Ideals generated by a -fold products of linear forms have linear graded free resolution
aa r X i v : . [ m a t h . A C ] A p r ZERO-DIMENSIONAL PROJECTIVE SCHEMES DEFINED BY a -FOLD PRODUCTS OFLINEAR FORMS RICARDO BURITY, S¸TEFAN O. TOH ˇANEANU AND YU XIEA
BSTRACT . Given Σ ⊂ R := K [ x , . . . , x k ] , any finite collection of linear forms, some possibly proportional,and any ≤ a ≤ | Σ | , it has been conjectured that I a (Σ) , the ideal generated by all a -fold products of Σ , haslinear graded free resolution. In this article we show the validity of this conjecture for the case when theseideals define a zero-dimensional projective scheme. As a consequence we validate the conjecture for the case k = 3 . This allows us to determine a generating set for the defining ideal of the Orlik-Terao algebra of thesecond order of a line arrangement in P K , and to conclude that for the case k = 3 , and Σ defining such aline arrangement, the ideal I | Σ |− (Σ) is of fiber type. We also show that the stable version of a Harbourne’sconjecture regarding to the containment of symbolic powers and regular powers is valid for ideals defined by aset of points giving a star configuration for s ≥ N + 1 hyperplanes in P N K .
1. I
NTRODUCTION
Let R := K [ x , . . . , x k ] be the ring of (homogeneous) polynomials with coefficients in a field K , withthe standard grading. Denote m := h x , . . . , x k i to be the irrelevant maximal ideal of R . Let L , . . . , L n belinear forms in R , some possibly proportional, and denote this collection by Σ = ( L , . . . , L n ) ⊂ R . For ℓ ∈ Σ , by Σ \ { ℓ } we will understand the collection of linear forms of Σ from which ℓ has been removed.Also, we denote | Σ | = n , and rk(Σ) := ht( h ℓ , . . . , ℓ n i ) .Let ≤ a ≤ n be an integer and define the ideal generated by a -fold products of Σ to be the ideal of RI a (Σ) := h{ L i · · · L i a | ≤ i < · · · < i a ≤ n }i . We also make the convention I (Σ) := R , and I b (Σ) = 0 , for all b > n . Also, if Σ = ∅ , I a (Σ) = 0 , for any a ≥ .A homogeneous ideal I ⊂ R generated in degree d is said to have linear (minimal) graded free resolution ,if one has the graded free resolution → R n b +1 ( − ( d + b )) → · · · → R n ( − ( d + 1)) → R n ( − d ) → R → R/I → , for some positive integer b . The integers n j ≥ are called the Betti numbers of R/I . By convention, thezero ideal has linear graded free resolution. Also we say that
R/I has linear graded free resolution if andonly if I has linear graded free resolution.[1, Conjecture 1] states that for any collection of linear forms Σ , and any ≤ a ≤ | Σ | , the ideals I a (Σ) (or R/I a (Σ) ) have linear graded free resolution. In [16] it is presented the current state of this conjecture,as well as it is shown that the conjecture is true whenever the support of Σ (i.e., the set of nonproportionalelements of Σ ) is generic. Mathematics Subject Classification.
Primary 13D02; Secondary 14N20, 52C35, 13A30, 94B27.
Key words and phrases. linear free resolution, ideals generated by products of linear forms, linear codes, Orlik-Terao algebra,ideal of fiber type.The first author was partially supported by CAPES - Brazil (grant: PVEX - 88881.336678/2019-01).Burity’s address: Departamento de Matemtica, Universidade Federal da Paraiba, J. Pessoa, Paraiba, 58051-900, Brazil, Email:[email protected]’s address: Department of Mathematics, University of Idaho, Moscow, Idaho 83844-1103, USA, Email: [email protected]’s address: Department of Mathematics, Widener University, Chester, Pennsylvania 19013, USA, Email: [email protected].
Suppose rk(Σ) = k . In the first part of this paper we check the validity of the conjecture whenever ht( I a (Σ)) = k − (see Theorem 2.1). We can handle this case because by [1, Proposition 2.3] we knowwhat is I a (Σ) sat , and then via Remark 1.1 below and the induction argument presented in [16], we willobtain the desired claim. As an immediate consequence of this case, we are able to prove the conjecturewhen k = 3 (see Theorem 2.3). These results generalize the main results in the unpublished preprint [14]. Remark 1.1.
Let J ⊂ R be an ideal generated in degree a . Then J ⊆ J sat ∩ m a . If R/J has linear gradedfree resolution (equivalently, reg(
R/J ) = a − ), since H m ( R/J ) = J sat /J , by [4, Theorem 4.3], we have ( J sat /J ) e = 0 , for any e ≥ a . This means that J sat ∩ m a ⊆ J , and therefore J = J sat ∩ m a . Theorem 2.1 allows us to analyze symbolic powers of some finite sets of points. We study (cid:0) sN (cid:1) points in P N K such that each point is the intersection of a subset of N of the s ≥ N + 1 hyperplanes H , . . . , H s withno N + 1 meet at a single point; we refer to such a set of points as a star configuration for s hyperplanes in P N K . We prove that the stable version of a Harbourne’s conjecture regarding to the containment of symbolicpowers and regular powers is valid for the defining ideals of such sets of points.In the second part of the paper we look at some elimination algebras associated to line arrangements in P K . This is the situation when k = 3 , and no two linear forms in Σ are proportional: i.e., Σ defines a linearrangement A in P K . Let s := |A| . In Remark 3.1, based on Theorem 2.3, we observe that the ideal I a ( A ) has linear powers for any a ∈ { , . . . , s } . Furthermore, when a = s − , if I := I s − ( A ) , in Theorem 3.4 wedetermine a set of generators for the defining ideal of the special fiber of I (also known as the Orlik-Teraoalgebra of the second order, see [13]), and in Theorem 3.5 we show that I is of fiber type. Both these resultsanswer affirmatively the two related conjectures stated in [13], for the case k = 3 .2. Z ERO - DIMENSIONAL SCHEMES WITH LINEAR FREE RESOLUTIONS
Some coding theory and duality.
