Equations of negative curves of blow-ups of Ehrhart rings of rational convex polygons
aa r X i v : . [ m a t h . A C ] J a n EQUATIONS OF NEGATIVE CURVES OF BLOW-UPS OFEHRHART RINGS OF RATIONAL CONVEX POLYGONS
KAZUHIKO KURANO
Abstract.
Finite generation of the symbolic Rees ring of a space monomial primeideal of a 3-dimensional weighted polynomial ring is a very interesting problem.Negative curves play important roles in finite generation of these rings. We areinterested in the structure of the negative curve. We shall prove that negativecurves are rational in many cases.We also see that the Cox ring of the blow-up of a toric variety at the point(1 , , . . . ,
1) coincides with the extended symbolic Rees ring of an ideal of a poly-nomial ring. For example, Roberts’ second counterexample to Cowsik’s question(and Hilbert’s 14th problem) coincides with the Cox ring of some normal projectivevariety (Remark 2.7). Introduction
Finite generation of symbolic Rees rings is one of very interesting problems incommutative ring theory. It is deeply related to Hilbert’s 14th problem or Kro-necker’s problem (Cowsik’s question [2]). It often happens that the Cox ring (or amulti-section ring) of an algebraic variety coincides with the extended symbolic Reesring of an ideal of a ring. Finite generation of these rings is also a very importantproblem in birational geometry.The simplest non-trivial examples of symbolic Rees rings are R s ( p a,b,c ) = M n ≥ p a,b,c ( n ) t n ⊂ k [ x, y, z, t ] , where p a,b,c is the kernel of the k -algebra homomorphism φ a,b,c : k [ x, y, z ] −→ k [ λ ]given by φ a,b,c ( x ) = λ a , φ a,b,c ( y ) = λ b , φ a,b,c ( z ) = λ c ( k is a field and a , b , c arepairwise coprime integers) and p a,b,c ( n ) = p a,b,cn k [ x, y, z ] p a,b,c ∩ k [ x, y, z ] is the n thsymbolic power. In this case, the extended symbolic Rees ring R s ( p a,b,c )[ t − ] isthe Cox ring of the blow-up Y ∆ a,b,c of the weighted projective surface X ∆ a,b,c =Proj( k [ x, y, z ]) at the point (1 ,
1) in the torus. Many commutative algebraists andalgebraic geometers studied them and gave many results ([13], [3], [9], [10], [6], [7],[17], [18] etc.). Finite generation of R s ( p a,b,c ) depends on a , b , c and the characteristicof k . There are many examples of finitely generated R s ( p a,b,c ). Examples of infinitelygenerated R s ( p a,b,c ) were first discovered by Goto-Nishida-Watanabe [10]. In thecase where the characteristic of k is positive, it is not known whether R s ( p a,b,c ) isfinitely generated for any a , b , c or not. We say that C is a negative curve if C is a curve in Y ∆ a,b,c with C < C is not the exceptional curve E of theblow-up π : Y ∆ a,b,c → X ∆ a,b,c . Since the Picard number of Y ∆ a,b,c is two, such acurve is unique if it exists. Cutkosky [3] proved that finite generation of the Coxring of Y ∆ a,b,c is equivalent to the existence of curves D and D in Y ∆ a,b,c such that D ∩ D = ∅ , D = E and D = E (the defining equations of π ( D ) and π ( D ) satisfyHuneke’s criterion [13] for finite generation). If R s ( p a,b,c ) is finitely generated with √ abc Z , we may assume that either D or D coincides with the negative curve C . If the negative curve does not exists, one can prove Nagata’s conjecture (for abc points) affirmatively as in [4]. Therefore the existence of the negative curve is a veryimportant question.The aim of this paper is to study the structure of the negative curve. In particular,the author is interested in the problem whether the negative curve is rational or not.If there exists a negative rational curve C , it is possible to estimate the degree ofthe curve D such that C and D satisfies Huneke’s criterion for finite generation inthe same way as in [18].Assume that the negative curve C in Y ∆ a,b,c exists. We shall study π ( C ) ∩ T in this paper, where T is the torus in X ∆ a,b,c . When C.E = r , the defining equationof π ( C ) ∩ T in T = Spec( k [ v ± , w ± ]) is an irreducible Laurent polynomial containedin ( v − , w − r k [ v ± , w ± ] and the Newton polygon has area less than r / r -nct in this paper (Defi-nition 3.1). Remark that C and π ( C ) ∩ T is birational. The author does not knowany example that C is not rational.We shall prove some basic properties on r -ncts in Proposition 3.4.It is proved that, for each r ≥
0, there exist essentially finitely many r -ncts. When r = 1, there exists essentially only one 1-nct ϕ = vw −
1. When r = 2, there existsessentially only one 2-nct ϕ = − v w − vw + 3 vw −
1. When r = 3, there existessentially two 3-ncts ϕ and ϕ ′ as in Example 3.2. Let P ϕ ′ be the Newton polygonof ϕ ′ . In the case where the characteristic of k is not 2, P ϕ ′ is a tetragon. However,in the case of characteristic 2, P ϕ ′ is a smaller triangle than this tetragon. This isa reason why p , , contains a negative curve in the case of characteristic 2. Inother characteristic, p , , does not contain a negative curve (see Example 3.5).The following is the main theorem of this paper. Theorem 1.1.
Let k be a field and ϕ be an r -nct over k , where r ≥
2. Let P ϕ bethe Newton polygon of ϕ . Consider the prime ideal p ϕ of the Ehrhart ring E ( P ϕ , λ ) = M d ≥ M ( α,β ) ∈ dP ϕ ∩ Z kv α w β λ d ⊂ k [ v ± , w ± ][ λ ]satisfying p ϕ = E ( P ϕ , λ ) ∩ ( v − , w − k [ v ± , w ± ][ λ ]. Let Y ∆ ϕ be the blow-up of X ∆ ϕ = Proj( E ( P ϕ , λ )) at the point p ϕ . Let C ϕ be the proper transform of the curve QUATIONS OF NEGATIVE CURVES 3 V + ( ϕλ ) in X ∆ ϕ . Let I ϕ (resp. B ϕ ) be the number of the interior lattice points (resp.the boundary lattice points) of P ϕ .Consider the following conditions:(1) − K Y ∆ ϕ is nef and big.(2) − K Y ∆ ϕ is nef.(3) ( − K Y ∆ ϕ ) > a , a , . . . , a n be the first lattice points of the 1-dimensional cones inthe fan corresponding to the toric variety X ∆ ϕ . We put P − K X ∆ ϕ = { x ∈ R | h x , a i i ≥ − i = 1 , , . . . , n ) } . Then | P − K X ∆ ϕ | > / − K Y ∆ ϕ is big.(6) The Cox ring Cox( Y ∆ ϕ ) is Noetherian. (7) B ϕ ≥ r .(8) H ( Y ∆ ϕ , O Y ∆ ϕ ( K Y ∆ ϕ + nC ϕ )) = 0 for n > I ϕ = r ( r − .(10) The extended symbolic Rees ring R ′ s ( p ϕ ) is Noetherian.(11) C ϕ ≃ P k is satisfied.Then we have the following:(a) (1) = ⇒ (2) ========== ⇒ (7) ⇓ ⇓ (3) = ⇒ (4) = ⇒ (5) = ⇒ (8) ⇐⇒ (9) ⇐ = (11) ⇓ ⇓ (6) = ⇒ (10)is satisfied. (b) If k is algebraically closed, then (9) is equivalent to (11).(c) If the characteristic of k is positive, then (10) is always satisfied.We shall prove this theorem in section 4. Using this theorem, we shall see thatthe negative curve C in Y ∆ a,b,c is rational in many cases in Remark 4.6. For instance,if r ≤
4, then C is rational. If r ≤ k ) = 0, C is rational.2. Preliminaries
All rings in this paper are commutative with unity. As we shall see in Remark 2.7, Cox( Y ∆ ϕ ) is a extended symbolic Rees ring of an ideal I overthe polynomial ring Cox( X ∆ ϕ ). If Cl( X ∆ ϕ ) is torsion-free, then I is a prime ideal. In the casewhere the characteristic of k is 0, it holds √ I = I . We do not have to restrict ourselves to the polygon P ϕ for (1) ⇒ (3) ⇒ (4) ⇒ (5) ⇒ (6) ⇒ (10). They hold for any integral convex polygon. KAZUHIKO KURANO
For a Noetherian ring A and a prime ideal Q of A , Q ( n ) = Q n A Q ∩ A is called the n th symbolic power of Q . When n <
0, we think Q ( n ) = A . We put R s ( Q ) = ⊕ n ≥ Q ( n ) t n ⊂ A [ t ] R ′ s ( Q ) = ⊕ n ∈ Z Q ( n ) t n ⊂ A [ t ± ]and call them the symbolic Rees ring of Q and the extended symbolic Rees ring of Q , respectively. We know that R s ( Q ) is Noetherian if and only if so is R ′ s ( Q ). Definition 2.1.
