Get 20M+ Full-Text Papers For Less Than $1.50/day. Start a 14-Day Trial for You or Your Team.

Learn More →

A Sequence of Cohen–Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially Deep Flaws

A Sequence of Cohen–Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially... Article A Sequence of Cohen–Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially Deep Flaws Mitsuhiro Miyazaki Department of Mathematics, Kyoto University of Education, 1 Fujinomori, Fukakusa, Fushimi-ku, Kyoto 612-8522, Japan; g53448@kyokyo-u.ac.jp Abstract: Let K be a field. In this paper, we construct a sequence of Cohen–Macaulay standard graded K-domains whose h-vectors are non-flawless and have exponentially deep flaws. Keywords: flawless O-sequence; stable-set polytope; t-perfect; Ehrhart ring MSC: 13H10; 52B20; 05E40; 05C17 1. Introduction In 1989, Hibi [1] made several conjectures on the h-vectors of Cohen–Macaulay standard graded algebras over a field. In particular, he conjectured that the h-vector of a standard graded Cohen–Macaulay domain is flawless ([1], Conjecture 1.4). The h- vector (h , h , . . . , h ), h 6= 0, of a Cohen–Macaulay standard graded algebra is flawless 0 1 s s if h  h for 0  i  bs/2c and h  h for 1  i  bs/2c. Niesi and Robbiano [2] i si i1 i disproved this conjecture by constructing a Cohen–Macaulay standard graded domain whose h-vector is (1, 3, 5, 4, 4, 1). Further, Hibi and Tsuchiya [3] showed that the Ehrhart rings of the stable-set polytopes of cycle graphs of length 9 and 11 have non-flawless h-vectors by computation using the software Normaliz [4]. Moreover, the present author showed that the Ehrhart ring of the stable-set polytope of any odd cycle graph whose Citation: Miyazaki, M. A Sequence of length is at least 9 has non-flawless h-vectors ([5], Theorem 5.2) by proving the conjecture Cohen–Macaulay Standard Graded of Hibi and Tsuchiya ([3], Conjecture 1). Domains Whose h-Vectors Have However, these examples have the slightest flaws, i.e., there exists i with 0  i Exponentially Deep Flaws. bs/2c and h = h + 1. In this paper, we construct a sequence of standard graded i si AppliedMath 2023, 3, 305–315. Cohen–Macaulay domains that have h-vectors with exponentially deep flaws, i.e., we show https://doi.org/10.3390/ the following. appliedmath3020017 Theorem 1. Let K be a field and ` an integer with `  2. Then, there exists a standard graded Academic Editor: Takayuki Hibi h`i h`i h`i Cohen–Macaulay domain A over K such that dim A = 8` 3, a( A ) = 4, and an Received: 21 February 2023 2`3 h2i h3i h-vector (h , h , . . . , h ), h 6= 0, with h = h + 2 . In particular, A , A , 0 1 s s bs /2c s bs /2c ` ` ` ` ` Revised: 21 March 2023 . . . is a sequence of Cohen–Macaulay standard graded domains over K who have exponentially deep Accepted: 22 March 2023 flaws. Published: 3 April 2023 This theorem is proved at the end of this paper. 