalization of Y.

Proof. By definition, the blowup is $\widetilde X = \textbf{Proj} \bigoplus _{d \geq 0} {\mathcal I}_C^d$ and the exceptional divisor can be written $E = \textbf{Proj} \bigoplus ({\mathcal I}_C^d \otimes {\mathcal O}_C)\text = \textbf{Proj} \bigoplus {\mathcal I}_C^d/{\mathcal I}_C^{d+1}$ . Since C is a local complete intersection, ${\mathcal I}_C/{\mathcal I}_C^2$ is locally free and $\textrm {Sym}^d {\mathcal I}_C/{\mathcal I}_C^2 = {\mathcal I}_C^d/{\mathcal I}_C^{d+1}$ for $d \geq 1$ [Reference Hartshorne21, II, Theorem 8.21A(e)] so that $E \cong {\mathbb P} ({\mathcal I}_C/{\mathcal I}_C^2)$ . Since ${\mathcal I}_E$ corresponds to ${\mathcal O} (1)$ under the $\textbf{Proj}$ construction of $\widetilde X$ , ${\mathcal I}_E \otimes {\mathcal O}_E$ also corresponds to ${\mathcal O} (1)$ under the $\textbf{Proj}$ construction for ${\mathbb P} ({\mathcal I}_C/{\mathcal I}_C^2)$ so that $\pi _* ({\mathcal I}_E^k\otimes {\mathcal O}_E) = {\mathcal I}_C^k/{\mathcal I}_C^{k+1}$ for $k \geq 0$ by [Reference Hartshorne21, II, Proposition 7.11(a)].

The case $k=0$ gives $\pi _* {\mathcal O}_E = {\mathcal O}_C$ . By Zariski’s main theorem, $\pi _* {\mathcal O}_{\widetilde X} = {\mathcal O}_X$ because $\pi $ is birational and X is normal; therefore, applying $\pi _*$ to $0 \to {\mathcal I}_E \to {\mathcal O}_{\widetilde X} \to {\mathcal O}_E \to 0$ shows that $\pi _* {\mathcal I}_E = {\mathcal I}_C$ . Inducting on k, we apply $\pi _*$ to the sequence $0 \to {\mathcal I}_E^{k+1} \to {\mathcal I}_E^k \to {\mathcal I}_E^k \otimes {\mathcal O}_E \to 0$ to see that $\pi _* {\mathcal I}_E^{k+1}$ is the kernel of the natural map ${\mathcal I}_C^k \to {\mathcal I}_C^k/{\mathcal I}_C^{k+1}$ , which is ${\mathcal I}_C^{k+1}$ .

Proof. By Remark 3.4(b), to apply Theorem 3.3, we need that $H^0 (K_{\widetilde X} + (d\widetilde H - mE)) \neq 0$ . If we choose $d_0$ such that $d_0\widetilde H - mE$ is very ample, then $K_{\widetilde X} + k(d_0\widetilde H-mE)$ has a nonzero section for sufficiently large k, and therefore $K_{\widetilde X} + kd_0\widetilde H - mE = K_{\widetilde X} + k(d_0\widetilde H-mE) + (km-m)E$ has a nonzero section, as well as $K_{\widetilde X} + (d\widetilde H - mE)$ for all $d\ge kd_0$ .

In more detail, the canonical bundle on $\widetilde X$ is $K_{\widetilde X} = \pi ^* K_X + E$ [Reference Hartshorne21, II, Exercise 8.5(b)], so we need the non-vanishing $H^0 (K_{\widetilde X} (d \widetilde H-mE)) = H^0 (\pi ^* K_X (dH-(m-1)E)) \neq 0$ . This condition is equivalent to $H^0 (\pi _* {\mathcal I}_E^{m-1} (K_X + dH)) \neq 0$ via the projection formula, which is equivalent to $H^0 ({\mathcal I}_C^{m-1} (K_X+dH)) \neq 0$ by Lemma 3.5.

