Proof. By [Reference Hartshorne20, Th. III.5.2], $H^0(X,\mathcal F)$ is a finitely generated R-module. By [Reference Hartshorne20, Th. III.5.2 and Cor. III.11.2], $H^1(X,\mathcal F)$ is an R-module of finite length and by [Reference Hartshorne20, Cor. III.11.2], $H^2(X,\mathcal F)=0$ since $\dim \pi ^{-1}(m_R)=1$ .

Proof. Write $D=\sum _{i=1}^r a_iF_i$ where the $F_i$ are distinct prime divisors on X and $a_i\in {\mathbb N}$ . Suppose that $F_j$ is not exceptional for $\pi $ . Let $q_j=\pi (F_j)$ , a height 1 prime ideal in R. Since $\pi $ is an isomorphism over $\mbox {Spec}(R)\setminus m_R$ , we have that $R_{q_j}={\mathcal O}_{X,F_j}$ , so

$$ \begin{align*}\begin{array}{lll} {\mathcal O}_X(-D)_{F_j}&=&(q_j^{a_j})_{q_j}=(I(\nu_j)_{a_j})_{q_j}=\Gamma(X,{\mathcal O}_X(-D))_{q_j}\\ &=&\Gamma(X,{\mathcal O}_X(-D )){\mathcal O}_{X,F_j}. \end{array} \end{align*} $$

Thus, $F_j$ is not in the support of F.

Proof. Let S be the set of exceptional prime divisors which are in the support of D. Write $D=B+\sum _{i=1}^ra_iE_i$ where B is an effective divisor which contains no exceptional prime divisors in its support and all $a_i\ge 0$ . For all $E_j$ , we have that

$$ \begin{align*}0\ge (D\cdot E_j)=(B\cdot E_j)+\sum_{i\ne j}a_i(E_i\cdot E_j)+a_j(E_j^2), \end{align*} $$

and so

(5) $$ \begin{align} -a_j(E_j^2)\ge (B\cdot E_j)+\sum_{i\ne j}a_i(E_i\cdot E_j)\ge 0. \end{align} $$

If B is nonzero, then there exists $E_{j}$ such that $(E_{j}\cdot B)>0$ and thus $a_{j}>0$ and so $E_j\in S$ . If $B=0$ , then there exists $E_j$ such that $(E_j\cdot D)<0$ since $D\ne 0$ and the intersection matrix $\left ((E_i\cdot E_j)\right )$ is nonsingular. Thus, S is nonempty. If $E_{j'}\in S$ and $E_j$ is such that $(E_j\cdot E_{j'})>0$ , then $E_j\in S$ by (5). The exceptional fiber $\pi ^{-1}(m_R)$ is connected as R is normal and $\pi $ is birational (by [Reference Hartshorne20, Cor. III.11.4]). Thus, S is the set of all exceptional prime divisors of X.

Proof. Since the intersection matrix $\left ((E_i\cdot E_j)\right )$ is negative definite, there exists an effective anti-ample ${\mathbb Q}$ -divisor A on X with exceptional support (by Proposition 2.3). Thus, $-dA$ is very ample for some $d\in {\mathbb Z}_{>0}$ . Let $I=\Gamma (X,{\mathcal O}_X(-dA))$ . The ideal I is $m_R$ -primary since the support of A is exceptional. The integral closure of $\sum _{n\ge 0}I^nt^n$ in $R[t]$ is

$$ \begin{align*}\sum_{n\ge 0}\overline{I^n}t^n=\sum_{n\ge 0}\Gamma(X,{\mathcal O}_X(-ndA))t^n. \end{align*} $$

Since R is excellent, $\sum _{n\ge 0}\overline {I^n}t^n$ is a finitely generated graded R-algebra. Thus, after replacing d with a higher power of d, we may assume that $I^n=\overline {I^n}=\Gamma (X,{\mathcal O}_X(-ndA))$ for all $n\in {\mathbb Z}_{>0}$ (as follows from [Reference Bourbaki4, Props. III.3.2 and III.3.3 on pages 158 and 159]).

Let $Y=\mbox {Proj}(\oplus _{n\ge 0}I^n)$ , which is normal since $\oplus _{n\ge 0}I^n$ is integrally closed. Since ${\mathcal O}_X(-dA)$ is generated by global sections, we have that $I{\mathcal O}_X={\mathcal O}_X(-dA)$ . By the universal property of blowing up (see [Reference Hartshorne20, Prop. II.7.14]), there exists a unique R-morphism $\phi :X\rightarrow Y$ such that $\phi ^*{\mathcal O}_Y(1)\cong {\mathcal O}_X(-dA)$ . $\phi $ is a birational morphism which is an isomorphism away from the preimage of $m_R$ . $\phi $ is of finite type since $X\rightarrow \mbox {Spec}(R)$ is. Since $(-A\cdot E)>0$ for all exceptional curves of X, we have that $\phi $ does not contract any curves of X and thus $\phi $ is quasi-finite. Let $p\in X$ and $q=\phi (p)$ . Let $A={\mathcal O}_{Y,q}$ and $B={\mathcal O}_{X,p}$ . The birational extension $A\rightarrow B$ satisfies $m_AB$ is $m_B$ -primary since $\phi $ is quasi-finite. Since A is normal and excellent, it is analytically irreducible by [Reference Grothendieck and Dieudonné18, Scholie IV.7.8.3(vii)]. Thus, by Zariski’s main theorem [Reference Abhyankar1, (10.7), p. 240] or [Reference Cutkosky10, Prop. 21.53], we have that $A=B$ and so $\phi $ is an isomorphism and X is the blowup of the $m_R$ -primary ideal I.

Proof. This follows from the proof on the last two lines of page 122 to the first half of page 123 of [Reference Hartshorne20, Th. II.5.19], along with the fact (observed in [Reference Hartshorne20, Rem. 5.19.2]) that the integral closure of a Nagata domain in its quotient field is a finite extension (by [Reference Matsumura26, Prop. 31.B]).

Proof. There exists an integer c such that $\deg (D_n)\ge c$ for all n. Let U be an effective divisor on E of degree larger than $2p_a(E)-2+c$ . By Serre duality (7),

$$ \begin{align*}h^1({\mathcal O}_{E}(D_n+Z))=h^0({\mathcal O}_{E}(K-(D_n+Z)), \end{align*} $$

where K is a cononical divisor on E. We have

$$ \begin{align*} \mbox{deg}(K-(Z+D_n))\le \deg(K-Z)-c. \end{align*} $$

If $\mbox {deg}(K-Z)-c<0$ , then certainly $h^0({\mathcal O}_{E}(K-(D_n+Z))=0$ . If $\mbox {deg}(K-Z)-c\ge 0$ , then $h^1({\mathcal O}_{E}(K-(D_n+Z)+U)=0$ by (8) and so

$$ \begin{align*}\begin{array}{lll} h^0({\mathcal O}_{E}(K-(D_n+Z))&\le& h^0({\mathcal O}_{E}(K-(D_n+Z)+U)\\ &=&\deg(K-(D_n+Z))+\deg(U)+1-p_a(E)\\ &\le& \deg(K-Z)-c+\mbox{deg}(U)+1-p_a(E). \end{array}\\[-40pt] \end{align*} $$

Proof. For $x=(x_1,\ldots ,x_r)\in {\mathbb R}^r$ , consider the inequalities

(9) $$ \begin{align} 0\le x_i \ \mbox{for} \ 1\le i\le r \end{align} $$

and

(10) $$ \begin{align} \left((D+\sum_{i=1}^rx_iE_i)\cdot E_j\right)\le 0 \ \mbox{for} \ 1\le j\le r. \end{align} $$

Since the matrix $((E_i\cdot E_j))$ is negative-definite and by Proposition 2.3, there exists an anti-ample, effective divisor $A=\sum _{i=1}^ra_iE_i$ on X. Thus, $a_i>0$ for all i (by Lemma 2.4) and after possibly replacing A with a positive multiple of A, $x=a=(a_1,\ldots ,a_r)$ satisfies (9) and (10). Let

