Properties of lexsegment ideals

Properties of lexsegment ideals

Viviana Ene and Anda Olteanu and Loredana Sorrenti Faculty of Mathematics and Computer Science, Ovidius University, Bd. Mamaia 124, 900527 Constanta, Romania, vivian@univ-ovidius.ro Faculty of Mathematics and Computer Science, Ovidius University, Bd. Mamaia 124, 900527 Constanta, Romania, olteanuandageorgiana@gmail.com DIMET University of Reggio Calabria, Faculty of Engineering, via Graziella (Feo di Vito), 89100 Reggio Calabria, Italy loredana.sorrenti@unirc.it.
Abstract.

We show that any lexsegment ideal with linear resolution has linear quotients with respect to a suitable ordering of its minimal monomial generators. For completely lexsegment ideals with linear resolution we show that the decomposition function is regular. For arbitrary lexsegment ideals we compute the depth and the dimension. As application we characterize the Cohen-Macaulay lexsegment ideals.

Keywords: lexsegment ideals, linear resolution, linear quotients, Cohen-Macaulay ideals.

MSC: 13D02, 13D05, 13C15.

The first author was supported by the grant CEX 05-D11-11/2005. The second author was supported by the CNCSIS grant TD 507/2007.

Introduction

Let be the polynomial ring in variables over a field . We order lexicographically the monomials of such that . Let be an integer and the set of monomials of degree . For two monomials , with , the set

is called a lexsegment. A lexsegment ideal in is a monomial ideal of which is generated by a lexsegment. Lexsegment ideals have been introduced by Hulett and Martin [HM]. Arbitrary lexsegment ideals have been studied by A. Aramova, E. De Negri, and J. Herzog in [ADH] and [DH]. They characterized the lexsegment ideals which have linear resolutions.

In this paper we show that any lexsegment ideal with linear resolution has linear quotients with respect to a suitable order of the generators.

Let be a monomial ideal and its minimal monomial set of generators. has linear quotients if there exists an ordering of the elements of such that for all , the colon ideals are generated by a subset of .

Lexsegment ideals which have linear quotients with respect to the lexicographical order of the generators have been characterized by the third author in [S].

In Section 1 we show that any completely lexsegment ideal with linear resolution has linear quotients with respect to the following order of the generators. Given two monomials of degree in , and we set if or and .

Let which define the completely lexsegment ideal with linear resolution. If , where , we show that has linear quotients with respect to this ordering of the generators. The non-completely lexsegment ideal will be separately studied in Section .

For the completely lexsegment ideals with linear resolution it will turn out that their decomposition function with respect to the ordering is regular . Therefore, one may apply the procedure developed in [HT] to get the explicit resolutions for this class of ideals.

In the last section of our paper we study the depth and the dimension of lexsegment ideals. Our results show that one may compute these invariants just looking at the ends of the lexsegment. As an application, we characterize the Cohen-Macaulay lexsegment ideals.

We acknowledge the support provided by the Computer Algebra Systems CoCoA [Co] and Singular [GPS] for the extensive experiments which helped us to obtain some of the results of this work.

1. Completely lexsegment ideals with linear resolutions

In the theory of Hilbert functions or in extremal combinatorics usually one considers initial lexsegment ideals, that is ideals generated by an initial lexsegment . Initial lexsegment ideals are stable in the sense of Eliahou and Kervaire ([EH], [AH]) and they have linear quotients with respect to lexicographical order [S, Proposition 2.1].

One may also define the final lexsegment . Final lexsegment ideals are generated by final lexsegments. They are also stable in the sense of Eliahou and Kervaire with respect to . Therefore they have linear quotients.

Throughout this paper we use the following notations. If is a monomial of , we denote by the exponent of the variable in , that is , . Also, we will denote

Hulett and Martin call a lexsegment completely lexsegment if all the iterated shadows of are again lexsegments. We recall that the shadow of a set of monomials is the set . The -th shadow is recursively defined as . The initial lexsegments have the property that their shadow is again an initial lexsegment, a fact which is not true for arbitrary lexsegments. An ideal spanned by a completely lexsegment is called a completely lexsegment ideal. All the completely lexsegment ideals with linear resolution are determined in [ADH]:

Theorem 1.1.