Proof. Since ${\mathcal I}_C (k-1)$ is globally generated, $k \widetilde H - E$ is very ample and embeds $\widetilde X$ into some ${\mathbb P}^N$ by [Reference Beltrametti and Sommese1, Proposition 2.1] and restriction gives an embedding $E \hookrightarrow {\mathbb P}^N$ . Composing with the m-uple embedding gives an embedding $E \hookrightarrow {\mathbb P}^{\binom {m+d}{d}-1}$ by the divisor $m k \widetilde H - mE$ and each curve in the image of E has degree a multiple of $m> 1$ . Now let $d\ge km$ . Note that $d \widetilde H-mE = mk \widetilde H - mE + (d-mk) \widetilde H$ is the sum of a very ample divisor and a base-point free divisor and hence is very ample [Reference Hartshorne21, Exercise 7.5(d)]. Any curve on E with respect to this embedding has degree at least $m>1$ , so with respect to the divisor $d \widetilde H -mE$ , the image of E is not ruled by lines, and therefore, the restriction $\textrm{Pic}\ E \to \textrm{Pic}\ \Gamma $ is injective for general $\Gamma \in |d \widetilde H - mE|$ by [Reference Lopez27, Proposition 2.2].

Proof. First, notice that ${\mathcal I}_C (\lfloor d/m \rfloor -1)$ globally generated implies that ${\mathcal I}_Z (d-1)$ is globally generated, so Hypothesis 3.1 hold. If we take $k = \lfloor d/m \rfloor $ , the hypothesis of Lemma 3.8 holds and hence its conclusion. Diagram (3.2) induces by pullback the diagram

(3.6)

The vertical map on the left is a split injection with cokernel the free abelian group generated by the class E of the exceptional divisor [Reference Hartshorne21, II, Exercise 8.5(a)]. Furthermore, the ruled surface $E \to C$ gives rise to a split injection $\textrm{Pic}\ C \to \textrm{Pic}\ E$ with cokernel a free abelian group generated by a section $C_0 \subset E$ [Reference Hartshorne21, V, Proposition 2.3]. Thus, Diagram (3.6) extends to the diagram

(3.7)

with short exact rows where the vertical map $\alpha $ sends $1$ to $E|_E$ modulo $\textrm{Pic}\ C$ .

Let $j: E \hookrightarrow \widetilde X$ denote the inclusion. Then $j^* (E)$ is the relative ${\mathcal O} (1)$ for the $\textbf{Proj}$ construction of E as a ruled surface over C [Reference Hartshorne21, II, Theorem 8.24(c)], but the section $C_0$ is a divisor corresponding to ${\mathcal O} (1)$ [Reference Hartshorne21, V, Proposition 2.8], which generates the cokernel of $\textrm{Pic}\ C \to \textrm{Pic}\ E$ , therefore $\alpha $ is an isomorphism, and $\textrm{Pic}\ X$ is the kernel of the induced map $\textrm{Pic}\ \widetilde X \to \textrm{Pic}\ E/\pi ^* \textrm{Pic}\ C$ .

Via the isomorphism $\textrm{Pic}\ S \cong \textrm{Pic}\ \widetilde X \cong \textrm{Pic}\ X \oplus E \mathbb Z$ , we see from the diagram

that the restriction map $\textrm{Pic}\ X \to \textrm{Pic}\ Y$ is injective, since the upper horizontal map is a split injection and the right vertical map is an isomorphism. More generally, the restriction map $\textrm{Pic}\ X \to \textrm{Pic}\ Y$ is injective for any integral ample divisor $Y \subset X$ by a well-known argument [Reference Grothendieck16, XII, Corollaire 3.6].

Now, we prove the theorem. Consider the sequence of maps:

$$\begin{align*}\textrm{Pic}\ X \stackrel{r}{\to} \textrm{Pic}\ Y \stackrel{\pi^*}{\to} \textrm{Pic}\ S \cong \textrm{Pic}\ \widetilde X \to \textrm{Pic}\ E/\pi^*\textrm{Pic}\ C \hookrightarrow \textrm{Pic}\ \Gamma/\pi^*\textrm{Pic}\ C. \end{align*}$$

In view of sequence (3.4), the composite map $\textrm{Pic}\ Y \to \textrm{Pic}\ \Gamma /\pi ^*\textrm{Pic}\ C$ is zero. By Lemma 3.8, the composite map $\textrm{Pic}\ Y \to \textrm{Pic}\ E/\pi ^*\textrm{Pic}\ C$ is also the zero map, but the kernel of the map $\textrm{Pic}\ \widetilde X \to \textrm{Pic}\ E/\pi ^*\textrm{Pic}\ C$ is exactly $\textrm{Pic}\ X \subset \textrm{Pic}\ \widetilde X$ by the argument following Diagram (3.7).