Let a , b , c be pairwise coprime integers. Let k be a field and S a,b,c = k [ x, y, z ] be a graded polynomial ring with deg( x ) = a , deg( y ) = b , deg( z ) = c . Let λ be a variable and consider the k -algebra homomorphism φ a,b,c : S a,b,c −→ k [ λ ]given by φ a,b,c ( x ) = λ a , φ a,b,c ( y ) = λ b , φ a,b,c ( z ) = λ c . Let p a,b,c be the kernel of φ a,b,c . Remark 2.2.
Let Y ∆ a,b,c be the blow-up of X ∆ a,b,c = Proj S a,b,c at V + ( p a,b,c ). (Here X ∆ a,b,c is a toric variety with a fan ∆ a,b,c . We shall define ∆ a,b,c in Remark 2.3.) Let E be the exceptional divisor and H be the pull back of O X ∆ a,b,c (1). Then { E, H } isa generating set of Cl( Y ∆ a,b,c ) ≃ Z . By Cutkosky [3], we have an identification H ( Y ∆ a,b,c , O Y ∆ a,b,c ( dH − rE )) = [ p a,b,c ( r ) ] d and an isomorphism of rings as follows:(2.1) Cox( Y ∆ a,b,c ) = M d,r ∈ Z H ( Y ∆ a,b,c , O Y ∆ a,b,c ( dH − rE )) = M r ∈ Z p a,b,c ( r ) t r = R ′ s ( p a,b,c ) Remark 2.3.
Consider S a,b,c , p a,b,c , X ∆ a,b,c , Y ∆ a,b,c as in Definition 2.1 and Re-mark 2.2.By Herzog [11], we have p a,b,c = I (cid:18) x s y t z u y t z u x s (cid:19) = ( x s − y t z u , y t − x s z u , z u − x s y t ) , where s = s + s , t = t + t , u = u + u , and I ( ) is the ideal generated by2 × v = x s z u /y t , w = x s y t /z u . Then S a,b,c [ x − , y − , z − ] is a Z -graded ring such that S a,b,c [ x − , y − , z − ] = k [ v ± , w ± ](cf. the proof of Lemma 3.6 in [18]).Take integers i , j satisfying i a + j b = 1. Putting λ = x i y j , we obtain S a,b,c ⊂ S a,b,c [ x − , y − , z − ] = (cid:0) S a,b,c [ x − , y − , z − ] (cid:1) [ λ ± ] = k [ v ± , w ± , λ ± ] . Here suppose v α w β λ d = x i y j z k ∈ S a,b,c . QUATIONS OF NEGATIVE CURVES 5
Then we have d = deg( x i y j z k ) = ia + jb + kc ≥ i = s α + s β + i d ≥ j = − tα + t β + j d ≥ k = u α − uβ ≥ . Consider the rational triangle P a,b,c = ( α, β ) ∈ R (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s α + s β + i ≥ − tα + t β + j ≥ u α − uβ ≥ . When s , t , u are positive integers , P a,b,c is the following triangle: P a,b,c ✘✘✘✘✘✘✘✘✘✘✘✘ u u tt − s s Thus we have the identification(2.2) ( S a,b,c ) d = M ( α,β ) ∈ dP a,b,c ∩ Z kv α w β λ d . Consider the Ehrhart ring of P a,b,c defined by E ( P a,b,c , λ ) = M d ≥ M ( α,β ) ∈ dP a,b,c ∩ Z kv α w β λ d ⊂ k [ v ± , w ± , λ ± ] . Thus E ( P a,b,c , λ ) is isomorphic to S a,b,c as a graded k -algebra. Let ∆ a,b,c be thecomplete fan in R with one dimensional cones R ≥ ( s , s ) , R ≥ ( − t, t ) , R ≥ ( u , − u ) . Then Proj( E ( P a,b,c , λ )) is the toric variety X ∆ a,b,c with the fan ∆ a,b,c .Furthermore we have p a,b,c = E ( P a,b,c , λ ) ∩ ( v − , w − k [ v ± , w ± , λ ± ] . If p a,b,c is not a complete intersection, then all of s , s , t , t , u , u are positive by Her-zog [11]. In the case where p a,b,c is a complete intersection, we can choose positive integers s , t , u after a suitable permutation of a , b , c . KAZUHIKO KURANO
It is easy to see(2.3) p a,b,c ( r ) = E ( P a,b,c , λ ) ∩ ( v − , w − r k [ v ± , w ± , λ ± ]for any r ≥
1. Therefore we have H ( Y ∆ a,b,c , O Y ∆ a,b,c ( dH − rE )) = p r ∩ M ( α,β ) ∈ dP a,b,c ∩ Z kv α w β λ d = [ p a,b,c ( r ) ] d , where p = ( v − , w − k [ v ± , w ± ]. Remark 2.4.
With notation as in Remark 2.3, we shall calculate the area of P a,b,c here.Suppose z u = v α w β λ cu . Then the bottom vertex of cuP a,b,c is ( α , β ). Since x s y t = wz u = v α w β +1 λ cu , the point ( α , β + 1) is on the upper edge of cuP a,b,c . ( α ,β )( α ,β +1) cuP a,b,c ✘✘✘✘✘✘✘✘✘✘✘✘ u u tt − s s •• Then the width of cuP a,b,c is1 tt − u u + 1 s s + u u = t ua + s ub = cu ab . Here recall that a = e (( λ a ) , k [ λ ]) = e (( x ) , S a,b,c / p a,b,c ) = ℓ S a,b,c ( S a,b,c / ( x ) + p a,b,c )= ℓ S a,b,c ( S a,b,c / ( x, y t z u , y t , z u )) = tu − t u ,b = su − s u = s u + s u and cu = s a + t b , where e ( ) is the multiplicity and ℓ ( ) is the length. Therefore the area of cuP a,b,c is cu ab . Thus we know that the areaof P a,b,c is abc . Definition 2.5.
If an irreducible polynomial f ∈ [ p a,b,c ( r ′ ) ] d ′ satisfies d ′ /r ′ < √ abc ,we say that f ∈ [ p a,b,c ( r ′ ) ] d ′ is a negative curve.If a negative curve f ∈ [ p a,b,c ( r ′ ) ] d ′ exists, then both r ′ and d ′ are uniquely deter-mined, and f is also unique up to a constant factor. We denote the proper transformof V + ( f ) by C ( ⊂ Y ∆ a,b,c ). Then C satisfies C = d ′ abc − r ′ < QUATIONS OF NEGATIVE CURVES 7
Lemma 2.6.
