On the Existence of Semi-Regular Sequences
aa r X i v : . [ m a t h . A C ] D ec ON THE EXISTENCE OF SEMI-REGULAR SEQUENCES
TIMOTHY J. HODGES, SERGIO D. MOLINA, AND JACOB SCHLATHER
Abstract.
Semi-regular sequences over F are sequences of homogeneous el-ements of the algebra B ( n ) = F [ X , ..., X n ] / ( X , ..., X n ), which have as fewrelations between them as possible. They were introduced in order to assessthe complexity of Gr¨obner basis algorithms such as F , F for the solutionof polynomial equations. Despite the experimental evidence that semi-regularsequences are common, it was unknown whether there existed semi-regular se-quences for all n , except in extremely trivial situations. We prove some resultson the existence and non-existence of semi-regular sequences. In particular, weshow that if an element of degree d in B ( n ) is semi-regular, then we must have n ≤ d . Also, we show that if d = 2 t and n = 3 d there exits a semi-regularelement of degree d establishing that the bound is sharp for infinitely many n .Finally, we generalize the result of non-existence of semi-regular elements tothe case of sequences of a fixed length m . Introduction
Semi-regular sequences over F are sequences of homogeneous elements of thealgebra B ( n ) = F [ X , ..., X n ] / ( X , ..., X n )which have as few relations between them as possible. They were introduced in[1, 2, 3, 4] in order to assess the complexity of Gr¨obner basis algorithms such as F , F for the solution of polynomial equations.Experimental evidence has shown that randomly generated sequences tend tobe semi-regular [3, Section 3]. On the other hand it has been observed than manysequences that arise in cryptography, such as those arising from the Hidden FieldEquation cryptosystems, are not semi-regular. Despite the experimental evidencethat semi-regular sequences are common, it was unknown whether there existedsemi-regular sequences for all n , except in extremely trivial situations.We prove here some results on the existence and non-existence of semi-regularsequences. We first look at the most elementary case, that of semi-regular elements(or sequences of length one). It was observed in [9, Lemma 3.12], that there are noquadratic semi-regular elements when n >
6. On the other hand it is trivial thatelements of degree n and n − n and d do there exist semi-regular elements of degree d in B ( n ) ? Mathematics Subject Classification.
Primary 11T55; Secondary 12E05, 12E20, 12Y05,13A02, 13D02, 13P15, 94A60.
Key words and phrases.
Semi-regularity, finite field.Corresponding author: Timothy Hodges, Department of Mathematical Sciences, University ofCincinnati, Cincinnati, OH 45221-0025, USA, email:[email protected] first author was partially supported by a grant from the Charles P. Taft Research Center.The second author was partially supported by a scholarship from COLFUTURO/Foundationfor the Future of Colombia and by the Maita Levine Fellowship, University of Cininnati, USA.
1N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 2
We prove here that if an element of degree d ≥ B ( n ) is semi-regular, then wemust have n ≤ d . We go somewhere towards understanding the sharpness of thisbound by determining when the symmetric polynomials σ d ( x , ..., x n ) = X ≤ j < ···
For a graded ring B = L Nk =0 B k , we define the index of B to be t if B t − = 0 and B k = 0 for all k ≥ t . We denote this number by Ind( B ). If λ , . . . , λ m is a set of homogeneous elements and I = ( λ , . . . , λ m ), then we defineInd( λ , . . . , λ m ) = Ind( B/I ). If B is strongly graded (that is, B i B j = B i + j for all i and j ), then Ind( B/I ) = min { d ≥ | I ∩ B d = B d } Definition 2.2.
Let λ , . . . , λ m be a sequence of homogeneous elements of B ( n ) ofpositive degree. The sequence λ , . . . , λ m is D -semi-regular if for all i = 1 , , . . . , m ,if µ is homogeneous and µλ i ∈ ( λ , . . . , λ i − ) and deg( µ ) + deg( λ i ) < D then µ ∈ ( λ , . . . , λ i ). A sequence of homogeneous polynomials λ , . . . , λ m is semi-regular if it is D -semi-regular for D = Ind( λ , . . . , λ m )Recall that the Hilbert function of a graded ring B is the function HF B ( k ) =dim B k and the Hilbert series is the series HS B ( z ) = P ∞ k =0 (dim B k ) z k . The Hilbertfunction and series for a graded ideal I of B are defined by HF I = HF B/I and HS I ( z ) = HS B/I ( z ) respectively. For any series a ( z ) = P i a i z i ∈ R [[ z ]] , we definethe index of a ( z ), Ind a ( z ), to be the first t for which a t ≤
0. If such a t doesnot exist define Ind a ( z ) = ∞ . For a series P i a i z i , we denote by (cid:2)P i a i z i (cid:3) t thetruncated series P t − i =0 a i z i and by (cid:2)P i a i z i (cid:3) the truncated series (cid:2)P i a i z i (cid:3) Ind( a ( z )) .It was asserted in [2] that a sequence λ , . . . , λ m is semi-regular if and only if HS ( λ ,...,λ m ) ( z ) = (cid:20) (1 + z ) n Q mi =1 (1 + z d i ) (cid:21) where d i = deg λ i . As noted in [5], the proofs in that article are incomplete. Webegin, therefore, by giving a complete proof. N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 3
For any d = ( d , . . . , d m ) ∈ N m , define T d ,n ( z ) = (1 + z ) n Q mi =1 (1 + z d i )and let t d ,n ( j ) be the coefficient of z j in T d ,n ( z ), so that T d ,n ( z ) = P ∞ i =0 t d ,n ( j ) z j .We begin with some observations on the way truncation behaves with respect tomultiplication. Lemma 2.3.
Let u ( z ) , v ( z ) , w ( z ) ∈ R [[ z ]] . Then(1) [ u ( z ) v ( z )] D = [[ u ( z )] D [ v ( z )] D ] D = [ u ( z ) [ v ( z )] D ] D (2) [ v ( z )] D = [ w ( z )] D = ⇒ [ u ( z ) v ( z )] D = [ u ( z ) w ( z )] D Proof. (1) First note that for any a ( z ) , c ( z ) ∈ R [[ z ]],[ a ( z ) + c ( z ) z D ] D = [ a ( z )] D Define u ′ ( z ) , v ′ ( z ) by u ( z ) = [ u ( z )] D + u ′ ( z ) z D and v ( z ) = [ v ( z )] D + v ′ ( z ) z D . Then[ u ( z ) v ( z )] D = (cid:2) ([ u ( z )] D + u ′ ( z ) z D )([ v ( z )] D + v ′ ( z ) z D ) (cid:3) D = (cid:2) [ u ( z )] D [ v ( z )] D + z D ( u ′ ( z )[ v ( z )] D + v ′ ( z )[ u ( z )] D + u ′ ( z ) v ′ ( z ) z D ) (cid:3) D = [[ u ( z )] D [ v ( z )] D ] D So [ u ( z ) [ v ( z )] D ] D = [[ u ( z )] D [ v ( z )] D ] D = [ u ( z ) v ( z )] D (2) If [ v ( z )] D = [ w ( z )] D , then[ u ( z ) v ( z )] D = [ u ( z )[ v ( z )] D ] D = [ u ( z )[ w ( z )] D ] D = [ u ( z ) w ( z )] D (cid:3) Since λ = 0 for any homogeneous element λ of positive degree, multiplicationby λ i is a well-defined map from B/ ( λ , . . . , λ i ) to B/ ( λ , . . . , λ i − ) whose image is( λ , . . . , λ i ) / ( λ , . . . , λ i − ). Let π i be the natural projection from B/ ( λ , . . . , λ i − )to B/ ( λ , . . . , λ i ). Thus we have an exact sequence, B/ ( λ , . . . , λ i ) λ i −→ B/ ( λ , . . . , λ i − ) π i −→ B/ ( λ , . . . , λ i ) → D -semi-regularity by saying thata sequence λ , . . . , λ m is D -semi-regular if and only if the sequence0 → ( B/ ( λ , . . . , λ i )) d − d i → ( B/ ( λ , . . . , λ i − )) d → ( B/ ( λ , . . . , λ i )) d → i = 1 , . . . , m and all d < D . Theorem 2.4.