Therefore, r and $\pi ^*$ above induce maps $\textrm{Pic}\ X \stackrel {r}{\to } \textrm{Pic}\ Y \stackrel {\pi ^*}{\to } \textrm{Pic}\ X$ whose composite is the identity and such that r is injective. Also, $\textrm{Ker}\pi ^*$ is finite, and thus torsion, by Theorem 2.1(b). By Theorem 2.1(a), r is an isomorphism on torsion, so any element of $\textrm{Ker} \pi ^*$ comes from an element of (the torsion subgroup of) $\textrm{Pic}\ X$ , and since $\pi ^*\circ r$ is the identity, this element must have already been $0$ . Thus, $\pi ^*:\textrm{Pic}\ Y \to \textrm{Pic}\ X$ is an isomorphism.

Proof. We verify that these conditions satisfy the hypotheses of Theorem 3.9. First, clearly $H^1({\mathcal O}_{{\mathbb P}^3}) = 0$ . Next, since $d\ge (s+1)m$ , $\lfloor \frac {d}{m}\rfloor \ge s+1$ , and since ${\mathcal I}_C(s)$ is generated by global sections, so is ${\mathcal I}_C(d-1)$ . Finally, the $(m-1)^{\mathrm {st}}$ power of the defining polynomial of the degree-r surface is a global section of ${\mathcal I}_C((m-1)r)$ , which is the same as ${\mathcal I}_C^{m-1}(K_{{\mathbb P}^3}(rm-r+4))$ , and thus, ${\mathcal I}_C^{m-1}(d)$ has a global section as well.

Proof. Observe that Z is a locally Cohen–Macaulay multiplicity- $r^2$ structure on the line $L=Z(x,y) = \textrm{Supp}\ Z$ . Let $\pi : \widetilde {\mathbb P}^3 \to {\mathbb P}^3$ be the blowup at Z with exceptional divisor $\widetilde E$ . The singular locus of $\widetilde {\mathbb P}^3$ is all of $\widetilde E_{\mathrm {red}}$ [Reference Nollet33, Proposition 2.1] because $r>1$ , so $\widetilde {\mathbb P}^3$ is not normal. Let $n: \overline {\mathbb P}^3 \to \widetilde {\mathbb P}^3$ be the normalization with $\overline E = \textrm{Supp}\ (n^{-1} \widetilde E)$ . We claim that $\overline {\mathbb P}^3$ is isomorphic to the blowup of ${\mathbb P}^3$ at L. With ${\mathbb P}^3 = \textrm{Proj}\ k[x,y,z,w]$ , the line $L = Z(x,y)$ is covered by open affines $w=1$ and $z=1$ . The affine $w=1$ is $U = \textrm{Spec}\ A$ , where $A = k[x,y,z]$ . Blowing up $I_Z = (x^r,y^r)$ yields $\textrm{Proj}\ A[u,v]/(u x^r - v y^r)$ , which is covered by affines $u=1$ and $v=1$ . The patch $u=1$ is $U_1 = \textrm{Spec}\ A[v]/(x^r - v y^r) = \textrm{Spec}\ A[x^r/y^r]$ . Clearly, $x/y$ is integral over $A[x^r/y^r]$ , hence the integral closure of $A[x^r/y^r]$ in the function field K contains $A [x/y]$ . On the other hand, since $A[x/y] \cong k[x,y,z,u]/(x-yu)$ is a regular ring, it is integrally closed, so the integral closure is exactly $A[x/y]$ . The maps on open affines glue to define a map from the blowup of ${\mathbb P}^3$ at L to $\widetilde {\mathbb P}^3$ , which therefore is the normalization map. so that $n^{-1} U_1 = \textrm{Spec}\ A [x/y]$ . We note that n is the map guaranteed by Moody’s result [Reference Moody31, Corollary 1], which characterizes maps that can occur between blowups. Therefore, $\overline {\mathbb P}^3$ is smooth, $\overline E \to \textrm{Supp}\ Z$ is a ${\mathbb P}^1$ -bundle, and $\textrm{Pic}\ \overline {\mathbb P}^3 \cong {\mathbb Z}^2$ generated by $\overline H$ and $\overline E$ [Reference Hartshorne21, II, 6.5]. Note that $n^* \widetilde E = r \overline E$ as Cartier divisors.