Let k be a field and P be a rational convex polygon (the convex hullof a finite subset of Q ) in R . Let A = k [ v ± , w ± ] be a Laurent polynomial ringwith two variables v , w . Consider the Ehrhart ring E ( P, λ ) = M d ≥ M ( α,β ) ∈ dP ∩ Z kv α w β λ d ⊂ A [ λ ] . Let d be a positive integer and put ϕ = P ( α,β ) ∈ d P ∩ Z c ( α,β ) v α w β where c ( α,β ) ∈ k and (2.4) N ϕ = { ( α, β ) ∈ Z | c ( α,β ) = 0 } . Assume that N ϕ contains at least two elements. Then the following two conditionsare equivalent: (1) ϕλ d is a prime element of E ( P, λ ) . (2) ϕ is irreducible in A and each edge of P contains an element in N ϕ .Proof. Before proving this lemma, we shall give some remarks here. Suppose thata lattice point ( α ′ , β ′ ) is in the interior of d ′ P for some d ′ >
0. Then we have E ( P, λ )[( v α ′ w β ′ λ d ′ ) − ] = A [ λ ± ]and(2.5) (cid:0) E ( P, λ ) / ( ϕλ d ) (cid:1) [( v α ′ w β ′ λ d ′ ) − ] = ( A/ϕA ) [ λ ± ] . Here remark that both sides are not 0 since N ϕ contains at least two elements. Let { E , . . . , E s } be the set of edges of P . Putting p i = M d> M ( α,β ) ∈ d ( P \ E i ) ∩ Z kv α w β λ d ⊂ E ( P, λ ) , p i is a prime ideal of E ( P, λ ) of height 1 for i = 1 , , . . . , s . It is easy to see(2.6) p ( v α ′ w β ′ λ d ′ ) = p ∩ · · · ∩ p s . First assume (1). Since ϕλ d is a prime element, ϕ is irreducible in A by (2.5). Ifsome E i does not meet N ϕ , ϕλ d ′ is contained in p i . Since ϕλ d is a prime element,we obtain p i = ( ϕλ d ). It is impossible since N ϕ contains at least two points.Next assume (2). Since ϕ is irreducible in A , the right-hand side of (2.5) is anintegral domain. In order to prove (1), it is enough to show that ϕλ d , v α ′ w β ′ λ d ′ isan E ( P, λ )-regular sequence. Since E ( P, λ ) is a normal domain, we knowAss E ( P,λ ) (cid:16) E ( P, λ ) / ( v α ′ w β ′ λ d ′ ) (cid:17) = { p , . . . , p s } by (2.6). Since none of p i ’s contains ϕλ d by (2), v α ′ w β ′ λ d ′ , ϕλ d is an E ( P, λ )-regularsequence. Since v α ′ w β ′ λ d ′ and ϕλ d are homogeneous elements, ϕλ d , v α ′ w β ′ λ d ′ is an E ( P, λ )-regular sequence. (cid:3)
KAZUHIKO KURANO
Remark 2.7.
As in Remark 2.2, the extended symbolic Rees ring R ′ s ( p a,b,c ) isidentified with the Cox ring of a blow-up of a toric surface. Here we generalize thisidentification.Let X ∆ be a d -dimensional toric variety with a fan ∆. Let { R ≥ a , R ≥ a , . . . , R ≥ a n } be the set of the 1-dimensional cones in ∆. We assume that each a i is in Z d suchthat the greatest common measure of the components of a i is 1, and P i R a i = R d .Here remark that a i = a j if i = j . Then we have the exact sequence0 ←− Cl( X ∆ ) ←− Z n a a ... a n ←− Z d ←− , where Cl( X ∆ ) is the divisor class group of X ∆ .The morphism of monoids(2.7) Cl( X ∆ ) ←− Z n ⊃ ( N ) n induces the morphism of semi-group rings k [Cl( X ∆ )] ←− A := k [ x , x , . . . , x n ] = k [( N ) n ] . Let I be the kernel of this ring homomorphism. By (2.7), A has a structure of aCl( X ∆ )-graded ring. Then we have an isomorphism A ≃ Cox( X ∆ ) as a Cl( X ∆ )-graded ring [1].Let { p , p , . . . , p ℓ } be the set of minimal prime ideals of I and put I ( r ) = I r A p ∩ · · · ∩ I r A p ℓ ∩ A. In the case where Cl( X ∆ ) is torsion-free, I is a prime ideal of A and I ( r ) is the r th symbolic power of I . If the characteristic of k is 0, then we have I = √ I and I ( r ) = p ( r )1 ∩ p ( r )2 ∩ · · · ∩ p ( r ) ℓ .Let Y ∆ be the blow-up of X ∆ at (1 , , . . . ,
1) in the torus ( k × ) n . Then we have anisomorphism(2.8) Cox( Y ∆ ) ≃ R ′ s ( I )as a Cl( X ∆ ) × Z -graded ring.In the case of X ∆ a,b,c and Y ∆ a,b,c in Definition 2.1, Remark 2.2 and Remark 2.3, weput a = ( s , s ), a = ( − t, t ), a = ( u , − u ). Then the morphism (2.7) of monoidscoincides with Z ( a b c ) ←− Z ⊃ ( N ) . Let I be the kernel of the k -algebra homomorphism k [ x , x , x ] → k [ y ± ] ( x y a , x y b , x y c ). Then I = p a,b,c and Cox( Y ∆ a,b,c ) ≃ R ′ s ( p a,b,c ) by (2.8). It is justthe isomorphism in Remark 2.2 given by Cutkosky. QUATIONS OF NEGATIVE CURVES 9
Suppose a = (2 , − a = ( − , − a = (0 , Z ⊕ Z / (2) ←− Z ⊃ ( N ) such that (1 , , (1 , , , (1 , , , (2 , I be the kernel of the k -algebra homomorphism k [ x , x , x ] → k [ y ± ,y ]( y − ( x y y , x y , x y y ), we have Cox( Y ∆ ) ≃ R ′ s ( I ). Consider the following k -algebrahomomorphisms f : k [ x , x , x ] → k [ y ] ( x y , x y , x y ) and f : k [ x , x , x ] → k [ y ] ( x
7→ − y , x y , x
7→ − y ). If the characteristic of k isnot 2, then we have I = Ker( f ) ∩ Ker( f ).Suppose t ≥ a a a a a a a = − t tt − tt t −
11 0 00 1 00 0 1 − − − . Then the morphism (2.7) of monoids is Z t + 1 0 0 t t + 1 0 t t + 1 t ←−−−−−−−−−−−−−−−−−−−−−−−−−−−− Z ⊃ ( N ) . Let I be the kernel of the k -algebra homomorphism k [ x , x , x , S, T, U, V ] → k [ x , x , x , W ] ( S x t +11 W , T x t +12 W , U x t +13 W , V ( x x x ) t W ). Then I is a prime ideal of height 3 and Cox( Y ∆ ) ≃ R ′ s ( I ). Roberts [23] proved that R ′ s ( I ) is not Noetherian and some pure subring of R ′ s ( I ) gives a counterexampleto Hilbert’s fourteenth problem if the characteristic of k is 0. On the other hand, R ′ s ( I ) is Noetherian if the characteristic of k is positive [16].Let E be the Ehrhart ring of the following tetragon • •• ••• • and put X ∆ = Proj( E ). In this case, the 1-dimensional cones are a a a a = − − − − − and the morphism (2.7) of monoids is Z − ←− Z ⊃ ( N ) . Let I be the kernel of the k -algebra homomorphism k [ x , x , x , x ] → k [ y ± , y ± ]( x y y , x y y − , x y y , x y ). Then I is a prime ideal called atoric ideal. By Example 4.2, we know that R ′ s ( I ) is a Noetherian ring. Let p be aprime ideal of the Ehrhart ring E ⊂ k [ v ± , w ± ][ λ ] satisfying p = E ∩ ( v − , w − k [ v ± , w ± ][ λ ]. Then the symbolic Rees ring R ′ s ( p ) is a pure subring of R ′ s ( I ).Therefore we know that R ′ s ( p ) is also a Noetherian ring.3. Definition of r -nct and basic properties We define an r -nct and give basic properties of it in this section. Definition 3.1.
