Finitely generated symbolic Rees rings of ideals defining certain finite sets of points in P^2
aa r X i v : . [ m a t h . A C ] A ug FINITELY GENERATED SYMBOLIC REES RINGS OF IDEALSDEFINING CERTAIN FINITE SETS OF POINTS IN P KEISUKE KAI AND KOJI NISHIDA
Abstract.
The purpose of this paper is to prove that the symbolic Rees rings of idealsdefining certain finite sets of points in the projective plane over an algebraically closedfield are finitely generated using a ring theoretical criterion which is known as Huneke’scriterion. Introduction
Let R be a commutative Noetherian ring and let a be a proper ideal of R . We denotethe set of minimal prime divisors of a by Min a . For any r ∈ Z , we define a ( r ) = \ p ∈ Min a ( p r R p ∩ R )and call it the r -th symbolic power of a . Moreover, taking an indeterminate t , we definethe symbolic Rees ring of a by R s ( a ) = X r ∈ N a ( r ) t r ⊂ R [ t ] , where N = { , , , . . . } . Although deciding whether the symbolic Rees rings of givenideals are finitely generated or not is an important problem in commutative algebra andalgebraic geometry, but usually it is a hard task. In this paper, we focus our attention ona ring theoretical criterion for finite generation of symbolic Rees rings which is known asHuneke’s criterion in a special situation described below.Let K be a field and let I be a proper homogeneous ideal of the polynomial ring S = K [ x, y, z ] which we regard as an N -graded ring by setting the degrees of x , y and z to suitable positive integers. We assume that S/I is a 1-dimensional reduced ring.Let m = ( x, y, z ) S . Because the symbolic powers of I are also homogeneous, we have S m ⊗ S R s ( I ) = R s ( IS m ), i.e., I ( r ) S m = ( IS m ) ( r ) for any r ∈ Z . On the other hand, if p ∈ Min I , we have IS p = p S p as √ I = I , and so S p ⊗ S R s ( I ) coincides with R ( S p ) = X r ∈ N p r S p · t r , which is the ordinary Rees ring of the 2-dimensional regular local ring S p . Here, let usrecall the following condition introduced in [4, Theorem 3.25] and [5, Proposition 2.1]. Definition 1.1.
Let < r i ∈ N and ξ i ∈ I ( r i ) for i = 1 , . We say that ξ and ξ satisfyHuneke’s condition on I if the following two equalities hold. Mathematics Subject Classification.
Primary: 13F20 ; Secondary: 13A02, 14N05.
Key words and phrases.
Symbolic power, Rees algebra, Symbolic Rees algebra. (a) IS m = p ( ξ , ξ ) S m . (b) G ( S p ) + = p ( ξ t r , ξ t r ) G ( S p ) for any p ∈ Min I ,where G ( S p ) = S p / p S p ⊗ R ( S p ) and G ( S p ) + denotes the ideal generated by the homoge-neous elements of positive degree. In both of these equalities, the right sides are obviouslycontained in the left sides, so the crucial requirement of the condition stated above is thatthe left sides are included in the right sides. Although the condition stated in Definition 1.1 is rather complicated, it is equivalent toan easy condition if the grading of S is ordinary and both of ξ and ξ are homogeneous. Proposition 1.2.
Suppose deg x = deg y = deg z = 1 . Let < r i , d i ∈ N and ξ i ∈ [ I ( r i ) ] d i for i = 1 , . Then ξ and ξ satisfy Huneke’s condition on I if and only if ht ( ξ , ξ ) S = 2 and d r · d r = e( S/I ) , where e( S/I ) denotes the multiplicity of S/I (cf. [1, Definition 4.1.5]) . Now, Huneke’s criterion can be described as follows.