2. Preliminaries Copyright: © 2023 by the author. In this section, we establish notation and terminology. For unexplained terminology Licensee MDPI, Basel, Switzerland. of commutative algebra and graphy theory we consult [6] and [7], respectively. This article is an open access article In this paper, all rings and algebras are assumed to be commutative with an identity distributed under the terms and element. Further, all graphs are assumed to be finite, simple and without loops. We denote conditions of the Creative Commons the set of non-negative integers, the set of integers, the set of rational numbers, the set of Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ real numbers and the set of non-negative real numbers by N, Z, Q, R and R , respectively. 4.0/). AppliedMath 2023, 3, 305–315. https://doi.org/10.3390/appliedmath3020017 https://www.mdpi.com/journal/appliedmath AppliedMath 2023, 3 306 For a set X, the cardinality of X is denoted by #X. For sets X and Y, we define Xn Y := fx 2 X j x 62 Yg. For non-empty sets X and Y, we denote the set of maps from X X X X to Y by Y . If Y is a subset of Y , then we treat Y as a subset of Y . If X is a finite set, 1 2 1 2 X #X X we identify R with the Euclidean space R . For f , f , f 2 R and a 2 R, we define 1 2 maps f  f and a f by ( f  f )(x) = f (x) f (x) and (a f )(x) = a( f (x)), for x 2 X. Let 2 2 2 1 1 1 A be a subset of X. We define the characteristic function c 2 R of A by c (x) = 1 for A A x 2 A and c (x) = 0 for x 2 Xn A. We denote the zero map, i.e., a map which sends all elements of X to 0, by 0. Further, if X is a subset of X, then we treat R as a coordinate X X X subspace of R , i.e., we identify R with f f 2 R j f (x) = 0 for any x 2 Xn X g. For a non-empty subset X of R , the convex hull (resp. affine span) of X is denoted by convX (resp. affX ). X + Definition 1. Let X be a finite set and x 2 R . For B  X, we set x (B) := x(b). b2B [n] For a field K, the polynomial ring with n variables over K is denoted by K . Let R = R be an N-graded ring. We say that R is a standard graded K-algebra if R = K n 0 n2N L L and R is generated by R as a K-algebra. Let R = R and S = S be N-graded 1 n2N n n2N n rings with R = S = K. We denote the Segre product R S of R and S by R#S. 0 0 n n n2N K Let Y be a finite set. Suppose that there is a family fT g of indeterminates indexed y y2Y f (y) Y f by Y. For f 2 Z , the Laurent monomial, Õ T , is denoted by T . A convex polyhe- y2Y Y Y dral cone in R is a set C of the form C = R a + + R a , where a , . . . , a 2 R . If 0 1 0 r 1 r one can take a , . . . , a 2 Q , we say that C is rational. 1 r Let C be a rational convex polyhedral cone. For a field K, we define K[Z \ C] by Y f Y K[Z \ C] := Y KT . By Gordon’s lemma, we see that K[Z \ C] is a finitely f2Z \C generated K-algebra. In particular, K[Z \ C] is Noetherian. Further, by the result of Hochster [8], we see that K[Z \ C] is normal and Cohen–Macaulay. Y Y A subspace W of R is rational if there is a basis of W contained in Q . Let W and W 1 2 be rational subspaces of R with W \ W = f0g and C be a rational convex polyhedral 1 2 i cone in W for i = 1, 2. Then, C + C is a rational convex polyhedral cone in R that i 1 2 Y Y is isomorphic to the Cartesian product C  C and K[Z \ (C + C )] K[Z \ C ] 1 2 1 2 1 K[Z \ C ]. Let X be a finite set and let P be a rational convex polytope in R , i.e., a convex X X polytope in R whose vertices are in Q . In addition, let ¥ be a new element that is not contained in X. We set X := X[f¥g. Further, we set C(P) := R f f 2 R j f (¥) = 1, fj 2 Pg. Then, C(P) is a rational convex polyhedral cone in R . We define the Ehrhart ring E [P] of P over a field K by E [P] := K[Z \ C(P)]. We K K define deg T = 1 and deg T = 0 for x 2 X. Then, E [P] is an N-graded K-algebra. ¥ x K Note that if W and W are rational subspaces of R with W \ W = f0g andP is a ra- 1 2 1 2 tional convex polytope in W for i = 1, 2, then P +P is a rational convex polytope in R i 1 2 that is isomorphic to the Cartesian product P P and E [P +P ] = E [P ]#E [P ]. 2 2 2 1 K 1 K 1 K It is known that dim E [P] = dimP + 1. Moreover, by the description of the canonical module of a normal affine semigroup ring by Stanley ([9], p. 82), we have the following. Lemma 1. The ideal KT f2Z \relint(C(P)) of E [P] is the canonical module of E [P], where relint(C(P)) denotes the interior of C(P) in K K the topological space aff(C(P)). The ideal of the above lemma is denoted by w and is called the canonical ideal E [P] of E [P]. Note that the a-invariant (cf. ([10], Definition 3.1.4)) , a(E [P]), of E [P] is K K K minf f (¥) j f 2 Z \ relint(C(P))g. AppliedMath 2023, 3 307 A stable set of a graph G = (V, E) is a subset S of V whose no two elements are adjacent. We treat the empty set as a stable set. Definition 2. The stable-set polytope STAB(G) of a graph G = (V, E) is convfc 2 R j S is a stable set of Gg. Note that c 2 STAB(G) for any v 2 V and c 2 STAB(G). In particular, fvg dim STAB(G) = #V. We set 0  f (x)  1 for any x 2 V, f (e)  1 for any e 2 E TSTAB(G) := f 2 R . + #C1 and f (C)  for any odd cycle C Then, TSTAB(G) is a rational convex polytope in R with TSTAB(G)  STAB(G). If TSTAB(G) = STAB(G), we say that G is t-perfect. Let G = (V, E) be an arbitrary graph and n 2 Z. Set K := fK  V j K is a clique and (n) #K  3g. We define tU (G) by 8 9 m(z)  n for any z 2 V, m (K) + n  m(¥) for > > < = any maximal element of K and m (C) + n  m(¥) (n) V tU (G) := m 2 Z . #C1 > > for any odd cycle C without chord and length at : ; least 5 (n) (n) We abbreviate tU (G) as tU if it is clear from the context. By the definition of E [TSTAB(G)], we see that E [TSTAB(G)] = KT . (0) m2tU V + Further, for m 2 Z , m 2 relint(C(E [TSTAB(G)])) if and only if m(z) > 0, m (K) < + #C1 m(¥) and m (C) < m(¥) , where z 2 V, K is a maximal element of K and C is an odd cycle without chords. However, since the values appearing in these inequalities are integers, these inequalities are equivalent to m(z)  1, m (K) + 1  m(¥) and + #C1 m (C) + 1  m(¥) , respectively. Therefore, by Lemma 1, we see that w = KT . E [TSTAB(G)] (1) m2tU 3. Construction Let K be a field. In this section, for each integer `  2, we construct a standard graded h`i Cohen–Macaulay K-algebra, A , which has a non-flawless h-vector. The flaw of the h-vector is computed in the next section. Let ` be an integer with `  2. We define a graph G = (V , E ) by the following way. ` ` ` Set AppliedMath 2023, 3 308 f1, 2, 3g 0  i  2` 4 I := , `,i f1g 2` 3  i  2` C := fc , c , . . . , c g, ` 0 1 2` B := fb j 0  i  2`, k 2 I g, ` ik `,i V := C [ B , ` ` ` E := ffc , c g j j i  1 (mod 2` + 1)g[ffc , b g j k 2 I g ` i j i ik `,i [ffc , b g j j i  1 (mod 2` + 1), k 2 I g ik `,i and G := (V , E ). ` ` ` The cases where ` = 3 and 4 are as follows. 