Let B = B ( n ) and let λ , . . . , λ m be a sequence of homogeneouselements of B with λ i being of degree d i . Set I = ( λ , . . . , λ m ) and d = ( d , . . . , d m ) .(1) If the sequence λ , . . . , λ m is D -semi-regular then [ HS I ( z )] D = [ T d ,n ( z )] D and HF I ( D ) ≥ t d ,n ( D ) .(2) If the sequence λ , . . . , λ m is D -semi-regular but not ( D + 1) -semi-regular,then HF I ( D ) > t d ,n ( D ) .(3) The sequence λ , . . . , λ m is semi-regular if and only if the Hilbert series of I is given by HS I ( z ) = [ T d ,n ( z )] N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 4 (4) If the sequence λ , . . . , λ m is D -semi-regular, then so is λ σ (1) , . . . , λ σ ( m ) forany permutation σ .Proof. Set d i = ( d , . . . , d i ) and denote t d i ,n ( d ) by t i ( d ). Note that(1 + z d j ) ∞ X d =0 t j ( d ) z d = ∞ X d =0 t j − ( d ) z d so t j − ( d ) = t j ( d ) + t j ( d − d j ) for all j and d .Set s i ( d ) = HF ( λ ,...,λ i ) ( d ) = dim( B/ ( λ , . . . , λ i )) d Let K i = ker (cid:16) B/ ( λ , . . . , λ i ) λ i −→ B/ ( λ , . . . , λ i − ) (cid:17) , let K i,d denote the subspace of degree d elements of K i and let k i ( d ) = dim K i,d .Note that λ , . . . , λ m is D -semi-regular if and only if k i ( d − d i ) = 0 for all d < D and all i = 1 , . . . , m .We have an exact sequence0 → K i → B/ ( λ , . . . , λ i ) λ i −→ B/ ( λ , . . . , λ i − ) → B/ ( λ , . . . , λ i ) → d → K i,d − d i → B/ ( λ , . . . , λ i ) d − d i λ i −→ B/ ( λ , . . . , λ i − ) d → B/ ( λ , . . . , λ i ) d → k i ( d − d i ) − s i ( d − d i ) + s i − ( d ) − s i ( d ) = 0We now prove the assertions in part (1) by induction on m using the case m = 0(the ”empty sequence”) as the base case. In this situation the assertions followfrom the fact that HS B ( z ) = (1 + z ) n . Now let m >
0. The hypothesis of D -semi-regularity implies that s m − ( d ) = s m ( d ) + s m ( d − d m ) for d = 0 , . . . , D −
1. Theinduction hypothesis implies that s m − ( d ) = t m − ( d ) for d < D . So[ HS ( λ ,...,λ m − ) ( z )] D = D − X d =0 s m − ( d ) z d = D − X d =0 s m ( d ) z d + D − X d =0 s m ( d − d m ) z d = (1 + z d m ) D − X d =0 s m ( d ) z d = (1 + z d m ) (cid:2) HS ( λ ,...,λ m ) ( z ) (cid:3) D Using Lemma 2.3 and induction on m yields[ HS ( λ ,...,λ m ) ( z )] D = (cid:20) z d m ) [ HS ( λ ,...,λ m − ) ( z )] D (cid:21) D = " z d m ) " (1 + z ) n Q m − j =1 (1 + z d j ) D D = " (1 + z ) n Q mj =1 (1 + z d j ) D N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 5 which proves the first assertion. For the second part we assume, by induction,that s m − ( D ) ≥ t m − ( D ) and observe that by semi-regularity and the first part, s m ( D − d m ) = t m ( D − d m ). Hence s m ( D ) = s m − ( D ) − s m ( D − d m ) + k m ( D − d m ) ≥ t m − ( D ) − t m ( D − d m ) = t m ( D )(2) Suppose that λ , . . . , λ m is D -semi-regular but not ( D +1)-semi-regular. Let u be the smallest integer such that λ , . . . , λ u is not semi-regular. Then k u ( D − d u ) >
0, so s u ( D ) = s u − ( D ) − s u ( D − d u ) + k u ( D − d u ) > t u − ( D ) − t u ( D − d u ) = t u ( D )Now suppose that s j ( D ) > t j ( D ) for some u ≤ j < m . Then s j +1 ( D ) = s j ( D ) − s j +1 ( D − d j +1 ) + k j +1 ( D − d j +1 ) > t j ( D ) − t j +1 ( D − d j +1 ) = t j +1 ( D )So by induction, s m ( D ) > t m ( D ).(3) Suppose now that the sequence λ , . . . , λ m is semi-regular, and set D =Ind( I ). Then [ HS I ( z )] D = [ T d ,n ( z )] D by part (1) because the sequence is D -semi-regular. Because D = Ind( I ), HS I ( z ) = [ HS I ( z )] D . By (1), t m ( d ) = s m ( d ) > d < D and t m ( D ) ≤ s m ( D ) = 0, so Ind( T d ,n ( z )) = D . Thus[ T d ,n ( z )] = [ T d ,n ( z )] D = [ HS I ( z )] D = HS I ( z )as required.Conversely, suppose that HS I ( z ) = [ T d ,n ( z )] and let D = Ind( T d ,n ( z )). Then bydefinition, D is the degree of regularity of the sequence λ , . . . , λ m . If the sequence λ , . . . , λ m is not D -semi-regular, then there exists a k < D such that it is k -semi-regular and not ( k + 1)-semi-regular. By part (2) we would then have that s m ( k ) > t m ( k )That is, the k -th coefficient of HS I ( z ) is strictly greater than the k -th coefficientof T d ,n ( z ), contradicting the fact that HS I ( z ) = [ T d ,n ( z )]. Thus the sequence is D -semi-regular and hence semi-regular.(4) follows immediately from (3) because the Hilbert series of B/I is independentof the order of the λ i . (cid:3) It is natural to expect that information about the semi-regular sequences shouldgive us information about arbitrary sequences. Since semi-regular sequences have asfew relations as possible, we expect the ideal generated by a semi-regular sequence( ν , . . . , ν m ) to grow at least as quickly as the ideal generated by an arbitrarysequence ( λ , . . . , λ m ). That is (if we use the notation P a i z i ≤ P b i z i ⇔ a i ≤ b i for all i ), HS ( λ ,...,λ m ) ( z ) ≥ HS ( ν ,...,ν m ) ( z )Thus it is tempting to expect for any sequence λ , . . . , λ m that HS ( λ ,...,λ m ) ( z ) ≥ [ T d ,n ( z )] . The following example shows that this is not true.
N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 6
Example 1.
Consider the element λ = x x + x x + x x + x x + x x + x x in B (12) and let I = ( λ ). Then, using [6, Theorem 2.1] we can calculate that HS I ( z ) = 1 + 12 z + 65 z + 208 z + 430 z + 584 z + 494 z + 208 z + 65 z + 12 z + z while (cid:20) (1 + z ) z (cid:21) = 1+12 z +65 z +208 z +430 z +584 z +494 z +208 z + z +12 z +65 z Note also that in this case Ind(( λ )) = Ind( T (12) ,n ( z )) but λ is not semi-regular.Thus the condition Ind( I ) = Ind( T d ,n ( z )) is not equivalent to semi-regularity.It would be interesting to know whether Ind(( λ , . . . , λ m )) ≥ Ind( T d ,n ( z )) for anarbitrary sequence λ , . . . , λ m . All known evidence points to this result being true.However, the failure of the inequality HS I ( z ) ≥ [ T d ,n ( z )] rules out the obvious wayof proving it.3. Conjectures and Questions on Semi-Regularity
It has long been conjectured that semi-regular sequences are in some sense“generic”. However very little progress has been made towards proving this conjec-ture. In fact even the question of the existence of semi-regular quadratic sequencesof length n in n of variables remains open. Let us begin by reviewing some of theconjectures made by Bardet et al. Conjecture 1. [1, 4] The proportion of semi-regular sequences tends to one as thenumber of variables tends to infinity.We prove this conjecture in Section 6 in the following precise sense. Recallthat the semi-regularity of a sequence depends only on the set of elements, noton their order, so we actually consider the proportion of homogeneous subsets of B ( n ) that are semi-regular. Let h ( n ) be the number of subsets of B ( n ) consistingof homogeneous elements of degree greater than or equal to one. Let s ( n ) be thenumber of such subsets that are semi-regular. We show in Theorem 6.4 thatlim n →∞ s ( n ) h ( n ) = 1Unfortunately this result does not give us the kind of information that we areinterested in. As the size of the set increases, so does the likelihood of it beingsemi-regular for trivial reasons (for instance any basis of the set of quadratic poly-nomials is trivially semi-regular). We show that the proportion of sequences thatare trivially semi-regular tends to one.A different formulation of the conjecture that most sequences are semi-regularis given in [3, Conjecture 2] Conjecture 2. [3] For any ( n, m, d , . . . , d m ) the proportion π ( n, m, d , . . . , d m ) ofsemi-regular sequences over F in the set E ( n, m, d , . . . , d m ) of algebraic systemsof m equations of degrees d , . . . , d m in n variables tends to 1 as n tends to ∞ .We show that this conjecture is false. In fact, the opposite is true. For a fixedchoice of ( m, d , . . . , d m ), Theorem 7.14 states thatlim n →∞ π ( n, m, d , . . . , d m ) = 0 . N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 7
Neither of these conjectures accurately addresses the observed fact that “most”quadratic sequences of length n in n variables are semi-regular. More generally wemake the following conjecture. Conjecture 3.
For any 1 ≤ d ≤ n define π ( n, d ) to be the proportion of sequencesof degree d and length n in n variables that are semi-regular. Thenlim n →∞ π ( n, d ) = 1 . In fact we expect much more to be true. Define π ( n, m, d ) to be the proportion ofsequences of degree d and length m in n variables that are semi-regular. The tablebelow gives estimates of π ( n, m,
2) for samples of 20 randomly chosen sequences ofquadratic homogeneous polynomials. n \ m . .
35 1 . . . . . . . .
95 1 . . . . . . . . . . . . . . . . . . . .
257 0 .
85 1 . .
95 1 18 . .
45 1 1 . . . . . . .
85 1 .
35 1 1 1 1 1 1 .
25 1 1 111 0 .
95 1 1 1 1 1 1 1 1 1 1 1 .
412 0 0 1 1 1 1 . .
45 1
Table 1.
Proportion of Samples of 20 Sets of m HomogeneousQuadratic Elements in n variables that are Semi-RegularTheorem 7.14 states that all columns of Table 1 eventually become zero. Weconjecture that the non-zero entries of the rows tend to one as n → ∞ . Oneformulation of this is the following conjecture. Conjecture 4.
Fix d >
1. Define π ( n, m, d ) to be the proportion of sequences ofdegree d and length m in n variables that are semi-regular. Then there exists an0 < η d ≤ / ǫ >
0, there exists an
N > π ( n, m, d ) > − ǫ for all n > N and all m > ηN. While we believe these conjectures to be true, it should be noted that the exis-tence question still remains largely open.
Question . For which pairs ( n, d = ( d , . . . , d m )) do there exist semi-regular se-quences λ , . . . , λ m with deg λ i = d i . Equivalently, for which n and d is T d ,n ( z ) = (1 + z ) n Q mi =1 (1 + z d i )the Hilbert series of an appropriate graded homomorphic image of the algebra B ( n ) ? N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 8
At both ends of the degree spectrum, the existence question is trivial. Sequencesof linear elements are semi-regular if and only if they are linearly independent.Likewise for sequences of degree n − n ). Also for sufficiently large m it iseasy to find sequences that are trivially semi-regular; for instance a basis of thespace of polynomials of degree d .4. The case m = 1 : semi-regularity of homogeneous polynomials In this section we give a complete answer to Question 1 in the case when thepolynomial is linear or quadratic. We show that the proportion of semi-regularelements of degree d in B ( n ) is zero when n > d . Also, we give a completedescription of the Hilbert Series and the index of a semi-regular element. Proposition 4.1.
Let λ ∈ B ( n )1 then λ is semi-regular and Ind( λ ) = n .Proof. Without loss of generality we can assume that λ = x ∈ B ( n ) . In this case B ( n ) / ( x ) ∼ = B ( n − which has Hilbert series (1 + z ) n − . On the otherhand T (1) ,n ( z ) = (1 + z ) n / (1 + z ) = (1 + z ) n − . So by Theorem 2.4, this element issemi-regular and Ind( λ ) = n . (cid:3) Theorem 4.2.
Let λ ∈ B ( n ) be homogeneous of degree d . If d = n or d = n − then λ is semi-regular .Proof. Let λ ∈ B ( n ) be homogeneous of degree d , with n − ≤ d ≤ n . Note thatInd( λ ) = n . The semi-regularity of λ follows trivially from the definition. (cid:3) Lemma 4.3.
Let λ ∈ B ( n ) be a monomial then Ann λ = (var( λ )) where var( λ ) isthe set of variables occurring in λ .Proof. The inclusion (var( λ )) ⊂ Ann λ is clear. Let ν ∈ Ann λ and write ν = ν + · · · + ν r , where the ν i are distinct monomials. Since λ is a monomial for i = j it follows that ν i λ = ν j λ unless ν i λ = ν j λ = 0. Thus λν j = 0 and so λ and ν j mustshare some x i . Hence ν j ∈ (var( λ )). (cid:3) Definition 4.4.