Theorem 1.3. R s ( I ) is finitely generated if and only if there exist elements in I ( r ) and I ( r ) satisfying Huneke’s condition on I for some < r , r ∈ N . Huneke’s criterion was originally proved by Huneke (cf. [4, Theorem 3.1, 3.2]) in thecase where I is a prime ideal, and the generalized version was given by Kurano and Nishida(cf. [5, Theorem 2.5]) so that it can be applied to radical ideals. The purpose of thispaper is to prove that the symbolic Rees rings of the ideals defining certain finite sets inthe projective plane P are finitely generated using Huneke’s criterion.Let K be an algebraically closed field and deg x = deg y = deg z = 1. For a point P = ( a : b : c ) ∈ P = P K , we denote by I P the ideal of S generated by the maximalminors of the matrix (cid:18) x y za b c (cid:19) , which is the defining ideal of P . Of course, I P is a prime ideal of S generated by a regularsequence. Moreover, for a set H = { P , P , . . . , P e } of e points in P , we set I H = I P ∩ I P ∩ · · · ∩ I P e . Then we have I ( r ) H = I rP ∩ I rP ∩ · · · ∩ I rP e for any r ∈ Z . As is well known, R s ( I H ) is finitely generated if and only if so is R ′ s ( I H ) = X r ∈ Z I ( r ) H t r ⊂ S [ t, t − ] , and the finite generation of these graded rings is related to that of the Cox ring ∆ H , whichis the subring X ( r ,...,r e ) ∈ Z e ( I r P ∩ · · · ∩ I r e P e ) t r · · · t r e e INITELY GENERATED SYMBOLIC REES RINGS 3 of S [ t ± , . . . , t ± e ], where t , . . . , t e are indeterminates. Since R ′ s ( I H ) coincides with thediagonal part of ∆ H , R ′ s ( I H ) is finitely generated if so is ∆ H . For example, in [2] Elizondo,Kurano and Watanabe proved that ∆ H is finitely generated if the points of H lie on a linein P . Moreover, in [7] Testa, Varilly-Alvarado and Velasco proved the finite generationof ∆ H for the following cases.(i) e ≤ e − H lie on a (possibly reducible) conic in P .(iii) H consists of 10 points of pairwise intersections of 5 general lines in P .(iv) There exist 3 distinct lines L , L and L in P such that H consists of pairwiseintersections of these lines and 2, 3 and 5 additional points on L , L and L ,respectively ( e = 13).Of course, R s ( I H ) can be finitely generated for wider classes of H . For example, thefollowing is known. Theorem 1.4.
Let n be a positive integer which is not a multiple of the characteristic of K and let θ be a primitive n -th root of unity. We set H = { (1 : 0 : 0) , (0 : 1 : 0) , (0 : 0 : 1) } ∪ { ( θ i : θ j : 1) | i, j = 1 , . . . , n } . Then R s ( I H ) is finitely generated. If n = 1 or 2, then the number of points in H stated in the above theorem is 4 or 7,and so the finite generation of R s ( I H ) follows from that of ∆ H . In [3], Harbourne andSeceleanu proved Theorem 1.4 in the case where n = 3, and the case where n ≥ I ( n ) H satisfying Huneke’s condition on I H . Although both of those elements arehomogeneous in the case where n = 3, but one of the two elements is not homogeneous if n ≥
4. Moreover, by a similar argument we prove that the following assertion holds.
Theorem 1.5.
Let f and g be homogeneous polynomials in S such that S/ ( f, g ) is a -dimensional reduced ring. We put deg f = m and deg g = n . Let us assume that f ∈ I mA , g ∈ I nB , f I B and g I A , where A and B are distinct two points in P . We set H = { A , B } ∪ { P ∈ P | ( f, g ) ⊆ I P } . Then R s ( I H ) is finitely generated. The above theorem will be proved in Section 4 showing that there exist linear forms f , f , . . . , f m ∈ [ I A ] and g , g , . . . , g n ∈ [ I B ] such that f = f f · · · f m , g = g g · · · g n ,f i I B for any i = 1 , , . . . , m and g j I A for any j = 1 , , . . . , n .Let P ij be the intersection point of the lines defined by f i and g j . Because S/ ( f, g ) isreduced, f i f k (i.e., f i /f k K ) if i = k , and g j g ℓ if j = ℓ . Consequently, we see H = { A, B } ∪ { P ij | i = 1 , . . . , m and j = 1 , . . . , n } K. KAI AND K. NISHIDA and ♯H = mn + 2 (Figure 1). We will prove that R s ( I H ) is finitely generated by findingelements in I ( mn ) H and I (2) H satisfying Huneke’s condition on I H . If m = n , then both ofthose elements are not homogeneous. A B f f f m g g g n Figure 1.
Theorem 1.5Setting f = y m − z m , g = z n − x n , A = (1 : 0 : 0) and (0 : 1 : 0) in Theorem 1.5, we getthe following example. Example 1.6.