Let k be a field and A = k [ v ± , w ± ] be a Laurent polynomial ringwith two variables v , w . Let r be a positive integer. An element ϕ of A is called an equation of a negative curve with multiplicity r over k (or simply an r -nct over k )if the following two conditions are satisfied:(1) ϕ is irreducible in A and ϕ ∈ ( v − , w − r A .(2) Let N ϕ be the set defined as in (2.4). Let P ϕ be the convex hull of N ϕ . Then2 | P ϕ | < r is satisfied, where | P ϕ | is the area of the polygon P ϕ . Example 3.2.
Polynomials v − w − ϕ = vw − k .The area of P ϕ is 0.The polynomial ϕ = − ϕ ( v − − v ( w − = − v w − vw + 3 vw − P ϕ is 3 / ϕ = − ϕ ( v −
1) + v ( w − = − vw − v w + v w − vw + v w + vw is a 3-nct since it is irreducible. The area of P ϕ is 4.The polynomial ϕ ′ = − ϕ ( vw − − ( v − ( w − vw = − vw − v w + v w − vw − v w + v w QUATIONS OF NEGATIVE CURVES 11 is also a 3-nct since it is irreducible. The area of P ϕ ′ is 4 if the characteristic of k is not 2. (The Newton polygon P ϕ ′ depends on the characteristic of the base field.See Example 3.5 (3).)If the characteristic of k is not 2, the Newton polygons P ϕ i ’s and P ϕ ′ are as follows: P ϕ ✲✻ • • P ϕ ✲✻ • •• • P ϕ ✲✻ • ••• •• • P ϕ ′ ✲✻ • •• ••• • We can inductively construct an r -nct ϕ r by ϕ r = − ϕ r − ( v −
1) + ( − r − v ( w − r for r ≥ r -ncts as follows: Proposition 3.3.
Let S a,b,c , p a,b,c be the ring and the ideal as in Definition 2.1.Let d and r be positive integers. Suppose that f ∈ ( S a,b,c ) d corresponds to ϕλ d ∈ (cid:16)L ( α,β ) ∈ d P a,b,c ∩ Z kv α w β (cid:17) λ d under the identification (2.2). Then f is irreducible in S a,b,c and contained in [ p a,b,c ( r ) ] d with d r < √ abc (thatis, f ∈ [ p a,b,c ( r ) ] d is a negative curve) if and only if the following four conditionsare satisfied: (1) ϕ is irreducible in A = k [ v ± , w ± ] . (2) There exists an element of N ϕ on each edge of d P a,b,c . (3) ϕ ∈ ( v − , w − r A . Remark that N ϕ is contained in d P a,b,c . (4) 2 | d P a,b,c | < r .When this is the case, ϕ is an r -nct.Proof. Recall that | P a,b,c | = 1 / abc as in Remark 2.4. Therefore the condition (4)is equivalent to d r < √ abc . The condition (3) is equivalent to f ∈ [ p a,b,c ( r ) ] d by(2.3).Recall r >
0. If either the condition (3) or f ∈ [ p a,b,c ( r ) ] d is satisfied, N ϕ containsat least two elements. Then, by Lemma 2.6, ϕλ d is irreducible in S a,b,c = E ( P a,b,c , λ )if and only if both (1) and (2) are satisfied. (cid:3) Here we state basic properties of r -ncts. Proposition 3.4.
Let k be a field and A = k [ v ± , w ± ] be a Laurent polynomial ringwith two variables v , w . We put p = ( v − , w − A . Let r be a positive integer. (1) A unit element of A is of the form cv α w β , where c ∈ k × , α, β ∈ Z . If u is aunit of A and ϕ is an r -nct, then uϕ is also an r -nct.Let ξ : A → A be a k -isomorphism such that ξ ( p ) = p . Then there exists ( a ij ) ∈ GL(2 , Z ) such that ξ ( v ) = v a w a , ξ ( w ) = v a w a . If ϕ is an r -nct,then so is ξ ( ϕ ) . (2) Assume that ϕ is an r -nct over k . (a) We have (3.1) p r ∩ M ( α,β ) ∈ P ϕ ∩ Z kv α w β = kϕ. (b) The number of P ϕ ∩ Z is at most r ( r +1)2 + 1 . (c) If ζ is a reducible element contained in p r , then P ϕ does not contain N ζ .In particular, if r ≥ , P ϕ does not contain r + 1 points on a line. (3) Assume that ϕ is an r -nct over k . Then there exists an element c ∈ k × suchthat all the coefficients of cϕ is in the prime field of k . (4) If ϕ is an r -nct over k , then ϕ p r +1 . (5) Assume that ϕ is an r -nct over k . Let L/k be a field extension. Then ϕ isalso an r -nct over L . (6) Let
N CT r be the set of r -ncts over k . Consider the equivalence relation ∼ on N CT r generated by ϕ ∼ uϕ and ϕ ∼ ξ ( ϕ ) as in (1). Then the quotientset N CT r / ∼ is a non-empty finite set for each r .Proof. It is easy to show (1). We omit a proof of it.We shall prove (2) (a). Assume the contrary. Let ϕ ′ be an element containedin the left-hand side but not in the right-hand one. Since ϕλ is a prime elementof E ( P ϕ , λ ) by Lemma 2.6, ϕλ , ϕ ′ λ is an E ( P ϕ , λ )-regular sequence. Take a ho-mogeneous element h ∈ E ( P ϕ , λ ) such that ϕλ , ϕ ′ λ , h is a homogeneous system ofparameters of E ( P ϕ , λ ). Since E ( P ϕ , λ ) is a 3-dimensional graded Cohen-Macaulay QUATIONS OF NEGATIVE CURVES 13 ring, we have ℓ E ( P ϕ ,λ ) ( E ( P ϕ , λ ) / ( ϕλ, ϕ ′ λ, h )) = e (( h ) , E ( P ϕ , λ ) / ( ϕλ, ϕ ′ λ )) ≥ ℓ E ( P ϕ ,λ ) p ϕ (cid:0) E ( P ϕ , λ ) p ϕ / ( ϕλ, ϕ ′ λ ) E ( P ϕ , λ ) p ϕ (cid:1) · e (( h ) , E ( P ϕ , λ ) / p ϕ ) ≥ r (deg h ) , where ℓ is the length, e is the multiplicity and p ϕ = E ( P ϕ , λ ) ∩ p A [ λ ]. Here we remarkthat the first inequality comes from the additive formula of multiplicities, and thesecond one comes from ϕλ, ϕ ′ λ ∈ p rϕ and e (( h ) , E ( P ϕ , λ ) / p ϕ )) = e (( λ deg h ) , k [ λ ]) =deg h . On the other hand, consider the Poincare series P ( E ( P ϕ , λ ) , s ) = X n ≥ dim k E ( P ϕ , λ ) n s n = f ( s )(1 − s ) , where we remark that E ( P ϕ , λ ) is generated by elements of degree 1 over k . Here f (1) is equal to the multiplicity of E ( P ϕ , λ ), which is 2 | P ϕ | by a theorem of Ehrhart(e.g. see Part II of [12]). Hence we have P ( E ( P ϕ , λ ) / ( ϕλ, ϕ ′ λ, h ) , s ) = f ( s )(1 − s ) (1 − s deg h )(1 − s ) since ϕλ , ϕ ′ λ , h is a regular sequence. Substituting 1 for s , we have ℓ E ( P ϕ ,λ ) ( E ( P ϕ , λ ) / ( ϕλ, ϕ ′ λ, h )) = 2 | P ϕ | (deg h ) < r (deg h ) . It is a contradiction. Thus the equality (3.1) is proved. The left-hand side of (3.1) is defined in L ( α,β ) ∈ P ϕ ∩ Z kv α w β by r ( r +1)2 linear equa-tions. Therefore (b) follows from (a).We shall prove (c). If ζ is a reducible element contained in p r , then kϕ doesnot contain ζ and P ϕ does not contain N ζ . Assume that P ϕ contains r + 1 pointson a line. Replacing ϕ by some uξ ( ϕ ) as in (1), we may assume that P ϕ contains(0 , , r, P ϕ is a convex polygon.) Then we have( v − r ∈ p r and N ( v − r ⊂ P ϕ . It is a contradiction since ( v − r is reducible for r ≥ F be the prime field of k . The assertion immediatelycomes from ( v − , w − r F [ v ± , w ± ] ∩ M ( α,β ) ∈ P ϕ ∩ Z F v α w β ⊗ F k = ( v − , w − r k [ v ± , w ± ] ∩ M ( α,β ) ∈ P ϕ ∩ Z kv α w β = kϕ. Next we shall prove (4). Suppose that an r -nct satisfies ϕ ∈ p r +1 . Multiplying