(11) $$ \begin{align} S=\{x\in {\mathbb R}^r\mid x_i\le a_i \ \mbox{for all} \ i \ \mbox{and the} \ 2r \ \mbox{inequalities} \ (9) \ \mbox{and} \ (10)\ \mbox{are satisfied}\}. \end{align} $$

The set S is nonempty and compact. Thus, there is at least one point in S such that $\sum _{i=1}^rx_i$ is minimized on S. Let $b=(b_1,\ldots ,b_r)$ be such a point. Let $B=b_1E_1+\cdots +b_rE_r$ and $\Delta =D+B$ . Then $\Delta $ is an effective, anti-nef ${\mathbb R}$ -divisor and B is an effective ${\mathbb R}$ -divisor with exceptional support. Let $E_j$ be a component of B. Since b minimizes $\sum x_i$ , $B-\epsilon E_j$ is effective and $\Delta -\epsilon E_j$ is not anti-nef for all $\epsilon>0$ sufficiently small. However, $\left ((\Delta -\epsilon E_j)\cdot E_i\right )\le 0$ for all $i\ne j$ , so we must have that $\left ((\Delta -\epsilon E_j)\cdot E_j\right )>0$ for all positive $\epsilon $ and thus $(\Delta \cdot E_j)=0$ since $\Delta $ is anti-nef. Thus, the decomposition $\Delta =D+B$ satisfies (1) and (2).

For $b=(b_1,\ldots ,b_r)$ and $b'=(b_1',\ldots ,b_r')\in {\mathbb R}^r$ , define

$$ \begin{align*}\mbox{min}(b,b')=(\mbox{min}(b_1,b_1'),\ldots,\mbox{min}(b_r,b_r')). \end{align*} $$

If b and $b'$ satisfy (9) and (10), then $\mbox {min}(b,b')$ also satisfies (9) and (10), as we now show. For a fixed j, we may assume that $\mbox {min}(b_j,b_j')=b_j$ (after possibly interchanging b and $b'$ ). Then, since $(E_i\cdot E_j)\ge 0$ if $i\ne j$ , we have that

$$ \begin{align*}((D+\sum_i\mbox{min}(b_i,b_i')E_i)\cdot E_j)\le ((D+\sum_ib_iE_i)\cdot E_j)\le 0. \end{align*} $$

Suppose that $B=\sum b_iE_i$ and $B'=\sum b_i'E_i$ are effective ${\mathbb R}$ -divisors such that $\Delta =D+B$ and $\Delta '=D=B'$ satisfy both (1) and (2). We will show that $B=B'$ and so $\Delta =\Delta '$ . Let $\mbox {min}(B,B')=\sum _i\mbox {min}(b_i,b^{\prime }_i)E_i$ . There exist $x_i\ge 0$ such that $\mbox {min}(B,B')=B-\sum _i x_iE_i$ . Since $D+\mbox {min}(B,B')$ is anti-nef, for each element $E_j$ of the support of B, we have

$$ \begin{align*}0\ge \left((D+\mbox{min}(B,B'))\cdot E_j\right)=\left((\Delta-\sum_i x_i E_i)\cdot E_j\right)=-\sum_i x_i(E_i\cdot E_j). \end{align*} $$

Thus, $\sum _i x_i(E_i\cdot E_j)\ge 0$ and so

$$ \begin{align*}\left((\sum_i x_iE_i)\cdot (\sum_j x_jE_j)\right)=\sum_i\sum_j x_ix_j(E_i\cdot E_j)\ge 0. \end{align*} $$

Since the matrix $\left ((E_i\cdot E_j)\right )$ is negative-definite, we have that $x_i=0$ for all i. Thus, $B=\mbox {min}(B,B')$ . Similarly, $B'=\mbox {min}(B,B')$ and so $B=B'$ . Thus, there is a unique effective ${\mathbb R}$ -divisor B with exceptional support such that B and $\Delta =D+B$ satisfy (1) and (2).

We now show that $\Delta $ is the unique minimal effective and anti-nef ${\mathbb R}$ -divisor on X such that $\Delta -D$ is effective with exceptional support. Let U be an effective anti-nef ${\mathbb R}$ -divisor on X such that $U-D$ is effective with exceptional support. Let $U'=D+\mbox {min}(\Delta -D,U-D)$ . As shown earlier in the proof, $U'\ge D$ is effective and anti-nef. Write $U'-D=\sum u_iE_i$ and $B=\Delta -D=\sum b_iE_i$ . We have $\sum u_i\le \sum b_i\le \sum a_i$ , so $U'-D\in S$ (defined in (11)). Since $\sum b_i$ is the minimum of $\sum x_i$ on S, we have that $u_i=b_i$ for all i and so $U'=\Delta $ . Thus, $\Delta \le U$ .

Now, suppose that D is an effective ${\mathbb Q}$ -divisor on X. Let $\Delta =D+B$ be the Zariski decomposition of D. After possibly reindexing the $E_1,\ldots ,E_r$ , we may assume that the support of B is $E_1\cup \cdots \cup E_s$ for some s with $1\le s\le r$ . Expand $D=F+\sum _{i=1}^rc_iE_i$ where F is an effective ${\mathbb Q}$ -divisor whose support does not contain any prime exceptional divisor and $c_1,\ldots ,c_r\in {\mathbb Q}_{\ge 0}$ . Then $\Delta =F+\sum _{i=1}^rd_iE_i$ with $c_i\le d_i$ for all i and $d_i=c_i$ for $s+1\le i\le r$ . Furthermore, for $1\le j\le s$ , we have $0=(\Delta \cdot E_j)=\sum _{i=1}^sd_i(E_i\cdot E_j)+g_j$ where $g_j=(F\cdot E_j)+\sum _{i=s+1}^rc_i(E_i\cdot E_j)\in {\mathbb Q}$ . Since the $s\times s$ matrix $\left ((E_i\cdot E_j)\right )_{1\le i,j\le s}$ is negative-definite, and thus is nonsingular, we have that $d_1,\ldots ,d_s\in {\mathbb Q}$ . Thus, $\Delta $ and B are ${\mathbb Q}$ -divisors.

Proof. Suppose that $f\in \Gamma (X,{\mathcal O}_X(-\lceil n\Delta \rceil ))$ . Then $(f)-\lceil n\Delta \rceil \ge 0$ . Writing $n\Delta =\lceil n\Delta \rceil -G$ with $G\ge 0$ , we have $-n\Delta =G-\lceil n\Delta \rceil $ . From

$$ \begin{align*}-nD=-n\Delta+nB=-\lceil n\Delta\rceil+(G+nB) \end{align*} $$

and the fact that $G+nB\ge 0$ , we have that $(f)-nD\ge 0$ so that $f\in \Gamma (X,{\mathcal O}_X(-\lceil nD\rceil ))$ .

Let S be the set of prime divisors in the support of B. Suppose that

$$ \begin{align*}f\in \Gamma(X,{\mathcal O}_X(-\lceil nD\rceil )). \end{align*} $$

Then $(f)-nD\ge 0$ . Write $(f)-nD=A+C$ where A and C are effective ${\mathbb R}$ -divisors on X, no components of A are in S and all components of C are in S. We have that $(f)-n\Delta =A+(C-nB)$ . If $E\in S$ , then

$$ \begin{align*}(E\cdot(A+(C-nB)))=(E\cdot((f)-n\Delta))=0, \end{align*} $$

which implies $(E\cdot (C-nB))=-(E\cdot A)\le 0$ . The intersection matrix of the curves in S is negative-definite since it is so for the set of all exceptional curves, so $C-nB\ge 0$ (by [Reference Zariski35, Lem. 7.1]). Thus, $(f)-n\Delta \ge 0$ , which implies $(f)-\lceil n\Delta \rceil \ge 0$ since $(f)$ is an integral divisor. Thus, $f\in \Gamma (X,{\mathcal O}_X(-\lceil n\Delta \rceil ))$ .