Let m, n be positive integers which are not multiples of the characteristicof K . Let θ m and θ n be primitive m -th and n -th root of unity, respectively. We set H = { (1 : 0 : 0) , (0 : 1 : 0) } ∪ { ( θ in : θ jm : 1) | i = 1 , . . . , n and j = 1 , . . . , m } . Then R s ( I H ) is finitely generated. Huneke’s condition
Let K be a field and let I be a proper homogeneous ideal of the polynomial ring S = K [ x, y, z ] which we regard as an N -graded ring setting the degrees of x , y and z to suitable positive integers. We assume that S/I is a 1-dimensional reduced ring. Let m = ( x, y, z ) and R = S m . The following result can be proved by the same argumentdeveloped in the proofs of [5, Proposition 2.1 and Lemma 2.2] replacing x with u . Theorem 2.1.
Suppose < r i ∈ N and ξ i ∈ I ( r i ) for i = 1 , . Let us take a homogeneouselement u of S so that uS + I is m -primary. Then we have ℓ R ( R/ ( u, ξ , ξ ) R ) ≥ r r · ℓ S ( S/uS + I ) and the following conditions are equivalent. (1) ℓ R ( R/ ( u, ξ , ξ ) R ) = r r · ℓ S ( S/uS + I ) . (2) ξ and ξ satisfy Huneke’s condition on I . As is described in Theorem 1.3, the finite generation of R s ( I ) can be characterizedby the existence of elements satisfying Huneke’s condition on I . Here, let us verify thatProposition 1.2 follows from the equivalence of the conditions (1) and (2) of Theorem 2.1.In the rest of this paper, we assume deg x = deg y = deg z = 1. Suppose ξ i ∈ [ I ( r i ) ] d i for INITELY GENERATED SYMBOLIC REES RINGS 5 i = 1 ,
2, where 0 < r i , d i ∈ N . If u is a linear form in S such that ℓ R ( R/ ( u, ξ , ξ ) R ) < ∞ ,then u, ξ , ξ is an S -regular sequence consisting of homogeneous polynomials of degrees1 , d , d , respectively, and so ℓ R ( R/ ( u, ξ , ξ ) R ) = ℓ S ( S/ ( u, ξ , ξ ) ) = d d . On the other hand, if u is a linear form of S whose image in the local ring R/IR generatesa reduction of the maximal ideal, we have ℓ S ( S/uS + I ) = ℓ R ( R/uR + IR ) = e uR ( R/IR ) = e m ( R/IR ) = e(
S/I ) . Consequently, if we choose a general linear form of x , y and z as u of Theorem 2.1, theequality of (1) holds if and only if d d = r r · e( S/I ). Thus we get Proposition 1.2.In order to explain how to use Proposition 1.2 and Theorem 1.3, let us verify thefollowing well known example.
Example 2.2.
Let H be a set of of distinct points P , P , P ∈ P . Then R s ( I H ) isfinitely generated.Proof . For i ∈ { , , } , we take a linear form f i of x , y and z which defines the line goingthrough P i and P i +1 , where P i +1 denotes P for i = 3. We set ξ = f f f and ξ = f f + f f + f f . Because I P = ( f , f ), I P = ( f , f ) and I P = ( f , f ), it follows thatMin ( ξ , ξ ) = { I P , I P , I P } , and so ht ( ξ , ξ ) = 2. On the other hand, as f i ∈ I P i ∩ I P i +1 for any i ∈ { , , } , we see ξ ∈ I P ∩ I P ∩ I P = I (2) H , and so ξ ∈ [ I (2) H ] . Similarly, we get ξ ∈ [ I H ] . Because32 ·
21 = 3 = ♯H = e( S/I H ) ,ξ and ξ satisfy Huneke’s condition on I H by Proposition 1.2. Therefore R s ( I H ) is finitelygenerated by Theorem 1.3. (cid:3) An alternative proof of Theorem 1.4
In the rest of this paper, K is an algebraically closed field and the grading of S = K [ x, y, z ] is ordinary. We put m = ( x, y, z ). As is well known, { p ∈ Spec S | p is homogeneous and dim S/ p = 1 } = { I P | P ∈ P } . For any P ∈ P , we denote the localization of S at I P and its maximal ideal by S P and m P , respectively. Let f and g be non-zero homogeneous polynomials of S such thatdeg f = m > g = n >
0. We set H f,g = { P ∈ P | ( f, g ) ⊆ I P } . Let us begin by verifying the following two lemmas, which may be well known.
Lemma 3.1.