ϕ by a unit of A , we may assume that ϕ is a Laurent polynomial over the prime field We can also prove (3.1) using intersection theory on the blow-up of Proj( E ( P ϕ , λ )) at (1 , and the origin is in N ϕ . Assume that the characteristic of k is a prime number p . (Inthe case of characteristic 0, we can prove the assertion easier.) If both α and β weredivided by p for any ( α, β ) ∈ N ϕ , then ϕ would be reducible since ϕ is a Laurentpolynomial over the prime field of characteristic p . Therefore we may assume thatthere exists ( α ′ , β ′ ) ∈ N ϕ such that p α ′ . Then we have0 = v ∂ϕ∂v ∈ p r ∩ M ( α,β ) ∈ P ϕ ∩ Z kv α w β = kϕ. Here remark that the constant term of v ∂ϕ∂v is 0. It is a contradiction since ϕ and v ∂ϕ∂v are linearly independent over k .We shall prove (5). We have only to show that ϕ is irreducible in L [ v ± , w ± ].Assume the contrary and suppose ϕ = ψ ψ in L [ v ± , w ± ]. Then we have P ϕ = P ψ + P ψ , where the right-hand side is the Minkowski sum, that is { a + b | a ∈ P ψ , b ∈ P ψ } .Suppose ψ ∈ ( v − , w − r L [ v ± , w ± ] \ ( v − , w − r +1 L [ v ± , w ± ] ,ψ ∈ ( v − , w − r L [ v ± , w ± ] \ ( v − , w − r +1 L [ v ± , w ± ] . Then, by (4), we have r = r + r . Since r > | P ϕ | , we have r √ r √ r √ > q | P ϕ | ≥ q | P ψ | + q | P ψ | . Here the second inequality is called the Brunn-Minkowski inequality. Thereforeeither r > | P ψ | or r > | P ψ | is satisfied. Hence we know that some irreducibledivisor ϕ ′ of ϕ is an r ′ -nct for some r ′ . By (3), we may assume that ϕ ′ is a Laurentpolynomial over the prime field. It is a contradiction since ϕ is irreducible over k .We shall prove (6). Since N CT r contains ϕ r in Example 3.2, the set N CT r / ∼ is not empty. Any 1-nct is equivalent to v −
1. Suppose r ≥
2. Let ϕ be an r -nct.If | P ϕ | = 0, then all the points of N ϕ are on a line and we may assume that ϕ isa polynomial in v . Then ( v − r divides ϕ . It contradicts to the irreducibility of ϕ . Therefore we have | P ϕ | >
0. Let Ω be the convex hull of the four points (0 , √ r , √ r , r ), (0 , r ). Let P be an integral convex polygon (a convexhull of a finite subset of Z with positive area). Then, by (2), r -ncts ϕ with P ϕ = P are equivalent to each others. Since Ω is bounded, there exist only finitely manyintegral convex polygons contained in Ω. Therefore there exists a finite subset F of N CT r such that any r -nct ϕ with P ϕ ⊂ Ω is equivalent to one of F . It is enoughto prove that any integral convex polygon with area less than r / ξ satisfying ξ ( Z ) = Z . Let P be an integralconvex polygon with area less than r /
2. By an affine transformation preserving thelattice, we may assume that (0 ,
0) and ( α ,
0) are adjacent vertices of P , where α is a positive integer. Since | P | < r /
2, we may assume that any point ( α, β ) in P satisfies 0 ≤ β < r . Let ( α , β ) be the other vertex adjacent to (0 , QUATIONS OF NEGATIVE CURVES 15 transformation of the form (cid:18) a (cid:19) for some a ∈ Z , we may assume β > α ≥ α, β ) in P satisfies 0 ≤ α . Suppose that P is not contained in Ω.Take ( α , β ) ∈ P \ Ω. Let ℓ be the line through (0 ,
0) and ( α , β ). The distance ofthe line ℓ and the point ( α , β ) is bigger than that of ℓ and ( √ r , | P | > (the area of the triangle (0 , α , β ), ( √ r , β √ r > r . It is a contradiction. Therefore we have P ⊂ Ω. (cid:3) Example 3.5. (1) The quotient set
N CR / ∼ consists of the equivalence classof ϕ in Example 3.2.(2) The quotient set N CR / ∼ consists of the equivalence class of ϕ in Exam-ple 3.2. In fact, suppose that ϕ is a 2-nct. By Proposition 3.4 (2) (c), P ϕ doesnot contain three points on a line. As in the proof of Proposition 3.4 (6),we may assume that P ϕ has three successive vertices ( α , β ), (0 , , ≤ α < β <
4. If ( α , β ) = (0 , P ϕ contains (1 , ζ = ( v − w − ∈ p , P ϕ contains N ζ . It contradicts to Proposition 3.4(2) (c). Then we know ( α , β ) = (2 , ϕ is equivalent to ϕ .In the case of r = 1 , N CR r / ∼ consists of the equivalence class of ϕ r and P ϕ r is independent of k . Therefore, for r = 1 ,
2, the existence of nega-tive curves contained in [ p a,b,c ( r ) ] d does not depend on the base field k (seeProposition 3.3).(3) One can prove that the quotient set N CT / ∼ consists of two elements, whichare represented by ϕ and ϕ ′ in Example 3.2, respectively. The polygon P ϕ is the triangle with vertices (0 , , , k is not 2, then P ϕ ′ is the tetragon as in Example 3.2.Assume that the characteristic of k is 2. Since the coefficient of vw in ϕ ′ is − ϕ ′ is a 3-nct such that P ϕ ′ is the triangle with vertices (0 , ,
1) and(2 , a, b, c ) = (9 , ,
13) and ch( k ) = 2, there exists anegative curve f ∈ [ p , , ] corresponding to a 3-nct which is equivalentto ϕ ′ as in the proof of Theorem 4.1 in [21]. There does not exist a negativecurve in p , , if ch( k ) = 2.4. The Cox ring of the blow-up of X ∆ ϕ and the symbolic Rees ringof the Ehrhart ring E ( P ϕ , λ )Let f ∈ [ p a,b,c ( r ) ] d be a negative curve and ϕ be an r -nct such that f = ϕλ d asin Proposition 3.3. Then we have | P ϕ | ≤ | d P a,b,c | < r / . From the results of Gonz´alezAnaya-Gonz´alez-Karu [7], it can be inferred that R ′ s ( p a,b,c ) tends to be finitely generated when | d P a,b,c | is close to | P ϕ | , and R ′ s ( p a,b,c )tends to be infinitely generated when | d P a,b,c | is close to r /
2. Therefore it is natural to ask: Is R ′ s ( p ϕ ) a Noetherian ring?Here p ϕ is a prime ideal of the Ehrhart ring E ( P ϕ , λ ) ⊂ k [ v ± , w ± ][ λ ] defined by p ϕ = E ( P ϕ , λ ) ∩ ( v − , w − k [ v ± , w ± ][ λ ].We put X ∆ a,b,c = Proj( E ( P a,b,c , λ )) and X ∆ ϕ = Proj( E ( P ϕ , λ )), where ∆ a,b,c and∆ ϕ are the fans corresponding to the toric varieties X ∆ a,b,c and X ∆ ϕ , respectively.Let Y ∆ a,b,c (resp. Y ∆ ϕ ) be the blow-up of X ∆ a,b,c (resp. X ∆ ϕ ) at the point (1 ,
1) inthe torus Spec k [ v ± , w ± ]. Another reason why we are studying R ′ s ( p ϕ ) is that Y ∆ ϕ contains a negative curve that is birational to the negative curve in Y ∆ a,b,c . We areinterested in the birational class of the negative curve in Y ∆ a,b,c . (The author doesnot know any example that the negative curve in Y ∆ a,b,c is not rational.)We shall prove Theorem 1.1 in this section. Before proving this theorem, we shallshow the following lemma: Lemma 4.1.