Since ${\mathcal I}_Z (r)$ is globally generated, [Reference Beltrametti and Sommese1, Theorem 1.3] tells us that $d \widetilde H - \widetilde E$ is very ample on $\widetilde {\mathbb P}^3$ for $d> r$ , giving an embedding $\sigma $ and a commuting diagram

(5.1)

where $\mathbb P^N = {\mathbb P} H^0 ({\mathcal I}_Z (d))^*$ .

If $\widetilde Y \subset \widetilde {\mathbb P}^3$ is a general hyperplane section of $\sigma $ , then $\widetilde Y$ maps isomorphically via $\pi $ to its image Y and $\widetilde E \cap \widetilde Y$ maps to Z under $\pi $ . This is because $\widetilde E$ is the relative ${\mathcal O} (1)$ for $\textbf{Proj}$ construction of the blowup, so the line bundle that embeds the ${\mathbb P}^1$ fibers under $\sigma $ takes those fibers to straight lines, and a hyperplane intersects those lines in schemes of length $\leq 1$ . In this situation, it is easy to prove that the resulting projection takes those hyperplane sections isomorphically to their images. Thus, the index $|\textrm{Pic}\ Y: {\mathbb Z} H\oplus {\mathbb Z} Z|$ is equal to $|\textrm{Pic}\ \widetilde Y: {\mathbb Z}\tilde H \oplus {\mathbb Z}(Y\cap \widetilde E)|$ . We will show that this latter index divides $(r, d)$ .

Proof. Since ${\mathcal O}_{\widetilde {\mathbb P}} (\widetilde H)$ is globally generated, it suffices to prove that $(n_* K_{\overline {\mathbb P}^3}) ((r+3)\widetilde H - \widetilde E)$ is globally generated. Since $K_{\overline {\mathbb P}^3} = -4 \overline H + \overline E$ [Reference Hartshorne21, II, Exercise 8.5] and $n^* \widetilde E = r \overline E$ , the projection formula gives

$$\begin{align*}(n_* K_{\overline {\mathbb P}^3}) ((r+3)\widetilde H - \widetilde E) = n_*( {\mathcal O}_{\overline {\mathbb P}^3} (- (r-1) \overline E + (r-1)\overline H). \end{align*}$$

To identify the global sections, we push forward to ${\mathbb P}^3$ by $\pi $ to obtain

$$\begin{align*}\pi_* n_* {\mathcal O} ((r-1)\overline H - (r-1) \overline E) = {\mathcal I}_L^{r-1} (r-1) \end{align*}$$

by Lemma 3.5 and the projection formula, and the latter sheaf has global sections spanned by the monomials $x^{r-1}, x^{r-2} y \dots , y^{r-1}$ . Since these sections generate ${\mathcal I}_L^{r-1} (r-1)$ on the open set $U = {\mathbb P}^3 - L$ , where the maps $\pi $ and n are isomorphisms, they generate on U.