12 b G : b 3 13 2 5 21 c 3 4 22 b b b 02 81 b 0 b 12 71 c c 1 8 c c 2 7 G : b 4 21 b b 22 61 c c 3 6 c c 4 5 b b 32 51 b b 41 b 43 In addition, set h`i h`i A := E [TSTAB(G )] and R := E [TSTAB(G(C ))], K ` K ` where G(C ) is the induced subgraph of G by C . ` ` ` In the following, up to the end of the proof of Lemma 5, we fix ` and write G , V , E , ` ` ` h`i h`i C , B , A , R and I as just G, V, E, C, B, A, R and I , respectively. Further, we consider ` ` `,i the subscripts of c , I and the first subscript of b modulo 2` + 1. For example, c = c , i i i,k 2`+1 0 I = I and b = b . 2 3,1 2`1 2`2,1 We set e := fc , c g and K := fc , c , b g i i i+1 i,k i i+1 i,k AppliedMath 2023, 3 309 for 0  i  2` and k 2 I . We also consider the subscript of e and the first subscript of K i i i,k modulo 2` + 1. We define m 2 Z for 0  i  2` and J  I by i2 1 z = c for some j with j i  0, 2, . . . , 2` 2 (mod 2` + m (z) = 1), z = b with k 2 J or z = ¥, i2,k 0 otherwise. J J (0) It is easily verified that m 2 tU . We also consider the subscript of m modulo i i J J + + 2` + 1. Note that (m ) (C) = ` and (m ) (K ) = 1 if j 6 i 2 (mod 2` + 1) or j  i 2 j,k i i (mod 2` + 1) and k 2 J. Otherwise, (m ) (K ) = 0. j,k First we show the following. Proposition 1. The ring A is a standard graded K-algebra. Proof. Since A = KT , (0) m2tU (0) (0) it is enough to show that for any m 2 tU with m(¥) = n > 0 there are m , . . . , m 2 tU 1 n with m (¥) = 1 for 1  i  n and m = m + + m (i.e., TSTAB(G) has the integer i 1 n decomposition property). We prove this fact by induction on n. The case where n = 1 is trivial. Suppose that n > 1. We first consider the case where m(c) > 0 for any c 2 C. Since 2` + + m (e ) = 2m (C)  2`n < (2` + 1)n, å i i=0 we see that there exists j with m (e ) < n. Set J = fk j m(b ) > 0g. Then, we claim that j j,k (0) m m 2 tU . j+2 First, since m(c) > 0 for any c 2 C by assumption and m(b ) > 0 for any k 2 J, we see j,k that (m m )(z)  0 for any z 2 V. j+2 Next let i be an integer with 0  i  2` and k 2 I . If i 6 j (mod 2` + 1) or i  j J J + + (mod 2` + 1) and k 2 J, then (m ) (K ) = 1. Thus, (m m ) (K )  m(¥) 1 = i,k i,k j+2 j+2 (m m )(¥). If i  j (mod 2` + 1) and k 62 J, then m (e ) < n and m(b ) = 0. i i,k j+2 J J + + Therefore, (m m ) (K ) = m (e )  n 1 = (m m )(¥). i,k j+2 j+2 J J J + + + Finally, (m m ) (C) = m (C) (m ) (C)  n` ` = `(m m )(¥). There- j+2 j+2 j+2 (0) fore, m m 2 tU . j+2 0 V Next, suppose that m(c ) = 0 for some i. Take i with m(c ) = 0. We define m 2 Z i 0 i by the following way. 0 0 First, we define m (c ) (0  j  2`) by induction on j. We define m (c ) = 0. i +j i +0 0 0 0 0 0 Suppose that 1  j  2` and for any â with 0  â  j 1, m (c 0) is defined so that i +â 0 0 0 0 m (c 0) 2 f0, 1g, m (c 0)  m(c 0) for 0  â  j 1, and m (c 0) = 1 implies i +â i +â i +â i +â 0 0 0 0 0 0 m (c 0 ) = 0 for 0  â  j 2 (these assumptions are trivially satisfied when j = 1). We i +â +1 set 1 if m (c ) = 0 and m(c ) > 0, i +j1 i +j 0 0 m (c ) = i +j 0 0 if m (c ) = 1 or m(c ) = 0. i +j1 i +j 0 0 AppliedMath 2023, 3 310 0 0 0 0 Then, m (c ) 2 f0, 1g, m (c )  m(c ) and m (c ) = 1 implies m (c ) = 0. i +j i +j i +j i +j1 i +j 0 0 0 0 0 Thus, we can continue the induction procedure up to j = 2`. We also set 0 0 1 if m (c ) = m (c ) = 0 and m(b ) > 0, 0 i i+1 i,k m (b ) = i,k 0 otherwise, m (¥) = 1 0 V 0 0 and we define m 2 Z . Note that Imm  f0, 1g. Note also that m (c ) = 1 implies 0 0 0 m (c ) = 0 and m (c ) = 0 implies m (c ) = 1 or m(c ) = 0, for any i 2 Z. i+1 i i1 i (0) Next we prove that m 2 tU . First since Imm  f0, 1g, we see that m(z)  0 for any z 2 V. Next we show that m (K )  1, for any i 2 Z and k 2 I . First consider the case where i,k i 0 0 0 0 m (c ) = 1. Then, m (c ) = 0. Further, m (b ) = 0 by the definition of m . Therefore, we i i+1 i,k 0 + 0 0 see that (m ) (K ) = 1. Next, consider the case where m (c ) = 0. Since m (b )  1 and i,k i i,k 0 0 0 + m (c ) = 1 implies that m (b ) = 0, we see that (m ) (K )  1. i+1 i,k i,k 0 0 0 Finally, since Imm  f0, 1g and m (c ) = 1 implies m (c ) = 0 for any i 2 Z, we see i i+1 0 + 0 0 (0) that (m ) (C)  ` = `m (¥). Thus, we see that m 2 tU . (0) Next, we prove that m m 2 tU . 0 0 First, by the definition of m , we see that m(z) = 0 implies m (z) = 0 for any z 2 V. 0 0 Since m (z) 2 f0, 1g for any z 2 V, we see that (m m )(z)  0 for any z 2 V. 0 + 0 0 + Next, we show that (m m ) (K )  (m m )(¥) for any i and k 2 I . If (m ) (K ) = i,k i i,k 0 + + 0 1, then (m m ) (K ) = m (K ) 1  m(¥) 1 = (m m )(¥). Assume that i,k i,k 0 + 0 0 0 (m ) (K ) = 0. Then, m (c ) = m (c ) = m (b ) = 0. Thus, we see that m(b ) = 0 by i,k i i+1 i,k i,k 0 0 0 0 the definition of m . Since m (c 0) = 0 implies m (c 0 ) = 1 or m(c 0) = 0, for any ı 2 Z, we ı ı 1 ı 0 + 0 see that m(c ) = 0. If m(c ) = 0, then (m m ) (K ) = 0  (m m )(¥). Suppose that i+1 i i,k 0 0 m(c ) > 0. Then, m (c ) = 1 by the property of m noted above. Therefore, m(c ) > 0. i i1 i1 Since m(c ) + m(c )  m (K )  m(¥), we see that m(c )  m(¥) 1. Therefore, i1 i i1,1 i 0 + 0 (m m ) (K ) = m(c )  (m m )(¥). i,k i 0 + 0 0 Finally we show that (m m ) (C)  `(m m )(¥). Since m(c ) = m (c ) = 0, we i i 0 0 see that 0 + 0 0 + (m m ) (C) = (m m )(c ) + (m m ) (e ) i å i +2j1 0 0 j=1 0 + = (m m ) (e ) å i +2j1 j=1 0 + (m m ) (K ) i +2j1,1 j=1 (m m )(¥) j=1 = `(m m )(¥). Remark 1. The functions m and m in the proof of Proposition 1 are the characteristic function j+2 of some stable set of G. Therefore, the above proof shows that G is a t-perfect graph. 4. Structure of the Canonical Module In this section, we study the generators and the structure of the canonical module of A. First, we set V + W := f f 2 R j f (C) = ` f (¥)g. AppliedMath 2023, 3 311 V B Then, W is a codimension 1 vector subspace of R with W  R . Further, we set (0) (0) (0) + tU = tU (G) := fm 2 tU (G) j m (C) = `m(¥)g, 0 0 (0) (0) + tU (G(C)) := fm 2 tU (G(C)) j m (C) = `m(¥)g and (0) m A := KT . (0) m2tU (0) h`i (0) Then, A is a K-subalgebra of A (we denote