Complete intersection vanishing ideals on degenerate tori over finite fields

We study the complete intersection property and the algebraic invariants (index of regularity, degree) of vanishing ideals on degenerate tori over finite fields. We establish a correspondence between vanishing ideals and toric ideals associated to numerical semigroups. This correspondence is shown to preserve the complete intersection property, and allows us to use some available algorithms to determine whether a given vanishing ideal is a complete intersection. We give formulae for the degree, and for the index of regularity of a complete intersection in terms of the Frobenius number and the generators of a numerical semigroup.


Introduction
Let K = F q be a finite field with q elements and let v 1 , . . ., v n be a sequence of positive integers.Consider the degenerate projective torus 1 , . . ., x vn n )] | x i ∈ K * for all i} ⊂ P n−1 , parameterized by the monomials x v 1 1 , . . ., x vn n , where K * = F q \ {0} and P n−1 is a projective space over the field K.This set is a multiplicative group under componentwise multiplication.If v i = 1 for all i, X is just a projective torus.
Let S = K[t 1 , . . ., t n ] = ⊕ ∞ d=0 S d be a polynomial ring over the field K with the standard grading.Recall that the vanishing ideal of X, denoted by I(X), is the ideal of S generated by the homogeneous polynomials that vanish on X.To study I(X), we will associate with this a semigroup S and a toric ideal P that depend on v 1 , . . ., v n and the multiplicative group of F q .
In what follows β denotes a generator of the cyclic group (K * , • ), d i denotes o(β v i ), the order of β v i for i = 1, . . ., n, and S denotes the semigroup Nd 1 + • • • + Nd n .If d 1 , . . ., d n are relatively prime, S is called a numerical semigroup.As is seen in Section 3, the algebra of I(X) is closely related to the algebra of the toric ideal of the semigroup ring , where K[y 1 ] is a polynomial ring.Recall that the toric ideal of K[S], denoted by P , is the kernel of the following epimorphism of K-algebras . ., y dn 1 ).Thus, S/P ≃ K[S].Since K[y 1 ] is integral over K[S] we have ht(P ) = n − 1.The ideal P is graded if one gives degree d i to variable t i .For n = 3, the first non-trivial case, this type of toric ideals were studied by Herzog [14].For n ≥ 4, these toric ideals have been studied by many authors [2,4,5,8,9,23].
In this paper, we relate some of the algebraic invariants and properties of I(X) with those of P and S. We are especially interested in the degree and the regularity index, and in the complete intersection property.
The most well known properties that P and I(X) have in common is that both are Cohen-Macaulay graded lattice ideals of dimension 1 [14,19].
The contents of this paper are as follows.In Section 2, we introduce some of the notions that will be needed throughout the paper.
A key fact that allows us to link the properties of P and I(X) is that the homogeneous lattices of these ideals are closely related (Proposition 3.2).If g 1 , . . ., g m is a set of generators for P consisting of binomials, then h 1 , . . ., h m is a set of generators for I(X), where h k is the binomial obtained from g k after substituting t i by t d i i for i = 1, . . ., n (Proposition 3.3).As a consequence if n = 3, then I(X) is minimally generated by 2 or 3 binomials (Corollary 3.4).If I(X) is a complete intersection, one of our main results show that a minimal generating set for I(X) consisting of binomials corresponds to a minimal generating set for P consisting of binomials and viceversa (Theorem 3.6).As a consequence I(X) is a complete intersection if and only if P is a complete intersection (Corollary 3.7).
We show a formula for the degree of S/I(X) (Lemma 3.11).The Frobenius number of a numerical semigroup is the largest integer not in the semigroup.For complete intersections, we give a formula that relates the index of regularity of S/I(X) with the Frobenius number of the numerical semigroup generated by o(β rv 1 ), . . ., o(β rvn ), where r is the greatest common divisor of d 1 , . . ., d n (Corollary 3.13).
The Frobenius number occurs in many branches of mathematics and is one of the most studied invariants in the theory of semigroups.A great deal of effort has been directed at the effective computation of this number, see the monograph of Ramírez-Alfonsín [18].