The following conditions are equivalent. (1) dim S/ ( f, g ) = 1 . K. KAI AND K. NISHIDA (2) Min ( f, g ) = { I P | P ∈ H f,g } . (3) H f,g is a finite set.When this is the case, S/ ( f, g ) is a Cohen-Macaulay ring.Proof . (1) ⇒ (2) Suppose dim S/ ( f, g ) = 1. Let us take any p ∈ Min ( f, g ). Then p ( m , and so 0 < dim S/ p ≤ dim S/ ( f, g ) = 1. Consequently, p is a homogeneous idealwith dim S/ p = 1, which means that p = I P for some P ∈ P . Conversely, if P ∈ H f,g ,we obviously have I P ∈ Min ( f, g ).(2) ⇒ (3) This implication holds since Min ( f, g ) is a finite subset of Spec S .(3) ⇒ (1) Suppose that H f,g is finite. If ht ( f, g ) = 1, there exists h ∈ S such that( f, g ) ⊆ hS , which is impossible since there exist infinitely many P ∈ P such that h ∈ I P .Thus we see ht ( f, g ) = 2, and so dim S/ ( f, g ) = 1. Then, as f, g is an S -regular sequence, S/ ( f, g ) is a Cohen-Macaulay ring. (cid:3) Lemma 3.2.
The following conditions are equivalent. (1) S/ ( f, g ) is a -dimensional reduced ring. (2) ♯H f,g = mn . (3) dim S/ ( f, g ) = 1 and ♯H f,g ≥ mn . (4) I H f,g = ( f, g ) .When this is the case, we have m P = ( f, g ) S P for any P ∈ H f,g and I ( r ) H f,g = ( f, g ) r forany r ∈ Z .Proof . (1) ⇒ (2) Suppose that S/ ( f, g ) is a 1-dimensional reduced ring. Becausedim S/ ( f, g ) = 1, we have Min ( f, g ) = { I P | P ∈ H f,g } by Lemma 3.1. Then, forany P ∈ H f,g , it follows that S P / ( f, g ) S P is a field since S/ ( f, g ) satisfies Serr’s condition(R ), which means m P = ( f, g ) S P . Here, let us choose a linear form u ∈ S generally sothat its image in the Cohen-Macaulay local ring R/ ( f, g ) R generates a reduction of themaximal ideal. Then u, f, g is a maximal R -regular sequence consisting of homogeneouspolynomials of degrees 1 , m, n , respectively, and we havee m ( R/ ( f, g ) R ) = e uR ( R/ ( f, g ) R )) = ℓ R ( R/ ( u, f, g ) R ) = ℓ S ( S/ ( u, f, g ) ) = mn . On the other hand, by the additive formula of multiplicity, we havee m ( R/ ( f, g ) R ) = X P ∈ H f,g ℓ S P ( S P / m P )e m P ( S P /I P ) = ♯H f,g . Thus we see that the condition (2) is satisfied.(2) ⇒ (3) We get this implication by (3) ⇒ (1) of Lemma 3.1.(3) ⇒ (4) Suppose dim S/ ( f, g ) = 1 and ♯H f,g ≥ mn . Again, let us take a linear form u ∈ S generally, then we havee( S/I H f,g ) = e m ( R/ ( I H f,g ) R ) = e uR ( R/ ( I H f,g ) R )= ℓ R ( R/uR + ( I H f,g ) R ) = ℓ S ( uS + I H f,g ) . On the other hand, we havee(
S/I H f,g ) = ♯H f,g ≥ mn = ℓ S ( S/ ( u, f, g ) ) . INITELY GENERATED SYMBOLIC REES RINGS 7
Consequently, we get ℓ S ( S/uS + I H f,g ) ≥ ℓ S ( S/ ( u, f, g ) ) . However, as the inclusion I H f,g ⊇ ( f, g ) holds obviously, it follows that the both sides ofthe above inequality are equal, and so uS + I H f,g = ( u, f, g ). Then I H f,g = ( u, f, g ) ∩ I H f,g = ( f, g ) + uS ∩ I H f,g = ( f, g ) + u · I H f,g . Therefore, by Nakayama’s lemma, we see I H f,g = ( f, g ).(4) ⇒ (1) This implication is obvious.Finally, we show I ( r ) H f,g = ( f, g ) r for any r ∈ Z when the equivalent conditions (1) - (4)are satisfied. Of course, we may assume r >