[ADH] Let , be monomials of degree with , and let be a completely lexsegment ideal. Then has linear resolution if and only if one of the following conditions holds:

  • for some

  • and for the largest monomial of degree one has

Theorem 1.2.

Let with and be monomials of degree with and let be a completely lexsegment ideal. Then has linear resolution if and only if has linear quotients.

Proof.

We have to prove that if has linear resolution then has linear quotients, since the other implication is known [H]. By Theorem 1.1, since has linear resolution, one of the conditions (a), (b), (c) holds.

We define on the set of the monomials of degree from the following total order: for

we set

Let

We will prove that has linear quotients with respect to this ordering of the generators.

Assume that satisfy the condition (a) and (the case is trivial). Then is isomorphic as –module with the ideal generated by the final lexsegment and the ordering of its minimal generators coincides with the lexicographical ordering The ideal is the initial ideal in defined by which has linear quotients with respect to Hence has linear quotients with respect to since it is obvious that the extension in the ring of a monomial ideal with linear quotients in has linear quotients too.

Next we assume that satisfy the condition (b) or (c).

By definition, has linear quotients with respect to the monomial generators if the colon ideals are generated by variables for all that is for all there exists an integer and an integer such that

In other words, for any we have to find a monomial such that

()

Let us fix and such that By the definition of the ordering we must have

Case : Let . One may find an integer , , such that for all and since, otherwise, which is impossible. We obviously have . If one may take which satisfies the condition (1.0) since the inequalities hold, and we will show that . This will imply that , hence .

The inequality is obviously fulfilled if or if and at least one of the inequalities for , is strict. If and for all , comparing the degrees of and it results It follows that that is This implies that and , that is

From now on, in the Case , we may assume that . We will show that at least one of the following monomials:

belongs to . It is clear that both monomials are strictly less than with respect to the ordering Therefore one of the monomials will satisfy the condition (1.0).

The following inequalities are fulfilled:

Let us assume, by contradiction, that and . Comparing the exponents of the variable , we obtain . Since the ideal generated by has linear resolution, we must have . Let be the largest monomial of degree such that . Then, by our assumption on , we also have the inequality .

Now we need the following

Lemma 1.3.

Let be two monomials of degree . If then .

Proof.

Let . Then there exists such that and . It is clear that . Comparing the degrees of and we get

If and , the required inequality is obvious.

Let and . Let us suppose, by contradiction, that This implies that , and, since , we get . Looking at the degree of we obtain that is . It follows that and , contradiction. ∎

Going back to the proof of our theorem, we apply the above lemma for the monomials and and we obtain which implies that By using condition (c) in the Theorem 1.1 it follows that On the other hand, Therefore, it results , which contradicts our assumption on .

Consequently, we have or , which proves that at least one of the monomials belongs to .

Case : Let and . Then there exists , , such that , for all and . If , then, looking at the degrees of and , we get contradiction. Therefore, . We proceed in a similar way as in the previous case. Namely, exactly as in the Case it results that at least one of the following two monomials belongs to . It is clear that both monomials are strictly less than with respect to the order . ∎

Example 1.4.

Let . We consider the monomials: and , and let be the monomial ideal generated by . The minimal system of generators of the ideal is

Since verifies the condition (c) in Theorem 1.1, it follows that is a completely lexsegment ideal with linear resolution. We denote the monomials from as follows: , so The colon ideal is not generated by a subset of . This shows that is not with linear quotients with respect to lexicographical order.

We consider now the order and check by direct computation that has linear quotients. We label the monomials from as follows: , so . Then .

We further study the decomposition function of a completely lexsegment ideal with linear resolution. The decomposition function of a monomial ideal was introduced by J. Herzog and Y. Takayama in [HT].

We recall the following notation. If is a monomial ideal with linear quotients with respect to the ordering of its minimal generators, then we denote

for .

Definition 1.5.