Let λ ∈ B ( n ) be a homogeneous polynomial of degree d . λ can bewritten in the form λ = X m ∈ µ d ǫ m m, where µ d is the set of monomials of degree d , and ǫ m is either 1 or 0. The support of λ is defined as Supp( λ ) = { m | ǫ m = 1 } . Proposition 4.5.
Let λ ∈ B ( n ) be homogeneous of degree d . Then Ind( λ ) > n − d .Proof. By renumbering we may assume x · · · x d ∈ Supp( λ ). We demonstrate x d +1 · · · x n / ∈ ( λ ). Suppose for the sake of contradiction that we have ν ∈ B ( n ) n − d such that νλ = x d +1 · · · x n . Writing ν and λ as polynomials in x , i.e. λ = x λ + λ and ν = x ν + ν then λ ν + x ( λ ν + λ ν ) = x d +1 · · · x n . N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 9 So λ ν + λ ν ∈ Ann x , but x / ∈ var( λ ν + λ ν ) therefore λ ν = λ ν . Inparticular x d +1 · · · x n λ = λ ν λ = λ ν = 0 , so λ ∈ Ann x d +1 · · · x n = ( x d +1 , · · · , x n ) but x · · · x d ∈ Supp λ , which is impos-sible. (cid:3) Now, we will use a result that appears in [8] about the first fall degree of ahomogeneous polynomial λ ∈ B ( n ) . Basically, the first fall degree of λ is the firstdegree at which non-trivial relations occur; trivial relation such as gλ = 0 where g ∈ ( λ ). In other words the first fall degree of λ is the first k such that there exists g in B ( n ) with the property that deg g + deg λ = k , gλ = 0 and g ( λ ). In [8] theauthors give a more detailed and general definition for the first fall degree. Definition 4.6.
Let λ be a homogeneous element of B ( n ) . The rank of λ is thesmallest integer s such that there exist µ , . . . , µ s ∈ B ( n )1 with λ ∈ F [ µ , . . . , µ s ].That is, s is the smallest number of linear elements required to generate λ . Theorem 4.7.
Let λ be an element of degree d > and rank s . Then D ff ( λ ) ≤ ( s + d + 2) / .Proof. See Theorem 4.9 in [8]. (cid:3)
This enables us to give a result on the non-existence of semi-regular elements ofdegree d ≥ n > d . Theorem 4.8.
Let λ , . . . , λ m be a sequence of homogeneous polynomials of de-grees d , . . . , d m in B ( n ) . If Ind( λ , . . . , λ m ) > D ff ( λ i ) for some ≤ i ≤ m then λ , . . . , λ m is not semi-regular.Proof. Suppose λ , . . . , λ m is a sequence of homogeneous polynomials such thatInd( λ , . . . , λ m ) > D ff ( λ i ) for some 1 ≤ i ≤ m . If λ , . . . , λ m is a semi-regular se-quence then by Theorem 2.4 any reordering of this sequence is also a semi-regularsequence. Thus, without loss of generality we can assume that Ind( λ , . . . , λ m ) > D ff ( λ ). By definition of first fall degree there exists g such that gλ = 0, deg g +deg λ = D ff ( λ ) < Ind( λ , . . . , λ m ) and g ( λ ). But it is not possible if λ , . . . , λ m is semi-regular. Therefore λ , . . . , λ m cannot be semi-regular. (cid:3) Corollary 4.9.
Let λ be homogeneous. If Ind( λ ) > D ff ( λ ) then λ is not semi-regular. Theorem 4.10.
There are no semi-regular elements of degree d ≥ for n > d .Proof. Let λ be a homogeneous element with deg λ = d > n > d . Then ( n + d ) / < n − d . Since the rank s of λ is less than or equal to n ,we have by Theorem 4.7 and Proposition 4.5 thatD ff ( λ ) ≤ s + d + 22 ≤ n + d + 22 < n − d + 1 ≤ Ind( λ ) . Since the first fall degree of λ is less than its index it cannot be semi-regular. (cid:3) Next theorem gives a complete description of the proportion of quadratic semi-regular elements.
N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 10
Theorem 4.11.
There are no semi-regular elements of degree 2 for n ≥ . Thatis, π ( n, ,
2) = 0 for n ≥ . For ≤ n ≤ the value of π ( n, , is given by thetable n π ( n, ,
2) 1 1 0 .
44 0 .
85 0 . Proof.
First part is just a consequence of Theorem 4.10.For the cases n = 2 , . . . ,
6, we can compute the Hilbert series of a polynomial ofa given rank using the specific case of x x , x x + x x and x x + x x + x x .The cases n = 2 and 3 follow from Theorem 4.2. When n = 4 the Hilbert series ofa rank two element is 1 + 4 z + 5 z + 2 z and that of a rank 4 element is 1 + 4 z + 5 z which is [ T (2) , ( z )]. Thus the rank four elements are semi-regular and the rank twoelements are not. There are 28 quadratic homogeneous elements of rank four and 35elements of rank two. Thus the proportion of semi-regular elements is 28 /
63 = 0 . n = 5, the Hilbert series of a rank two element is 1 + 5 z + 9 z + 7 z + 2 z and that of a rank 4 element is 1 + 5 z + 9 z + 5 z + z = [ T (2) , ( z )]. The number ofelements of ranks two and four is respectively 155 and 868 yielding a proportion ofsemi-regular elements of 0 .
85. When n = 6 the Hilbert series of a rank two elementis 1+6 z +14 z +16 z +9 z +2 z , that of a rank 4 element is 1+6 z +14 z +14 z +5 z and that of a rank six element is 1 + 6 z + 14 z + 14 z + z = [ T (2) , ( z )]. The numberof elements of ranks two, four and six is respectively 651, 18 ,
228 and 13 ,
888 yieldinga proportion of semi-regular elements of 0 . (cid:3) In her thesis [1], Bardet asserts that the element P ≤ i
3. The table below gives some experimental data for samples of 20homogeneous polynomials of degree d in n variables. n \ d . . .
45 1 .
95 1 17 0 1 0 1 18 1 .
25 1 .
25 1 19 0 1 .
65 1 0 1 110 . . Table 2.
Proportion of Samples of 20 Homogeneous Elements ofDegree d in n variables that are Semi-RegularNote that the ones on the upper two diagonals reflect that fact that all elementsof degree n − n are semi-regular, whereas the ones on the other diagonals reflectonly a high probability of semi-regularity since a monomial of degree less than n − d = 2 t and n = 3 d then σ n,d is semi-regular, thus establishing that the bound d ≤ n/ n . N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 11
Now, we will give a complete description of the truncated series (cid:20) (1 + z ) n z d (cid:21) when n ≤ d . First, note that(1 + z ) n z d = (1 + z ) n (1 − z d + z d + · · · + ( − j z jd + · · · ) . Therefore, (1 + z ) n z d = ∞ X k =0 γ ( n, k, d ) z k (4.1)where γ ( n, k, d ) = ⌊ k/d ⌋ X j =0 ( − j (cid:18) nk − jd (cid:19) . Lemma 4.12.
Let n , and d be two natural numbers.(a) If k is a non-negative integer number such that k < ⌈ ( n + d ) / ⌉ then (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) > (b) If n + d is odd and k = ⌈ ( n + d ) / ⌉ = ( n + d + 1) / then (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) + (cid:18) n (cid:19) ≤ Proof.
Suppose k is a non-negative integer number such that k < ⌈ ( n + d ) / ⌉ . Notethat ⌈ ( n + d ) / ⌉ = ( n + d ) / ⌈ ( n + d ) / ⌉ = ( n + d + 1) /
2. In any case since k isinteger we have that k < ( n + d ) /
2. Also note that (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) > k ≤ n/
2. Now suppose that n/ ≤ k < ( n + d ) /
2. Then k − d < n − k ≤ n/ (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) = (cid:18) nn − k (cid:19) − (cid:18) nk − d (cid:19) > ⌈ ( n + d ) / ⌉ = ( n + d + 1) / k . Let us provethat (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) + (cid:18) n (cid:19) ≤ . Since ( n + d + 1) / k then ( n + d ) / < k , thus n − k < k − d . Also, since k = ( n + d + 1) / ≤ ( n + 2 d ) / k − d ≤ n/
2. Therefore, (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) = (cid:18) nn − k (cid:19) − (cid:18) nk − d (cid:19) < (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) + (cid:18) n (cid:19) ≤ . (cid:3) N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 12
Theorem 4.13.
Let n , and d be two natural numbers. If n < d then (cid:20) (1 + z ) n z d (cid:21) = D − X k =0 (cid:20)(cid:18) nk (cid:19) − (cid:18) nk − d (cid:19)(cid:21) z k where D = ⌈ ( n + d ) / ⌉ .Proof. Since (1 + z ) n z d = ∞ X k =0 γ ( n, k, d ) z k we want to show that γ ( n, k, d ) > k < ⌈ ( n + d ) / ⌉ and that γ ( n, k, d ) ≤ k = ⌈ ( n + d ) / ⌉ . Since n < d then ( n + d ) / < d , thus ⌈ ( n + d ) / ⌉ ≤ d . Let k be a non-negative integer such that k < ⌈ ( n + d ) / ⌉ ≤ d . By Lemma 4.12 we havethat γ ( n, k, d ) = (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) > . Suppose now that k = ⌈ ( n + d ) / ⌉ . If n + d is even, then k = ( n + d ) / < (3 d + d ) / d and n − k = k − d . So γ ( n, k, d ) = (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) = (cid:18) nk (cid:19) − (cid:18) nn − k (cid:19) = 0 . If n + d is odd then k = ( n + d + 1) / k ≤ d . Hence γ ( n, k, d ) ≤ (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) + (cid:18) n (cid:19) ≤ (cid:3) Lemma 4.14.
Let d ≥ . Then (3 d )!(2 d + 1)!( d + 1)! ≥ Proof.
For d = 2 we have that (3 d )!(2 d + 1)!( d + 1)! = 6!5!3! = 1Suppose the result is true for d let us prove it for d + 1. By induction we have that(3( d + 1))!(2( d + 1) + 1)!( d + 2)! = (3 d )!(2 d + 1)!( d + 1)! (3 d + 1)(3 d + 2)(3 d + 3)(2 d + 2)(2 d + 3)( d + 2) ≥ (3 d + 1)(3 d + 2)(3 d + 3)(2 d + 2)(2 d + 3)( d + 2)To show that (3 d + 1)(3 d + 2)(3 d + 3)(2 d + 2)(2 d + 3)( d + 2) ≥ d + 36 d + 7 d − ≥ . The last inequality is true since for d ≥ d + 36 d + 7 d ≥ (cid:3) Lemma 4.15.
Let d ≥ . Then (cid:18) d d + 1 (cid:19) − (cid:18) dd + 1 (cid:19) + (cid:18) d (cid:19) < N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 13
Proof.