With notation as in Theorem 1.1, the following conditions are equiv-alent: (i) H ( Y ∆ ϕ , O Y ∆ ϕ ( K Y ∆ ϕ + nC ϕ )) = 0 for any n > , (ii) H ( Y ∆ ϕ , O Y ∆ ϕ ( K Y ∆ ϕ + C ϕ )) = 0 , (iii) C ϕ . ( K Y ∆ ϕ + C ϕ ) = − , (iv) C ϕ . ( K Y ∆ ϕ + C ϕ ) < , (v) I ϕ = r ( r − , (vi) I ϕ ≤ r ( r − .If C ϕ ≃ P k , then (ii) is satisfied. In the case where k is algebraically closed, theconverse is also true.Proof. (i) ⇒ (ii), (iii) ⇒ (iv), (v) ⇒ (vi) are obvious.We shall prove (iv) ⇒ (ii). Assume that there exists an effective Weil divisor D on Y ∆ ϕ that is linearly equivalent to K Y ∆ ϕ + C ϕ . By (iv), there exists an effectiveWeil divisor D ′ such that D = C ϕ + D ′ . Let π ϕ : Y ∆ ϕ → X ∆ ϕ be the blow-up at thepoint (1 , K X ∆ ϕ + ( π ϕ ) ∗ ( C ϕ ) ∼ ( π ϕ ) ∗ ( D ) = ( π ϕ ) ∗ ( C ϕ ) + ( π ϕ ) ∗ ( D ′ ) , we have ( π ϕ ) ∗ ( D ′ ) − K X ∆ ϕ ∼ . It contradicts to the fact that the left-hand side is a non-zero effective divisor.We shall prove (ii) ⇒ (vi). We have0 = H ( Y ∆ ϕ , O Y ∆ ϕ ( K Y ∆ ϕ + C ϕ )) = p r − ∩ M ( α,β ) ∈ P ϕ ◦ ∩ Z kv α w β , where P ϕ ◦ is the interior of P ϕ and p = ( v − , w − k [ v ± , w ± ]. Since p r − isdefined by r ( r − linear equations in k [ v ± , w ± ], the number of P ϕ ◦ ∩ Z is less thanor equal to r ( r − . QUATIONS OF NEGATIVE CURVES 17
Next we shall prove (iii) ⇔ (v) and (vi) ⇒ (iv). By Pick’s theorem, we have2 | P ϕ | = B ϕ + 2 I ϕ − . Since C ϕ = 2 | P ϕ | − r and C ϕ . ( − K Y ∆ ϕ ) = B ϕ − r , we have C ϕ . ( K Y ∆ ϕ + C ϕ ) + 2 = − B ϕ + r + 2 | P ϕ | − r + 2 = 2 I ϕ − r ( r − . It is easy to check (iii) ⇔ (v) and (vi) ⇒ (iv).We shall prove (ii) ⇒ (i). As we have already seen, (ii) is equivalent to (iv).Therefore we have C ϕ . ( K X ∆ ϕ + nC ϕ ) < n >
0. Since any effective divisorthat is linearly equivalent to K Y ∆ ϕ + nC ϕ has C ϕ as a component, the multiplicationby C ϕ H ( Y ∆ ϕ , O Y ∆ ϕ ( K Y ∆ ϕ + ( n − C ϕ )) → H ( Y ∆ ϕ , O Y ∆ ϕ ( K Y ∆ ϕ + nC ϕ ))is bijective. Therefore (ii) implies (i).Next we shall prove (vi) ⇒ (v). We may assume that k is algebraically closed.Remark that(4.1) h ( O Y ∆ ϕ ( K Y ∆ ϕ + C ϕ )) = h ( O Y ∆ ϕ ( − C ϕ )) = h ( O C ϕ ) . Since (vi) is equivalent to (ii), we have h ( O C ϕ ) = 0 and C ϕ ≃ P k . By definition, C ϕ does not meet the singular points of Y ∆ ϕ . Then, by the adjunction formula, wehave ω Y ∆ ϕ ( C ϕ ) | C ϕ = ω C ϕ . Since (iii) is satisfied, (v) holds.If C ϕ ≃ P k , (ii) is satisfied by (4.1). (ii) implies C ϕ ≃ P k when k is algebraicallyclosed. (cid:3) Now we start to prove Theorem 1.1. (1) ⇒ (2) is trivially true.We shall prove (1) ⇒ (3) ⇒ (4) ⇒ (5) ⇒ (6) ⇒ (10) for any integral convexpolygon P and the corresponding complete 2-dimensional fan ∆.(1) ⇒ (3) is a basic fact. (Any nef and big Q -Cartier divisor D on a normalprojective surface satisfies D > ⇒ (4). Recall that − K X ∆ = D + · · · + D s is a Q -Cartier divisor,where each D i is a toric prime divisor corresponding to each edge of P . Take a posi-tive integer q such that qP − K X ∆ is an integral convex polygon and − qK X ∆ is a Cartierdivisor on X ∆ . By the Riemann-Roch theorem, we know that χ ( O X ∆ ( − nqK X ∆ )) isa polynomial in n of degree 2 such that the coefficient of n is ( − qK X ∆ ) /
2. On theother hand, h ( O X ∆ ( − nqK X ∆ )) = ( nqP − K X ∆ ∩ Z ) is a polynomial of degree 2 for n ≥ n is | qP − K X ∆ | .Since χ ( O X ∆ ( − nqK X ∆ )) = h ( O X ∆ ( − nqK X ∆ )) − h ( O X ∆ ( − nqK X ∆ ))+ h ( O X ∆ ( − nqK X ∆ ))and h ( O X ∆ ( − nqK X ∆ )) = h ( O X ∆ ( K X ∆ + nqK X ∆ )) = 0 for n ≥
0, we obtain( − K X ∆ ) / ≤ | P − K X ∆ | . Let E be the exceptional divisor of the blow-up Y ∆ → X ∆ at (1 ,
1) in the torus.Since ( − K Y ∆ ) = ( − K X ∆ ) + E = ( − K X ∆ ) −
1, we obtain | P − K X ∆ | > . We shall prove (4) ⇒ (5). We have h ( O X ∆ ( − nqK X ∆ )) = ( nqP − K X ∆ ∩ Z ). Bydefinition, H ( Y ∆ , O Y ∆ ( − nqK Y ∆ )) is a k -vector subspace of H ( X ∆ , O X ∆ ( − nqK X ∆ ))defined by nq ( nq + 1) / ( nqP − K X ∆ ∩ Z ) − nq ( nq + 1)2 ≤ h ( O Y ∆ ( − nqK Y ∆ )) . Since ( nqP − K X ∆ ∩ Z ) is a polynomial of degree 2 for n ≥ n is | qP − K X ∆ | , we know that − K Y ∆ is big if | P − K X ∆ | > .Next we shall prove (5) ⇒ (6). We may assume that k is an algebraically closedfield.First we shall construct a refinement ∆ ′ of ∆ such that • X ∆ ′ is a smooth toric variety, and • P − K X ∆ = P − K X ∆ ′ .Let { R ≥ a , R ≥ a , . . . , R ≥ a n } be the set of the 1-dimensional cones in ∆. We assume that each a i is the shortestinteger vector in the cone R ≥ a i . Assume that a , a , . . . , a n are arranged counter-clockwise around the origin. We think that each a i is a row vector of length 2. Weshall construct ∆ ′ using induction on(4.2) (cid:12)(cid:12)(cid:12)(cid:12) det (cid:18) a a (cid:19) × det (cid:18) a a (cid:19) × · · · × det (cid:18) a n − a n (cid:19) × det (cid:18) a n a (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) . Suppose that X ∆ is not smooth. After a linear transformation in SL(2 , Z ), we mayassume that a = (1 ,
0) and a = ( a, b ), where b > a >