Proof. By Lemma 2.2, E is exceptional. We will assume that $(\Delta \cdot E)<0$ and derive a contradiction. Since $\Gamma (X,{\mathcal O}_X(-\Delta ))\ne 0$ , there exists an effective divisor D on X such that $D\sim -\Delta $ . Write $D = U+F_1+\cdots +F_s$ where U is an effective divisor with no exceptional divisors in its support and $F_1=E,F_2,\ldots ,F_s$ are prime exceptional divisors. Let $\Delta _i=U+ F_1+\cdots +F_i$ for $0\le i\le s$ .

We have short exact sequences

$$ \begin{align*}0\rightarrow {\mathcal O}_X(nD-\Delta_0)\rightarrow {\mathcal O}_X(nD)\rightarrow {\mathcal O}_X(nD)\otimes{\mathcal O}_{\Delta_0}\rightarrow 0. \end{align*} $$

There exists a very ample effective divisor H on X which contains no exceptional prime divisors in its support and whose support is disjoint from $\Delta _0$ by [Reference Hartshorne20, Th. III.5.2] since $\Delta _0$ intersects $\pi ^{-1}(m_R)$ in only a finite number of closed points and so $\Delta _0$ is a closed subscheme of the affine scheme $X\setminus V(H)$ and thus $\Delta _0$ is an affine scheme. We thus have that $H^1(\Delta _0,{\mathcal O}_X(-nD)\otimes {\mathcal O}_{\Delta _0})=0$ for all n and so

(12) $$ \begin{align} h^1({\mathcal O}_X(nD))\le h^1({\mathcal O}_X(nD-\Delta_0)) \end{align} $$

for all $n\in {\mathbb N}$ .

For $i<s$ and $n\in {\mathbb N}$ , we have short exact sequences

$$ \begin{align*}0\rightarrow {\mathcal O}_X(nD-\Delta_i-F_{i+1})\rightarrow {\mathcal O}_X(nD-\Delta_i)\rightarrow {\mathcal O}_X(nD-\Delta_i)\otimes {\mathcal O}_{F_{i+1}}\rightarrow 0. \end{align*} $$

Thus,

$$ \begin{align*}h^1({\mathcal O}_X(nD-\Delta_i))\le h^1({\mathcal O}_X(nD-\Delta_{i+1})+h^1(F_{i+1},O_X(nD-\Delta_i)\otimes {\mathcal O}_{F_{i+1}}). \end{align*} $$

$(D\cdot F_{i+1})=(-\Delta \cdot F_{i+1})\ge 0$ implies that there exists $\sigma _i>0$ such that $h^1(F_{i+1},O_X(nD-\Delta _i)\otimes {\mathcal O}_{F_{i+1}})\le \sigma _i$ for all $n\in {\mathbb N}$ by Lemma 3.1, so

(13) $$ \begin{align} h^1({\mathcal O}_X(nD-\Delta_i))\le h^1({\mathcal O}_X(nD-\Delta_{i+1}))+\sigma_i \end{align} $$

for all $i\ge 0$ and $n\in {\mathbb N}$ .

Now, consider the exact sequences

$$ \begin{align*}0\rightarrow {\mathcal O}_X(nD-\Delta_0-F_1)\rightarrow {\mathcal O}_X(nD-\Delta_0)\rightarrow {\mathcal O}_X(nD-\Delta_0)\otimes {\mathcal O}_{F_1}\rightarrow 0 \end{align*} $$

for $n\in {\mathbb N}$ . Since $(F_1\cdot D)=(F_1\cdot -\Delta )>0$ , we have that $H^1(F_1,{\mathcal O}_X(nD-\Delta _i)\otimes {\mathcal O}_{F_1})=0$ for $n\gg 0$ by (8). From the natural inclusion ${\mathcal O}_X(nD-\Delta _0)\rightarrow {\mathcal O}_X(nD)$ , we deduce that $F_1$ is in the support of the fixed locus of $nD-\Delta _0$ if $F_1$ is in the support of the fixed locus of $-n\Delta $ . Thus, for n such that $F_1$ is a component of the base locus $B_n$ of $-n\Delta $ , the image of $H^0(X,{\mathcal O}_X(nD-\Delta _0))$ in $H^0(F_1,{\mathcal O}_X(nD-\Delta _i)\otimes {\mathcal O}_{F_1})$ is zero. Thus,

$$ \begin{align*}h^1({\mathcal O}_X(nD-\Delta_0))=h^1({\mathcal O}_X(nD-\Delta_0-F_1))-\chi({\mathcal O}_{F_1}(nD-\Delta_0)\otimes{\mathcal O}_{F_1}) \end{align*} $$

so that by the Riemann Roch theorem (6),

(14) $$ \begin{align} h^1({\mathcal O}_X(nD-\Delta_0))=h^1({\mathcal O}_X(nD-\Delta_0-F_1))+n(\Delta\cdot F_1)+(\Delta_0\cdot F_1)+p_a(F_1)-1. \end{align} $$

As explained before the statement of Lemma 2.2, there exists a positive integer h such that for $n\gg 0$ , $F_1$ is a component of $B_n$ if $h\nmid n$ .

By (12) and (13), there exists a constant $c>0$ such that

$$ \begin{align*}h^1({\mathcal O}_X(nD))\le h^1({\mathcal O}_X((n-1)D))+c \end{align*} $$

for all $n\in {\mathbb Z}_{>0}$ and for all $n\gg 0$ such that $h\nmid n$ we have by (12)–(14) that

$$ \begin{align*}h^1({\mathcal O}_X(nD))\le h^1({\mathcal O}_X((n-1)D)+n(\Delta\cdot F_1)+c. \end{align*} $$

Thus, we have $h^1({\mathcal O}_X(nD))<0$ for $n\gg 0$ since we have assumed that $(\Delta \cdot F_1)<0$ . However, this is impossible, giving a contradiction and so $(\Delta \cdot F_1)=0$ .

Proof. The set of base points

$$ \begin{align*}\Omega=\{p\in X\mid {\mathcal O}_X(-\Gamma)_p \ \mbox{is not generated by global sections}\} \end{align*} $$

of $\Gamma (X,{\mathcal O}_X(-\Gamma ))$ is a finite set of closed points, which are necessarily contained in the exceptional fiber of $\pi $ . Let $C\ge 0$ be an effective divisor on X such that $-C$ is very ample for $\pi $ . There exists an integer $m>0$ such that there exists an effective divisor $H\sim -mC$ with no exceptional components in its support and such that $\Omega $ is disjoint from its support (by [Reference Hartshorne20, Th. III.5.2]). After replacing C with this multiple $mC$ , we may assume that $H\sim -C$ . Let $f\in K$ , the quotient field of R, be such that $(f)-C=H$ . We may regard the effective divisor H as a closed subscheme of X.

We have a short exact sequence

$$ \begin{align*}0\rightarrow {\mathcal O}_X(C)\stackrel{f}{\rightarrow} {\mathcal O}_X\rightarrow {\mathcal O}_H\rightarrow 0, \end{align*} $$

and tensoring with ${\mathcal O}_X(-i\Gamma -jC)$ , we have short exact sequences

(15) $$ \begin{align} 0\rightarrow {\mathcal O}_X(-i\Gamma-(j-1)C)\stackrel{f}{\rightarrow}{\mathcal O}_X(-i\Gamma-jC)\rightarrow {\mathcal O}_X(-i\Gamma-jC)\otimes{\mathcal O}_H\rightarrow 0. \end{align} $$

For $i,j\ge 0$ , let $A_{i,j}$ be the natural image of $\Gamma (X,\mathcal O_X(-i\Gamma -jC))$ in

$$ \begin{align*}\Gamma(H,{\mathcal O}_X(-i\Gamma-jC)\otimes {\mathcal O}_H), \end{align*} $$

upon taking global sections of (15). Since the base points of $\Gamma (X,\mathcal O_X(-i\Gamma -jC))$ are a subset of $\Omega $ and so are disjoint from H, we have that, for all $i,j\ge 0$ , $A_{i,j}{\mathcal O}_{H,q}={\mathcal O}_X(-i\Gamma -jC)_q$ for all $q\in H$ .