0. Because I ( r ) H f,g ⊇ ( f, g ) r holds obviously, itis enough to show I ( r ) H f,g S p = ( f, g ) r S p , where p is any associated prime ideal of S/ ( f, g ) r .In fact, as S/ ( f, g ) r is a 1-dimensional Cohen-Macaulay ring, we have p ∈ Min ( f, g ), andso there exists P ∈ H f,g such that p = I P . Then, m P = ( f, g ) S P as is proved in the proofof (1) ⇒ (2). Hence we have I ( r ) H f,g S P = I rP S P = m rP = ( f, g ) r S P , and so the proof is complete as S p = S P . (cid:3) Now, we are ready to give an alternative proof for Theorem 1.4 using Huneke’s criterion.In the rest of this section, let n be a positive integer which is not a multiple of thecharacteristic of K . We take a primitive n -th root θ of unity, and set H = { (1 : 0 : 0) , (0 : 1 : 0) , (0 : 0 : 1) } ∪ { P ij | i, j = 1 , . . . , n } ⊂ P , where P ij = ( θ i : θ j : 1). Let f = y n − z n , g = z n − x n and h = x n − y n . Then, as f + g + h = 0, we have(3.1) ( f, g ) = ( g, h ) = ( h, f ) and H f,g = H g,h = H h,f . Moreover, it is easy to see that(3.2) f , g and h are elements of I P ij for any i, j = 1 , . . . , n ,which means { P ij } i,j ⊆ H f,g . Because dim S/ ( f, g ) = 1 and ♯ { P ij } i,j = n , by Lemma 3.2we see(3.3) H f,g = { P ij } i,j , I H f,g = ( f, g ) and m P ij = ( f, g ) S P ij for any i, j .Because I (1 : 0 : 0) = ( y, z ), I (0 : 1 : 0) = ( z, x ) and I (0 : 0 : 1) = ( x, y ), we get the followingassertions by Lemma 3.2, (3.1), (3.2) and (3.3).(3.4) I ( r ) H = ( y, z ) r ∩ ( z, x ) r ∩ ( x, y ) r ∩ ( f, g ) r for any r ∈ Z .(3.5) xf , yg and zh are elements of I H .If n = 1 or 2, then ♯H = 4 or 7, and so R s ( I H ) is finitely generated as is mentioned inIntroduction. Hence, we may assume n ≥ K. KAI AND K. NISHIDA
First, let us consider the case where n = 3. In this case, we set ξ = f gh and ξ = xf · yg + yg · zh + zh · xf . By (3.4) and (3.5), we have ξ ∈ [ I (3) H ] and ξ ∈ [ I H ] ⊆ [ I (2) H ] . Let p be any prime idealof S containing ξ and ξ . Because ξ ∈ p , one of f , g and h belongs to p . If f ∈ p , then yg · zh ∈ p as ξ ∈ p , and so ht p ≥ p includes one of ( f, y ), ( f, z ) or ( f, g ) (= ( f, h )).Similarly, we get ht p ≥ g ∈ p or h ∈ p . Consequently, we have ht ( ξ , ξ ) = 2. Hence,by Proposition 1.2 it follows that ξ and ξ satisfy HC on I H since93 ·
82 = 12 = ♯H = e( S/I H ) . Therefore R s ( I H ) is finitely generated by Theorem 1.3.In the rest of this section, we assume n ≥
4. In this case, taking an element α ∈ K sothat α = 0 ,
1, we set ξ = f gh · ( αf + g ) n − and ξ = ( xf ) · ( yg ) n − + ( yg ) · ( zh ) n − + ( zh ) · ( xf ) n − + f n − gh . Let us notice that ξ is not homogeneous although so is ξ . By (3.4) and (3.5) we caneasily verify that(3.6) ξ and ξ belongs to I ( n ) H .We aim to show that ξ and ξ satisfy Huneke’s condition on I H .First, let us verify I H R = p ( ξ , ξ ) R , where R = S m . As is noticed in Definition 1.1,the crucial point is to prove that the right side includes the left side. For that purpose, itis enough to see that the following assertion is true by (3.4). Claim 3.3.
Let p be a prime ideal of S such that ( ξ , ξ ) ⊆ p ⊆ m . Then p includes oneof ( x, y ) , ( y, z ) , ( z, x ) or ( f, g ) . In fact, as ξ ∈ p , one of f , g , h or αf + g belongs to p . If f ∈ p , then ( yg ) · ( zh ) n − ∈ p as ξ ∈ p , and so p includes ( y, z ) or ( f, g ) since p ( f, y ) = p ( f, z ) = ( y, z ) and ( f, g ) = ( h, f )by (3.1). Similarly, we see that p includes one of ( x, y ), ( z, x ) or ( f, g ) if g ∈ p or h ∈ p .So, let us consider the case where αf + g ∈ p . Then, as g ≡ − αf mod p and h = − ( f + g ) ≡ ( α − f mod p , it follows that ξ ≡ f n η mod p , where η = ( − α ) n − x y n − + α ( α − n − y z n − + ( α − z x n − − α ( α − . Because α ( α − = 0, we have η m , which means η p . Hence, we get f ∈ p as ξ ∈ p ,and so p includes ( αf + g, f ) = ( f, g ). Thus we have seen Claim 3.3.Next, we verify G ( S P ) + = p ( ξ t n , ξ t n ) G ( S P ) for any P ∈ H . Again, the crucialpoint is to prove that the right side includes the left side, which is deduced from the nextassertion. Claim 3.4.