To prove that these sections also generate ${\mathcal I}_L^{r-1} (r-1)$ on $\pi ^{-1} L$ , we use local calculations. $n_*( {\mathcal O}_{\overline {\mathbb P}^3} (- (r-1) \overline E)$ is an ideal sheaf on $\widetilde {\mathbb P}^3$ , where we can interpret the monomial sections and show generation at each stalk. Again considering a representative open affine corresponding to the map $ \textrm{Spec}\ A[x^r/y^r] \to \textrm{Spec}\ A [x/y]$ , we see that With ${\mathbb P}^3 = \textrm{Proj}\ k[x,y,z,w]$ , the line $L = Z(x,y)$ is covered by open affines $w=1$ and $z=1$ . The line bundle $-(p-1) \overline E = {\mathcal I}_{\overline E, \overline {\mathbb P}^3}^{p-1}$ on $\overline {\mathbb P}^3$ is the ideal sheaf generated locally by $y^{r-1} \in A[x/y]$ . Observe that $y \in A[x^r/y^r]$ . Since $A[x/y]$ is generated by $1, x/y, (x/y)^2, \dots , (x/y)^{r-1}$ as a module over $A[x^r/y^r]$ and the products $y^{r-1}, (x/y) y^{r-1}, \dots (x^{r-1}/y^{r-1}) y^{r-1}$ lie in the subring $A[x^r/y^r]$ , it follows that the principal ideal $(y^{r-1}) \subset A[x/y]$ is in fact an ideal in the subring $A[x^r/y^r]$ as well. Thus, $n_* {\mathcal O}_{\overline {\mathbb P}^3} (-(r-1) \overline E )$ is an ideal sheaf on $\widetilde {\mathbb P}^3$ which is locally generated on $U_1$ by $y^{r-1}, x y^{r-2}, \dots , x^{r-1}$ . Reversing the roles of x and y, we see that these same elements generate the sheaf on the open affine $v=1$ as well. Reversing the roles of w and z gives global generation on the affine $z=1$ , so the sections $y^{r-1}, x y^{r-2}, \dots , x^{r-1}$ generate $n_* {\mathcal O} ((r-1)H - (r-1) \overline E)$ at every point in $\widetilde {\mathbb P}^3$ .

We now return to the proof of Proposition 5.2. Lemma 5.3 shows that, for large d, $n_* K_{\overline {\mathbb P}^3} (d\widetilde H - \widetilde E)$ is globally generated and so gives a map $f: \overline {\mathbb P}^3 \to {\mathbb P}^N$ , for which the general hyperplane section $\overline Y$ yields an isomorphism $\textrm{Cl}\ \overline {\mathbb P}^3 \to \textrm{Cl}\ \overline Y$ by the theorem of Ravindra and Srinivas in [Reference Ravindra and Srinivas34]. Moreover, $\overline {\mathbb P}^3$ is smooth, hence so is $\overline Y$ by Bertini’s theorem, so we get the same isomorphism on Picard groups, thus $\textrm{Pic}\ Y \cong {\mathbb Z}^2$ , generated by $\overline H$ and $\overline E \cap \overline Y$ . Moreover, $\overline Y$ is the normalization of $\widetilde Y$ [Reference Manaresi29, Corollary II.8], so Theorem 2.1(a) gives an inclusion $n^*: \textrm{Pic}\ \widetilde Y \to \textrm{Pic}\ \overline Y$ ; in order to determine the index $|\textrm{Pic}\ \widetilde Y : {\mathbb Z}\widetilde H \oplus {\mathbb Z} (\widetilde E \cap \widetilde Y)|$ , we identify the image of this pullback map.

For compactness of notation, denote by $\widetilde E, \overline E$ their respective intersections with $\widetilde Y, \overline Y$ . Note that $\widetilde H$ and $\widetilde E$ are Cartier on $\widetilde Y$ and that $n^* \widetilde E = r \overline E$ , so the inclusion $n^*: \textrm{Pic}\ \widetilde Y \to \textrm{Pic}\ \overline Y$ has cokernel generated by the image of $\overline E$ . Identifying $\textrm{Pic}\ \tilde Y$ with its image under $n^*$ gives a chain of subgroups $ {\mathbb Z}\widetilde H \oplus {\mathbb Z}\widetilde E \le \textrm{Pic}\ \widetilde Y \le \textrm{Pic}\ \overline Y$ , and $|\textrm{Pic}\ \overline Y : {\mathbb Z}\widetilde H \oplus {\mathbb Z}\widetilde E| = r$ . Let k be the smallest integer such that $k\overline E\in \textrm{Im} n^*$ ; then, clearly $k\mid m$ ; moreover, $\overline E$ clearly generates $\textrm{Pic}\ \overline Y$ over $ {\mathbb Z}\widetilde H \oplus {\mathbb Z}\widetilde E$ , so $|Pic\ \overline Y: \textrm{Pic}\ \widetilde Y| = k$ and thus $|\textrm{Pic}\ \widetilde Y : {\mathbb Z}\widetilde H \oplus {\mathbb Z}\widetilde E| = \frac {r}{k}$ . We will show that $r\mid dk$ , which will imply that $r\mid (r,d)k$ and therefore that $|\textrm{Pic}\ \widetilde Y : {\mathbb Z}\widetilde H \oplus {\mathbb Z}\widetilde E| = \frac {r}{k} \mid (r,d)$ .