The complete intersection property of P has been nicely characterized, using the notion of a binary tree [2,4] and the notion of suites distinguées [5].For n = 3, there is a classical result of [14] showing an algorithm to construct a generating set for P .Thus, using our results, one can obtain various classifications of the complete intersection property of I(X).Furthermore, in [2] an effective algorithm is given to determine whether P is a complete intersection.This algorithm has been implemented in the distributed library cimonom.lib[3] of Singular [11].Thus, using our results, one can use this algorithm to determine whether I(X) is a complete intersection (see Example 3.14).If I(X) is a complete intersection, this algorithm returns the generators of P and the Frobenius number.As a byproduct, we can construct interesting examples of complete intersection vanishing ideals (see Example 3.16).
We show how to compute the vanishing ideal I(X) using the notion of saturation of an ideal with respect to a polynomial (Proposition 3.18).
It is worth mentioning that our results could be applied to coding theory.The algebraic invariants of S/I(X) occur in algebraic coding theory as we now briefly explain.An evaluation code over X is a linear code obtained by evaluating the linear space of homogeneous d-forms of S on the set of points X ⊂ P n−1 .A linear code obtained in this way, denoted by C X (d), has length |X| and dimension dim K (S/I(X)) d .The computation of the index of regularity of S/I(X) is important for applications to coding theory: for d ≥ reg S/I(X) the code C X (d) coincides with the underlying vector space K |X| and has, accordingly, minimum distance equal to 1. Thus, potentially good codes C X (d) can occur only if 1 ≤ d < reg(S/I(X)).The length, dimension and minimum distance of evaluation codes C X (d) arising from complete intersections have been studied in [6,10,12,15,16,20,21].
For all unexplained terminology and additional information, we refer to [7] (for the theory of lattice ideals), [22,24] (for commutative algebra and the theory of Hilbert functions).

Preliminaries
We continue to use the notation and definitions used in Section 1.In this section, we introduce the notions of degree and regularity via Hilbert functions, and the notion of a lattice ideal.
The Hilbert function of S/I(X) is given by H X (d) := dim K (S d /I(X) ∩ S d ), and the Krulldimension of S/I(X) is denoted by dim(S/I(X)).The unique polynomial for d ≥ |X| − 1.Thus, |X| is the degree of S/I(X).
Definition 2.1.The index of regularity of S/I(X), denoted by reg(S/I(X)), is the least integer The index of regularity of S/I(X) is equal to the Castelnuovo Mumford regularity of S/I(X) because this ring is Cohen-Macaulay of dimension 1.
Remark 2.2.The Hilbert series of S/I(X) can be written as where h 0 , . . ., h r are positive integers.The number r is the regularity index of S/I(X) and h 0 + • • • + h r is the degree of S/I(X) (see [24,Corollary 4.1.12]).The same observation holds for any graded Cohen-Macaulay ideal I ⊂ S of height n − 1.
Recall that a binomial in S is a polynomial of the form t a − t b , where a, b ∈ N n and where, if The vector c can be written as c = c + − c − , where c + and c − are two nonnegative vectors with disjoint support.If for some lattice L in Z n .A lattice L is called homogeneous if there is an integral vector ω with positive entries such that ω, a = 0 for a ∈ L. Definition 2.4.An ideal I ⊂ S is called a complete intersection if there exists g 1 , . . ., g m such that I = (g 1 , . . ., g m ), where m is the height of I.
Remark 2.5.A graded binomial ideal I ⊂ S is a complete intersection if and only if I is generated by a set of homogeneous binomials g 1 , . . ., g m , with m = ht(I), and any such set of homogeneous generators is already a regular sequence (see [24, Proposition 1.3.17,Lemma 1.3.18]).
Lemma 2.6.Let S = K[t 1 , . . ., t n ] be a polynomial ring with the standard grading.If I is a graded ideal of S generated by a homogeneous regular sequence f 1 , . . ., f n−1 , then Proof.We set δ i = deg(f i ).By [24, p. 104], the Hilbert series of S/I is given by (2.1) Thus, by Remark 2.2, reg(

Complete intersections and algebraic invariants
We continue to use the notation and definitions used in Sections 1 and 2. In this section, we study vanishing ideals over degenerate projective tori.We study the complete intersection property and the algebraic invariants of vanishing ideals.We will establish a correspondence between vanishing ideals and toric ideals associated to semigroups of N.