[HT] Let be a monomial ideal with linear quotients with respect to the sequence of minimal monomial generators and set , for . Let be the set of all monomials in . The map defined as: , where is the smallest number such that , is called the decomposition function of .

We say that the decomposition function is regular if for all and .

We show in the sequel that completely lexsegment ideals which have linear quotients with respect to have also regular decomposition functions.

In order to do this, we need some preparatory notations and results.

For an arbitrary lexsegment with the elements ordered by , we denote by , the ideal generated by all the monomials with . will be the ideal generated by all the monomials with .

Lemma 1.6.

Let be a lexsegment ideal which has linear quotients with respect to the order of the generators. Then, for any , .

Proof.

Let us assume that , that is . It follows that there exists , , and a variable such that . Obviously, we have . But this equality shows that , which is impossible since . ∎

Lemma 1.7.

Let be a completely lexsegment ideal which has linear quotients with respect to the ordering of the generators. If and , then

Proof.

Let and .

In the first place we consider

Since, by Lemma 1.6, we have , the above inequality shows that . We have to show that , that is . It is clear that , hence . Let such that . We have to show that . Let such that , for some variable . Then by the definition of our ordering . This implies that .

Now we have to consider the second inequality,

(1.1)

Since , we have , that is there exists , , and a variable , such that

(1.2)

If , then , contradiction. Hence . We also note that since , thus . The following inequalities hold:

(1.3)

If , we obviously get . Let . From the inequality (1.1) we obtain .

If then and by Theorem 1.1. Since , by using Lemma 1.3, we have the last inequality being true by Lemma 1.6. Therefore,

If then the condition (c) in Theorem 1.1 holds. Let be the largest monomial with respect to the lexicographical order such that . Since by hypothesis, we also have . By Lemma 1.3 we obtain Next we apply the condition (c) from Theorem 1.1 and get the following inequalities:

(1.4)

From the equality (1.2) we have . As and the inequality gives that is which implies that . This shows that . Now looking at the inequalities (1.4), we have

(1.5)

From (1.5) and (1.3) we obtain .

It remains to show that Let We obviously have By the choice of we have

for some variable

If we get which is impossible since Therefore, Then so . It follows that . If we have which implies that contradiction. Therefore and, moreover, , the inequality being true by the definition of the ordering . This yields Therefore we have proved that . ∎

After this preparation, we prove the following

Theorem 1.8.

Let , with , and be a completely lexsegment ideal which has linear resolution. Then the decomposition function associated to the ordering of the generators from is regular.

Proof.

Let and . We have to show that .

Let . In order to prove that , that is , we will consider the following two cases:

Case 1: Let . By Lemma 1.7, . Since , we have

so there exists , , and a variable , such that that is

(1.6)

By Lemma 1.6, and, since , we have . Note also that since . If then and .

Now let . If , we have which implies that . The following inequalities hold: the first one being true since , so . These inequalities show that . But we also have , hence .

To finish this case we only need to treat the case . We are going to show that at least one of the monomials or belongs to In any case this will lead to the conclusion that by using (1.6).

From the equality (1.6), we have , hence , and . Since and , we get

(1.7)

which gives

If the inequality

(1.8)

holds, then we get . We also note that and (by (1.7)). Therefore and we may write This implies that .

Now we look at the monomial for which we have , so If the inequality

(1.9)

holds, we obtain . Obviously we have . By using (1.6), we may write which shows that .

To finish the proof in the Case we need to consider the situation when both inequalities (1.8) and (1.9) fail. Hence, let

We will show that this inequalities cannot hold simultaneously. Comparing the exponents of in the monomials involved in the above inequalities, we obtain . Since, by hypothesis, we have On the other hand, implies that So and satisfies the condition (c) in Theorem 1.1. Let, as usually, be the largest monomial with respect to the lexicographical order such that .

Since , we have . By Lemma 1.3 and using the condition , we obtain: But our assumption was that Therefore, combining the last two inequalities, after cancellation, one obtains that This leads to the inequality and, since , we get , which is impossible.

Case 2: Let