Since d ≥
2, by above lemma we have (cid:18) d d + 1 (cid:19) − (cid:18) dd + 1 (cid:19) + (cid:18) d (cid:19) = (3 d )!(2 d + 1)!( d − − (3 d )!( d + 1)!(2 d − d = (3 d )!( d ( d + 1))(2 d + 1)!( d + 1)! − (3 d )!(2 d (2 d + 1))( d + 1)!(2 d + 1)! + 3 d = (3 d )!(2 d + 1)!( d + 1)! ( − d − d ) + 3 d ≤ − d − d + 3 d = − d + 2 d = d ( − d + 2) < (cid:3) Theorem 4.16.
Let n , and d be two natural numbers. If n = 3 d and d ≥ then (cid:20) (1 + z ) n z d (cid:21) = d − X k =0 (cid:20)(cid:18) nk (cid:19) − (cid:18) nk − d (cid:19)(cid:21) z k + z d Proof.
We know that (1 + z ) n z d = ∞ X k =0 γ ( n, k, d ) z k . Suppose n = 3 d . Then for k < d , γ ( n, k, d ) = (cid:18) dk (cid:19) − (cid:18) dk − d (cid:19) > γ ( n, d, d ) = (cid:18) d d (cid:19) − (cid:18) dd (cid:19) + (cid:18) d (cid:19) = 1and by Lemma 4.15 c d +1 = (cid:18) d d + 1 (cid:19) − (cid:18) dd + 1 (cid:19) + (cid:18) d (cid:19) < . It proves the result. (cid:3)
Theorem 4.17.
Suppose that λ is a semi-regular homogeneous element of degree d > . Then n ≤ d and Ind( λ ) = ( ⌈ ( n + d ) / ⌉ if n < d ( n + d + 2) / d + 1 if n = 3 d Proof. If λ is semi-regular, then by Theorem 2.4 HS ( λ ) ( z ) = (cid:20) (1 + z ) n z d (cid:21) thus Ind( λ ) = Ind(1 + z ) n / (1 + z d ) so the result follows from Theorem 4.13 andTheorem 4.16. (cid:3) To finish this section, we present a theorem that will be used in the follow-ing section where we prove some results about semi-regularity of the elementarysymmetric polynomials.
N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 14
Lemma 4.18.
Let λ ∈ B ( n ) be a homogeneous element of degree d . Suppose thatfor k < n the map B ( n ) k λ −→ B ( n ) k + d multiplication by λ , is injective. Then the map B ( n ) k − λ −→ B ( n ) k − d is injective.Proof. Suppose that for k < n the map B ( n ) k λ −→ B ( n ) k + d is injective. Suppose that there exists α = 0 such that α ∈ B ( n ) k − such that αλ = 0.Since k < n we have that there exits x i such that x i ∤ α . So x i α = 0 and x i α ∈ B ( n ) k satisfies that x i αλ = 0, which is a contradiction. (cid:3) Theorem 4.19.
Let λ ∈ B ( n ) be a homogeneous element of degree d . Suppose d = deg( λ ) ≥ n/ .(a) If d > n/ then λ is semi-regular if and only if for D = ⌈ ( n + d ) / ⌉ the map B ( n ) D − − d λ −→ B ( n ) D − is injective, and the map B ( n ) D − d λ −→ B ( n ) D is surjective.(b) If d = n/ then λ is semi-regular if and only if the maps B ( n ) d − λ −→ B ( n )2 d − and B ( n ) d λB ( n )0 λ −→ B ( n )2 d are injective and the map B ( n ) d +1 λ −→ B ( n )2 d +1 is surjective.Proof. Suppose n = 3 d . Suppose λ is semi-regular. By Theorem 2.4 and Theorem4.16 we have HS ( λ ) ( z ) = ∞ X k =0 dim (cid:16) B ( n ) k /λB ( n ) k − d (cid:17) z k = (cid:20) (1 + z ) n z d (cid:21) = d − X k =0 (cid:20)(cid:18) nk (cid:19) − (cid:18) nk − d (cid:19)(cid:21) z k + z d . Thus dim( B ( n )2 d − ) − dim( λB ( n ) d − ) = dim (cid:16) B ( n )2 d − /λB ( n ) d − (cid:17) = (cid:18) n d − (cid:19) − (cid:18) nd − (cid:19) = dim( B ( n )2 d − ) − dim( B ( n ) d − ) . N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 15
So dim( λB ( n ) d − ) = dim( B ( n ) d − ), then we have that the map B ( n ) d − λ −→ B ( n )2 d − is injective. Now, since n = 3 d note that dim B ( n )2 d = dim B ( n ) d anddim( B ( n )2 d ) − dim( λB ( n ) d ) = dim( B ( n )2 d /λB ( n ) d )= 1= dim( B ( n )2 d ) − (dim( B ( n ) d ) − . So dim( λB ( n ) d ) = dim( B ( n ) d ) −
1, then the kernel of the map B ( n ) d λ −→ B ( n )2 d has dimension 1, however λ ∈ B ( n ) d is in this kernel. Therefore the kernel should be λB ( n )0 . So, the map B ( n ) d λB ( n )0 λ −→ B ( n )2 d is injective. Finally, dim( B ( n )2 d +1 /λB ( n ) d +1 ) = 0 then we have that B ( n ) d +1 λ −→ B ( n )2 d +1 is surjective. Conversely, suppose that the maps B ( n ) d − λ −→ B ( n )2 d − and B ( n ) d λB ( n )0 λ −→ B ( n )2 d are injective and the map B ( n ) d +1 λ −→ B ( n )2 d +1 is surjective. Since the map B ( n ) d − λ −→ B ( n )2 d − is injective, then by Lemma 4.18 we have that B ( n ) k − d λ −→ B ( n ) k is injective, for all k < d . Thus, by Lemma 4.12 we have that for all k < d dim (cid:16) B ( n ) k /λB ( n ) k − d (cid:17) = dim( B ( n ) k ) − dim( λB ( n ) k − d )= dim( B ( n ) k ) − dim( B ( n ) k − d )= (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) > . Since the map B ( n ) d λB ( n )0 λ −→ B ( n )2 d is injective and dim( B ( n )2 d ) = dim( B ( n ) d ) then dim( B ( n )2 d /λB ( n ) d ) = 1. Finally, sincethe map B ( n ) d +1 λ −→ B ( n )2 d +1N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 16 is surjective then dim( B ( n )2 d +1 / ( λB ( n ) d +1 )) = 0. Putting together this information wehave that HS ( λ ) ( z ) = ∞ X k =0 dim (cid:16) B ( n ) k /λB ( n ) k − d (cid:17) z k = d − X k =0 (cid:20)(cid:18) nk (cid:19) − (cid:18) nk − d (cid:19)(cid:21) z k + z d . By Theorem 4.16 (cid:20) (1 + z ) n z d (cid:21) = d − X k =0 (cid:20)(cid:18) nk (cid:19) − (cid:18) nk − d (cid:19)(cid:21) z k + z d . Thus, HS ( λ ) ( z ) = (cid:20) (1 + z ) n z d (cid:21) . So λ is semi-regular. It proves (b). The proof of (a) is similar. (cid:3) Semi-regularity of elementary symmetric polynomials
Consider the ring of polynomials over F in n variables, F [ X , ..., X n ]. In thisring we have the elementary symmetric polynomial of degree d which is defined as σ d ( X , ..., X n ) = X ≤ i < ···
Let n , d , and k be natural numbers such that ≤ k ≤ n then σ d ( x , ..., x n ) = d X i =0 σ d − i ( x , ..., x k ) σ i ( x k +1 , ..., x n ) (5.1) Proof.
Note that n X j =1 σ j ( x , ..., x n ) t j = n Y i =1 (1 + tx i )= k Y i =1 (1 + tx i ) n − k Y j =1 (1 + tx j )= k X i =1 σ i ( x , ..., x k ) t i n − k X j =1 σ j ( x k +1 , ..., x n ) t j (cid:3) N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 17
Lemma 5.2. σ a,n σ b,n = (cid:18) a + ba (cid:19) σ a + b,n where k denotes the image of k in F .Proof. Let M be a monomial in x , . . . , x n of degree a + b . Then M will occur oncein σ a,n σ b,n for each occurrence of a sub-monomial of M of degree a in σ a,n . Thereare precisely (cid:0) a + ba (cid:1) such sub-monomials. (cid:3) Corollary 5.3. σ ,n σ k,n = ( if k is odd σ k,n if k is even Theorem 5.4. If d is odd, then σ d,n is semi-regular if and only if d = n − , n .Proof. Suppose that σ d,n is semi-regular. By above corollary, the first fall degreeof σ d,n is less than or equal to d + 1. On the other hand, we know thatInd( σ d,n ) ≥ (cid:24) n + d (cid:25) . Since σ d,n is semi-regular by Corollary 4.9 we have that Ind( σ d,n ) ≤ D ff ( σ d,n ), thus ⌈ ( n + d ) / ⌉ ≤ d +1. This implies that n + d ≤ d +2 from which we obtain d ≥ n − d = n −
2, then Ind( σ d,n ) = ⌈ ( n + d ) / ⌉ = n −
1. Thus, dim( B ( n ) n − /σ d,n B ( n )1 ) = 0then multiplication by σ d,n would be an surjective map from B ( n )1 to B ( n ) n − ; however,dim B ( n ) n − = dim B ( n )1 = n , then multiplication by σ d,n would be an injective mapfrom B ( n )1 to B ( n ) n − ; but σ ,n σ d,n = 0 by above corollary so this does not happen.Hence we must have that d = n − n . Conversely, any element λ ∈ B ( n ) n ∪ B ( n ) n − is semi-regular as we show in Theorem 4.2. (cid:3) Lemma 5.5.
Suppose that for k < n the map B ( n ) k σ d,n −−−→ B ( n ) k + d is injective. Then the map B ( n +1) k σ d,n +1 −−−−→ B ( n +1) k + d is injective.Proof. Suppose that the map B ( n ) k σ d,n −−−→ B ( n ) k + d is injective. Suppose that there exists α ∈ B ( n +1) k such that ασ d,n +1 = 0. Byequation (5 .
1) we have that σ d,n +1 = σ d,n + x n +1 σ d − ,n , also notice that α = α ′ + x n +1 β with α ′ ∈ B ( n ) k and β ∈ B ( n ) k − . Thus 0 = ασ d,n +1 = α ′ σ d,n + x n +1 ( α ′ σ d − ,n + βσ d,n ). Since α ′ σ d,n ∈ B ( n ) we have that α ′ σ d,n = 0 so by injectivity α ′ = 0. Thus,0 = x n +1 ( βσ d,n ), but βσ d,n ∈ B ( n ) , so βσ d,n = 0. Therefore by Lemma 4 .
18 wehave that β = 0. So α = 0. (cid:3) N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 18
Theorem 5.6.
