Unmixedness and arithmetic properties of matroidal ideals

Let R=k[x1,…,xn]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$R=k[x_1,\ldots ,x_n]$$\end{document} be the polynomial ring in n variables over a field k and let I be a matroidal ideal of degree d. In this paper, we study the unmixedness properties and the arithmetical rank of I. Moreover, we show that ara(I)=n-d+1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ara(I)=n-d+1$$\end{document}. This answers the conjecture made by Chiang-Hsieh (Comm Algebra 38:944–952, 2010, Conjecture).

a problem the classification of all unmixed polymatroidal ideals. After that, Vlǎdoiu [16] studied the unmixed polymatroidal ideals and he showed that an ideal of Veronese type is unmixed if and only if it is Cohen-Macaulay. The arithmetical rank of I is defined as follows: ara(I) := min{t ∈ N : there exist a 1 , . . . , a t ∈ R such that (a 1 , . . . , a t ) = √ I}.
Note that ideals with the same radical have the same arithmetical rank. It is well-known that ht(I) ≤ ara(I) ≤ µ(I), where µ(I) is the minimal number of generators of I and ht is the height of I. If I is a square-free monomial ideal in R, then by using [12] and the Auslander-Buchsbaum formula, we have the following well-known inequalities: ht(I) ≤ pd(R/I) ≤ ara(I) ≤ µ(I), where pd is the projective dimension of I. In particular, I is Cohen-Macaulay if and only if ht(I) = pd(R/I). An ideal I is called a set-theoretic complete intersection when ara(I) = ht(I).
Kimura et al. [11] raised the following question: Let I be a square-free monomial ideal in R. When does ara(I) = pd(R/I) hold? In particular, suppose that R/I is Cohen-Macaulay. When is I a set-theoretic complete intersection? A considerable number of studies have been made on this question (see, for example, [3,4,11], and [5]). The above question does not always hold as was shown by [17]. Chiang-Hsieh [5] proved that if I is a matroidal ideal of degree d, then ara(I) = pd(R/I) provided that one of the following conditions holds: In the end of her article, she proposed the following conjecture: Conjecture: Let I be a full-supported matroidal ideal of degree d. Then ara(I) = n − d + 1.
The main purpose of this note is to study the unmixed properties and arithmetical rank of matroidal ideals. Also, we give an affirmative answer to the above conjecture. For any unexplained notion or terminology, we refer the reader to [8,15]. Several explicit examples were performed with help of the computer algebra systems Macaulay2 [7].
1 · · · x an n , a i = 0}. Throughout this paper, we assume that all polymatroidal ideals are full-supported, that is, supp(I) = [n].
Proof. Since m / ∈ Ass(R/I), we have I = (I : m). Therefore there exists m ≤ n . Thus (ii) holds. If xy ∈ I, then by definition of S i , it is clear that x ∈ S i and y ∈ S j for i = j. Conversely, let x ∈ S i and y ∈ S j for i = j. Then xx i , yx j / ∈ I for i = j and so (I : In the following, we assume that m is the number of S i as we use in Theorem 1.2. and |S i | = n − |G(I : x i )| for all i, we have n < m(n − ht(I)) and this is a contradiction. Therefore I is unmixed. If n is a prime number, then ht(I) = n − 1. Since depth(R/I) > 0, by using the Auslander-Buchsbaum formula, it follows that ht(I) = pd(R/I). Therefore I is Cohen-Macaulay.
It is know that for n = 6, there is a counter-example which is an unmixed matroidal ideal but which is not Cohen-Macaulay. For n = 4, we consider I = (x 1 x 3 , x 1 x 4 , x 2 x 3 , x 2 x 4 ) such that I is an unmixed matroidal ideal but it is not Cohen-Macaulay.

Arithmetical rank of matroidal ideals.
We start this section by the following lemma which is proved by Schmitt and Vogel [13]. Lemma 2.1. Let P be a finite subset of R and let P 0 , P 1 , . . . , P r be subsets of P such that the following conditions hold: (a) ∪ r i=0 P i = P ; (b) P 0 has exactly one element; (c) If p and p are different elements of P i (0 < i ≤ r), then there is an integer j with 0 ≤ j < i and an element p ∈ P j such that pp ∈ (p ).
Let q i = p∈Pi p. Then (P ) = (q 0 , . . . , q r ).  The following result answers the conjecture made by H.J. Chiang-Hsieh in [5,Conjecture]. Proof. If I is a set-theoretic complete intersection, then I is Cohen-Macaulay and so by [9,Theorem 4.2], I is square-free Veronese. Conversely, let I be square-free Veronese. Then I is Cohen-Macaulay and so ht(I) = pd(R/I). By Theorem 2.3, we have ht(I) = ara(I), as required.