Let D be the non-singular matrix D = diag(d 1 , . . ., d n ).Consider the homomorphisms of Z-modules: There are two homogeneous lattices that will play a role here: The map D induces a Z-isomorphism between L 1 and L. It is well known [24] that the toric ideal P is the lattice ideal of L 1 .Below, we show that I(X) is the lattice ideal of L.
Lemma 3.1.The map t a − t b → t D(a) − t D(b) induces a bijection between the binomials t a − t b of P whose terms t a , t b have disjoint support and the binomials is homogeneous in the standard grading of S. As (β v i ) d i = 1 for all i, it is seen that g vanishes at all points of X.Hence, g ∈ I(X) and the map is well defined.
The map is clearly injective.To show that the map is onto, take a binomial vanishes at all points of X.Hence, since the order of Proof.As mentioned above, the first equality is well known [24].Since I(X) is a lattice ideal [19], it is generated by binomials of the form t a + − t a − (this follows using that t i is a non-zero divisor of S/I(X) for all i).To show the second equality, take t a + − t a − in I(X).Then, by Lemma 3.1, a + − a − ∈ L and t a + − t a − is in I(L).Thus, I(X) ⊂ I(L).Conversely, take f = t a + − t a − in I(L) with a + − a − in L.Then, there is c ∈ L 1 such that a + − a − = D(c + − c − ).Then, t c + − t c − is in P and maps, under the map of Lemma 3.1, to f .Thus, f ∈ I(X).This proves that I(L) ⊂ I(X).
Proof.We set Notice that h i is equal to g i (t d 1 , . . ., t dn ), the evaluation of g i at (t d 1 1 , . . ., t dn n ).By Lemma 3.1, one has the inclusion (h 1 , . . ., h m ) ⊂ I(X).To show the reverse inclusion take a binomial 0 = f ∈ I(X).We may assume that f = t a + − t a − .Then, by Lemma 3.1, there is g . By hypothesis we can write g = m i=1 f i g i for some f 1 , . . ., f m in S.Then, evaluating both sides of this equality at (t d 1 1 , . . ., t dn n ), we get where is minimally generated by at most 3 binomials.
Proof.By a classical theorem of Herzog [14], P is generated by at most 3 binomials.Hence, by Proposition 3.3, I(X) is generated by at most 3 binomials.
Given a subset I ⊂ S, its variety, denoted by V (I), is the set of all a ∈ A n K such that f (a) = 0 for all f ∈ I, where A n K is the affine space over K. Given a binomial g = t a − t b , we set g = a − b.If B is a subset of Z n , B denotes the subgroup of Z n generated by B. Proposition 3.5.[4, Proposition 2.5] Let B = {g 1 , . . ., g n−1 } be a set of binomials in P .Then, P = (B) if and only if the following two conditions hold: We come to the main result of this section.Theorem 3.6.(a) If I(X) is a complete intersection generated by binomials h 1 , . . ., h n−1 , then P is a complete intersection generated by binomials g 1 , . . ., g n−1 such that h i is equal to g i (t d 1 1 , . . ., t dn n ) for all i.(b) If P is a complete intersection generated by binomials g 1 , . . ., g n−1 , then I(X) is a complete intersection generated by binomials h 1 , . . ., h n−1 , where h i is equal to g i (t d 1 1 , . . ., t dn n ) for all i.
Proof.(a) Since t k is a non-zero divisor of S/I(X) for all k, it is not hard to see that the monomials of h i have disjoint support for all i, i.e., we can write h i = t a + i −t a − i for i = 1, . . ., n−1.We claim that the following two conditions hold.
(i) L = a 1 , . . ., a n−1 , where a i = a + i − a − i and L is the lattice that defines I(X).