Let n and k be non-negative integers. Then (cid:18) nk (cid:19) = ( if n is even and k is odd (cid:0) ⌊ n/ ⌋⌊ k/ ⌋ (cid:1) otherwisewhere a denotes the image of a in F .Proof. See Theorem 4.1.10 in [7]. (cid:3)
Lemma 5.7.
Let d = 2 n l . Then for all ≤ k ≤ n − we have that (cid:18) n l + jk (cid:19) = ( if j = k if ≤ j ≤ k − where a denotes the image of a in F .Proof. If n = 1, then k = 1 and clearly we have that (cid:0) l +11 (cid:1) = 1 and (cid:0) l (cid:1) = 0.Suppose by induction that the result is true for n , let us prove it for n + 1. For k = 1 we have that (cid:0) n +1 l +11 (cid:1) = 1 and (cid:0) n +1 l (cid:1) = 0. Suppose 2 ≤ k ≤ n +1 − ≤ j ≤ k −
1, then 1 ≤ ⌊ k/ ⌋ ≤ n −
1, and 0 ≤ ⌊ j/ ⌋ ≤ ⌊ k/ ⌋ −
1. Suppose firstthat k is even. Then by Theorem 5 . (cid:18) n +1 l + jk (cid:19) = (cid:18) n l + ⌊ j/ ⌋⌊ k/ ⌋ (cid:19) = ( j = k ≤ j ≤ k − k is odd. Note that 2 n +1 l + k is odd then by Theorem 5 . (cid:18) n +1 l + kk (cid:19) = (cid:18) n l + ⌊ k/ ⌋⌊ k/ ⌋ (cid:19) = 1 . Now if 0 ≤ j ≤ k − . (cid:18) n +1 l + jk (cid:19) = 0 . If 0 ≤ j ≤ k − . (cid:18) n +1 l + jk (cid:19) = (cid:18) n l + ⌊ j/ ⌋⌊ k/ ⌋ (cid:19) = 0 . (cid:3) Lemma 5.8.
The map B ( n ) k − d σ d,n −−−→ B ( n ) k is surjective if and only if there exits α ∈ B ( n ) k − d such that ασ d,n = x · · · x k Proof.
One way is trivial. In the other way suppose that there exits α ∈ B ( n ) k − d suchthat ασ d,n = x · · · x k . Note that the set { x i · · · x i k | ≤ i < · · · < i k ≤ n } is abasis for the vector space B ( n ) k . By the natural action of the group of permutations N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 19 of n -elements Σ n on B ( n ) , given g ∈ Σ n we have that x g (1) · · · x g ( k ) = g ∗ ( x · · · x k )= g ∗ ( ασ d,n )= ( g ∗ α )( g ∗ σ d,n )= ( g ∗ α ) σ d,n (cid:3) The next theorem is one of the key results of this section.
Theorem 5.9.
Let d = 2 m l . Then σ d,n is semi-regular for all n = d + i , with ≤ i ≤ m +1 − .Proof. For 0 ≤ i ≤ n = d + i with 0 ≤ i ≤ m +1 − n < d . Thus, by Theorem 4 .
19 weneed to prove that for D = ⌈ ( n + d ) / ⌉ the map B ( n ) D − − d σ d,n −−−→ B ( n ) D − is injective and the map B ( n ) D − d σ d,n −−−→ B ( n ) D is surjective. Let us prove by induction over i that the result is true for all 2 ≤ i ≤ m +1 −
1. If i = 2 we have that n = d + 2 thus D = d + 1 therefore we need toshow that B ( d +2)1 σ d,d +2 −−−−→ B ( d +2) d +1 is surjective. By equation (5.1) and Lemma 5.2 we have that σ ,d +1 σ d,d +2 = σ ,d +1 ( σ d,d +1 + σ d − ,d +1 x d +2 )= (cid:18) d + 11 (cid:19) σ d +1 ,d +1 + (cid:18) d (cid:19) σ d,d +1 x d +2 = σ d +1 ,d +1 = x · · · x d +1 . Therefore by Lemma 5 . d + i with 2 ≤ i ≤ m +1 −
2. Let us see that σ d,n is semi-regular for n = d + i + 1.Suppose that i + 1 is even, so 4 ≤ i + 1 ≤ m +1 − k = ( i + 1) / ≤ k ≤ m − n = d + 2 k and D = (cid:24) d + i + 1 + d (cid:25) = (cid:24) d + i + d (cid:25) = d + k. By Theorem 4.19 we want to show that the map B ( n ) D − − d σ d,n −−−→ B ( n ) D − is injective and the map B ( n ) D − d σ d,n −−−→ B ( n ) D is surjective. By induction we have that σ d,n − is semi-regular. Since ⌈ ( n − i ) / ⌉ = ⌈ ( d + i + d ) / ⌉ = D we have that the map B ( n − D − − d σ d,n − −−−−→ B ( n − D − is injective. By Lemma 5.5 we have that B ( n ) D − − d σ d,n −−−→ B ( n ) D −
1N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 20 is injective. Now, want to show that B ( n ) D − d σ d,n −−−→ B ( n ) D is surjective. Since D = d + k , the we need to show that B ( n ) k σ d,n −−−→ B ( n ) d + k is surjective. By equation (5.1) and Lemma 5.2 we have that σ k,d + k σ d,n = σ k,d + k σ d,d +2 k = σ k,d + k d X j =0 σ d − j,d + k σ j ( x d + k +1 , ..., x d +2 k )= d X j =0 (cid:18) d + k − jk (cid:19) σ d + k − j,d + k σ j ( x d + k +1 , ..., x d +2 k ) . Since 2 ≤ k ≤ m − σ k,d + k σ d,n = σ d + k,d + k = x · · · x d + k . By Lemma 5 . i + 1 is odd. Since 2 ≤ i ≤ m +1 − k = i/ ≤ k ≤ m , n = d + 2 k − D = (cid:24) d + i + 1 + d (cid:25) = (cid:24) d + i + d (cid:25) + 1 = d + k. We want to show that the map B ( n ) D − − d σ d,n −−−→ B ( n ) D − is injective and the map B ( n ) D − d σ d,n −−−→ B ( n ) D is surjective. By induction we have that σ d,n − is semi-regular. Since ⌈ ( n − i ) / ⌉ = ⌈ ( d + i + d ) / ⌉ = d + k − D − B ( n − D − − d σ d,n − −−−−→ B ( n − D − is injective and the map B ( n − D − − d σ d,n − −−−−→ B ( n − D − is surjective. But notice thatdim B ( n − D − − d = dim B ( n − k − = (cid:18) n − k − (cid:19) = (cid:18) d + ii/ (cid:19) = (cid:18) d + id + i/ (cid:19) = dim B ( n − d + k − = dim B ( n − D − . Therefore the map B ( n − D − − d σ d,n − −−−−→ B ( n − D −
1N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 21 is injective. By Lemma 5.5 we have that B ( n ) D − − d σ d,n −−−→ B ( n ) D − is injective. Now, we want to show that B ( n ) D − d σ d,n −−−→ B ( n ) D is surjective. Since k = D − d then we want to show that B ( n ) k σ d,n −−−→ B ( n ) d + k is surjective. By equation (5.1) and Lemma 5.2 we have that σ k − ,d + k σ d,n = σ k − ,d + k σ d,d +2 k − = σ k − ,d + k d X j =0 σ d − j,d + k σ j ( x d + k +1 , ..., x d +2 k − )= d X j =0 (cid:18) d + k − − jk − (cid:19) σ d + k − − j,d + k σ j ( x d + k +1 , ..., x d +2 k − ) . Since 2 ≤ k ≤ m , then 1 ≤ k − ≤ m −
1. Thus, by Lemma 5.7 we have that σ k − ,d + k σ d,n = σ d + k − ,d + k . Therefore, x σ k − ,d + k ∈ B ( n ) k and x σ k − ,d + k σ d,n = x σ d + k − ,d + k = σ d + k,d + k = x · · · x d + k . By Lemma 5 . (cid:3) Lemma 5.10.
Let m be a positive integer (cid:18) j m (cid:19) = ( if j = 2 m +1 if m ≤ j ≤ m +1 − where k denotes the image of k in F .Proof. If m = 1 we have that (cid:18) (cid:19) = 0 , and (cid:18) (cid:19) = (cid:18) (cid:19) = 1 . Suppose the result is true for m ≥ m + 1. Notethat by Theorem 5.6 and induction (cid:18) m +2 m +1 (cid:19) = (cid:18) m +1 m (cid:19) = 0 . Now, if 2 m +1 ≤ j ≤ m +2 −
1, then 2 m ≤ ⌊ j/ ⌋ ≤ m +1 − (cid:18) j m +1 (cid:19) = (cid:18) ⌊ j/ ⌋ m (cid:19) = 1 . (cid:3) Lemma 5.11.
Let d = 2 m and n = 3 d = 2 m +1 + 2 m . Then σ d ( x i , . . . , x i m +1 ) σ d,n = x i · · · x i m +1 + σ d,n N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 22
Proof.
Let µ ∈ Supp( σ d,n ). Let j be the number of common variables between µ and x i · · · x i m +1 . Note that 2 m ≤ j ≤ m +1 . µ will occur once in the product σ d ( x i , . . . , x i m +1 ) σ d,n for each occurrence of a sub-monomial of µ of degree d = 2 m in σ d ( x i , . . . , x i m +1 ). There are exactly (cid:0) j m (cid:1) such sub-monomials. If j = 2 m +1 then µ = x i · · · x i m +1 and by the above lemma this element appears an evennumber of times. So x i · · · x i m +1 Supp( σ d ( x i , . . . , x i m +1 ) σ d,n ). If 2 m ≤ j ≤ m +1 − µ ∈ Supp( σ d ( x i , . . . , x i m +1 ) σ d,n ). Thus, σ d ( x i , . . . , x i m +1 ) σ d,n = x i · · · x i m +1 + σ d,n . (cid:3) Lemma 5.12.
Let d = 2 m and n = 3 d . Then the map B ( n ) d +1 σ d −→ B ( n )2 d +1 is surjective.Proof. First, let us see that σ d − , d +1 σ d,n = σ d − , d +1 . By equation (5.1), Lemma 5.2 and Lemma 5.7 we have σ d − , d +1 σ d,n = σ k,d + k d X j =0 σ d − j, d +1 σ j ( x d +2 , ..., x d )= d X j =0 (cid:18) d + ( d − − j ) d − (cid:19) σ d − − j, d +1 σ j ( x d +2 , ..., x d )= d X j =0 (cid:18) m + (2 m − − j )2 m − (cid:19) σ d − − j, d +1 σ j ( x d +2 , ..., x d )= σ d − , d +1 . Note that σ d − , d +1 = x σ d − ( x , ..., x d +1 ) + x σ d − ( x , x , ..., x d − )+ x x σ d − ( x , ..., x d − ) + x · · · x d +1 . Therefore x x σ d − , d +1 σ d,n = x x σ d − , d +1 = x x x · · · x d +1 . Since x x σ d − , d +1 ∈ B ( n ) d +1 then by Lemma 5 . (cid:3) Theorem 5.13.