A linear code C of dimension k , and length n , is the image of a K -linear map K k −→ K n , given by a (generating) matrix G of size k × n ; most of the time one supposesthat G has no zero column, and we suppose this as well. The minimum distance, d ( C ) , is the minimumnumber of nonzero entries in a nonzero vector in C . The numbers dim( C ) = rk( G ) , n , and d ( C ) are calledthe parameters of C , and they are invariant under rescaling and permutation of the columns of G . The left-multiplication of G by any k × k invertible matrix gives the same linear code C . Because of these properties,one has the following duality:To each column ( c , . . . , c k ) T of G consider the dual linear form c x + · · · + c k x k in R = K [ x , . . . , x k ] .Consequently, to G we can associate a collection of linear forms Σ = ( ℓ , . . . , ℓ n ) ⊂ R . Also, this processis reversible: to any collection of linear forms we can associate a generating matrix of a linear code C . Thedimension of C equals the rank of the collection of linear forms dual to some (any) generating matrix of thecode. Because of this duality, we will replace Σ by its dual code C .Suppose rk( G ) = k . For any ≤ r ≤ k one defines the r -th generalized Hamming weight , d r ( C ) , in thefollowing way. Let D ⊆ C be a subcode (i.e., subspace of C ). The support of D is Supp ( D ) := { i : ∃ ( x , . . . , x n ) ∈ D with x i = 0 } . Let m ( D ) := | Supp ( D ) | be the cardinality of the support of D . Then, d r ( C ) := min D⊆C , dim D = r m ( D ) . By convention, d ( C ) = 0 .By classical results in coding theory (see for example [1, Corollary 1.3 and Proposition 1.7]) this invarianthas the following description: n − d r ( C ) is the maximum number of columns of G that span a k − r dimensional vector space. For example, n − d k ( C ) = 0 , since we assumed that G has no zero columns. Or, n − d k − ( C ) is the maximum number of columns of G that are proportional to each-other. If I ⊂ R is a homogeneous ideal, by definition I sat := { f ∈ R | ∃ n ( f ) ≥ such that m n ( f ) · f ⊂ I } . ERO-DIMENSIONAL PROJECTIVE SCHEMES DEFINED BY a -FOLD PRODUCTS OF LINEAR FORMS 3 The r -th generalized Hamming weights help determine the heights of ideals generated by a -fold productsof linear forms. From [1, Proposition 2.2], for r = 1 , . . . , k , for any d r − ( C ) < a ≤ d r ( C ) , one has ht( I a ( C )) = k + 1 − r. For example, if ≤ a ≤ d ( C ) , then ht( I a ( C )) = k , the maximum possible value. From [12, Theorem 3.1],we have that for any ≤ a ≤ d ( C ) , I a ( C ) = m a . This has been the starting point of [1, Conjecture 1], since powers of the irrelevant ideal have linear gradedfree resolution (see for example [6, Corollary 1.5]).For d ( C ) < a ≤ d ( C ) , we have that I a ( C ) defines a zero-dimensional projective scheme. These will bethe main object of our study in this paper.Let Σ = ( ℓ , . . . , ℓ | {z } m , . . . , ℓ s , . . . , ℓ s | {z } m s ) be a collection of linear forms in R := K [ x , . . . , x k ] , with gcd( ℓ i , ℓ j ) = 1 if i = j . The support of Σ is Supp(Σ) := { ℓ , . . . , ℓ s } . Suppose rk(Σ) = k . Let n := m + · · · + m s . Let C be the linear code dual to Σ . Then, by the discussions above, ht( I a (Σ)) = k − if and only if d ( C ) < a ≤ d ( C ) . Suppose d ( C ) < a ≤ d ( C ) . Then, by [1, Proposition 2.3], we have the primary decomposition I a (Σ) = p a − n + ν Σ ( p )1 ∩ · · · ∩ p a − n + ν Σ ( p u ) u ∩ K, where K is an m -primary ideal, { p , . . . , p u } is the set of minimal primes over I a (Σ) , and for ≤ j ≤ u , ν Σ ( p j ) denotes the number of linear forms from Σ that belong to the ideal p j .As discussed in [12] (see also [16]), because p j is a minimal prime, then it is a linear prime generated byat least n − a + 1 elements of Σ (counted with multiplicity). So if p j = h ℓ i , . . . , ℓ i c i , with ℓ i , . . . , ℓ i c ∈ Supp(Σ) , then ν Σ ( p j ) = m i + · · · + m i c which is ≥ n − a + 1 . Also ht( p j ) = k − , so we mustobviously have c ≥ k − . As discussed, n − d ( C ) is the maximum number of linear forms in Σ , countedwith multiplicity, that generate a height k − linear prime ideal. So ν Σ ( p j ) ≤ n − d ( C ) , which gives a − d ( C ) ≥ a − n + ν Σ ( p j ) ≥ . To conclude, for d ( C ) < a ≤ d ( C ) , denote Γ k − (Σ) to be the set of all linear primes of height k − ,generated by at least n − a + 1 elements of Σ counted with multiplicity. Then we have I a (Σ) sat = \ p ∈ Γ k − (Σ) p a − n + ν Σ ( p ) . Theorem 2.1.
Let
Σ = ( ℓ , . . . , ℓ | {z } m , . . . , ℓ s , . . . , ℓ s | {z } m s ) be a collection of linear forms in R := K [ x , . . . , x k ] ,with gcd( ℓ i , ℓ j ) = 1 if i = j . Let C be the linear code dual to Σ . Then, for d ( C ) < a ≤ d ( C ) , the ideal I a (Σ) has linear graded free resolution.Proof. Similar to the proof of [16, Theorem 2.3], we will prove the result by induction on the pairs ( | Σ | , rk(Σ)) , with | Σ | ≥ rk(Σ) ≥ , Base cases. If rk(Σ) = 2 , then, modulo an embedding, [13, Theorem 2.2] proves this case. If | Σ | = rk(Σ) , then after a change of variables we can assume that Σ = ( x , . . . , x k ) . This is the Booleanarrangement, which leads to a particular case of star configurations, and therefore I a (Σ) has graded linearfree resolution for any ≤ a ≤ | Σ | (see part (3) in the Introduction of [13]). Inductive step.
Suppose | Σ | > rk(Σ) ≥ . Let C be the linear code dual to Σ , and consider a ∈{ d ( C ) + 1 , . . . , d ( C ) } . Denote n := | Σ | . After a change of variables, and possibly embedding in a smallerring, we may suppose rk(Σ) = k .Let ℓ ∈ Σ , Σ ′ := Σ \ { ℓ } , and let C ′ be the linear code dual to Σ ′ . Denote n ′ := n − | Σ ′ | . We firstshow the following claim: RICARDO BURITY, S¸TEFAN O. TOH ˇANEANU AND YU XIE
CLAIM: There exists ℓ ∈ Σ , such that I a (Σ) : ℓ = I a − (Σ ′ ) . Proof.