Let P ∈ H and P be a prime ideal of G ( S P ) containing ( ξ t n , ξ t n ) G ( S P ) .Then we have P = G ( S P ) + . INITELY GENERATED SYMBOLIC REES RINGS 9 If P ∈ H and η ∈ m rP ∩ S for r ∈ N , we denote by ηt r the image of ( η/ t r ∈ m rP t r under the homomorphism R ( S P ) → G ( S P ).Let us start the proof of Claim 3.4 with checking the case where P ∈ H f,g . In this case,we have m P = ( f, g ) S P = ( g, h ) S P = ( h, f ) S P by (3.1) and Lemma 3.2, and none of x , y and z belongs to I P , which means that x , y and z are unites of G ( S P ). We set U = f t , V = gt and W = ht . Then we have U + V + W = 0 and G ( S P ) + is generated by any two elements of { U, V, W } as an ideal of G ( S P ). Moreover, we have the equalities ξ t n = U V W ( α · U + V ) n − and ξ t n = xy n − · U V n − + y z n − · V W n − + z x n − · W U n − + U n − V W in G ( S P ). Because ξ t n ∈ P , one of U , V , W and α · U + V belongs to P . If U ∈ P ,then y z n − · V W n − ∈ P as ξ t n ∈ P , and so P includes G ( S P ) + = ( U, V ) = (
U, W )as y z n − is a unit of G ( S P ). Similarly, we can see that P includes G ( S P ) + if V ∈ P or W ∈ P . So, let us consider the case where α · U + V ∈ P . Then, as V ≡ − α · U mod P and W = − ( U + V ) ≡ α − · U mod P , it follows that ξ t n ≡ η · U n mod P , where η is the element stated in the proof of Claim 3.3. Because η m , η is a unit of G ( S P ). Hence we get U ∈ P as ξ t n ∈ P . Thus we see that P includes α · U + V and U , which means P = G ( S P ) + .Next, let us consider the case where P = (0 : 0 : 1). Then, none of f , g and z belongsto I P = ( x, y ). Moreover, αf + g I P since αf + g ≡ (1 − α ) z n mod I P and α = 1. Hence f , g , z and αf + g are units of G ( S P ). On the other hand, m P = ( x, y ) S P . So, we set X = xt and Y = yt . Then G ( S P ) = ( X, Y ), and we have the equalities ξ t n = f g ( αf + g ) n − · ( X n − Y n ) and ξ t n = f g n − · X Y n − + f n − g · ( X n − Y n )in G ( S P ), where the second equality holds since ( yg ) · ( zh ) n − and ( zh ) · ( xf ) n − areincluded in I n +1 P . Because ξ t n ∈ P and f g ( αf + g ) n − is a unit of G ( S P ), we have X n − Y n ∈ P . Then we get X Y n − ∈ P since ξ t n ∈ P and f g n − is a unit of G ( S P ).Consequently, we see that X and Y belong to P , which means P = G ( S P ) + .If P = (0 : 1 : 0), we can prove P = G ( S P ) + similarly as the above case.Finally, we suppose P = (1 : 0 : 0). In this case, I P = ( y, z ) and g , h , x and αf + g areunits in G ( S P ). On the other hand, m P = ( y, z ) S P . So, we set Y = yt and Z = zt . Then G ( S P ) + = ( Y, Z ), and we have the equalities ξ t n = gh ( αf + g ) n − · ( Y n − Z n ) and ξ t n = g h n − · Y Z n − in G ( S P ), where the second equality holds since ( xf ) · ( yg ) n − , ( zh ) · ( xf ) n − and f n − gh are included in I n +1 P . Because ξ t n ∈ P and gh ( αf + g ) n − is a unit of G ( S P ), we get Y n − Z n ∈ P . On the other hand, we get Y Z n − ∈ P since ξ t n ∈ P and g h n − is a unit of G ( S P ). Consequently, we see that Y and Z belong to P , which means P = G ( S P ) + . Thus the proof of Theorem 1.4 is complete. Remark 3.5.