Let d = 2 m . If n = d + 2 m +1 then σ d,n is semi-regular.Proof. Suppose n = d + 2 m +1 then n = 3 d . By Theorem 4.19 we want to show thatthe maps B ( n ) d − σ d,n −−−→ B ( n )2 d − and B ( n ) d σ d,n B ( n )0 σ d,n −−−→ B ( n )2 d N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 23 are injective and the map B ( n ) d +1 σ d,n −−−→ B ( n )2 d +1 is surjective. By Theorem 5.9 we have that σ d,n − is semi-regular. Since n − < n = 3 d then ⌈ ( n − d ) / ⌉ = 2 d , thus the map B ( n − d − σ d,n − −−−−→ B ( n − d − is injective. So, by Lemma 5.5 the map B ( n ) d − σ d,n −−−→ B ( n )2 d − is injective. Also, by Lemma 5.12 the map B ( n ) d +1 σ d,n −−−→ B ( n )2 d +1 is surjective. Finally, by Lemma 5.11 we have that σ d ( x i , . . . , x i d ) σ d,n = x i · · · x i d + σ d,n . Consider the set S = { x i · · · x i d + σ d,n | ≤ i ≤ · · · ≤ i d ≤ n } Note that the set S \ { x · · · x m +1 + σ d,n } ⊂ B ( n )2 d is linearly independent. Andthis set has (cid:0) n d (cid:1) − B ( n )2 d − B ( n ) d σ d B ( n )0 σ d,n −−−→ B ( n )2 d is injective. By all the above, σ d,n is semi-regular. (cid:3) The following theorem is our main result of this section.
Theorem 5.14.
Let d ≥ , where d = 2 m l with l an odd number, and m a non-negative integer. Then(a) If l > , σ d,n is semi-regular if and only if n = d, d + 1 , ..., d + 2 m +1 − .(b) If l = 1 , σ d,n is semi-regular if and only if n = d, d + 1 , ..., d + 2 m +1 .Proof. Suppose first that m = 0. Thus, d is an odd number. By Theorem 5.4 wehave that (a) is true in the case m = 0. Suppose now that m ≥
1. Suppose that σ d,n is semi-regular. By Theorem 4.10 we have that n ≤ d . By Corollary 4.17Ind( σ d,n ) ≥ (cid:24) n + d (cid:25) ≥ n + d . Note that σ m ,n σ d,n = (cid:18) m l + 2 m m (cid:19) σ m + d,n = (cid:18) l + 11 (cid:19) σ m + d,n = 0 . Thus D ff ( σ d,n ) ≤ d + 2 m . Since σ d,n is semi-regular we have that Ind( σ d,n ) ≤ D ff ( σ d,n ). So n + d ≤ d + 2 m +1 from which we obtain n ≤ d + 2 m +1 . Suppose l = 1, by Theorem 5.9 and Theorem 5.13 we have that σ d,n is semi-regular for n = d, d + 1 , ..., d + 2 m +1 . So (b) is proved. Suppose l >
1. We already know that n ≤ d + 2 m +1 . Suppose n = d + 2 m +1 = 2 m +1 l + 2 m +1 note that since l > n < d . Thus, if σ d,n is semi-regular then Ind( σ d,n ) = ⌈ ( n + d ) / ⌉ = d + 2 m . ByTheorem 4.19 the map B ( n )2 m σ d,n −−−→ B ( n ) d +2 m N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 24 is surjective. However, dim B ( n )2 m = (cid:18) n m (cid:19) = (cid:18) d + 2 m +1 m (cid:19) = (cid:18) d + 2 m +1 d + 2 m (cid:19) = dim B ( n ) d +2 m . So the map B ( n )2 m σ d,n −−−→ B ( n ) d +2 m is injective. But that is not possible since σ m ,n σ d,m = 0. Hence we must have n ≤ d + 2 m +1 −
1. Conversely, by Theorem 5.9 σ d,n is semi-regular for n = d, d + 1 , ..., d + 2 m +1 −
1. It proves (a). (cid:3)
The following table gives a visual interpretation of Theorem 5.14. n \ d x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Table 3.
Semi-Regularity of σ d,n . The values when σ d,n is semi-regular are marked with an x Most homogeneous sequences are semi-regular
In her thesis [1, § F , dans le sens ou la proportion desuites semi-r´eguli`eres tend vers 1 quand n tend vers l’infini” (We conjecture nonethe less that a sequence ‘chosen at random’ will be F -semi-regular in the sense thatthe proportion of sequences that are semi-regular tends to 1 as n tends to infinity).We prove this result in its broadest interpretation: namely that the proportion ofhomogeneous subsets of B ( n ) that are semi-regular tends to 1 as n tends to infinity. Lemma 6.1.
Let k be a positive integer and suppose that { λ , . . . , λ m } spans B k .Then λ , . . . , λ m is a semi-regular sequence.Proof. Notice that deg( λ i ) = k so ( λ , . . . , λ m ) ∩ B j = { } for all j < k . On theother hand, ( λ , . . . , λ m ) ∩ B k = B k , so Ind( λ , . . . , λ m ) = k . For any homogeneous f ∈ B we have deg( f ) + deg( λ i ) ≥ k so k -semi-regularity is trivially satisfied. (cid:3) N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 25
Lemma 6.2.
Let { λ , . . . , λ m } be a semi-regular sequence of homogeneous elementsof B and let D = Ind( λ , . . . , λ m ) . If ν ∈ B is a homogeneous element of degreegreater than or equal to D then λ , . . . , λ m , ν is also semi-regular.Proof. Since deg( ν ) ≥ D we have( λ , . . . , λ m ) = ( λ , . . . , λ m , ν )so the degree of regularity does not change and again we see for any homogeneous f ∈ B that deg( f )+deg( ν ) ≥ D . So D -semi-regularity is again trivially verified. (cid:3) Proposition 6.3.
Let V n be an n -dimensional F -vector space and let P ( V n \{ } ) be the set of subsets of V n \{ } . Let S n be the number of subsets of V n \{ } thatspan V n . Then lim n →∞ S n |P ( V n \{ } ) | = 1 Equivalently the probability of randomly picked subset of V n being a spanning setgoes to as n → ∞ .Proof. We find an upper bound on the probability that a randomly picked subset A does not span V n . Note that if a subset does not span V n then it must be containedin some n − V n . The probability that A is contained ina particular n − n − / n = 1 / n − . There are 2 n − n − / n − . Taking a limit we havelim n →∞ n − n − = 0as desired. (cid:3) Theorem 6.4.
Let d be a positive integer and let P d ( n ) be the proportion of homo-geneous sequences of degree greater than or equal to d in B ( n ) that are semi-regular.Then lim n →∞ P d ( n ) = 1 Proof.
Let h d ( n ) be the number of homogeneous elements of B = B ( n ) of degreegreater than or equal to d . Then there are 2 h d ( n ) possible homogeneous sequences(note that we are ignoring order in the sequence). From Proposition 3.2 and Propo-sition 3.3 we know that if we pick a spanning set of B d and add any elements from B i where i > d then this is a semi-regular sequence. The number of homogeneouselements of B of degree greater than d is h d +1 ( n ) = h d ( n ) − ( | B d | − . Let S d ( n ) be the number of spanning sets of B d . Then a lower bound on theproportion of semi-regular sequences is given by S d ( n )2 h d +1 ( n ) h d ( n ) = S d ( n )2 | B d |− Since d ≥
1, dim B d → ∞ as n → ∞ and so Proposition 6.3, implies thatlim n →∞ S d ( n )2 | B d |− = 1 . (cid:3) N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 26 Non-Existence of Semi-Regular Sequences over F In this section we prove the following theorem.
Theorem 7.1.
Let d , . . . , d m be a sequence of integers with d i ≥ for some ≤ i ≤ m . Then there exists an N such that for all n ≥ N , there cannot be asemi-regular sequence λ , . . . , λ m of homogeneous polynomials of degrees d , . . . , d m . This theorem implies that Conjecture 2 in [3] is false.The idea of the proof is the following. For d = ( d , . . . , d m ), we define thefunction τ d ( n ) = Ind (1 + z ) n Q mi =1 (1 + z d i ) . (7.1)We show that this function is bounded below by a linear function g ( n ) = rn + c ,with r > /
2. Suppose that for some j we have that d j ≥
2. Since r > / N such that for all n ≥ Nτ d ( n ) > n d j . Suppose λ , . . . , λ m is a semi-regular sequence of homogeneous polynomials of de-grees d , . . . , d m in B ( n ) , n ≥ N . Thus, by Theorem 2.4, Ind( λ , . . . , λ m ) = τ d ( n ) > ( n + d j + 2) /
2. Also, by Theorem 4.7 we have that D ff ( λ j ) ≤ n + d j + 22 . Therefore, D ff ( λ j ) < Ind( λ , . . . , λ m ), but Theorem4.8 tells us that this is notpossible for a semi-regular sequence. Lemma 7.2.
Let f : N → R be a non-decreasing function. If there exist n , N ∈ N ,and A ∈ R , such that for all n ≥ n we have f ( n + N ) ≥ f ( n ) + A then there exists a constant c such that f ( n ) ≥ ( A/N ) n + c for all natural number n .Proof. Consider the function g ( n ) = f ( n ) + ( A/N )( n − ( n + N )). Let us showthat for all n ≥ n we have that f ( n ) > g ( n ). Let m ≥ n . Write m − n = lN + b ,where b is an integer b < N . By hypothesis we have that f ( m ) = f ( n + b + lN ) ≥ f ( n + b ) + lA. Since f is non-decreasing we have that f ( m ) ≥ f ( n ) + lA . Now, g ( m ) = g ( n + b + lN )= f ( n ) + ( A/N )( n + b + lN − n − N )= f ( n ) + A ( l −
1) + (
A/N ) b. But b < N , so g ( m ) < f ( n ) + Al ≤ f ( m ). Thus, for all n ≥ n we have that f ( n ) > g ( n ). Note that g is defined as g ( n ) = ( A/N ) n + k, where k = f ( n ) − ( A/N )( n + N ). So, for all n ≥ n we have that f ( n ) > ( A/N ) n + k . Since we have a finite number natural numbers less that n then N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 27 for an appropriate choice of a constant c we have that for all natural number nf ( n ) > ( A/N ) n + c . (cid:3) Lemma 7.3.
For any u between and n , and any a j ∈ R n X j =0 a j (cid:18) nj (cid:19) = u − d X j =0 γ ( n, j, d )( a j + a j + d ) + u X j = u − d +1 γ ( n, j, d ) a j + n X j = u +1 (cid:18) nj (cid:19) a j Proof.