As I(X) is generated by h 1 , . . ., h n−1 , by [17, Lemma 2.5], condition (i) holds.The binomial t q−1 i − t q−1 n is in I(X) for all i because F * q is a group of order q − 1.Thus, V (I(X), t i ) = {0} for all i.From the equality (h 1 , . . ., h n−1 , t i ) = (I(X), t i ), we get Thus, (ii) holds.This completes the proof of the claim.
By (i) and Proposition 3.2, there are b 1 , . . ., b n−1 in L 1 = ker(ψ) such that a i = D(b i ) for all i.Accordingly a Clearly, all the g i 's are in P and h i is equal to g i (t d 1 1 , . . ., t dn n ) for all i.Next, we prove that P is generated by g 1 , . . ., g n−1 .By Proposition 3.5 it suffices to show that the following two conditions hold: First we show (i ′ ).Since b 1 , . . ., b n−1 are in L 1 , we need only show the inclusion "⊂".Take γ ∈ ker(ψ), then D(γ) ∈ L, and by (i) it follows that γ ∈ b 1 , . . ., b n−1 .
Next we show (ii ′ ).For simplicity of notation, we may assume that i = n.Take c in the variety V (g 1 , . . ., g n−1 , t n ) and write c = (c 1 , . . ., c n ).Then, c n = 0 and g i (c) = c b + i − c b − i = 0 for all i, were c b + i means to evaluate the monomial t b + i at the point c.Let i be a fixed but arbitrary integer in {1, . . ., n − 1}.We can write Then We claim that h i (c v 1 1 , . . ., c vn n ) = 0. To show this we consider two cases.Case (I): b + in > 0.Then, as Hence, there is j such that b − ij > 0 and c j = 0. Thus, by Eq. (3.1), Hence, there is k such that c k = 0 and b + ik > 0. Thus, by Eq. (3.1), h i (c v 1 1 , . . ., c vn n ) = 0. We may now assume that c j = 0 if b + ij > 0, and Let β be a generator of the cyclic group (F * , • ).Any c j = 0 has the form c j = β j ℓ .Thus, using that (β v j ) d j = 1, we get that (c Hence, by Eq. (3.1), h i (c v 1 1 , . . ., c vn n ) = 0, as required.This completes the proof of the claim.
. By (ii), the point c ′ es zero.Hence, c = 0 as required.This completes the proof of (ii ′ ).Hence, P is a complete intersection generated by g 1 , . . ., g n−1 .
(b) It follows from Proposition 3.3.
Using the notion of a binary tree, a criterion for complete intersection toric ideals of affine monomial curves is given in [4].In [2] an effective algorithm is given to determine whether P is a complete intersection.If P is a complete intersection, this algorithm returns the generators of P and the Frobenius number.
In our situation, the next result allows us to: (A) use the results of [4,5,14] to give criteria for complete intersection vanishing ideals over a finite field, (B) use the effective algorithms of [2] to recognize complete intersection vanishing ideals over finite fields and to compute its invariants (see Example 3.14).
Corollary 3.7.I(X) is a complete intersection if and only if P is a complete intersection.
Proof.Assume that I(X) is a complete intersection.By Remark 2.5, there are binomials h 1 , . . ., h n−1 that generate I(X).Hence, P is a complete intersection by Theorem 3.6.The converse follows by similar reasons.
Proof.It follows readily by recalling that o( In what follows X ′ will denote the degenerate torus in P n−1 parameterized by , where v ′ i = rv i and r = gcd(d 1 , . . ., d n ).Below, we relate I(X) and I(X ′ ).Proposition 3.9.The vanishing ideal I(X) is a complete intersection if and only if I(X ′ ) is a complete intersection.