There exists $n\in {\mathbb Z}_{>0}$ such that there exists an effective divisor G on X such that $G\sim -nC$ , the support of G contains no exceptional components of $\pi $ and $\mbox {sup}(H)\cap \mbox {sup}(G)\cap \mbox {sup}(\pi ^{-1}(m_R))=\emptyset $ (by [Reference Hartshorne20, Th. III.5.2]). We may regard G as a closed subscheme of X. Thus, H is a closed subscheme of the affine scheme $X\setminus G$ and so H is affine, say $H=\mbox {Spec}(S)$ . The restriction of $\pi $ to H is determined by a ring homomorphism $R\rightarrow S$ . Now, $S=\Gamma (H,{\mathcal O}_H)$ is a finitely generated R-module since $\pi $ is a projective morphism (by [Reference Hartshorne20, Cor. II.5.20]). As explained in [Reference Hartshorne20, Cor. II.5.5], since S is Noetherian, the functor $M\rightarrow \tilde M$ gives an equivalence of categories between the category of finitely generated S-modules and the category of coherent ${\mathcal O}_{\mbox {Spec}(S)}$ -modules, with inverse $\mathcal F\mapsto \Gamma (\mbox {Spec}(S),\mathcal F)$ .

In particular, letting $B_{i,j}=\Gamma (H,{\mathcal O}_X(-i\Gamma -jC)\otimes {\mathcal O}_H)$ for $i,j\ge 0$ , we have that ${\mathcal O}_X(-i\Gamma -jC)\otimes {\mathcal O}_H=\widetilde {B_{i,j}}$ . We also have that $B_{i,j}$ is the tensor product over S of i copies of $B_{1,0}$ and j copies of $B_{0,1}$ (see [Reference Hartshorne20, Prop. II.5.2]).

We have that the ring $A_{0,0}$ is a quotient of $\Gamma (X,{\mathcal O}_X)=R$ since $\pi $ is proper birational and R is normal. Let $A_{0,0}[t_1,t_2]$ be a polynomial ring over $A_{0,0}$ , which is bigraded by specifying that $\deg (a)=(0,0)$ if $a\in A_{0,0}$ , $\deg (t_1)=(1,0)$ , and $\deg (t_2)=(0,1)$ . Let M be the bigraded $A_{0,0}$ -subalgebra $M:=\sum _{i,j\ge 0} A_{i,j}t_1^it_2^j$ of $A_{0,0}[t_1,t_2]$ . Similarly, let B be the bigraded S-subalgebra $B:=\bigoplus _{i,j\ge 0}B_{i,j}t_1^it_2^j$ of $S[t_1,t_2]$ .

We have a natural inclusion of graded rings $M\rightarrow B$ .

Since H is disjoint from $\Omega $ , we have that

$$ \begin{align*}A_{1,0}^iA_{0,1}^jS_q=A_{ij}S_q={\mathcal O}_X(-i\Gamma-jA)\otimes{\mathcal O}_{H,q}=(B_{i,j})_q \end{align*} $$

for all $q\in H$ and $i,j\ge 0$ . Thus,

(16) $$ \begin{align} A_{1,0}^iA_{0,1}^jS=B_{i,j} \ \mbox{for all} \ i,j\ge 0. \end{align} $$

Let A be the bigraded $A_{0,0}$ -subalgebra $A:= A_{0,0}[A_{1,0}t_1,A_{0,1}t_2]$ of M. Now, we have a natural surjection $A_{1,0}^iA_{0,1}^j\otimes _RS\rightarrow B_{i,j}$ for all $i,j\ge 0$ by (16). Thus, the natural homomorphism $A\otimes _RS\rightarrow B$ is surjective. Since S is a finitely generated R-module, we have that B is a finitely generated bigraded A-module. Since $A\subset M\subset B$ and A is Noetherian, we have that M is also a finitely generated A-module.

By [Reference Zariski35, Lem. 4.3], since A is generated in bidegrees $(1,0)$ and $(0,1)$ , and M is a finitely generated bigraded R-module, there exists $N\in {\mathbb Z}_{>0}$ such that

(17) $$ \begin{align} A_{i,j}=A_{i,j-1}A_{0,1} \ \mbox{whenever} \ j\ge N\ \mbox{and} \ i\ge 0 \ \mbox{is arbitrary} \end{align} $$

and

(18) $$ \begin{align} A_{i,j}=A_{i-1,j}A_{1,0} \ \mbox{whenever} \ i\ge N \ \mbox{and} \ j\ge 0 \ \mbox{is arbitrary.} \end{align} $$

Thus, taking global sections in the short exact sequences (15), and applying (18), we have that if $i\ge N$ and $j\ge 0$ , then

(19) $$ \begin{align} &\Gamma(X,{\mathcal O}_X(-i\Gamma-jC))\nonumber\\&\quad=\Gamma(X,{\mathcal O}_X(-i\Gamma-(j-1)C))f+\Gamma(X,{\mathcal O}_X(-(i-1)\Gamma-jC))\Gamma(X,{\mathcal O}_X(-\Gamma)). \end{align} $$

Since $-C$ is ample, for fixed i, ${\mathcal O}_X(-i\Gamma -jC)$ is generated by global sections for all $j\gg 0$ (by [Reference Hartshorne20, Th. II.5.17]). Let i be a fixed integer $\ge N$ and let $j>0$ be such that ${\mathcal O}_X(-i\Gamma -jC)$ is generated by global sections.

The only points $q\in X$ where it is possible for ${\mathcal O}_X(-i\Gamma -(j-1)C)_q$ to not be generated by global sections are the points of $\Omega $ . Suppose that $q\in \Omega $ . Thus, q is not in the support of $H=(f)-C$ , and so $f=0$ is a local equation of C at q and $f{\mathcal O}_{X,q}={\mathcal O}_X(-C)_q$ . Furthermore, since $q\in \Omega $ , $\Gamma (X,{\mathcal O}_X(-\Gamma ))O_{X,q}\subset m_q{\mathcal O}_{X}(-\Gamma )$ where $m_q$ is the maximal ideal of ${\mathcal O}_{X,q}$ , equation (19), and Nakayama’s lemma show that

$$ \begin{align*}\begin{array}{lll} {\mathcal O}_X(-i\Gamma-jC)_q&=&\Gamma(X,{\mathcal O}_X(-i\Gamma-jC)){\mathcal O}_{X,q}\\ &=&\Gamma(X,{\mathcal O}_X(-i\Gamma-(j-1)C))f{\mathcal O}_{X,q}\\ &&+\Gamma(X,{\mathcal O}_X(-(i-1)\Gamma-jC){\mathcal O}_X(-\Gamma)m_q\\ &=&\Gamma(X,{\mathcal O}_X(-i\Gamma-(j-1)C)){\mathcal O}_X(-C)_q. \end{array} \end{align*} $$

Thus, $\Gamma (X,{\mathcal O}_X(-i\Gamma -(j-1)C)){\mathcal O}_{X,q}= {\mathcal O}_X(-i\Gamma -(j-1)C)_q$ , and since this is true for all $q\in \Omega $ , ${\mathcal O}_X(-i\Gamma -(j-1)C)$ is generated by global sections.

By descending induction on j, we obtain that ${\mathcal O}_X(-i\Gamma )$ is generated by global sections for all $i\ge N$ .

We now prove the second statement of the proposition. Let $g_0,\ldots ,g_r\in \Gamma (X,{\mathcal O}_X(-N\Gamma ))$ generate $\Gamma (X,{\mathcal O}_X(-N\Gamma ))$ as an R-module. Then $g_0,\ldots ,g_r$ induce a proper R-morphism $\phi :X\rightarrow {\mathbb P}_R^r$ such that $\phi ^*{\mathcal O}_{{\mathbb P}_R^r}(1)\cong {\mathcal O}_X(-N\Gamma )$ (by [Reference Hartshorne20, Th. II.7.1 and Cor. II.4.8]). In fact, $\phi $ is projective, by [Reference Grothendieck and Dieudonné17, Prop. II.5.5(v)] or [34, Lem. 29.43.15 and Tag 01W7] and [34, Lem. 29.43.16(1) and Tag 01W7]. Let Z be the image of $\phi $ in ${\mathbb P}_R^r$ (which is closed since $\phi $ is proper), and let ${\mathcal O}_Z(1)={\mathcal O}_{{\mathbb P}^r_R}(1)\otimes {\mathcal O}_Z$ . Let $\overline \phi :X\rightarrow Z$ be the induced projective R-morphism. By [Reference Hartshorne20, Cor. III.11.2], for $s\in {\mathbb Z}$ , the support of $R^1\overline \phi *{\mathcal O}_X(-s\Gamma )$ is contained in the finite set of closed points of Z which are the images of curves contracted by $\overline \phi $ (the prime exceptional divisors E of $\pi $ such that $(E\cdot -\Gamma )=0$ ). By [Reference Hartshorne20, Th. II.5.19], $\Gamma (Z,R^1\overline \phi *{\mathcal O}_X(-s\Gamma ))$ is a finitely generated R-module. Since its support is the maximal ideal of R, the length of $\Gamma (Z,R^1\overline \phi *{\mathcal O}_X(-s\Gamma ))$ as an R-module is finite.