Note that by definition of γ ( n, j, d ), we have that (cid:18) nj (cid:19) = γ ( n, j, d ) + γ ( n, j − d, d ) . Thus u X j =0 a j (cid:18) nj (cid:19) = u X j =0 ( γ ( n, j, d ) + γ ( n, j − d, d )) a j = u X j =0 γ ( n, j, d ) a j + u X j =0 γ ( n, j − d, d ) a j = u X j =0 γ ( n, j, d ) a j + u − d X j =0 γ ( n, j, d ) a j + d = u − d X j =0 γ ( n, j, d )( a j + a j + d ) + u X j = u − d +1 γ ( n, j, d ) a j (cid:3) Lemma 7.4.
Let N , d be natural numbers. Let β ( z ) = ∞ X j =0 b j z j . Suppose that
Ind β ( z ) ≥ and b i + b i − d ≥ for all ≤ i ≤ Ind β ( z ) + Ind (1 + z ) N z d − d − . Then
Ind(1 + z ) N β ( z ) ≥ Ind β ( z ) + Ind (1 + z ) N z d − d. Proof.
Let (1 + z ) N β ( z ) = X c i z i and let l = Ind β ( z ) and s = Ind(1 + z ) N / (1 + z d ). Suppose that l ≥ b i + b i − d ≥ ≤ i ≤ l + s − d − . N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 28
We want to show that Ind(1 + z ) N β ( z ) ≥ l + s − d. That is, c i > ≤ i ≤ l + s − d −
1. Clearly Ind(1 + z ) N β ( z ) ≥ l . It remainsto show that c l + i >
0, for i = 0 , . . . , s − d −
1. For 0 ≤ i ≤ s − d − c l + i = N X j =0 b l + i − j (cid:18) Nj (cid:19) = s − − d X j =0 γ ( N, j, d )( b l + i − j + b l + i − j − d )+ s − X j = s − d γ ( N, j, d ) b l + i − j + N X j = s (cid:18) Nj (cid:19) b l + i − j . For j = 0 , . . . , s −
1, we have that γ ( N, j, d ) >
0, since s = Ind(1 + z ) N / (1 + z d ).Also, b l + i − j + b l + i − j − d ≥
0, since l + i − j ≤ l + s − d −
1. So s − − d X j =0 γ ( N, j, d )( b l + i − j + b l + i − j − d ) ≥ . Finally, if j ≥ s − d , then l + i − j ≤ l + ( s − d − − ( s − d ) = l −
1, so b l + i − j > s − X j = s − d γ ( N, j, d ) b l + i − j + N X j = s (cid:18) Nj (cid:19) b l + i − j > . Thus, we have shown that c i > ≤ i ≤ l + s − d −
1. SoInd(1 + z ) N β ( z ) ≥ l + s − d. (cid:3) Theorem 7.5. If Ind α ( z ) ≥ and Ind α ( z ) ≥ Ind α ( z )1 + z d + Ind (1 + z ) N z d − d then Ind (1 + z ) N α ( z )1 + z d ≥ Ind α ( z )1 + z d + Ind (1 + z ) N z d − d Proof.
Let α ( z ) = X a i z i , α ( z )1 + z d = X b i z i = β ( z )and let l = Ind β ( z ) and s = Ind(1 + z ) N / (1 + z d ). Suppose thatInd α ( z ) ≥ Ind α ( z )1 + z d + Ind (1 + z ) N z d − d with Ind α ( z ) ≥
1. In other words, Ind α ( z ) ≥ α ( z ) ≥ l + s − d. We want to show that Ind(1 + z ) N β ( z ) ≥ l + s − d. N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 29
Since Ind α ( z ) ≥ β ( z ) ≥
1. Also, note that b i + b i − d = a i > ≤ i ≤ l + s − d −
1, since Ind α ( z ) ≥ l + s − d . Thus, by Lemma 7.4 wehave that Ind(1 + z ) N β ( z ) ≥ l + s − d. (cid:3) Lemma 7.6.
Let d , N be natural numbers. Then Ind (1 + z ) N (1 + z ) n z d ≥ Ind (1 + z ) n z d + Ind (1 + z ) N z d − d for all natural number n .Proof. Consider (1 + z ) n z d = ∞ X i =0 b i z i = β ( z ) . We want to show thatInd(1 + z ) N β ( z ) ≥ Ind β ( z ) + Ind (1 + z ) N z d − d. Clearly Ind β ( z ) ≥
1. Also, we have that b i + b i − d = (cid:18) ni (cid:19) ≥ i . Thus, the result follows from Lemma 7.4. (cid:3) Theorem 7.7.
Suppose that r ∈ (cid:26) n (cid:18) Ind (1 + z ) n z d − d (cid:19) | n ≥ (cid:27) then there exists a c such that Ind (1 + z ) n z d ≥ rn + c for all n .Proof. Let r = 1 N (cid:18) Ind (1 + z ) N z d − d (cid:19) . Consider the function τ ( d ) ( k ) = Ind (1 + z ) k z d . By Lemma 7.6 we have τ ( d ) ( n + N ) = Ind (1 + z ) n (1 + z ) N z d ≥ Ind (1 + z ) n z d + Ind (1 + z ) N z d − d = τ ( d ) ( n ) + ( τ ( d ) ( N ) − d ) . By Lemma 7.2 there exists c such that τ ( d ) ( n ) ≥ N ( τ ( d ) ( N ) − d ) n + c N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 30 for all n . In other words, Ind (1 + z ) n z d ≥ rn + c for all n . (cid:3) Theorem 7.8.
Let d = ( d , . . . , d m ) and let d ′ = ( d , . . . , d m , d ) . Suppose that r = 1 N (cid:18) Ind (1 + z ) N z d − d (cid:19) for some positive integer N . Consider the function τ d ( n ) as defined in (7.1). If τ d ( n ) ≥ τ d ′ ( n ) + rN then τ d ′ ( n + N ) ≥ τ d ′ ( n ) + rN Proof.
Consider α ( z ) = (1 + z ) n Q mi =1 (1 + z d i ) . In this case τ d ( n ) = Ind α ( z ) τ d ′ ( n ) = Ind (1 + z ) n Q mi =1 (1 + z d i )(1 + z d ) = Ind α ( z )1 + z d τ d ′ ( n + N ) = Ind (1 + z ) n + N Q mi =1 (1 + z d i )(1 + z d ) = Ind α ( z )(1 + z ) N z d . Also, rN = Ind (1 + z ) N z d − d. Suppose τ d ( n ) ≥ τ d ′ ( n ) + rN. Thus, Ind α ( z ) ≥ Ind α ( z )1 + z d + Ind (1 + z ) N z d − d. By Theorem 7.5Ind α ( z )(1 + z ) N z d ≥ Ind α ( z )1 + z d + Ind (1 + z ) N z d − d. Therefore, τ d ′ ( n + N ) ≥ τ d ′ ( n ) + rN. (cid:3) Theorem 7.9.
Let d = ( d , . . . , d m ) and let d ′ = ( d , . . . , d m , d ) . Suppose that s = 1 N (cid:18) Ind (1 + z ) N z d − d (cid:19) for some positive integer N . Suppose that there exist r ≥ s and c such that τ d ( n ) ≥ rn + c, for all n . Then there exists c ′ such that τ d ′ ( n ) ≥ sn + c ′ , N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 31 for all n .Proof. Let d = ( d , . . . , d m ) and let d ′ = ( d , . . . , d m , d ). Suppose that s = 1 N (cid:18) Ind (1 + z ) N z d − d (cid:19) for some positive integer N . Suppose that there exist r ≥ s and c such that τ d ( n ) ≥ rn + c, for all n . Let us prove that for c ′ = min { c − sN, − sN, } we have τ d ′ ( n ) ≥ sn + c ′ , for all n . If s ≤
0, the Theorem is true since c ′ ≥ τ d ′ ( n ) ≥
0. Suppose that s >
0. Let n be any natural number. We want to show that τ d ′ ( n ) ≥ sn + c ′ . Let k be the largest positive integer less than or equal to n such that τ d ′ ( k ) ≥ sk + ( c − sN ), and set n = k . If no such positive integer exists, set n = 0. If n = n , the assertion is true so assume that n < n . Write n − n − hN + b where b is an integer b < N . Let m = n + 1 + jN where 0 ≤ j ≤ h . Then τ d ′ ( m ) < sm + ( c − sN ) ≤ rm + ( c − sN ) . Hence τ d ( m ) ≥ τ d ′ ( m ) + sN . By Theorem 7 . τ d ′ ( m + N ) ≥ τ d ′ ( m ) + sN . So by iterating this argument, τ d ′ ( m ) ≥ τ d ′ ( n + 1) + jsN ≥ τ d ′ ( n ) + jsN Hence, τ d ′ ( n ) ≥ τ d ′ ( n + 1 + hN ) ≥ τ d ′ ( n ) + hsN. If n = k we have that τ d ′ ( n ) ≥ τ d ′ ( n ) + hsN ≥ s ( n ) + ( c − sN ) + hsN = s ( n + hN ) + ( c − sN )= s ( n − − b ) + ( c − sN )= sn − s (1 + b ) + ( c − sN ) ≥ sn − sN + ( c − sN ) ≥ sn + c ′ . If n = 0 we have τ d ′ ( n ) ≥ τ d ′ ( n ) + hsN ≥ hsN = s ( n − − b )= sn − s (1 + b ) ≥ sn − sN ≥ sn + c ′ . (cid:3) N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 32
Theorem 7.10.
Let d = ( d , . . . , d m ) . Suppose that r is such that for all i thereexits an n i such that r ≤ n i (cid:18) Ind (1 + z ) n i z d i − d i (cid:19) . Then there exists a c such that τ d ( n ) = Ind (1 + z ) n Q mi =1 (1 + z d i ) ≥ rn + c for all n .Proof. Let r i = 1 n i (cid:18) Ind (1 + z ) n i z d i − d i (cid:19) . Reordering we can suppose that r ≥ r ≥ · · · ≥ r m ≥ r . By Theorem 7 . c such that τ ( d ) ( n ) ≥ r n + c , for all n . By Theorem 7 . c such that τ ( d ,d ) ( n ) ≥ r n + c , for all n . By iterating this argument we have that there exists c such that τ d ( n ) ≥ rn + c, for all n . (cid:3) Lemma 7.11.
Let d be a natural number. Then there exists M such that for all n ≥ M (cid:18) nn + d − j (cid:19) − (cid:18) nn − j (cid:19) + (cid:18) nn − j − d (cid:19) − (cid:18) nn − j − d (cid:19) > for all ≤ j ≤ d − ⌊ d/ ⌋ − . Proof.