0. Here we put b = (1 , (cid:12)(cid:12)(cid:12)(cid:12) det (cid:18) a b (cid:19) × det (cid:18) ba (cid:19) × det (cid:18) a a (cid:19) × · · · × det (cid:18) a n − a n (cid:19) × det (cid:18) a n a (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) is strictly less than (4.2). Let ¯∆ be the complete fan in R with 1-dimensional cones { R ≥ a , R ≥ b , R ≥ a , . . . , R ≥ a n } . Here one can check P − K X ∆ = P − K X ¯∆ . Repeating this process, we can construct ∆ ′ satisfying the required conditions.Let Y ∆ ′ be the blow-up of X ∆ ′ at the point (1 ,
1) in the torus. Then Y ∆ ′ → Y ∆ isa resolution of singularities. Then, since P − K X ∆ = P − K X ∆ ′ , we have H ( Y ∆ , O Y ∆ ( − nK Y ∆ )) = p n ∩ M ( α,β ) ∈ nP − KX ∆ ∩ Z kv α w β = H ( Y ∆ ′ , O Y ∆ ′ ( − nK Y ∆ ′ )) , where p = ( v − , w − k [ v ± , w ± ]. Hence, if − K Y ∆ is big, so is − K Y ∆ ′ . Then,by Theorem 1 in Testa-VarillyAlvarado-Velasco [25], Cox( Y ∆ ′ ) is finitely generated.Since Y ∆ ′ → Y ∆ is surjective, Cox( Y ∆ ) is also finitely generated by Theorem 1.1 inOkawa [22]. We have completed the proof of (5) ⇒ (6). QUATIONS OF NEGATIVE CURVES 19
We shall prove (6) ⇒ (10). Let E be the exceptional divisor of Y ∆ → X ∆ =Proj( E ( P, λ )). Let H be the pullback of O X ∆ (1). We put p = E ( P, λ ) ∩ p k [ v ± , w ± , λ ]. Then the extended symbolic Rees ring R ′ s ( p ) coincides with themultisection ring M r,d ∈ Z H ( Y ∆ , O Y ∆ ( dH − rE )) , which is a pure subring of Cox( Y ∆ ). Therefore, if Cox( Y ∆ ) is finitely generated, sois R ′ s ( p ).In the rest of this proof, we consider an r -nct ϕ , the integral convex polygon P ϕ and the corresponding fan ∆ ϕ .We shall prove (5) ⇒ (8). Suppose that there exists an effective divisor D that islinearly equivalent to K Y ∆ ϕ + nC ϕ . Then we have nC ϕ ∼ D − K Y ∆ ϕ and C ϕ is big.It contradicts to C ϕ = 2 | P ϕ | − r < . (2) ⇒ (7) follows from ( − K Y ∆ ϕ ) .C ϕ = B ϕ − r .We shall prove (7) ⇒ (8). Since 0 ≤ B ϕ − r = ( − K ∆ ϕ ) .C ϕ and C ϕ <
0, (iv) inLemma 4.1 is satisfied. Therefore (i) in Lemma 4.1 is satisfied.(8) ⇔ (9) is nothing but (i) ⇔ (v) in Lemma 4.1.(11) ⇒ (9) and (b) follow from Lemma 4.1.In the rest of the proof, we prove (8) ⇒ (10) and (c).When we prove (c), we may assume that k is a finite field. Then we can prove(10) in the same way as Theorem 1 in Cutkosky [3].Next we shall prove (8) ⇒ (10). By (c), we may assume that k is a field ofcharacteristic 0. Furthermore, we may assume that k is the field of complex numbers C . Since 0 = h ( O Y ∆ ϕ ( K Y ∆ ϕ + nC ϕ )) = h ( O Y ∆ ϕ ( − nC ϕ )) = h ( O nC ϕ )for any n >
0, we can contract C ϕ , that is, there exists a birational morphism ξ : Y ∆ ϕ → Z such that ξ ( C ϕ ) is a point, where Z is a normal projective surface withat most rational singularity. Here put H = ( π ϕ ) ∗ ( O X ∆ ϕ (1)), where π ϕ : Y ∆ ϕ → X ∆ ϕ is the blow-up at (1 ,
1) in the torus. Let E be the exceptional divisor of π ϕ . Let i and j be positive integers such that ( iH − jE ) .C ϕ = 0. Then there exists n > nξ ∗ ( iH − jE ) is a very ample Cartier divisor on Z . Then one can provethat n ( iH − jE ) is a semi-ample Cartier divisor on Y ∆ ϕ . It is easy to verify that R ′ s ( p ϕ ) is Noetherian. (cid:3) Example 4.2.
Suppose that the characteristic of k is 0. In Gonz´alezAnaya-Gonz´alez-Karu [7], they constructed two distinct r -ncts ϕ r and ϕ ′ r over k for each r ≥
3. Here P ϕ r is a triangle with vertices ( − , − r − ,
0) and (0 , r − P ϕ ′ r is a tetragon with vertices ( − , − r − , r − , r − r − , r − r is odd, and ( − , − r − , r , r − r − , r −
1) in thecase where r is even. Both P ϕ r and P ϕ ′ r contain r ( r +1)2 + 1 lattice points. Both Y ∆ ϕr and Y ∆ ϕ ′ r satisfy the condition (1) in Theorem 1.1. We shall give an outline of the proof of the above assertions for ϕ ′ r . (One canprove the same assertions for ϕ r in the same way.) Consider the tetragon P withfour vertices as above. Obviously it contains r ( r +1)2 + 1 lattice points. Put p =( v − , w − k [ v ± , w ± ]. Since(4.3) p r ∩ M ( α,β ) ∈ P ∩ Z kv α w β is defined by r ( r +1)2 linear equations in (cid:16)L ( α,β ) ∈ P ∩ Z kv α w β (cid:17) , (4.3) is not 0. Let ϕ ′ r be a non-zero element in (4.3). Using Lemma 4.3 below, we have p r ∩ M ( α,β ) ∈ P ∩ Z ( α,β ) =( − , − kv α w β = 0 . Therefore the coefficient of v − w − in ϕ ′ r is not zero. In the same way, we knowthat the coefficients of monomials corresponding to the vertices of P are not zero.Thus we obtain P ϕ ′ r = P . By Lemma 2.1 in [7], we know ϕ ′ r is irreducible. Since | P | = r − , we know that ϕ ′ r is an r -nct.Next we shall prove that − K Y ∆ ϕ ′ r is nef and big. It is easy to see that | P − K X ∆ ϕ ′ r | > . Therefore − K Y ∆ ϕ ′ r is big by Theorem 1.1. Let V be the closure ofSpec( k [ v ± , w ± ] / ( w − X ∆ ϕ ′ r . Let D be the toric prime divisor on X ∆ ϕ ′ r corresponding to the bottomedge of P ϕ ′ r . Let ˜ V and ˜ D be the proper transforms of V and D , respectively. Thenwe know − K Y ∆ ϕ ′ r is linearly equivalent to ˜ V + ˜ D . Since V > D >
0, weobtain ˜ V > D >
0. Thus we know − K Y ∆ ϕ ′ r is nef. Lemma 4.3.
Let k be a field of characteristic and n be a positive integer. For asubset U of Z , we put k h U i = { X ( α,β ) ∈ U c ( α,β ) v α w β | c ( α,β ) ∈ k } . Let L be a line in R such that ( L ∩ U ) = n . Put U ′ = U \ ( L ∩ U ) . Then we havean isomorphism of k -vector spaces k h U i ∩ ( v − , w − n k [ v ± , w ± ] ≃ k h U ′ i ∩ ( v − , w − n − k [ v ± , w ± ] . We omit a proof of Lemma 4.3. We can prove this lemma in the same way as theproof of Lemma 4.5 in [18].
Remark 4.4.