From the Leray spectral sequence, we obtain exact sequences (see [Reference Rotman32, Th. 11.2]) for $m\in {\mathbb Z}$ ,

$$ \begin{align*}0\rightarrow H^1(Z,\overline \phi_*{\mathcal O}_X(-m\Gamma))\rightarrow H^1(X,{\mathcal O}_X(-m\Gamma)) \rightarrow H^0(Z,R^1\overline\phi_*{\mathcal O}_X(-m\Gamma)). \end{align*} $$

For $m\in {\mathbb N}$ , write $m=nN+s$ with $0\le s<N$ . Then ${\mathcal O}_X(-m\Gamma )\cong \overline \phi ^*{\mathcal O}_Z(n)\otimes {\mathcal O}_X(-s\Gamma )$ . Then, by the projection formula (see [Reference Hartshorne20, Exer. III.8.3]), we obtain exact sequences for $n,s\in {\mathbb Z}$

(20) $$ \begin{align} \begin{array}{l} 0\rightarrow H^1(Z,{\mathcal O}_Z(n)\otimes \overline\phi_*{\mathcal O}_X(-s\Gamma)) \rightarrow H^1(X,\overline\phi^*{\mathcal O}_Z(n)\otimes{\mathcal O}_X(-s\Gamma))\\ \rightarrow H^0(Z,(R^1\overline\phi_*{\mathcal O}_X(-s\Gamma))\otimes{\mathcal O}_Z(n)). \end{array} \end{align} $$

Let

$$ \begin{align*}s_1=\max\{\lambda_R\Gamma(Z,R^1\overline\phi_*{\mathcal O}_X(-s\Gamma))\mid 0\le s<N\}. \end{align*} $$

We have that $H^1(Z,{\mathcal O}_Z(n)\otimes \overline \phi _*{\mathcal O}_X(-s\Gamma ))=0$ for all $0\le s<N$ and $n\gg 0$ (see [Reference Hartshorne20, Th. III.5.2]). Let

$$ \begin{align*}s_2=\max\{\lambda_RH^1(Z,{\mathcal O}_Z(n) \otimes \overline\phi_*{\mathcal O}_X(-s\Gamma)) \mid 0\le s<N \mbox{ and }n\in {\mathbb N}\} \end{align*} $$

$s_2$ is finite by [Reference Hartshorne20, Prop. III.8.5, Th. III.8.8, and Cor. III.11.2]. By (20), we have that $\lambda _RH^1(X,{\mathcal O}_X(-m\Gamma ))\le s_1+s_2$ for all $m\in {\mathbb N}$ .

Proof. To prove the proposition, it suffices to prove it for it for some positive multiple d of $\Delta $ , since for $n\in {\mathbb N}$ , writing $n=md+s$ with $0\le s<d$ , we have $B_n\le B_{md}+B_s$ .

Write $-\Delta =\sum _{i=1}^t a_iF_i$ . Let

$$ \begin{align*}M_i=\{n\in {\mathbb N}\mid F_i\mbox{ is not a component of }B_n\}. \end{align*} $$

$M_i$ is a numerical semigroup, so if $M_i$ is nonzero, there exists $h_i\in {\mathbb Z}_{>0}$ such that for $n\gg 0$ , $n\in M_i$ if and only if $h_i$ divides n. Let

$$ \begin{align*}\mathcal B(D)=\{F_i\mid F_i\mbox{ is a component of } \ B_n \ \mbox{for infinitely many} \ n\}. \end{align*} $$

By Proposition 5.1, $F_i\in \mathcal B(D)$ implies $(F_i\cdot \Delta )=0$ and $F_i$ is exceptional for $\pi $ . After possibly reindexing the $F_i$ , we may assume that the support of $\mathcal B(D)$ is $\cup _{i=1}^sF_i$ , for some $s\le t$ . We have that $M_i=0$ or $h_i>1$ for $1\le i\le s$ . Thus, the support of $B_n$ is $\cup _{i=1}^sF_i$ if $n\gg 0$ and $h_i\nmid n$ for all i such that $1\le i\le s$ and $M_i$ is nonzero.

If we replace $\Delta $ with $n_0\Delta $ for some $n_0\gg 0$ , we have that the support of $B_1$ is $\mathcal B(D)$ . By Proposition 5.2, there exists $s_0\in {\mathbb N}$ such that the effective divisor $\Gamma =\Delta +B_1$ satisfies the condition that $h^1({\mathcal O}_X(-n\Gamma ))\le s_0$ for all $n\ge 1$ since $-\Gamma $ has no fixed component.

For a given $n\in {\mathbb Z}_{>0}$ , consider the following conditions on a divisor $Z_n$ .

1. (a) $n\Gamma \ge Z_n\ge n\Delta $ .

2. (b) $-Z_n$ has no fixed component.

3. (c) $h^1({\mathcal O}_X(-Z_n))\le s_0$ .

Let $C_n$ be a minimal element in the set of divisors satisfying (a), (b), and (c). Let $B_n'=C_n-n\Delta $ . Then $nB_1\ge B_n'\ge B_n$ (since $-n\Delta =-n\Gamma +nB_1=-C_n+B_n'$ and $C_n\le n\Gamma $ ). Thus, it suffices to show that the $B_n'$ are bounded from above.

For $1\le i\le s$ , we have short exact sequences

$$ \begin{align*}0\rightarrow O_X(-C_n)\rightarrow {\mathcal O}_X(-C_n+F_i)\rightarrow {\mathcal O}_X(-C_n+F_i)\otimes {\mathcal O}_{F_i}\rightarrow 0, \end{align*} $$

giving exact sequences

$$ \begin{align*}\begin{array}{l} 0\rightarrow H^0(X,{\mathcal O}_X(-C_n))\rightarrow H^0(X,{\mathcal O}_X(-C_n+F_i))\rightarrow H^0(F_i,{\mathcal O}_X(-C_n+F_i)\otimes {\mathcal O}_{F_i})\\ \rightarrow H^1(X,{\mathcal O}_X(-C_n))\rightarrow H^1(X,{\mathcal O}_X(-C_n+F_i)) \rightarrow H^1(F_i,{\mathcal O}_X(-C_n+F_i)\otimes {\mathcal O}_{F_i})\rightarrow 0. \end{array} \end{align*} $$

We will show that

(21) $$ \begin{align} -(C_n\cdot F_i)\le \mbox{max}\{s_0-(F_i^2)-1+p_a(F_i),2p_a(F_i)-2-(F_i^2),0\} \end{align} $$

for all n and $1\le i\le s$ .

First assume that $F_i$ is not a component of $B_n'$ . Then $(B_n'\cdot F_i)\ge 0$ . Since $(F_i\cdot \Delta )=0$ by Proposition 5.1, we have that $(C_n\cdot F_i)\ge 0$ and so (21) holds.