Suppose n ≥ . Then, I ( n ) H has no reduction generated by two homogeneouspolynomials by [6, Proposition 5.1] . However, by the argument stated in the proof of [5,Theorem 2.5] , we can prove that ( ξ , ξ ) R is a reduction of I ( n ) H R . Proof of Theorem 1.5
In this section, let f and g be homogeneous polynomials of S having positive degrees m and n , respectively. We assume f ∈ I mA and g ∈ I nB , where A and B are points of P .Let us take linear forms u, v ∈ S so that I A = ( u, v ). Because f ∈ I mA , we can express f = m X j =0 a j u j v m − j ( a j ∈ S ) . However, as f is a homogeneous polynomial of degree m , we can choose a , a , . . . , a m from K . Then fv m = m X j =0 a j · ( uv ) j ∈ K [ uv ] . Because K is algebraically closed, we can express fv m = m Y i =1 ( α i · uv − β i ) ( α i , β i ∈ K ) . Then, setting f i = α i u − β i v ∈ [ I A ] for i = 1 , , . . . , m , we have f = f f · · · f m . Similarly, there exist linear forms g , g , . . . , g n ∈ [ I B ] such that g = g g · · · g n . In the rest of this section, we assume A = B , f I B and g I A . Then we have(4.1)
A, B H f,g . Moreover, for any i = 1 , . . . , m and j = 1 , . . . , n , we have f i I B and g j I A , and so f i g j , which means that f i and g j define distinct two lines in P intersecting at thepoint P ij with I P ij = ( f i , g j ). Of course P ij ∈ H f,g for any i.j .Let us assume furthermore that S/ ( f, g ) is a 1-dimensional reduced ring. Then thefollowing assertions hold by Lemma 3.2.(4.2) ♯H f,g = mn . (4.3) m P = ( f, g ) S P for any P ∈ H f,g . INITELY GENERATED SYMBOLIC REES RINGS 11 (4.4) I ( r ) H f,g = ( f, g ) r for any r ∈ Z .Moreover, we have f i f k if i = k and g j g ℓ if j = ℓ . Here we suppose P ij = P kℓ .Then f k ∈ I P kℓ = I P ij . Hence, if i = k , we have I P ij = ( f i , f k ) = I A as f i f k , whichcontradicts to (4.1). Thus we get i = k . Similarly, we get also j = ℓ . Consequently, wesee P ij = P kℓ if i = k or j = ℓ , and so ♯ { P ij } i,j = mn . Hence the following assertion isdeduced by (4.2).(4.5) H f,g = { P ij | i = 1 , . . . , m and j = i, . . . , n } . Let h be a linear form in S defining a line going through A and B , i.e., h ∈ [ I A ∩ I B ] .For any i = 1 , . . . , m , we have f i h since f i I B and h ∈ I B . Hence we see(4.6) I A = ( f i , h ) for any i = 1 , . . . , m .As a consequence, we get(4.7) ( f, h ) ⊆ p ∈ Spec S ⇒ I A ⊆ p . The following two assertions can be verified similarly as (4.6) and (4.7).(4.8) I B = ( g j , h ) for any j = 1 , . . . , n .(4.9) ( g, h ) ⊆ p ∈ Spec S ⇒ I B ⊆ p . Let us take any P ∈ H f,g . If h ∈ I P , then I A = I P by (4.7), which contradicts to (4.1).Hence we have(4.10) h I P for any P ∈ H f,g .We set H = { A, B } ∪ H f,g . By (4.1) and (4.4), we have(4.11) I ( r ) H = I rA ∩ I rB ∩ ( f, g ) r for any r ∈ Z .Similarly as in Section 3, if P ∈ H and η ∈ m rP ∩ S for r ∈ N , we denote by ηt r the imageof ( η/ t r ∈ m rP t r under the homomorphism R ( S P ) → G ( S P ). Here we want to show thefollowing assertion.(4.12) f t m , ht is an sop for G ( S A ) .It is enough to show that G ( S A ) + is the unique prime ideal of G ( S A ) containing f t m and ht . So, let us take any P ∈ Spec G ( S A ) containing f t m and ht . Because the factorization f t m = m Y i =1 f i t holds in G ( S A ), we can choose i = 1 , . . . , m so that f i t ∈ P . Then, we have P = G ( S A ) + since m A = ( f i , h ) S A by (4.6). Similarly, the following assertion holds.(4.13) gt n , ht is an sop for G ( S B ) .Now, we are ready to prove Theorem 1.5. If m = 1 or n = 1, then all the points of H except for just one point lie on a line, and so the Cox ring ∆ H is finitely generated by theresult due to Testa, Varilly-Alvarado and Verasco (The case (ii) stated in Introductioncan be applied). So, in the rest, we assume m ≥ n ≥
2. We set ξ = f n g m ( f + g ) mn − m − n and ξ = f g + ( f + g ) h . Because f ∈ I mA , g ∈ I nB and h ∈ I A ∩ I B , we have ξ ∈ I ( mn ) H and ξ ∈ I (2) H by (4.11). We aim to show that ξ and ξ satisfy Huneke’s condition on I H .First, let us verify I H = p ( ξ , ξ ), which implies I H R = p ( ξ , ξ ) R . For that purpose,it is enough to see that the following assertion is true by (4.7), (4.9) and (4.11). Claim 4.1.
