Hilbert series of generic ideals in products of projective spaces
aa r X i v : . [ m a t h . A C ] F e b Hilbert series of generic ideals in products of projective spaces
Ralf Fr¨oberg
Abstract If k [ x , . . . , x n ] /I = R = P i ≥ R i , k a field, is a standard graded algebra, theHilbert series of R is the formal power series P i ≥ dim k R i t i . It is known already sinceMacaulay which power series are Hilbert series of graded algebras [12]. A much harderquestion is which series are Hilbert series if we fix the number of generators of I andtheir degrees, say for ideals I = ( f , . . . , f r ), deg f i = d i , i = 1 , . . . , r . In some sense”most” ideals with fixed degrees of their generators have the same Hilbert series. Thereis a conjecture for the Hilbert series of those ”generic” ideals, see below. In this paperwe make a conjecture, and prove it in some cases, in the case of generic ideals of fixeddegrees in the coordinate ring of P × P , which might be easier to prove.KEYWORDS: P × P , Hilbert series, generic idealsAMS subject classification: 14N10, 13D40 The conjecture and the results in section 2 below are inspired by the corresponding con-jecture and results in the singly graded case.
Conjecture 1 ([5])
Let I = ( f , . . . , f r ) ⊂ k [ x , . . . , x n ] , k an infinite field, be an idealgenerated by generic forms f i with deg( f i ) = d i , and let R = k [ x , . . . , x n ] /I . Then R ( u ) = [ r Y i =1 (1 − u d i ) / (1 − u ) n ] + here [ P a i u i ] + = P b i u i , where b i = a i if for all j ≤ i we have a j > , and b i = 0 otherwise. We first comment on the use of the word ”generic”. A polynomial of degree d in k [ x , . . . , x n ]is a linear combination of (cid:0) n + d − d (cid:1) monomials. Thus an ideal ( f , . . . , f r ), deg( f i ) = d i , canbe considered as a point in A = A N , N = P ri =1 (cid:0) n + d i − d i (cid:1) . There is a Zariski-open sub-set of A , for which the Hilbert series is constant. Ideals corresponding to points in thatZariski-open set are what we call generic, see [8].The conjecture is proved for r ≤ n (trivial), for n ≤ n = 3 [1], for r = n + 1[16]. There are partial results in [10], [7], [2], [13], [3], [15], [14].1f J = ( l d , . . . , l d r r ), where l i are generic linear forms. Sometimes, but not always, theHilbert series of k [ x , . . . , x n ] /J equals the one in the conjecture. There is a conjecture onwhen it does, [11, 4]. P × P We are considering homogeneous ideals in the coordinate ring of P × P . Thus, let k be aninfinite field, S = k [ x , , x , y , y ] be bigraded, deg( x i ) = (1 , y j ) = (0 , I be a bihomogeneous ideal, so generated by bihomogeneous elements. Hence R = S/I isbigraded, R = ⊕ i,j ≥ R i,j . The Hilbert series of R is defined as R ( u, v ) = P dim k R i,j u i v j .We are interested in the case when the ideal is generated by ”generic” elements. Given asequence of degrees d , . . . , d r , we denote the space of ideals I = ( f , . . . , f r ) where deg f i = d i by I d ,..., d r . An element of degree ( d, e ) is a linear combination of ( d +1)( e +1) monomials.Thus an ideal in I d ,..., d r can be considered as a point in A Nk , N = P ri =1 ( d i +1)( e i +1) where d i = ( d i , e i ). We partially order Hilbert series termwise, so that P a ij u i v j ≥ P b ij u i v j if a ij ≥ b ij for all i, j . Theorem 2
There are only a finite number of possibilities for Hilbert series of ideals in I d ,..., d r . There is a nonempty Zariski open part of of I d ,..., d r where the Hilbert series isconstant. This constant Hilbert series is the smallest possible for ideals in I d ,..., d r . Proof
The corresponding theorems in the singly graded case, [8, Theorem 1] , [5, Theoremp.120], and [6, Proposition 1] are easily adapted. We call points in this nonempty Zariskiopen set generic.We define ( d, e ) ≤ ( f, g ) if d ≤ f and e ≤ g , and ( d, e ) ≥ ( f, g ) if d ≥ f and e ≥ g .Furthermore [ P i,j a ij u i v j ] + = P i,j b ij u i v j , where b ij = a ij if a kl > k, l ) ≤ ( i, j )and b i,j = 0 otherwise. Lemma 3
Let R = S/I be bigraded and f ∈ R i,j . Then ( R/f )( u, v ) ≥ [(1 − t i u j )( R ( u, v )] + , Proof
Consider the map f · : R d − i,e − j → R d,e . The image is largest if the map is of maximalrank, i.e., either injective or surjective, so dim k ( R/f ) d,e ≥ max { , dim R d,e − dim R i,j } . Ifdim( R/f ) d,e = 0, then dim( R/f ) d + f,e + g = 0 for all ( f, g ) ≥ Lemma 4 [(1 − u i v j )[ P i,j a ij u i v j ] + ] + = [(1 − u i v j ) P i,j a ij u i v j ] + . Proof
Easy calculation.These two lemmas give the following.
Theorem 5
Let I = ( f , . . . , f r ) ⊂ k [ x , x , y , y ] = S , deg f i = ( d i , e i ) . Then S/I ( u, v ) ≥ [ Q ri =1 (1 − u d i v e i ) / ((1 − u ) (1 − v ) )] + .
2e now give a conjecture in the case when the f i ’s are generic, c.f. [9]. To prove theconjecture for some fixed ( d , . . . , d r ) it suffices to give one example with the conjecturedseries. If the conjecture is true for these parameters, then almost all ideals have theconjectured series, so we must be very unlucky if we miss the series with a random choiceof coefficients. Conjecture 6
Let I = ( f , . . . , f r ) ⊂ k [ x , x , y , y ] = S , deg f i = ( d i , e i ) generic. Then ( S/I )( u, v ) = [ Q ri =1 (1 − u d i v e i ) / ((1 − u ) (1 − v ) )] + . We have checked that the conjecture is true in the following cases. Some of thesewere checked by Alessandro Oneto. Except for the first class, we have used computercalculations.1. For small r the concepts of ideal generated by generic forms and complete intersectionagrees. It is well known that the conjecture is true for complete intersections.2. deg f i = (1 ,
1) for all i , any r .3. Some f i of degree (1,1), some of degree (1,2), any r .4. deg f i = (1 ,
2) for all i , any r .5. deg f i = (2 ,
2) for all i , any r .We also checked that the corresponing conjecture is true for deg f i = (1 , ,
1) for all i ,any r , in P × P × P .On the other hand, the corresponding conjecture in P × P cannot be true. For fourgeneric forms of degree (1,1) the conjecture would give that R d,d = 0 if d >>
0. Thecorrect statement is that dim k R d,d = 6 if d >> Question
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