Proof.Let P and P ′ be the toric ideals of K[y d 1 1 , . . ., y dn 1 ] and K[y It is not hard to see that P = P ′ .Then, by Theorem 3.6, P is a complete intersection if and only if I(X) is a complete intersection and P ′ is a complete intersection if and only if I(X ′ ) is a complete intersection.Thus, I(X) is a complete intersection if and only if I(X ′ ) is a complete intersection.Proof.Let S i = β v i be the cyclic group generated by β v i .The set X * is equal to the cartesian product S 1 × • • • × S n .Hence, to show the first equality, it suffices to recall that |S i | is o(β v i ), the order of β v i .Notice that any element of X * can be written as ((β i 1 ) v 1 , . . ., (β in ) vn ) for some integers i 1 , . . ., i n .The kernel of the epimorphism of groups Theorem 4,p. 4]).Thus, the second equality follows.Definition 3.12.If S is a numerical semigroup of N, the Frobenius number of S, denoted by g(S), is the largest integer not in S. Proof.Part (i) follows at once from Lemma 3.11.Next, we prove (ii).Let P and P ′ be as in the proof of Proposition 3.9.With the notation above, by Lemma 3.8, we get that d i = rd ′ i for all i.The toric ideals P and P ′ are equal but they are graded differently.Recall that P and P ′ are graded with respect to the gradings induced by assigning deg(t i ) = d i and deg(t i ) = d ′ i for all i, respectively.Let g 1 , . . ., g n−1 be a generating set of P = P ′ consisting of binomials.Then, by Theorem 3.6, I(X) is generated by h 1 , . . ., h n−1 , where h i is g i (t d 1 1 , . . ., t dn n ) for all i.Accordingly,

Consider the semigroup S
As P ′ is a complete intersection generated by g 1 , . . ., g n−1 and deg P ′ (g i ) = D ′ i for all i, using [4, Remark 4.5], we get Therefore, using the equality reg S/I(X) = n−1 i=1 (D i − 1) (see Lemma 2.6), the formula for the regularity follows.
Example 3.14.To illustrate how to use the algorithm of [2] we consider the degenerate torus X, over the field F q , parameterized by x v 1 1 , . . ., x v 5 5 , where v 1 = 1500, v 2 = 1000, v 3 = 432, v 4 = 360, v 5 = 240, and q = 54001.In this case, one has Using [2, Algorithm CI, p. 981], we get that P is a complete intersection generated by the binomials g 1 = t , the index of regularity and degree of S/I(X) are 1379 and 8201250000, respectively.
The next example is interesting because if one chooses v 1 , . . ., v n at random, it is likely that I(X) will be generated by binomials of the form t m i − t m j .Example 3.15.Let F q be the field with q = 211 elements.Consider the sequence v 1 = 42, v 2 = 35, v 3 = 30.In this case, one has d 1 = 5, d 2 = 6, d 3 = 7.By a well known result of Herzog [14], one has P = (t do not generate I(X).
The next example was found using Theorem 3.6.Without using this theorem it is very difficult to construct examples of complete intersection vanishing ideals not generated by binomials of the form t m i − t m j .
The computation of the vanishing ideal.In this part we show how to compute the vanishing ideal using the notion of saturation of an ideal with respect to a polynomial.
The next lemma is easy to show.
Lemma 3.17.If c ij := lcm{d i , d j } = lcm{o(β v i ), o(β v j )}, then t The set of toric relations T = {t 1 ≤ i, j ≤ n} do not generate I(X), as is seen in Example 3.15.If v i = 1 for all i, then c ij = q − 1 for all i, j and I(X) is generated by T .
For an ideal I ⊂ S and a polynomial h ∈ S the saturation of I with respect to h is the ideal Thus, the claim follows from the equality L = D(L 1 ).The inclusion "⊃" follows readily using that t i is a non-zero divisor of S/I(X) for all i because I(X) is a lattice ideal containing I ′ (see Lemma 3.17).To show the inclusion "⊂", take a binomial f = t a − t b ∈ I(X).By Proposition 3.2, I(X) = I(L).Thus, a − b ∈ L. Using the previous claim and [17, Lemma 2.3], there is δ ∈ N n such that t δ f ∈ I ′ .Hence, f ∈ (I ′ : (t 1 • • • t n ) ∞ ).
homogeneous, one has |a ′ | = |b ′ |.It follows readily that c ∈ L 1 and a ′ − b ′ = D(c).We can write c = c + − c − .As a ′ and b ′ have disjoint support, we get a ′ = D(c + ) and b ′ = D(c − ).Thus, the binomial f = t c + − t c − is in P and maps to t a ′ − t b ′ .Proposition 3.2.P = I(L 1 ) and I(X) = I(L).