Let ℓ ∈ Σ .(i) If rk(Σ ′ ) = k − , then the multiplicity of ℓ must be one, and after a change of variables we cansuppose ℓ = x k , and Σ ′ ⊂ K [ x , . . . , x k − ] . Since I a (Σ) = ℓ · I a − (Σ ′ ) + I a (Σ ′ ) , and since ℓ is anonzero divisor in K [ x , . . . , x k ] /I a (Σ ′ ) , we obtain I a (Σ) : ℓ ⊆ I a − (Σ ′ ) . Since the other inclusionis obvious, the claim is shown for this situation (for any ≤ a ≤ n ).(ii) If d ( C ) = 1 , after a change of coordinates we can take ℓ as in part (i) above: the generating matrixof C will have its last row consisting of all zeroes, except in the position corresponding to the columndual to ℓ .(iii) Suppose d ( C ) ≥ . Then rk(Σ ′ ) = k , for any ℓ we choose. This is because otherwise, all the n − columns of the generating matrix of C ′ will span a k − dimensional vector space, and hence d ( C ) = n − ( n −
1) = 1 .Suppose we are in situation (iii) above. Some basic coding theory (see [9]) gives that for any ≤ r ≤ k , d r ( C ′ ) ≤ d r ( C ) ≤ d r ( C ′ ) + 1 . Here is why: we know that d r ( C ) is n minus the max number of columns of G that span a k − r dimensionalvector space, and similarly, d r ( C ′ ) is n − minus the max number of columns of G ′ that span a k − r dimensional vector space. • Since G ′ is obtained from G by deleting a column from it, we have n − − d r ( C ′ ) ≤ n − d r ( C ) ;hence the second inequality. • Consider n − d r ( C ) columns of G that span a k − r dimensional vector space. At least all of thesecolumns except one are columns of G ′ . So n − d r ( C ) − ≤ ( n − − d r ( C ′ ) ; hence the firstinequality.We have d ( C ) + 1 ≤ a ≤ d ( C ) , which leads to d ( C ′ ) ≤ d ( C ) ≤ a − ≤ d ( C ) − ≤ d ( C ′ ) . If a − d ( C ′ ) , then by [12, Theorem 3.1], we have I a − (Σ ′ ) = m a − ; remember that dim( C ′ ) = k ,and m = h x , . . . , x k i . But I a (Σ) is generated in degree a , so I a (Σ) : ℓ ⊆ m a − . So we obtain again that I a (Σ) : ℓ = I a − (Σ ′ ) .If d ( C ′ ) < a − ≤ d ( C ′ ) , then by inductive hypotheses I a − (Σ ′ ) has linear graded free resolution. ByRemark 1.1, we have I a − (Σ ′ ) = I a − (Σ ′ ) sat ∩ m a − . Also, by previous discussions we have I a − (Σ ′ ) sat = \ q ∈ Γ k − (Σ ′ ) q ( a − − n ′ + ν Σ ′ ( q ) . We have I a (Σ) ⊆ I a (Σ) sat ∩ m a , and consequently I a (Σ) : ℓ ⊆ ( I a (Σ) sat : ℓ ) ∩ ( m a : ℓ ) | {z } m a − . Recall Γ k − (Σ ′ ) consists of all the linear primes of height k − generated by at least n ′ − ( a −
1) + 1 = n − a + 1 elements of Σ ′ . Since Σ ′ ⊂ Σ , we have Γ k − (Σ ′ ) ⊆ Γ k − (Σ) . Therefore, I a (Σ) sat = \ p ∈ Γ k − (Σ) p a − n + ν Σ ( p ) ⊆ \ q ∈ Γ k − (Σ ′ ) q a − n + ν Σ ( q ) , and consequently, Here, G and G ′ are the generating matrices of the codes C and C ′ , respectively. ERO-DIMENSIONAL PROJECTIVE SCHEMES DEFINED BY a -FOLD PRODUCTS OF LINEAR FORMS 5 I a (Σ) sat : ℓ ⊆ \ q ∈ Γ k − (Σ ′ ) ( q a − n + ν Σ ( q ) : ℓ ) . Now, let q ∈ Γ k − (Σ ′ ) .(1) If ℓ / ∈ q , then ν Σ ′ ( q ) = ν Σ ( q ) . So q a − n + ν Σ ( q ) : ℓ = q a − n + ν Σ ( q ) = q ( a − − n ′ + ν Σ ′ ( q ) . (2) If ℓ ∈ q , then ν Σ ′ ( q ) = ν Σ ( q ) − , so q a − n + ν Σ ( q ) : ℓ = q a − n + ν Σ ( q ) − = q ( a − − n ′ + ν Σ ′ ( q ) . Therefore, from items (1) and (2) above, we obtain I a (Σ) sat : ℓ ⊆ I a − (Σ ′ ) sat , and therefore I a (Σ) : ℓ ⊆ I a − (Σ ′ ) . Since the other inclusion is obvious, this way we conclude the proof of the CLAIM. (cid:3) Let ℓ ∈ Σ satisfying the CLAIM. Because I a (Σ) : ℓ = I a − (Σ ′ ) we have the short exact sequence ofgraded R -modules: −→ R ( − I a − (Σ ′ ) −→ RI a (Σ) −→ R h ℓ, I a (Σ) i −→ . First we deal with the leftmost nonzero module.(a) If d ( C ′ ) < a − ≤ d ( C ′ ) , then by inductive hypotheses, I a − (Σ ′ ) has linear graded free resolution.(b) If d ( C ′ ) = a − , and dim( C ′ ) = rk(Σ ′ ) = k , then I a − (Σ ′ ) = m a − , which has linear graded freeresolution.(c) If d ( C ′ ) = a − , and dim( C ′ ) = rk(Σ ′ ) = k − , then d ( C ) = 1 , and therefore d ( C ′ ) = 1 aswell. So a = 2 , and after a change of variables, I a − (Σ ′ ) = h x , . . . , x k − i , which is a linear primein R , so it has linear graded free resolution.In conclusion, reg (cid:18) R ( − I a − (Σ ′ ) (cid:19) = ( a −
2) + 1 = a − . Now we deal with the rightmost nonzero module. We can suppose, after a change of variables, that ℓ = ℓ s = x k . Suppose that for all ≤ i ≤ s − , ℓ i = ¯ ℓ i + c i x k , c i ∈ K , where ¯ ℓ i are linear forms invariables x , . . . , x k − .Consider ¯ C the linear code dual to ¯Σ := (¯ ℓ , . . . , ¯ ℓ | {z } m , . . . , ¯ ℓ s − , . . . , ¯ ℓ s − | {z } m s − ) ⊂ ¯ R := K [ x , . . . , x k − ] . Let ¯ G be the corresponding generating matrix. Since we assumed that rk(Σ) = k , then rk( ¯Σ) = rk( ¯ G ) =dim( ¯ C ) = k − . The length of ¯ C is u := m + · · · + m s − < n .For r = 1 , , we have that v := u − d r ( ¯ C ) is the maximum number of columns of ¯ G that span an ( k − − r = k − r − dimensional vector space. Without loss of generality, suppose these are the first v columns of ¯ G . Then ht( h ¯ ℓ , . . . , ¯ ℓ v i ) = k − r − . Since x k is a nonzero divisor mod h x , . . . , x k − i , wehave ht( h ¯ ℓ , . . . , ¯ ℓ v , x k i ) = k − r . But h ¯ ℓ , . . . , ¯ ℓ v , x k i = h ℓ , . . . , ℓ v , x k i = h ℓ , . . . , ℓ v , ℓ s , . . . , ℓ s | {z } m s i . So the first v and the last m s = n − u columns of G span a k − r dimensional vector space. Therefore n − d r ( C ) ≥ v + n − u = n − d r ( ¯ C ) , leading to d r ( ¯ C ) ≥ d r ( C ) .We have d ( C ) + 1 ≤ a ≤ d ( C ) ≤ d ( ¯ C ) . RICARDO BURITY, S¸TEFAN O. TOH ˇANEANU AND YU XIE (1) If d ( ¯ C ) + 1 ≤ a , then by inductive hypotheses, I a ( ¯Σ) has a linear graded free resolution.(2) If a ≤ d ( ¯ C ) , then I a ( ¯Σ) = h x , . . . , x k − i a , which has linear graded free resolution.To sum up we got reg (cid:18) ¯ RI a ( ¯Σ) (cid:19) = a − . But R/ h ℓ, I a (Σ) i and ¯ R/I a ( ¯Σ) are isomorphic as R -modules, so they have the same regularity (see [4,Corollary 4.6]).Applying the inequalities of regularity under short exact sequence (see [5, Corollary 20.19 b.]), we canconclude that reg( R/I a (Σ)) ≤ a − , and therefore that I a (Σ) has linear graded free resolution. (cid:3) Symbolic powers of star configurations.