Let ϕ be an r -nct. By Proposition 3.4 (2) (b), we have ( P ϕ ∩ Z ) ≤ r ( r +1)2 + 1. In many cases ( P ϕ ∩ Z ) = r ( r +1)2 + 1 is satisfied. Here we show that QUATIONS OF NEGATIVE CURVES 21 it is equivalent to B ϕ = r + 1. If B ϕ = r + 1 is satisfied, we know I ϕ = r ( r − byTheorem 1.1. Thus we obtain ( P ϕ ∩ Z ) = B ϕ + I ϕ = r ( r + 1)2 + 1 . Conversely assume B ϕ + I ϕ = r ( r +1)2 + 1. Since0 < r − | P ϕ | = r − I ϕ − B ϕ + 2 = r − I ϕ + B ϕ ) + B ϕ + 2 = B ϕ − r by Pick’s theorem, we know I ϕ = r ( r − by Theorem 1.1. Therefore we have B ϕ = ( P ϕ ∩ Z ) − I ϕ = r + 1.In particular, the condition (7) in Theorem 1.1 is satisfied if ( P ϕ ∩ Z ) = r ( r +1)2 +1. Remark 4.5.
Assume that there exists a negative curve f ∈ [ p a,b,c ( r ) ] d for somepairwise coprime positive integers a , b , c . Then we know dim k [ p a,b,c ( r ) ] d = 1 bythe same reason as Proposition 3.4 (2) (a). Since [ p a,b,c ( r ) ] d is defined in [ S a,b,c ] d by r ( r +1)2 linear equations, we know dim k [ S a,b,c ] d ≤ r ( r +1)2 + 1. Let ϕ be the r -nctcorresponding to f as in Proposition 3.3. Let P be the convex hull of dP a,b,c ∩ Z .Then we have P ϕ ⊂ P ⊂ dP a,b,c . Usually it is very difficult to determine P ϕ .Here assume ( dP a,b,c ∩ Z ) = dim k [ S a,b,c ] d = r ( r +1)2 + 1. Then P contains just r ( r +1)2 + 1 lattice points. Since | P | ≤ | dP a,b,c | < r /
2, we know the number of latticepoints in the boundary of P is bigger than or equal to r + 1 and that in the interiorof P is less than or equal to r ( r − / P ϕ is contained in that of P , the condition (9) in Theorem 1.1 is satisfied for ϕ byLemma 4.1. Remark 4.6.
Assume that k is algebraically closed and there exists a negativecurve f ∈ [ p a,b,c ( r ) ] d for some pairwise coprime positive integers a , b , c . Let ϕ be the r -nct corresponding to f as in Proposition 3.3. Then the negative curve C in Y ∆ a,b,c is birational to C ϕ in Y ∆ ϕ . If the condition (9) in Theorem 1.1 is satisfied for ϕ , weknow that C is a rational curve.If r = 1, it is easy to check C ≃ P k (e.g. Lemma 3.2 in [18]).Suppose r = 2. Since the unique 2-nct satisfies (1) in Theorem 1.1 (cf. Example 3.5(2), Example 4.2), C is a rational curve. It is easy to see that C is singular ifand only if ( dP a,b,c ◦ ∩ Z ) >
1. There are many examples such that C ≃ P k (e.g. ( a, b, c ) = (3 , , , , , . . . ). There are also many examples such that C is singular (e.g. ( a, b, c ) = (5 , , , , , . . . ). In the case where( a, b, c ) = (3 , , , , R ′ s ( p a,b,c ) is Noetherian. In the case where ( a, b, c ) =(16 , , , , R ′ s ( p a,b,c ) is not Noetherian.Assume that r = 3. In this case any 3-nct satisfies (1) in Theorem 1.1 over anyfield k . In fact, when ch( k ) = 2, we can prove it in the same way as Example 4.2.Assume that ch( k ) = 2. For ϕ in Example 3.2, we can also prove it in the sameway as Example 4.2. Consider ϕ ′ in Example 3.2. Remark that P ϕ ′ is a triangle and the Picard number of Y ∆ ϕ ′ is 2. Since C ϕ ′ < − K Y ∆ ϕ ′ .C ϕ ′ = B ϕ ′ − − K Y ∆ ϕ ′ is nef and big.If r = 4, then (9) in Theorem 1.1 is satisfied over any field k . If r = 4 andch( k ) = 0, then (7) in Theorem 1.1 is satisfied.If r = 5 and ch( k ) = 0, then one can prove that (9) in Theorem 1.1 is satisfied.The author does not know any example that the condition (7) in Theorem 1.1 isnot satisfied in the case where ch( k ) = 0.From the above, if r ≤
4, then C is rational. If r = 5 and ch( k ) = 0, then C isrational.In many case dim k [ S a,b,c ] d = r ( r +1)2 + 1 is satisfied. When this is the case, C isrational since (9) in Theorem 1.1 is satisfied (cf. Remark 4.5). The author knows afew examples that dim k [ S a,b,c ] d < r ( r +1)2 + 1. See the next example. Example 4.7.
Suppose that k is of characteristic 0 and ( a, b, c ) = (8 , , f ∈ [ p a,b,c (9) ] . Let ϕ bethe corresponding 9-nct as in Proposition 3.3. Then P ϕ is the following pentagon: • • ••• ••• ••••• ••••• ••••••• •••••••• •••••• •••• •• It satisfies B ϕ + I ϕ = r ( r +1)2 = 45, B ϕ = r = 9, I ϕ = r ( r − = 36. In this case, P a,b,c is the triangle with three edges having slopes 1 / − / /
2. Since a + b + c is less than d/r in this case, − K Y ∆ a,b,c is neither nef nor big. Since we have abirational surjective map Y ∆ ϕ → Y ∆ a,b,c , we know that − K Y ∆ ϕ is neither nef norbig. The author does not know whether Cox( Y ∆ ϕ ) is Noetherian or not. Since thecondition (7) in Theorem 1.1 is satisfied, the extended symbolic Rees ring R ′ s ( p ϕ ) isNoetherian. In this case, generators of one-dimensional cones of the fan ∆ ϕ is a a a a a = − − − − . QUATIONS OF NEGATIVE CURVES 23
Therefore the morphism of monoids (2.7) is Z − − − ←− Z ⊃ ( N ) . Let I be the kernel of the k -algebra homomorphism k [ x , x , x , x , x ] → k [ y ± , y ± , y ± ]( x y y y − , x y − y − y , x y , x y , x y ). Then we haveCox( Y ∆ ϕ ) = R ′ s ( I ) as in Remark 2.7.Using a computer, we know that there exists a negative curve f ∈ [ P (18) ] in the case where ( a, b, c ) = (5 , ,
49) over a field k of characteristic 0. Let ϕ be the corresponding 18-nct as in Proposition 3.3. Let P be the convex hull of1617 P , , ∩ Z as in Remark 4.5. The number of lattice points in the boundary of P is r = 18 and that in the interior of P is r ( r − / × /
2. Since P ϕ is containedin P , the number of lattice points in the interior of P ϕ is r ( r − / × / ϕ . In thiscase, − K Y ∆ ϕ is neither nef nor big. The author does not know whether Cox( Y ∆ ϕ ) isNoetherian or not. Remark 4.8.
Let ϕ be an r -nct. Then we have h ( Y ∆ ϕ , O Y ∆ ϕ ( − K Y ∆ ϕ + C ϕ )) = p a ( C ϕ ), h ( Y ∆ ϕ , O Y ∆ ϕ ( − K Y ∆ ϕ + C ϕ )) = 0 and I ϕ = r ( r − + p a ( C ϕ ).Let f ∈ [ p a,b,c ( r ) ] d be the negative curve. Then we have h ( Y ∆ a,b,c , O Y ∆ a,b,c ( − K Y ∆ a,b,c + C )) = p a ( C ), h ( Y ∆ a,b,c , O Y ∆ a,b,c ( − K Y ∆ a,b,c + C )) = 0 and the number of lattice pointsin the interior of dP a,b,c is r ( r − + p a ( C ). References [1]
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