Let p be a prime ideal of S containing ξ and ξ . Then p includes one of ( f, h ) , ( g, h ) or ( f, g ) . In fact, as ξ ∈ p , one of f , g or f + g belongs to p . If f ∈ p , then ( f + g ) h ∈ p as ξ ∈ p , and so p includes ( f, f + g ) = ( f, g ) or ( f, h ). Similarly, we see that p includes( f, g ) or ( g, h ) if g ∈ p . If f + g ∈ p , then f g ∈ p as ξ ∈ p , and so p includes ( f, g ) as f ∈ p or g ∈ p . Thus we have seen Claim 4.1.Next, we verify G ( S P ) + = p ( ξ t mn , ξ t ) G ( S P ) for any P ∈ H , which is deduced fromthe next assertion. Claim 4.2.
Let P ∈ H and P be a prime ideal of G ( S P ) containing ξ t mn and ξ t .Then we have P = G ( S P ) + . Let us start the proof of the above assertion with checking the case where P ∈ H f,g . Inthis case, we have m P = ( f, g ) S P by (4.3) and h is a unit of G ( S P ) by (4.10). We set U = f t and V = gt . Then G ( S P ) + = ( U, V ) = (
U, U + V ) = ( U + V, V ) and we have the equalities ξ t mn = U n V m ( U + V ) mn − m − n and ξ t = U V + ( U + V ) · h in G ( S P ). Because ξ t mn ∈ P , one of U , V or U + V belongs to P . If U ∈ P , then U + V ∈ P since ξ t ∈ P and h = ( h ) is a unit, and so P = G ( S P ) + . Similarly,we see P = G ( S P ) + if V ∈ P . If U + V ∈ P , then U V ∈ P as ξ t ∈ P , and so P = G ( S P ) + as P contains U or V .Next, we consider the case where P = A . Let us notice that the equalities ξ t mn = ( f t m ) n · g m ( f + g ) mn − m − n and ξ t = f t · g + ( f + g ) · ( ht ) hold in G ( S A ). Because I A does not include g and f + g , it follows that g m ( f + g ) mn − m − n , g and ( f + g ) are units in G ( S A ). Hence we have f t m ∈ P as ξ t mn ∈ P . Then f t also belongs to P as it vanishes if m ≥
3, and so ht ∈ P as ξ t ∈ P . Therefore we see P = G ( S A ) + by (4.12).Finally, the case where P = B can be verified as above using (4.13). Thus we have seenClaim 4.2, and the proof of Theorem 1.5 is complete. References [1]
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C. Huneke , Hilbert functions and symbolic powers , Michigan Math. J. (1987), 293–318.[5] K. Kurano and K. Nishida , Infinitely generated symbolic Rees rings of space monomial curveshaving negative curves , Michigan Math. J. (1987), 293-318.[6] U. Nagel and A. Seceleanu , Ordinary and symbolic Rees algebras for ideals of Fermat pointconfigurations , J. Algebra (2016), 80–102.[7]
D. Testa, A. Varilly-Alvarado and M. Velasco , Big rational surfaces , Math. Ann. (2011),95–107.(K. Kai)
Department of Mathematics and Informatics, Graduate School of Science,Chiba University, Yayoi-cho 1-33, Inage-ku, Chiba 263-8522, Japan
E-mail address : k k04 [email protected] (K. Nishida, corresponding author) Department of Mathematics and Informatics, GraduateSchool of Science, Chiba University, Yayoi-cho 1-33, Inage-ku, Chiba 263-8522, Japan
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