Let A = { H , . . . , H s } be a collection of s ≥ N + 1 hyper-planes in P N K , and suppose ℓ , . . . , ℓ s ∈ R = K [ x , . . . , x N ] are defining linear forms of these hyperplanes:i.e., H i = V ( ℓ i ) , i = 1 , . . . , s . Suppose A is generic, meaning that any N + 1 of the defining linear formsare linearly independent.Define the star configuration with support A to be V ( A ) := [ ≤ j < ··· Let I = I ( V ( A )) be the ideal of star configurations of s ≥ N + 1 hyperplanes in P N K .Then for m ≥ s/N , one has I ( Nm − N +1) ⊆ I m . Proof. By Theorem 2.1 and Remark 1.1, or by [15, Theorem 3.2], we have for m ≥ , I m = I m ( s − N +1) ( ℓ m · · · ℓ ms ) = \ ≤ j < ··· 1) = m ( s − N + 1) . This yields that I ( Nm − N +1) ⊆ M ( s − N +1) m . (cid:3) Multi line arrangements in P . Let Σ = ( ℓ , . . . , ℓ | {z } m , . . . , ℓ s , . . . , ℓ s | {z } m s ) , s ≥ be a collection of linearforms in R := K [ x, y, z ] , with gcd( ℓ i , ℓ j ) = 1 if i = j . Let n := m + · · · + m s , and suppose rk(Σ) = 3 .Also, let C be the linear code dual to Σ .In this section we will show that for any ≤ a ≤ n , the ideal I a (Σ) has linear graded free resolution.This is known to be true in various instances:(a) If a = n , then I n (Σ) = h ℓ m · · · ℓ m s s i , which has linear graded free resolution.(b) If a = n − , then, [13, Section 2.1] shows that I n − (Σ) has linear graded free resolution.(c) If a = n − , then [13, Theorem 2.5] shows that I n − (Σ) has linear graded free resolution.(d) If ≤ a ≤ d ( C ) , then [12, Theorem 3.1] gives that I a (Σ) = h x, y, z i a , which has linear gradedfree resolution.(e) If d ( C ) + 1 ≤ a ≤ d ( C ) , then, by Theorem 2.1, I a (Σ) has linear graded free resolution. Theorem 2.3. Let Σ be as above. Then for any ≤ a ≤ n , I a (Σ) has linear graded free resolution.Proof. From items (d) and (e) above, we have to show the theorem when d ( C ) + 1 ≤ a ≤ n . This is theinstance when ht( I a (Σ)) = 1 .Let M := max { m , . . . , m s } . Since n − d ( C ) is the maximum number of columns of a generatingmatrix that spans a − dimensional vector space, we have that d ( C ) = n − M . Suppose M = m and m ≥ m ≥ · · · ≥ m s ≥ .We prove the theorem by induction on n ≥ s . Base Case. If n = s , then m = · · · = m s = 1 , and so d ( C ) = n − . From what we talked previously,we only have to check the claim for a = ( n − 1) + 1 = n , which is true by item (a) above. Inductive Step. Suppose n > s (and so M = m ≥ ). Because a ≥ n − m + 1 = m + · · · + m s + 1 ,then every a -fold product of elements of Σ will have at least one factor that is ℓ . So I a (Σ) = ℓ · I a − (Σ ′ ) , where Σ ′ = ( ℓ , . . . , ℓ | {z } m − , . . . , ℓ s , . . . , ℓ s | {z } m s ) .By inductive hypotheses, I a − (Σ ′ ) has linear graded free resolution, and therefore, so does I a (Σ) . (cid:3) As a first consequence, we can now prove equality in the statement of [16, Lemma 2.1], when k = 3 . Proposition 2.4. Let Σ = ( ℓ , . . . , ℓ | {z } m , . . . , ℓ s , . . . , ℓ s | {z } m s ) , s ≥ be a collection of linear forms in R := K [ x, y, z ] , with gcd( ℓ i , ℓ j ) = 1 if i = j . Let n := m + · · · + m s , and suppose rk(Σ) = 3 . For ≤ a ≤ n ,let I := I a (Σ) , and let Γ( I ) be the set of all linear primes containing I of the form h ℓ i , . . . , ℓ i c i for ≤ i < · · · < i c ≤ s . Then, I = \ p ∈ Γ( I ) p a − n + ν Σ ( p ) . RICARDO BURITY, S¸TEFAN O. TOH ˇANEANU AND YU XIE Proof. For ≤ a ≤ d ( C ) , we have I = h x, y, z i a . In this situation, Γ( I ) = {h x, y, z i} . Obviously wehave ν Σ ( h x, y, z i ) = n , so the claim is true.For d ( C ) + 1 ≤ a ≤ d ( C ) , we have Γ( I ) = Γ ( I ) ∪ {h x, y, z i} . From Theorem 2.1 and Remark 1.1,we have I = I sat ∩ h x, y, z i a . So the discussions before Theorem 2.1 will prove the claim in this case aswell.Suppose d ( C ) + 1 ≤ a ≤ n .If n = s , then m = · · · = m s = 1 , so d ( C ) = n − . So a = n = s , and therefore I = h ℓ · · · ℓ s i = h ℓ i ∩ · · · ∩ h ℓ s i , which is the claim for this case (we have gcd( ℓ i , ℓ j ) = 1 , and ν Σ ( h ℓ i i ) = m i = 1 ).If n > s , then M := max { m , . . . , m s } ≥ . Suppose m , . . . , m t ≥ n − a + 1 , and m t +1 , . . . , m s ≤ n − a . By [16, Lemma 2.1], we have I ⊆ h ℓ i a − n + m ∩ · · · ∩ h ℓ t i a − n + m t ∩ K ∩ h x, y, z i a = h ℓ a − n + m · · · ℓ a − n + m t t i ∩ K ∩ h x, y, z i a , where K = \ p ∈ Γ( I ) , ht( p )=2 p a − n + ν Σ ( p ) . Let i ∈ { , . . . , t } . Any a -fold product of elements of Σ has at least a factor ℓ a − ( n − m i ) i . So we have I = ℓ a − n + m · · · ℓ a − n + m t t I b (Σ ′ ) , where Σ ′ consists of ℓ , . . . , ℓ t each of multiplicity n − a , and ℓ t +1 , . . . , ℓ s each of multiplicity m t +1 , . . . , m s , and b := a − P ti =1 ( a − n + m i ) = | Σ ′ | − ( n − a ) .If a = n , everything is clear. So suppose a ≤ n − . Then rk(Σ ′ ) = 3 . Let C ′ be the linear code dualto Σ ′ . Let M ′ := max { n − a, m t +1 , . . . , m s } , which is equal to n − a . So d ( C ′ ) = | Σ ′ | − M ′ = b . Thisimplies that ht( I b (Σ ′ )) = 2 , and from previous discussions, J := I b (Σ ′ ) = \ q ∈ Γ ( J ) q b −| Σ ′ | + ν Σ ′ ( q ) ∩ h x, y, z i b = \ q ∈ Γ ( J ) q a − n + ν Σ ′ ( q ) ∩ h x, y, z i b . If q ∈ Γ ( J ) , then ht( q ) = 2 , and ν Σ ′ ( q ) ≥ | Σ ′ | − b + 1 = n − a + 1 . Since Σ ′ ⊂ Σ , we have that q ∈ Γ( I ) and ht( q ) = 2 . Because ν Σ ( q ) ≥ ν Σ ′ ( q ) , we have K ⊆ \ q ∈ Γ ( J ) q a − n + ν Σ ( q ) =: L, and therefore I ⊆ h f i ∩ L ∩ h x, y, z i a , where f := ℓ a − n + m · · · ℓ a − n + m t t .We want to show the other inclusion. Let g be an element in the intersection at the right. Then g = f · h ,with h ∈ L : f , and h ∈ h x, y, z i a : f = h x, y, z i b .Let q ∈ Γ ( J ) , and suppose ℓ , . . . , ℓ v ∈ q , but ℓ v +1 , . . . , ℓ t / ∈ q . Then ν Σ ( q ) = m + · · · + m v + α ,and ν Σ ′ ( q ) = v ( n − a ) + α , for some α , an integer counting the multiplicities of other linear forms from Σ (and also from Σ ′ ) that belong to q . Then, q a − n + ν Σ ( q ) : f = q a − n + ν Σ ( q ) − P vj =1 ( a − n + m j ) = q a − n + ν Σ ′ ( q ) . This way we obtained that h ∈ J . So g ∈ f · J , which we saw that equals I .This concludes