Now, assume that $F_i$ is a component of $B_n'$ . We have that either

(22) $$ \begin{align} H^0(X,{\mathcal O}_X(-C_n+F_i))=H^0(X,{\mathcal O}_X(-C_n)) \end{align} $$

or

(23) $$ \begin{align} h^1({\mathcal O}_X(-C_n+F_i))>s_0. \end{align} $$

If (22) holds, then $h^0({\mathcal O}_X(-C_n+F_i)\otimes O_{F_i})\le s_0$ . Thus,

$$ \begin{align*}s_0\ge h^0({\mathcal O}_X(-C_n+F_i)\otimes{\mathcal O}_{F_i})\ge ((-C_n+F_i)\cdot F_i)+1-p_a(F_i) \end{align*} $$

by the Riemann–Roch formula (6), and so (21) holds.

Suppose that (23) holds. Then $h^1(F_i,{\mathcal O}_X(-C_n+F_i)\otimes {\mathcal O}_{F_i})>0$ , and so

$$ \begin{align*}((-C_n+F_i)\cdot F_i)<2p_a(F_i)-2 \end{align*} $$

by (8). Thus, (21) holds.

For i with $1\le i\le s$ , let $\sigma _i=\mbox {max}\{s_0-(F_i^2)-1+p_a(F_i),2p_a(F_i)-2-(F_i^2),0\}$ . Since $(F_i\cdot \Delta )=0$ for $1\le i\le s$ by Proposition 5.1, and by (21), we have that

$$ \begin{align*}(B_n'\cdot F_i)=((C_n-n\Delta)\cdot F_i)=(C_n\cdot F_i)\ge -\sigma_i. \end{align*} $$

In particular, $\sigma _i\ge -(B_n'\cdot F_i)$ .

Since the intersection matrix $((F_i\cdot F_j))$ for $1\le i,j\le s$ is negative-definite, and thus is nonsingular, there exists a ${\mathbb Q}$ -divisor $\mathcal E=c_1F_1+\cdots +c_s F_s$ such that $(\mathcal E\cdot F_i)=-\sigma _i$ for $1\le i\le s$ . Then

$$ \begin{align*}((\mathcal E-B_n')\cdot F_i)=-\sigma_i-(B_n'\cdot F_i)\le 0 \end{align*} $$

for all i implies $\mathcal E\ge B_n'$ by [Reference Zariski35, Lem. 7.1], since the intersection matrix is negative-definite. Thus, the $B_n'$ are bounded from above.

Proof. Statement (1) follows from Lemma 2.2. If $\Delta $ is an integral divisor, then Statement (2) follows from Proposition 5.3.

Now, assume that $\Delta $ is a ${\mathbb Q}$ -divisor. Write $\Delta =\sum \frac {b_i}{d}F_i$ with $d\in {\mathbb Z}_{>0}$ and $b_i\in {\mathbb N}$ , where the $F_i$ are distinct prime divisors on X. Since $d\Delta $ is an integral divisor, there exists an effective integral divisor C with exceptional support such that $B_{nd}\le C$ for all $n\in {\mathbb N}$ . Let $n\in {\mathbb N}$ , and write $n=md-c$ with $m\in {\mathbb N}$ and $0\le c<d$ . Then ${\mathcal O}_X(-\lceil n\Delta \rceil )={\mathcal O}_X(-md\Delta +\lfloor c\Delta \rfloor )$ . Thus, $B_n\le B_{md}+\lfloor c\Delta \rceil \le C+d\Delta $ .

Proof. By induction on the number of components of Z, we may assume that $h^1({\mathcal O}_X(D_n+Z-F))$ is bounded where F is a prime component of Z. We have a short exact sequence

$$ \begin{align*}0\rightarrow {\mathcal O}_X(-F)\rightarrow {\mathcal O}_X\rightarrow {\mathcal O}_{F}\rightarrow 0, \end{align*} $$

giving exact sequences

$$ \begin{align*}H^1(X,{\mathcal O}_X(D_n+Z-F))\rightarrow H^1(X,{\mathcal O}_X(D_n+Z))\rightarrow H^1(F,{\mathcal O}_{X}(D_n+Z)\otimes{\mathcal O}_{F}). \end{align*} $$

If F is exceptional, there exists $s\in {\mathbb Z}_{>0}$ such that $h^1(F,{\mathcal O}_X(D_n+Z)\otimes {\mathcal O}_{F})\le s$ for all $n\ge 0$ by Lemma 3.1, so $h^1({\mathcal O}_X(D_n+Z))$ is bounded from above. If F is not exceptional, then F is affine and so $H^1(F,{\mathcal O}_X(D_m+Z))\otimes {\mathcal O}_{F})=0$ for all m, so again $h^1({\mathcal O}_X(D_n+Z))$ is bounded from above.

Proof. Let $C_n$ be the effective divisors of the proof of Proposition 5.3, so that $B_n'=C_n-n\Delta $ are effective divisors and there exists an effective divisor G with exceptional support such that $B_n'\le G$ for all $n\in {\mathbb N}$ . Since $-\Delta $ is nef, we have that $(-C_n\cdot E)$ is bounded from below for each prime exceptional component E of G. Furthermore, we have (by the proof of Proposition 5.3) that $h^1({\mathcal O}_X(-C_n))\le s_0$ for all $n\in {\mathbb N}$ . For each effective divisor $Z\le G$ , Proposition 5.5 gives us an upper bound for $h^1({\mathcal O}_X(-C_n+Z))$ over $n\in {\mathbb N}$ . The maximum of these bounds is an upper bound for $h^1({\mathcal O}_X(-n\Delta ))$ over $n\in {\mathbb N}$ .

Proof. There exists an effective anti-ample divisor A on X with exceptional support by Proposition 2.3. There exists $n_0\in {\mathbb Z}_{>0}$ such that $\mathcal F\otimes {\mathcal O}(-n_0A)$ is generated by global sections, so there is a surjection ${\mathcal O}_X^s\rightarrow \mathcal F\otimes {\mathcal O}_X(-n_0A)$ for some s, giving a short exact sequence of coherent sheaves

$$ \begin{align*}0\rightarrow \mathcal K\rightarrow {\mathcal O}_X(n_0A)^s\rightarrow \mathcal F\rightarrow 0 \end{align*} $$

and surjections

$$ \begin{align*} H^1(X,{\mathcal O}_X(-n\Delta+n_0A))^s\rightarrow H^1(X,{\mathcal O}_X(-n\Delta)\otimes\mathcal F). \end{align*} $$

Thus, $h^1({\mathcal O}_X(-n\Delta )\otimes \mathcal F)$ is bounded above for $n\in {\mathbb N}$ since $-\Delta $ is nef, and by Lemma 5.5 and Proposition 5.6.

Proof. Write $\Delta _1=\sum a_iF_i$ and $\Delta _2=\sum b_iF_i$ where the $F_i$ are distinct prime divisors on X. We have $b_i\ge a_i$ for all i and $b_j>a_j$ for some j. If $F_j$ is not exceptional, then certainly $\Gamma (X,{\mathcal O}_X(-\lceil n\Delta _2\rceil ))\ne \Gamma (X,{\mathcal O}_X(-\lceil n\Delta _1\rceil ))$ for n sufficiently large by Lemma 2.2.

Now, suppose that $F_j$ is exceptional. By (2) of Lemma 5.4, there exists an effective exceptional divisor $H=\sum c_iF_i$ such that the fixed component $B_n$ of $\Gamma (X,{\mathcal O}_X(-\lceil n\Delta _1\rceil ))$ satisfies $B_n\le H$ for all $n\in {\mathbb N}$ . Observe that $g\in \Gamma (X,{\mathcal O}_X(-\lceil n\Delta _2\rceil ))$ implies $\nu _j(g)\ge \lceil n b_j\rceil $ . By definition of $B_n$ , for $n\in {\mathbb Z}_{>0}$ , there exists $f_n\in \Gamma (X,{\mathcal O}_X(-\lceil n\Delta _1\rceil ))$ such that $(f_n)-\lceil n\Delta _1\rceil =A_n+B_n$ where $F_j$ is not a component of the effective divisor $A_n$ . Thus, $\nu _j(f_n)=\lceil na_j\rceil +\delta $ with $\delta \le c_j$ . We have that $n>\frac {c_j+1}{b_j-a_j}$ implies $\lceil na_j\rceil +\delta <\lceil nb_j\rceil $ . Thus, $\nu _j(f_n)<\lceil nb_j\rceil $ so that $f_n\not \in \Gamma (X,{\mathcal O}_X(-\lceil n\Delta _2\rceil ))$ .