Let $L \subset \widetilde Y$ be the conductor for n with scheme-theoretic inverse image $\Gamma = n^{-1} (L)$ . Since $\widetilde {\mathbb P}^3$ has discrete Picard group, $H^1 ({\mathcal O}_{\widetilde {\mathbb P}^3})=0$ , so Diagram (2.1) applies. The proof of Theorem 2.1c(i) gives $H^0 (\mathcal N)=0$ in that diagram, so taking long exact cohomology gives the commutative diagram

(5.2)

with the right vertical map an isomorphism. From this diagram, it follows that the image of the class of $k\overline E \in \textrm{Pic}\ \Gamma $ is in the image of $\textrm{Pic}\ L$ .

Let $C = \textrm{Supp}\ L$ and $D = \textrm{Supp}\ \Gamma $ . C is supported on the non-normal locus of $\widetilde Y$ , which for general sections of $\sigma $ is $\widetilde Y \cap \widetilde E$ . We will think of C as a curve on the ruled surface $\textrm{Supp}\ \widetilde E$ . Since $n^{-1} \textrm{Supp}\ \widetilde E = \overline E$ and the map $\overline E \to \widetilde E$ is a finite map of degree r on the ${\mathbb P}^1$ fibers for the ruled structures (see the proof of Proposition 5.3), it follows that $n: D \to C$ is a degree-r map of curves. Since $\overline E$ and $\textrm{Supp}\ \widetilde E$ are ruled surfaces over $Z (x,y) \cong {\mathbb P}^1$ and hence smooth, it follows from Bertini’s theorem for the induced maps to ${\mathbb P}^N$ that D and C are smooth curves and $C \cong {\mathbb P}^1$ .

By [Reference Hartshorne and Polini24, Proposition 5.9], we have a diagram of exact sequences

Since $k\overline E \in \textrm{Pic}\ \Gamma $ is in the image of $\textrm{Pic}\ L$ , it follows from this diagram that the class of $k\overline E \in \textrm{Pic}\ D$ is in the image of $\textrm{Pic}\ C$ . Since $D \to C$ is a degree-r map of integral curves, all divisors in the image of $\textrm{Pic}\ C \to \textrm{Pic}\ D$ have degree a multiple of r. We can compute the degree of the divisor $\overline E \in \textrm{Pic}\ D$ using intersection theory. The map $\widetilde E \to {\mathbb P}^N$ is defined by the line bundle $d \widetilde H - \widetilde E$ , hence the map $\overline E \to {\mathbb P}^N$ is defined by the line bundle $d \overline H - r \overline E$ , which gives the class of $D \subset \overline E$ . The degree of $\overline E$ restricted to the curve D is the intersection number $\overline E \cdot (d \overline H - r \overline E)$ calculated on $\overline E$ , or working in the ambient space $\overline {\mathbb P}^3$ , it is given by the intersection number $\overline E \cdot \overline E \cdot (d \overline H - r \overline E)$ . The same calculations from [Reference Nollet32] used in Example 3.7 give

$$\begin{align*}\overline E \cdot \overline E \cdot (d \overline H - r \overline E) = \overline E^2 (d\overline H - r\overline E) = d \overline H.\overline E^2 - r \overline E^3 = -d + 2r. \end{align*}$$

Therefore, the degree of $k\overline E$ restricted to D, namely, $k( -d + 2r)$ , must be a multiple of r, so $r\mid dk$ as desired.

Question 5.8. What are the one-dimensional subschemes $Z\subseteq {\mathbb P}^3$ such that Z is Cartier on the very general surface Y of high degree containing Z?