the proof since we showed I ⊆ h f i ∩ K ∩ h x, y, z i a ⊆ h f i ∩ L ∩ h x, y, z i a ⊆ I. (cid:3) 3. R EES ALGEBRAS OF LINE ARRANGEMENTS In [13, Section 3] there were conjectured some results concerning the generators of some eliminationalgebras, such as the special fiber, and now we can shed more light when the ambient ring is the polynomialring in three variables. ERO-DIMENSIONAL PROJECTIVE SCHEMES DEFINED BY a -FOLD PRODUCTS OF LINEAR FORMS 9 Basic definitions and concepts. The Rees algebra of an ideal I in a ring R is the homogeneous R -subalgebra of the standard graded polynomial R [ t ] in one variable over R , generated by the elements at, a ∈ I , of degree . Fixing a set of generators of I determines a surjective homomorphism of R -algebrasfrom a polynomial ring over R to R [ It ] . The kernel of such a map is called a presentation ideal of R [ It ] .If R = K [ x , . . . , x k ] is a standard graded polynomial ring over a field K and I = h g , . . . , g n i is an idealgenerated by forms g , . . . , g n of the same degree, consider T = R [ y , . . . , y n ] = K [ x , . . . , x k ; y , . . . , y n ] ,a standard bigraded K -algebra with deg x i = (1 , and deg y j = (0 , . Using the given generators toobtain an R -algebra homomorphism ϕ : T = R [ y , . . . , y n ] −→ R [ It ] , y i g i t, yields a presentation ideal I which is bihomogeneous in the bigrading of T . In this polynomial setup, one defines the special fiber F ( I ) := R [ It ] ⊗ R R/ m ≃ ⊕ s ≥ I s / m I s , where m = h x , . . . , x k i ⊂ R . The Krull dimension of the special fiber ℓ ( I ) := dim F ( I ) is called the analyticspread of I .As noted before, the presentation ideal of R [ It ] I = M ( u,v ) ∈ N × N I ( u,v ) , is a bihomogeneous ideal in the standard bigrading of T . Two basic subideals of I are hI (0 , − ) i and hI ( − , i ,and they have significant importance in the theory: • hI (0 , − ) i is the homogeneous defining ideal of the special fiber. • hI ( − , i coincides with the ideal of T generated by the biforms s y + · · · + s n y n ∈ T , whenever ( s , . . . , s n ) is a syzygy of g , . . . , g n of certain degree in R . Therefore, T / hI ( − , i is a presentationof the symmetric algebra S ( I ) of I .The ideal I is said to be of linear type provided I = hI ( − , i , and it is said to be of fiber type if I = hI ( − , i + hI (0 , − ) i .3.2. The case of line arrangements in P . Let A ⊂ P be a rank 3 arrangement of s lines, defined bythe linear forms ℓ , . . . , ℓ s ∈ R := K [ x, y, z ] . Let C be the linear code dual to the linear forms defining A . If m denotes the maximum number of concurrent lines in A , then d ( C ) = s − m . Also, because m = · · · = m s = 1 , we have d ( C ) = s − . Let a ∈ { s − m + 1 , . . . , s − } , and consider I := I a ( A ) .For this range of a , we have ht( I ) = 2 , and the goal is to analyze R [ It ] . When a is outside of this range,things are more or less clear: when a = s , I is a principal ideal, and when ≤ a ≤ s − m , then I = h x, y, z i a (and the Rees algebra is well understood in this case; see for example, the discussions in [13, Section 3.3]).The case when a = s − has been treated extensively in [7], so we will assume a ≤ s − .In terms of generators for I , we pick the standard generators ℓ i · · · ℓ i a , for all ≤ i < · · · < i a ≤ s . Ofcourse, this set of generators is not minimal. In fact, [1, Proposition 2.10] gives the formula µ ( I ) = min { ,s − a } X u =0 c − u,s − a − u , where c i,j are coefficients occurring in the Tutte polynomial of the matroid of A .Denote I ( A , a ) the presentation ideal of R [ It ] for the above chosen set of generators, where a is in therange s − m + 1 ≤ a ≤ s − . Remark 3.1. For any integer e ≥ , we have I e = I ea ( A ( e )) , where A ( e ) := ( ℓ , . . . , ℓ | {z } e , . . . , ℓ s , . . . , ℓ s | {z } e ) .So, from Theorem 2.3, we have that I e has linear graded free resolution. But this translates into I having Here we’ll be talking about the presentation ideal of R [ It ] by fixing a particular set of homogeneous generators of I of thesame degree. linear powers , which, by [3, Theorem 2.5], is equivalent to reg (1 , ( R [ It ]) = 0 . In particular, what thismeans is that I ( A , a ) doesn’t have any minimal generators in bidegree ( u, v ) , with u ≥ .[3, Example 2.6] gives an example of an ideal with linear powers that is not of fiber type. We areconjecturing that the ideals generated by a -fold products of linear forms are of fiber type.3.3. The case a = s − . In this situation [13, Proposition 3.6] presents the generators for the symmetricideal of I := I s − ( A ) , i.e., sym( I ) := hI ( − , i . Also, in the same paper we started presenting the generatorsof the presentation ideal of the special fiber, i.e., ∂ ( I ) := hI (0 , − ) i . Next we review these results, and thenotations.Let f i,j := ℓ · · · ℓ s ℓ i ℓ j , ≤ i < j ≤ s be the (standard) generators of I . By [13, Theorem 2.4] we have µ ( I ) = (cid:18) s (cid:19) − t X j =1 (cid:18) n j − (cid:19) , where Sing ( A ) := { P , . . . , P t } is the set of all intersection points of thelines of A , and n j is the number of lines of A intersecting at the point P j (indeed, n j = ν A ( I ( P j )) ). TheRees ideal, denoted here simply by I , corresponding to this set of generators is the kernel of the map: T := R [ . . . , t i,j , . . . ] −→ R [ It ] , t i,j f i,j t. Recall [13, Lemma 3.2 and Proposition 3.6] give the following information about important elements in I . (I) If { i , i , i } , ≤ i < i < i ≤ s , is a circuit in the matroid of A , then it gives a dependency c i ℓ i + c i ℓ i + c i ℓ i = 0 . This in turn gives the following element of I : L i ,i ,i := c i t i ,i + c i t i ,i + c i t i ,i . (II) For any ≤ a < b < c ≤ s the followings are elements of I : A a,b,c := ℓ a t a,b − ℓ c t b,c ,B a,b,c := ℓ a t a,c − ℓ b t b,c ,C a,b,c := ℓ b t a,b − ℓ c t a,c . (III) If s ≥ , for any ≤ a < b < c < d ≤ s , the followings are elements of I : Q a,b,c,d := t a,b t c,d − t a,c t b,d ,Q a,b,c,d := t a,b t c,d − t a,d t b,c . We have that sym( I ) is generated by the sets (I) and (II). Also the sets (I) and (III) belong to ∂ ( I ) .There is another set of elements that belong to ∂ ( I ) . Since the rank of A is 3, then any four of thedefining linear forms are linearly dependent. By (I), we are left to analyze the circuits of size four; forexample { , , , } , where any subset of three elements of this circuit is independent. This circuit comeswith the linear dependency d ℓ + d ℓ + d ℓ + d ℓ = 0 , where all the coefficients are not zero. Now wefollow the ideas in [13], on how to obtain elements of ∂ ( I ) from the elements of the Orlik-Terao ideal. Ourcircuit of size four leads to the following element of the Orlik-Terao ideal G := d y y y + d y y y + d y y y + d y y y ∈ S := K [ y , . . . , y s ] . Multiplying this by any y k , after pairing two y ’s with different indices, via the preimage of the map in [13,Proposition 3.3], we obtain, modulo elements of type (III), the following elements of ∂ ( I ) : P , , , := d t , t , + d t , t , + d t , t , + d t , t , ,P , , , := d t , t , + d t , t , + d t , t , + d t , t , ,P , , , := d t , t , + d t , t , + d t , t , + d t , t , ,P , , , := d t , t , + d t , t , + d t , t , + d t , t , ,R k , , , := d t , t ,k + d t , t ,k + d t , t ,k + d t , t ,k , ≤ k ≤ s. ERO-DIMENSIONAL PROJECTIVE SCHEMES DEFINED BY a -FOLD PRODUCTS OF LINEAR FORMS 11 Denote the set of all these elements with (IV), for all circuits { j , j , j , j } , ≤ j < j < j < j ≤ s . Remark 3.2. In [13, Subsection 3.2.1] it is shown how one can obtain canonically (i.e., via Sylvesterforms) the elements of types (I) and (III), and some elements of type (IV), from elements of type (II). Butafter modulo elements of type (III), we also have t , P , , , = t , P , , , and t ,k P , , , = t , R k , , , .So via Sylvester forms we can obtain all elements of (IV) from elements of type (II).In [13, Example 3.5] it is obtained a minimal generator of ∂ ( I ) that is not of any of the types (I), (III),nor (IV): F := t , t , + t , t , − t , t , . That generator was obtained from the dependency · ℓ + 1 · ℓ + ( − · ℓ + 0 · ℓ = 0 . This gave the element of the Orlik-Terao ideal G := y y y + y y y − y y y (observe that we will notsimplify by y ; we could simplify if we want because ∂ ( I ) is a prime ideal not containing the variables, andwe would get the standard minimal generator of the Orlik-Terao ideal corresponding to the circuit { , , } ).Multiplying G in order by the variables y , y , y , y , modulo (III) we obtain: P , , , = t , L , , ,P , , , = t , L , , ,P , , , = t , L , , ,P , , , = F. From now on we can include the elements similar to F into the type (IV) ones, by allowing the set { , , , } to be dependent (not necessarily minimally dependent, i.e., a circuit). To sum up, below weshow how via Sylvester forms we can obtain all the elements of types (I), (III), and (IV), from elements oftype (II).Suppose we have the dependency a ℓ + a ℓ + a ℓ + a ℓ = 0 , where a , a , a = 0 , but a may equalto zero, with ℓ , ℓ , ℓ linearly independent. Suppose also that a = − . Consider the following elementsof type (II): A , , = ℓ t , − ℓ t , , B , , = ℓ t , − ℓ t , , A , , = ℓ t , − ( a ℓ + a ℓ + a ℓ ) t , . We have A , , B , , A , , = t , − t , t , − t , t , − a t , − a t , − a t , · ℓ ℓ ℓ . Taking the determinant of the × content matrix we obtain t , [ a t , t , + a t , t , − t , t , + a t , t , ] = t , [ P , , , − a Q , , , ] ∈ I . By primality, P , , , ∈ ∂ ( I ) . Furthermore, if a = 0 , then P , , , = t , L , , , and so L , , ∈ ∂ ( I ) . Sinceany other element of type (IV) corresponding to the dependent set { , , , } can be obtained from P , , , ,the conclusion follows.As we observed above, for any ≤ j < j < j < j ≤ s , the (not necessarily minimal) dependent set { j , j , j , j } leads to the canonical construction of elements of type (I), (III), (IV) of ∂ ( I ) . The questionis if there are any other elements of ∂ ( I ) that cannot be generated by elements of type (I), (III), and (IV).We claim that there aren’t any. Remark 3.3. As it is explained in [13], any generator of ∂ ( I ) is obtained by pairing variables y ’s withdifferent indices in M · ( d j y j y j y j + d j y j y j y j + d j y j y j y j + d j y j y j y j ) =: M G j ,...,j , where d j ℓ j + d j ℓ j + d j ℓ j + d j ℓ j = 0 , and M is a monomial in S := K [ y , . . . , y s ] of odd degree. We can suppose j i = i for i = 1 , . . . , . Let M = y m · · · y m n n , where m + · · · + m n = m . Sincewe need to pair y i ’s in M G ,..., , we have a natural restriction about the degrees: m a ≤ than the sum ofthe exponents of all the other variables in that term if the variable y a shows up in a term of M G ,..., ,consequently we have m l ≤ m + 1 , for l = 1 , . . . , m l ≤ m + 3 , for l = 5 , . . . , n. A monomial M satisfying these inequalities will be said to satisfy the exponents restrictions ; and the pairingsof the variables y i ’s that pull back to an element of ∂ ( I ) will be called valid pairings . Theorem 3.4. Using the above notations, ∂ ( I ) is generated by elements of types (I), (III), and (IV).Proof. As discussed in Remark 3.3, we can suppose that any generator of ∂ ( I ) is obtained by pairing vari-ables y ’s with different indices in M · ( d y y y + d y y y + d y y y + d y y y ) =: M G , , , , where d ℓ + d ℓ + d ℓ + d ℓ = 0 , and M is a monomial in K [ y , . . . , y s ] of odd degree satisfying theexponent restrictions.The first idea is to reduce the worked case to deg( M ) ≤ .So suppose deg( M ) = m ≥ . The goal is to show that for any valid pairings of M G ,..., , there exist i = j such that M = y i y j M ′ , where M ′ = y m ′ · · · y m ′ n n and m ′ = deg( M ′ ) = m − , and M ′ satisfies theexponent restrictions: m ′ l ≤ m ′ + 1 , for l = 1 , . . . , m ′ l ≤ m ′ + 3 , for l = 5 , . . . , n. With this at hand, modulo elements of type (III), the valid parings of M G ,..., will “transfer” to some validpairings of M ′ G ,..., . But by induction, this can be generated by elements of types (I), (III), and (IV), andthe pullback of the pairing y i y j will be just the variable t i,j . Case . Assume m l < m + 1 , for all l = 1 , . . . , m l < m + 3 , for all l = 5 , . . . , s . Since m is odd, then for any choice of i and j , i = j , we have m ′ i = m i − and m ′ i ≤ m ′ + 1 and 2 m ′ i ≤ m ′ + 3 ,depending if i ∈ { , . . . , } or i ∈ { , . . . , n } (the same for m ′ j = m j − ). So M ′ satisfies the exponentrestrictions: m ′ l ≤ m ′ + 1 , for l = 1 , . . . , m ′ l ≤ m ′ + 3 , for l = 5 , . . . , n. This is saying that no matter what the valid pairings we chose for M G ,..., , we can write M = y i y j M ′ forsome i = j , with M ′ satisfying the exponents restrictions , and Case . Suppose there are k, l ∈ { , . . . , } such that m k = m + 1 = 2 m l . Then m k + 2 m l =2( m + 1) . Impossible. Case . Suppose there are k, l ∈ { , . . . , n } such that m k = m + 3 = 2 m l . Then m k + 2 m l =2( m + 3) . Impossible. Case . Suppose there is k ∈ { , . . . , } such that m k = m + 1 , and there is l ∈ { , . . . , n } such that m l = m + 3 . Then m l + 2 m k = 2( m + 2) . Impossible. Case . Suppose m k = m + 1 ( k ∈ { , . . . , } ) and m r < m + 1 , for all r ∈ { , . . . , } \ { k } , and m l < m + 3 for all l ∈ { , . . . , n } . Then we can choose i = k and j any, and M ′ satisfies the exponentsrestrictions. Since m ≥ , then m k ≥ . So in any valid pairings of M G , , , , there will be at least a y k not paired with any of the y ’s showing in the expansion of G , , , . But this y i = y k must pair with another y j from M . Case . Suppose m i < m + 1 for all i ∈ { , . . . , } , m l = m + 3 for some l ∈ { , . . . , n } , and m r < m + 3 , for all r ∈ { , . . . , n } \ { l } . Then we can choose i any, and j = l , and therefore M ′ satisfiesthe exponents restrictions. Same as in the previous case, since m ≥ , then m l ≥ . So in any valid pairings If M = y mk , since m ≥ and deg( G , , , ) = 3 , any valid pairing of MG , , , must lead to a valid pairing in y m − k , m − ≥ , which is impossible. So there are i = j with m i ≥ and m j ≥ . ERO-DIMENSIONAL PROJECTIVE SCHEMES DEFINED BY a -FOLD PRODUCTS OF LINEAR FORMS 13 of M G ,..., , there will be at least a y l not paired with any of the y ’s showing in the expansion of G ,..., .But this y j = y l must pair with another y i from M .Above we showed how recursively we can drop the degree of the monomial M by 2, if m ≥ . If m = 1 ,then we recover elements of type (IV). So we need to focus on the case m = 3 . We divide this case insubcases:(1) y i y j , i = j .(a) i ∈ { , . . . , } ; suppose i = 1 . Then, by looking at the last three terms of y y j ( d y y y + d y y y + d y y y + d y y y ) , we have only one valid pairings possible in those terms. The pullback looks d A + d t , t , t ,j + d t , t , t ,j + d t , t , t ,j , where we have various options for the pairings that give A . But modulo elements of type (III) ,we can write the above as t ,j ( d t , t , + d t , t , + d t , t , + d t , t , ) = t ,j P , , , . (b) i ∈ { , . . . , n } , and j ∈ { , . . . , } ; suppose j = 1 . Then, by looking at the last three terms of y i y ( d y y y + d y y y + d y y y + d y y y ) , we have only one valid pairings possible in those terms. The pullback looks d A + d t , t ,i t ,i + d t , t ,i t ,i + d t , t ,i t ,i , where we have various options for the pairings that give A . But modulo elements of type (III) ,we can write the above as t ,i ( d t , t ,i + d t , t ,i + d t , t ,i + d t , t ,i ) = t ,i R i , , , . Now, suppose j ∈ { , . . . , n } and i = j . So, we have y i y j ( d y y y + d y y y + d y y y + d y y y ) . Suppose that the pullback of some valid pairing is d t ,i t ,i t ,j + d t ,i t ,i t ,j + d t ,i t ,i t ,j + d t ,i t ,i t ,j . But t ,i t ,j = t , t i,j , t ,i t ,j = t , t i,j and t ,i t ,j = t , t i,j modulo elements of type (III),then we can rewrite t i,j ( d t ,i t , + d t ,i t , + d t ,i t , + d t ,i t , ) . Again t ,i t , = t , t ,i , t ,i t , = t , t ,i , t ,i t , = t , t ,i and t ,i t , = t , t ,i moduloelements of type (III), so we can rewrite t i,j ( d t , t ,i + d t , t ,i + d t , t ,i + d t , t ,i ) = t i,j R i , , , . (2) y i y j y k . A similar computations modulo elements of type (III) leads to the pullback of any validpairings of y i y j y k G , , , rewritten as t u,v F with F of type (IV). As an example, suppose i =1 , j = 2 , k = 3 . We must do valid pairings in y y y ( d y y y + d y y y + d y y y + d y y y ) . The pullback of the last term can only be d t , t , t , . For each of the other three terms, thereare three possible valid pairings; for example the first term can be d t , t , , d t , t , t , , or d t , t , t , . Suppose we have d t , t , + d t , t , + d t , t , t , + d t , t , t , . For example A = t , t , t ,j , we use the fact that t , t ,j ≡ t , t ,j . For example A = t , t ,i t ,i , we use the facts that t , t ,i ≡ t ,i t , , and t , t ,i ≡ t , t ,i . As t , t , ≡ t , t , , modulo elements of type (III), we can rewrite t , ( d t , t , + d t , t , + d t , t , + d t , t , ) . Since t , t , ≡ t , t , , modulo elements of type (III), we obtain t , P , , , . (3) y i , i ∈ { , . . . , n } ; suppose i = 5 . From the relation d ℓ + d ℓ + d ℓ + d ℓ = 0 , we have y ( d y y y + d y y y + d y y y + d y y y ) . Then using the remark, we have only one option: P := d t , t , t , + d t , t , t , + d t , t , t , + d t , t , t , . From the relation f ℓ + f ℓ + f ℓ + f ℓ = 0 (we can assume f = d ), we can associate theelement in Orlik-Terao ideal d y y y + f y y y + f y y y + f y y y . Multiplying by y y , the pullback of a valid pairings is: P := d t , t , t , + f t , t , t , + f t , t , t , + f t , t , t , . Note that P is already treated in the case (1) (a). From the two relations above we have the thirdrelation: e ℓ + e ℓ + e ℓ + e ℓ = 0 with e = d − f , e = d − f , e = d and e = − f .This relation gives us another element in Orlik-Terao ideal e y y y + e y y y + e y y y + e y y y . Multiplying by y y , the pullback of a valid pairing is: P := e t , t , t , + e t , t , t , + e t , t , t , + e t , t , t , . Note that P is already treated in the case (1) (a). But we obviously have P = P + P . So the case m = 3 is also completely analysed. (cid:3) Theorem 3.5. Using the above notations, the ideal I := I s − ( A ) is of fiber type.Proof. Let F ( . . . , t i,j , . . . ) ∈ I a generator of degree d + 1 of the Rees ideal of I . If d = 0 , then bydefinition, F is a linear form in t i,j , with constant coefficients, so it is and element of ∂ ( I ) (i.e., a linearcombination of elements of type (I)).Suppose d ≥ . By the Remark 3.1 we can suppose F ( . . . , t i,j , . . . ) = X n i ,i + ··· + n iu,iu +1 = d L i ,i ,...,i u ,i u +1 t n i ,i i ,i · · · t n iu,iu +1 i u ,i u +1 , with L i ,i ,...,i u ,i u +1 ∈ K [ x, y, z ] a linear form.Since F ( . . . , t i,j , . . . ) ∈ I , we have F ( . . . , f i,j , . . . ) = 0 , that is, F ( . . . , fℓ i ℓ j , . . . ) = 0 , with f = ℓ · · · ℓ s . Multiplying by f d , we have f d F ( . . . , fℓ i ℓ j , . . . ) = F ( . . . , fℓ i fℓ j , . . . ) = 0 . Then we can consider F ( . . . , y i y j , . . . ) ∈ I ( A , s − ⊂ K [ x, y, z, y , . . . , y s ] the presentation ideal of the Rees algebra of theideal I s − ( A ) = h ℓ · · · ℓ s , . . . , ℓ · · · ℓ s − i . By [7, Theorem 4.2], we know that I s − ( A ) is of fiber type, that is, I ( A , s − 1) = h sym( I s − ( A )) , ∂ ( I s − ( A )) i , where ∂ ( I s − ( A )) = hI ( A , s − (0 , − ) i the Orlik-Terao ideal of A . By [7, Lemma 3.1(b)] and [10], sets of generators of these ideals are: sym( I s − ( A )) = h ℓ i y i − ℓ i +1 y i +1 | ≤ i ≤ s − i , and ERO-DIMENSIONAL PROJECTIVE SCHEMES DEFINED BY a -FOLD PRODUCTS OF LINEAR FORMS 15 ∂ ( I s − ( A )) = h G j ,...,j | d j ℓ j + d j ℓ j + d j ℓ j + d j ℓ j = 0 i , where G j ,...,j is described in the Remark 3.3. So, we can write F ( . . . , y i y j , . . . ) = X n i ,i + ··· + n iu,iu +1 = d L i ,i ,...,i u ,i u +1 ( y i y i ) n i ,i · · · ( y i u y i u +1 ) n iu,iu +1 = s − X i =1 B i ( ℓ i y i − ℓ i +1 y i +1 ) + X C j ,...,j G j ,...,j (1)with B i , C j ,...,j ∈ K [ x, y, z, y , . . . , y s ] . About these polynomials we have the following properties:(*) All the monomials showing up in expression (1) are constructed from pairings y i y j ’s, and thesepairings must be valid pairings (as we say in Remark 3.3) in order to pull back to variables t i,j inthe expression of F . Below, we’ll use “ b ” to denote this pull back.(**) As F ( . . . , y i y j , . . . ) has degree d in variables y i ’s and degree in variables x, y, z , then B i musthave degree d − in variables y i ’s and degree in variables x, y, z . About C j ,...,j , it must havedegree in variables x, y, z and have either degree d − , or d − , in variables y i ’s.Here, and below, by “monomial” we will understand a monomial in variables y i ’s.So, using the statements above and the fact that d ≥ , we can suppose that in each monomial in B i and C j ,...,j there is a variable y r for some r ∈ { , . . . , s } . If d = 1 , then the degree of C j ,...,j in variables y i ’scan only be 0, so the pull back (i.e., pairing of y i y j ↔ t i,j ) of this term will give a combination of generatorsof type (I) with coefficients polynomials in variables x, y, z , so an element of ∂ ( I ) . Claim: In regard to B i , for each i ∈ { , . . . , s } , we can suppose r = i, i + 1 . Proof of Claim. Suppose that we have a monomial in B i of the form L y m i i y m i +1 i +1 with m i + m i +1 = 2 d − and L ∈ K [ x, y, z ] . If m i +1 = 0 (or m i = 0 ), then we will have L y d − i ( ℓ i y i − ℓ i +1 y i +1 ) showing up inexpression (1), but this contradicts (*). If m i , m i +1 > then we have L y m i i y m i +1 i +1 ( ℓ i y i − ℓ i +1 y i +1 ) . For (*) to happen (i.e., valid pairings), in the first monomial above we need to have m i = d − and m i +1 = d , and in the second monomial we need to have m i = d and m i +1 = d − ; an obvious contradiction.And Claim is shown.So, from the Claim above, for each monomial M of each B i there is Q ∈ K [ x, y, z, y , . . . , y s ] d − suchthat M = Qy r ( ℓ i y i − ℓ i +1 y i +1 ) = Q ( ℓ i y i y r − ℓ i +1 y i +1 y r ) . If r > i + 1 , then the pull back looks c M = b QB i,i +1 ,r and if r < i , then the pullback looks c M = b QC r,i,i +1 ,where B i,i +1 ,r , C r,i,i +1 are elements of the type (II).Now we analyse each monomial of C j ,...,j only in the variables y ′ i s (we disregard the “coefficients”which are linear forms in variables x, y, z since they do not affect the pull back to variables t i,j ).Suppose ( j , . . . , j ) = (1 , . . . , . Let N = y n y n y n y n y n . . . y n s s be such a monomial. In this casewe have only to discuss the case n + · · · + n s = 2 d − . The condition (*) (or the exponent restrictionsin Remark 3.3) applied to N G ,..., leads to n i ≤ d − , that is, n i ≤ d − for each i ∈ { , . . . , s } . By symmetry, we only need to analyse two cases: if n ≤ n ≤ · · · ≤ n s or n s ≤ n s − ≤ · · · ≤ n .Suppose we are in the first case. Then we can organize the term N G ,..., in the following way: N G ,..., = ( y n y n y n y n y n · · · y n s − s ) y s ( d y y y + d y y y + d y y y + d y y y ) . (2)Note that we can pair all the variables in the element y n y n y n y n y n . . . y n s − s . For that we need to havethe exponent restrictions n i ≤ n + · · · b n i + · · · + n s − for each i ∈ { , . . . , s − } and n s − ≤ n · · · + n s − .But by the hypothesis we already have it for i ≤ s − . If we suppose n s − > n + · · · + n s − , then we If deg( G ,..., ) = 2 (i.e., one of d , d , d , d is 0), then G ,..., pulls back to an element of type (I), and also, it is not difficultto see that we must have valid pairings of the y i ’s in N . have n s − > d − − n s , that is, n s > d − , equivalently n s > d − , which it is a contradiction. So,the pull back of the expression (2) looks b N R s , , , , with R s , , , element of the type (IV).If we suppose n s ≤ n s − ≤ · · · ≤ n we can organize in the same way and the pull back of the expression(2) looks b N P , , , , with P , , , element of the type (IV).So, pulling back F ( . . . , y i y j , . . . ) to F ( . . . , t i,j , . . . ) , we can see that it belongs to sym( I ) + ∂ ( I ) , hence I is of fiber type. (cid:3) Acknowledgment. Ricardo Burity is grateful to CAPES for funding his one year stay at University of Idaho.R EFERENCES [1] B. Anzis, M. Garrousian and S¸. 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