Proof. Proposition 6.1 proves the essential implication (1) implies (3). The directions (3) implies (2) and (2) implies (1) are immediate.

Proof. It follows from Lemma 2.4 that $E=F_j$ for some j with $a_j>0$ .

Let $D_1=\sum _{i\ne j}a_iF_i$ , so that $D_1\le \Delta $ . Let $\Delta _1=D_1+B_1$ be the Zariski decomposition of $D_1$ . We have that $\Delta _1\le \Delta $ by Remark 4.2, and so $0\le \Delta -\Delta _1=a_jF_j-B_1$ so that $0\le B_1\le a_jF_j$ . Thus, $\Delta _1=\Delta -\lambda F_j$ with $0\le \lambda \le a_j$ .

If $\Delta _1\ne \Delta $ , then $\lambda>0$ , and so

(24) $$ \begin{align} (F_j\cdot\Delta)=(F_j\cdot\Delta_1)+\lambda(F_j^2)<0. \end{align} $$

If $\Delta _1=\Delta $ , then $B_1=a_jF_j$ . Since $a_j>0$ , we have that

(25) $$ \begin{align} 0=(\Delta_1\cdot F_j)=(\Delta\cdot F_j) \end{align} $$

by (2) of Lemma 4.1.

Suppose that (1) holds. Then $\Delta _1\ne \Delta $ so that $(F_j\cdot \Delta )<0$ by (24), so that (1) implies (3) holds. Certainly, (2) implies (1) is true, so we are reduced to proving (3) implies (2). Now, (3) implies $\Delta _1\ne \Delta $ by (24) and (25). If (2) does not hold, then there exist infinitely many $n\in {\mathbb Z}_{>0}$ such that $\Gamma (X,{\mathcal O}_X(-\lceil n\Delta \rceil ))=\Gamma (X,{\mathcal O}_X(-\lceil n\Delta _1\rceil ))$ so that $\Delta _1=\Delta _2$ by Corollary 6.2, giving a contradiction.

Proof. Write $\Delta =\sum _{i=1}^sa_iF_i$ , so that $I(n\Delta )=\cap _{i=1}^sI(\nu _{F_i})_{na_i}$ . For a fixed n, we have that $m_R\in \mbox {Ass}(R/\cap _{i=1}^sI(\nu _{F_i})_{na_i})$ if and only if

$$ \begin{align*}\cap_{i=1}^sI(\nu_{F_i})_{na_i}\ne \cap_{F_i}I(\nu_{F_i})_{na_i}, \end{align*} $$

where the second intersection is over the $F_i$ which are not exceptional. This condition occurs if and only if there exists j such that $F_j$ is exceptional and

$$ \begin{align*}\cap_{i=1}^sI(\nu_{F_i})_{na_i}\ne \cap_{i\ne j}I(\nu_{F_i})_{na_i}. \end{align*} $$

Thus, by Proposition 6.3, the three conditions of the corollary are equivalent.

Proof. Suppose that for some $m>0$ there exists $f\in \Gamma (X,{\mathcal O}_X(-\lceil m\Delta \rceil ))$ such that its class $\overline f$ in $\Gamma (X,{\mathcal O}_X(-\lceil m\Delta \rceil ))/\Gamma (X,{\mathcal O}_X(-\lceil m\Delta \rceil -E_j))$ is nonzero. Then

$$ \begin{align*}\overline ft^m\in \sum_{n=0}^{\infty}\Gamma(X,{\mathcal O}_X(-\lceil n\Delta\rceil))/\Gamma(X,{\mathcal O}_X(-\lceil n\Delta\rceil-E_j))t^n=R[\Delta]/P_j \end{align*} $$

is nonzero. The element $\overline ft^m$ is not a unit since it is homogeneous of positive degree and it is not nilpotent since $R[\Delta ]/P_j$ is an integral domain. Thus, $\dim R[\Delta ]/P_j>0$ . Thus, by (26), $\dim R[\Delta ]/P_j=0$ implies $R[\Delta ]/P_j=R/m_R$ .

Proof. We have that $\sqrt {m_RR[\Delta ]}\subset \cap _{i=1}^rP_i$ by (27).

Let $h\in \cap _{i=1}^rP_i$ . We will show that $h^n\in m_RR[\Delta ]$ for some $n\in {\mathbb Z}_{>0}$ , which will establish the proposition. We may assume that h is homogeneous, so that

$$ \begin{align*}h\in \cap_{i=1}^r\Gamma(X,{\mathcal O}_X(-\lceil a\Delta\rceil-E_i))= \Gamma(X,{\mathcal O}_X(-\lceil a \Delta\rceil-E_1-\cdots-E_r) \end{align*} $$

for some $a\in {\mathbb N}$ . We must show that $h^n\in m_R\Gamma (X,{\mathcal O}_X(-\lceil an\Delta \rceil ))$ for some $n\in {\mathbb Z}_{>0}$ .

First, suppose that $a=0$ . We have that $\Gamma (X,{\mathcal O}_X(-E_1-\cdots -E_r)=m_R$ , so we already have that $h\in m_R\Gamma (X,{\mathcal O}_X)=m_R$ .

Now, suppose that $a>0$ . After replacing $\Delta $ with a positive multiple of $\Delta $ and h with a power of h, we may assume that $\Delta $ is an integral divisor and that $h\in \Gamma (X,{\mathcal O}_X(-\Delta -\sum _{i=1}^rE_i))$ . By Lemma 2.5, there exists an $m_R$ -primary ideal I in R such that X is the blowup of I, so that $X=\mbox {Proj}(\bigoplus _{n\ge 0}I^n)$ and $I{\mathcal O}_X={\mathcal O}_X(-C)$ is very ample, where C is an effective divisor whose support is the union of all exceptional prime divisors $E_1,\ldots , E_r$ . The graded ring $\bigoplus _{n\ge 0}\Gamma (X,I^n{\mathcal O}_X)$ is a finite $\bigoplus _{n\ge 0}I^n$ -module and there exists $n_0\in {\mathbb Z}_{>0}$ such that the R-ideal $\Gamma (X,I^n{\mathcal O}_X)=I^n$ for $n\ge n_0$ by Lemma 2.6. Since R and X are normal, $\Gamma (X,I^n{\mathcal O}_X)=\overline {I^n}$ for all $n\ge 0$ .

After possibly replacing I with a positive power of I, we may assume that $\Gamma (X,I^n{\mathcal O}_X)=I^n$ for all $n\in {\mathbb N}$ and that there exists an effective divisor $H\sim -C$ on X with no exceptional prime divisors in its support. Let $f\in \Gamma (X,{\mathcal O}_X(-C))=I$ be such that $(f)-C=H$ . We have a short exact sequence

$$ \begin{align*}0\rightarrow {\mathcal O}_X(C)\stackrel{f}{\rightarrow}{\mathcal O}_X\rightarrow {\mathcal O}_H\rightarrow 0. \end{align*} $$

There exists $\alpha \in {\mathbb Q}_{>0}$ such that $F:=\sum _{i=1}^rE_i-\alpha C\ge 0$ . There exists $e\in {\mathbb Z}_{>0}$ such that $e\alpha C$ is an integral divisor and so $eF$ is an integral divisor. Thus, for $n\in {\mathbb Z}_{>0}$ , we have that

$$ \begin{align*}\begin{array}{lll} h^{n2e}&\in& \Gamma(X,{\mathcal O}_X(-n2e\Delta-n2e(\sum_{i=1}^rE_i))=\Gamma(X,{\mathcal O}_X(-n2e\Delta-n2e\alpha C-n2eF)\\ &\subset& \Gamma(X,{\mathcal O}_X(-n2e\Delta-n2e\alpha C))=\Gamma(X,{\mathcal O}_X(-n2e(\Delta+\frac{\alpha}{2}C)-n2e\frac{\alpha}{2}C)). \end{array} \end{align*} $$