Let p = ⌊ d/ ⌋ . Note that for 0 ≤ j ≤ d − p − (cid:18) nn + d − j (cid:19) − (cid:18) nn − j (cid:19) + (cid:18) nn − j − d (cid:19) − (cid:18) nn − j − d (cid:19) ≥ (cid:18) nn + d (cid:19) − (cid:18) nn (cid:19) + (cid:18) nn − d + p + 1 (cid:19) − (cid:18) nn − d (cid:19) . Now, (cid:18) nn + d (cid:19) − (cid:18) nn (cid:19) + (cid:18) nn − d + p + 1 (cid:19) − (cid:18) nn − d (cid:19) = (2 n )!( n + d )!( n − d )! − (2 n )! n ! n ! + (2 n )!( n − d + p + 1)!( n + 2 d − p − − (2 n )!( n − d )!( n + 2 d )! . N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 33
Note that 1( n + d )!( n − d )! − n ! n ! + 1( n − d + p + 1)!( n + 2 d − p − − n − d )!( n + 2 d )!= q ( n )( n + 2 d − p − n + 2 d )! , where q ( n ) = d Y i =1 ( n + d + i ) d − p − Y i =1 ( n − d + i ) − d Y i =1 ( n + i ) d − p − Y i =1 ( n + i )+ d − p − Y i =1 ( n − d + p + 1 + i ) − d − p − Y i =1 ( n − d + i ) . Clearly q ( n ) is a polynomial in n of degree at most 4 d − p −
1. The coefficient of n d − p − is easily seen to be zero and that of n d − p − is (cid:18) d + d ( d + 1)2 + (3 d − p − − d ) + (3 d − p − d − p )2 (cid:19) − (cid:18) (2 d )(2 d + 1)2 + (2 d − p − d − p )2 (cid:19) + (cid:18) (4 d − p − − d + p + 1) + (4 d − p − d − p )2 (cid:19) − (cid:18) (4 d − p − − d ) + (4 d − p − d − p )2 (cid:19) = 4 dp − d − p + 4 d − p − , which is positive for all d ≥
1. Thus the leading coefficient of q ( n ) is positive and q ( n ) is positive for all n ≥ M , for some M . (cid:3) Lemma 7.12.
Let n , d be natural numbers. Then γ (2 n, k, d ) > for all ≤ k ≤ n + ⌊ d/ ⌋ .Proof. By definition we have γ (2 n, k, d ) = ⌊ k/d ⌋ X j =0 ( − j (cid:18) nk − jd (cid:19) . We know that (cid:0) nj (cid:1) is strictly increasing when 0 ≤ j ≤ n , therefore γ (2 n, k, d ) > ≤ k ≤ n . Now, for n ≤ k ≤ n + ⌊ d/ ⌋ we have γ (2 n, k, d ) = (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) + γ (2 n, k − d, d ) . If k ≤ n + ⌊ d/ ⌋ , then k − d ≤ n . Thus γ (2 n, k − d, d ) > N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 34 for all n ≤ k ≤ n + ⌊ d/ ⌋ . In order to finish let us show that (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) ≥ n ≤ k ≤ n + ⌊ d/ ⌋ . If n ≤ k ≤ n + ⌊ d/ ⌋ then n − d ≤ k − d ≤ n + ⌊ d/ ⌋ − d ≤ n − ⌊ d/ ⌋ and n − ⌊ d/ ⌋ ≤ n − k ≤ n. So, for n ≤ k ≤ n + ⌊ d/ ⌋ we have that (cid:18) nn − d (cid:19) ≤ (cid:18) nk − d (cid:19) ≤ (cid:18) nn − ⌊ d/ ⌋ (cid:19) and (cid:18) nn − ⌊ d/ ⌋ (cid:19) ≤ (cid:18) n n − k (cid:19) ≤ (cid:18) nn (cid:19) . Thus, for n ≤ k ≤ n + ⌊ d/ ⌋ (cid:18) nk (cid:19) − (cid:18) nk − d (cid:19) = (cid:18) n n − k (cid:19) − (cid:18) nk − d (cid:19) ≥ (cid:18) nn − ⌊ d/ ⌋ (cid:19) − (cid:18) nn − ⌊ d/ ⌋ (cid:19) = 0 . The result is proved. (cid:3)
Theorem 7.13.
Let d be a natural number. There exists K such that for all n ≥ K we have Ind (1 + z ) n z d > n d Proof.
First let us prove that there exists M such that for all n ≥ M we haveInd (1 + z ) n z d > n d By equation (4.1), we need to show that there exists M such that for all n ≥ M wehave γ (2 n, k, d ) > , for all 0 ≤ k ≤ n + d . By Lemma 7.12 we have that for any nγ (2 n, k, d ) > , for all 0 ≤ k ≤ n + ⌊ d/ ⌋ . (7.2)It remains to show that there exists M such that for all n ≥ Mγ (2 n, n + d − j, d ) > ≤ j ≤ d − ⌊ d/ ⌋ −
1. For all 0 ≤ j ≤ d − ⌊ d/ ⌋ −
1, we have that γ (2 n, n + d − j, d ) = (cid:18) nn + d − j (cid:19) − (cid:18) nn − j (cid:19) + (cid:18) nn − j − d (cid:19) − (cid:18) nn − j − d (cid:19) + γ (2 n, n − j − d, d ) . N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 35
By Lemma 7 .
11, we have that there exist M such that for all n ≥ M (cid:18) nn + d − j (cid:19) − (cid:18) nn − j (cid:19) + (cid:18) nn − j − d (cid:19) − (cid:18) nn − j − d (cid:19) > ≤ j ≤ d − ⌊ d/ ⌋ −
1. Also, by (7.2) we have that γ (2 n, n − j − d, d ) > , for all 0 ≤ j ≤ d − ⌊ d/ ⌋ − n ≥ Mγ (2 n, k, d ) > , for all n + ⌊ d/ ⌋ + 1 ≤ k ≤ n + d . (7.3)Thus from (7.2) and (7.3) we have that for all n ≥ Mγ (2 n, k, d ) > , for all 0 ≤ k ≤ n + d . In other words, for all n ≥ M Ind (1 + z ) n z d ≥ n d + 1 > n d. (7.4)So, for all n ≥ M Ind (1 + z ) n +1 z d ≥ Ind (1 + z ) n z d ≥ n d + 1 > n + 12 + d. (7.5)From (7.4) and (7.5) we have thatInd (1 + z ) n z d > n d for all n ≥ K , where K = 2 M + 1. (cid:3) Now, let us prove the main theorem of this section.
Theorem 7.14.
Let d = ( d , . . . , d m ) , with d j ≥ for some ≤ j ≤ m . Thenthere exists an N such that for all n ≥ N , there are no semi-regular sequences oftype d in B ( n ) .Proof. By Theorem 7.13, for all 0 ≤ i ≤ m there exists an n i such thatInd (1 + z ) n i z d i > n i d i . Set r i = 1 n i (cid:18) Ind (1 + z ) n i z d i − d i (cid:19) > r = min r i . By Theorem 7.10 there exists c such that τ d ( n ) ≥ rn + c, for all n. Suppose that for some j we have that d j ≥
2. Since r > /
2, there exists N suchthat for all n ≥ N τ d ( n ) > n d j . Suppose λ , . . . , λ m is a semi-regular sequence of homogeneous polynomials of de-grees d , . . . , d m in B ( n ) , n ≥ N . Thus, by Theorem 2.4, Ind( λ , . . . , λ m ) = τ d ( n ) > ( n + d j + 2) /
2. Since d j ≥
2, Theorem 4.7 tells us that D ff ( λ j ) ≤ n + d j + 22 . This would imply that D ff ( λ j ) < Ind( λ , . . . , λ m ), but by Theorem 4.8 this is notpossible for a semi-regular sequence. (cid:3) N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 36 Conclusion
Since the introduction of the concept of a semi-regular sequence over F , it hasbeen conjectured that such sequences are in some sense “generic”. However littleconcrete progress has been made towards proving this conjecture. In fact evenin one of the simplest and most important cases, that of quadratic sequences oflength n in n variables, the question of the existence of semi-regular sequences forall n remains open. In this paper we established four results about the existence ofsemi-regular sequences(1) The proportion of sequences of homogeneous elements in n variables thatare semi-regular tends to one as n tends to infinity.(2) A homogeneous element of degree d can only be semi-regular if n ≤ d .(3) We established precisely when the symmetric element σ d,n = X ≤ i < ··· ǫ . This appears to be a hard problem. There do appear to besporadic values of ( n, m ) for which the proportion of semi-regular elements is low(such as ( n, m ) = (10 , , (11 ,
15) and (15 , n, m ) for which the coefficient of (1 + z ) n / (1 + z ) m is zeroat the index. If this phenomenon can occur for arbitrarily large values of n and m ,then it is possible that Conjecture 4 will be false. References [1] M. Bardet, ´Etude des syst`emes alg´ebriques surd´etermin´es. Applications aux codes cor-rectuers et la cryptograhie . PhD thesis, Universit´e Paris V!, D´ecembre 2004.[2] M. Bardet, J.-C. Faug`ere, and B. Salvy,
On the complexity of gr¨obner basis computa-tion of semi-regular overdetermined algebraic equations , in: Proc. ICPSS InternationalConference on Polynomial System Solving Paris, 2004, 71-75.[3] M. Bardet, J-C. Faugre, B. Salvy,
Complexity of Gr¨obner basis computation for Semi-regular Overdetermined sequences over F with solutions in F , in: INRIA ResearchReport 5049, 2003.[4] M. Bardet, J.-C. Faug`ere, B. Salvy and B.-Y. Yang, Asymptotic Expansion of the Degreeof Regularity for Semi-Regular Systems of Equations , MEGA 2005 Sardinia (Italy) ,[5] C. Diem,
Bounded Regularity , J. of Algebra, vol. 423, 2015, pg 1143-1160.[6] J. Ding, T. J. Hodges, V. Kruglov, D. Schmidt, S. Tohaneanu,
Growth of the ideal gener-ated by a multivariate quadratic function over GF(3) , J. of Algebra and Its Applications,12 (2013), 1250219-1 to 23.[7] J. L. Gross,
Combinatorial Methods with Computer Applications , CRC Press, Nov. 16,2007.
N THE EXISTENCE OF SEMI-REGULAR SEQUENCES 37 [8] T. J. Hodges, C. Petit and J. Schlather,
First Fall Degree and Weil Descent , Finite Fieldsand Their Applications 30 (2014), 155-177.[9] V. Kruglov,
Growth of the ideal generated by a quadratic multivariate function , PhD.Thesis, University of Cincinnati, USA, 2010.
E-mail address , Timothy Hodges: [email protected]
University of Cincinnati, Cincinnati, OH 45221-0025, USA
E-mail address , Sergio Molina: [email protected]
University of Cincinnati, Cincinnati, OH 45221-0025, USA
E-mail address , Jacob Schlather: [email protected]@gmail.com