Now, the effective integral divisor $2e(\Delta +\frac {\alpha }{2}C)$ is anti-ample by Proposition 2.3, since $\Delta $ is anti-nef. Thus, there exists $n_0\in {\mathbb Z}_{>0}$ such that ${\mathcal O}_X(-n2e(\Delta +\frac {\alpha }{2}C))$ is generated by global sections for all $n\ge n_0$ . Let $\Gamma =n_02e(\Delta +\frac {\alpha }{2}C)$ . By the argument of the proof of Proposition 5.2, applying (17), there exists $N>0$ such that

$$ \begin{align*}&\Gamma(X,{\mathcal O}_X(-i\Gamma-jC))\\&\quad=\Gamma(X,{\mathcal O}_X(-i\Gamma-(j-1)C)\Gamma(X,{\mathcal O}_X(-C))+f\Gamma(X,{\mathcal O}_X(-i\Gamma-(j-1)C) \end{align*} $$

whenever $j\ge N$ and $i\ge 0$ . Since $f\in \Gamma (X,{\mathcal O}_X(-C))$ , we have that

$$ \begin{align*}\begin{array}{lll} \Gamma(X,{\mathcal O}_X(-i\Gamma-jC))&=&\Gamma(X,{\mathcal O}_X(-i\Gamma-(j-1)C))\Gamma(X,{\mathcal O}_X(-C))\\ &=&I\Gamma(X,{\mathcal O}_X(-i\Gamma-(j-1)C))\subset I\Gamma(X,{\mathcal O}_X(-in_02e\Delta)). \end{array} \end{align*} $$

Thus,

$$ \begin{align*}h^{nn_02e}\in \Gamma(X,{\mathcal O}_X(-n\Gamma-\frac{nn_02e\alpha}{2}C))\subset I\Gamma(X,{\mathcal O}_X(-nn_02e\Delta)) \subset m_R\Gamma(X,{\mathcal O}_X(-nn_02e\Delta))\end{align*} $$

whenever n is so large that $n\ge \frac {N}{n_0e\alpha }$ .

Proof. By Proposition 6.7, we have that $\dim R[\Delta ]/m_RR[\Delta ]=0$ if and only if $\dim R[\Delta ]/P_j=0$ for all j, and this second condition holds if and only if $R[\Delta ]/P_j=R/m_R$ for all j by Remark 6.6.

Proof. Suppose that $(\Delta \cdot E_j)<0$ . We have short exact sequences

$$ \begin{align*}0\rightarrow {\mathcal O}_X(-\lceil n\Delta\rceil-E_j)\rightarrow {\mathcal O}_X(-\lceil n\Delta\rceil )\rightarrow {\mathcal O}_X(-\lceil n\Delta\rceil)\otimes{\mathcal O}_{E_j}\rightarrow 0. \end{align*} $$

Taking global sections, we have short exact sequences

$$ \begin{align*}\begin{array}{l} 0\rightarrow \Gamma(X,{\mathcal O}_X(-\lceil n\Delta\rceil-E_j)) \rightarrow \Gamma(X,{\mathcal O}_X(-\lceil n\Delta\rceil))\\\rightarrow \Gamma(E_j,O_X(-\lceil n\Delta\rceil)\otimes{\mathcal O}_{E_j}) \rightarrow H^1(X,{\mathcal O}_X(-\lceil n\Delta\rceil-E_j)). \end{array} \end{align*} $$

There exists $d\in {\mathbb Z}_{>0}$ such that $d\Delta $ is an integral divisor. By Corollary 5.7, applied to $d\Delta $ and the coherent sheaves ${\mathcal O}_X(-\lceil s\Delta \rceil -E_j)$ for $0\le s<d$ , we have that

$$ \begin{align*}h^1(X,{\mathcal O}_X(-\lceil n\Delta\rceil -E_j)) \end{align*} $$

is bounded for positive n. Since $(-\Delta \cdot E_j)>0$ , we have (by the Riemann–Roch theorem (6)) that there exists $c'>0$ such that

$$ \begin{align*}h^0({\mathcal O}_X(-\lceil n\Delta\rceil )\otimes{\mathcal O}_{E_j})>c'n \end{align*} $$

for $n\gg 0$ . Thus, there exists $c>0$ such that the image $A_n:=\mbox {Im}(\Gamma (X,{\mathcal O}_X(-\lceil n\Delta \rceil ))$ in $B_n:=\Gamma (E_j,{\mathcal O}_X(-\lceil n\Delta \rceil )\otimes {\mathcal O}_{E_j})$ satisfies

(28) $$ \begin{align} \lambda_R(\Gamma(X,{\mathcal O}_X(-\lceil n\Delta\rceil))/\Gamma(X,{\mathcal O}_X(-\lceil n\Delta\rceil -E_j))= \lambda_R(A_n)=\dim_kA_n\ge cn \end{align} $$

for $n\gg 0$ .

Let $A=\oplus _{n\ge 0}A_n$ . We have that $B_{0}$ is a finite field extension of $k=R/m_R=A_0$ . Now, ${\mathcal O}_X(-d\Delta )\otimes {\mathcal O}_{E_j}$ is ample on the projective curve $E_j$ , so there exists $e\in {\mathbb Z}_{>0}$ which is divisible by d such that ${\mathcal O}_X(-e\Delta )\otimes {\mathcal O}_{E_j}$ is very ample and $\overline B=\oplus _{m\ge 0}B_{me}$ is a finitely generated $B_{0}$ -algebra which is generated by its terms of the lowest positive degree $me$ (see [Reference Hartshorne20, Th. II.5.19 and Exer. II.5.14]). Thus, $\overline B$ is the coordinate ring of a projective embedding of the curve $E_j$ in a projective space over $B_0$ , determined by a $B_0$ -basis of $\Gamma (E_j,{\mathcal O}_X(-e\Delta )\otimes {\mathcal O}_{E_j})$ . Thus, $\overline B$ has dimension 2. Let $\overline A=\bigoplus _{m\ge 0}A_{me}$ .

By (28), for $n\gg 0$ , there exists $F\in A_{ne}$ such that $0\ne F$ . The ring $\overline B_{(F)}$ of elements of degree zero in the localization $\overline B_F$ is such that $\mbox {Spec}(\overline B_{(F)})$ is the affine variety $E_j\setminus V(F)$ , with maximal ideals in $\overline B_{(F)}$ corresponding to height 1 homogeneous prime ideals in $\mbox {Proj}(\overline B)$ which do not contain F (by [Reference Hartshorne20, Prop. II.2.5]). Thus, there exists a homogeneous height 1 prime ideal $Q=\oplus _{n> 0}Q_{ne}$ in $\overline B$ which does not contain F.

Let $P=\overline A\cap Q$ , where $P=\oplus _{n> 0}P_{ne}$ with $P_{ne}=Q_{ne}\cap A_{ne}$ . $\dim \overline B/Q=1$ implies that there exists $d\in {\mathbb Z}_{>0}$ such that $\dim _k(B_{ne}/Q_{ne})<d$ for all n (by [Reference Bruns and Herzog5, Th. 4.1.3]). Thus, by (28), we have that $P\ne 0$ . P is not the graded maximal ideal $\oplus _{n\ge 0}A_{ne}$ of $\overline A$ since $F\not \in P$ .

We have constructed a chain of distinct homogeneous prime ideals $0\subset P\subset \bigoplus _{n>0}A_{ne}$ in $\overline A$ and thus $\overline A$ has dimension $\ge \ 2$ . The extension $\overline A\rightarrow A$ is integral, so $\dim A\ge 2$ by the going up theorem (see [Reference Atiyah and MacDonald2, Th. 5.11]). We have that $m_R\Gamma (X,{\mathcal O}_X(-\lceil n\Delta \rceil ))\subset \Gamma (X,{\mathcal O}_X(-\lceil n\Delta \rceil -E_j))$ for all $n\ge 0$ by (27). We thus have a surjection $R[\Delta ]/m_RR[\Delta ]\rightarrow A$ and so $\dim A\le \dim R[\Delta ]/m_RR[\Delta ]$ . However, $\dim R[\Delta ]/m_RR[\Delta ]\le 2$ by [Reference Cutkosky and Srinivas14, Lem. 3.6], so that $\dim A=2$ .