this ring by ( A ) when it is necessary to express `). Further, since (0) (0) m 2 tU (G) () mj 2 tU (G(C)), 0 0 (0) for m 2 tU , we see that (0) m R\ A = KT . (0) m2tU (G(C)) J (0) (0) We denote this ring by R . Note that m 2 tU for any i 2 Z and J  I . By ([5], i2 i 0 Lemma 4.3) and the argument following the proof of it, we see the following. Æ Æ Æ V Theorem 2. The elements m , m , . . . , m of R are linearly independent and 0 1 2` Æ Æ Æ (0) m m m 0 1 2` R = K[T , T , . . . , T ]. Further, we see the following. Lemma 2. It holds that 2` (0) m V J A = K[T j 0  i  2`, J  I ] = K[Z \ R m ]. i2 0 å å i=0 J I i2 Proof. It is clear that 2` m V (0) K[T j 0  i  2`, J  I ]  K[Z \ R m ]  A . i2 å å i=0 J I i2 (0) m In order to prove the inclusion A  K[T j 0  i  2`, J  I ], it is enough to i2 (0) m m show that for any m 2 tU , T 2 K[T j 0  i  2`, J  I ]. We prove this fact by i2 induction on m(¥). (0) The case where m(¥) = 0 is trivial. Let m be an arbitrary element of tU with m(¥) > 0. By the proof of Lemma 4.3 in [5], we see that there is i with (m m )(c)  0 Æ + Æ for any c 2 C and (m m ) (e )  (m m )(¥) for any j. Set J = fk j m(b ) > 0g. j i2,k i i (0) Then, it holds that m m 2 tU . i 0 In fact, (m m )(z)  0 for any z 2 V by the choice of i and the definition of J. If j 6 i 2 (mod 2` + 1) or j  i 2 (mod 2` + 1) and k 2 J, then (m ) (K ) = 1. Thus, j,k J J + + (m m ) (K ) = m (K ) 1  m(¥) 1 = (m m )(¥). If j  i 2 (mod 2` + 1) j,k j,k i i and k 62 J, then m(b ) = m (b ) = 0 by the definition of J. Therefore, by the choice of i2,k i2,k J J + + Æ + Æ i, we see that (m m ) (K ) = (m m ) (e ) = (m m ) (e )  (m m )(¥) = j,k i2 i2 i i i i J J J J + + + (m m )(¥). Finally, (m m ) (C) = m (C) (m ) (C) = `m(¥) ` = `(m m )(C). i i i i J (0) Thus, we see that m m 2 tU . i 0 AppliedMath 2023, 3 312 Since (m m )(¥) = m(¥) 1, we see, by the induction hypothesis, that J J mm m 0 i i T 2 K[T j 0  i  2`, J  I ]. i2 Thus, we see that J J J m m mm m 0 i i i T = T T 2 K[T j 0  i  2`, J  I ]. i2 Since V C B B R = R  R and W  R , we see that C B W = (R \ W) R . C C C Thus, R \ W is a codimension 1 vector subspace of R . Since dimR = #C = Æ C Æ Æ Æ 2` + 2 and m 2 R \ W for any 0  i  2`, we see, by Theorem 2, that m , m , . . . , m is i 1 2` a basis of R \ W. Set 0 Æ 0 W = Rc and W = Rm  W å fb g i i i i i2,k k2 I i2 for 0  i  2`. Then, B 0 0 0 R = W  W  W , W = W  W  W and W = Rm 0 1 1 2` i å 2` i J I i2 for 0  i  2`. Set C = R m i å 0 J I i2 for 0  i  2`. Then, by Lemma 2, we see that 2` (0) V V V A = K[Z \ C ] = K[Z \ C ] K[Z \ C ]. å i 0 2` i=0 It is easily verified that K[Z \ C ] is isomorphic to the Ehrhart ring of the unit cube for 2  i  2` 2. Therefore, V [2] [2] [2] K[Z \ C ] K #K #K for 2  i  2` 2. Further, it is easily verified that V [2] K[Z \ C ] K for i = 0, 1, 2` 1 and 2`. Thus, we see that (0) [2] [2] [2] 2`3 [8] A (K #K #K ) K . It is verified by a direct computation, or by Theorem 2.1 in [11], that the Hilbert series [2] [2] [2] 1+4l+l (0) of K #K #K is . Therefore, the Hilbert series of A is (1l) 2 2`3 (1 + 4l + l ) 8`4 (1 l) AppliedMath 2023, 3 313 For each integer k with 1  k  2` 1, we define h 2 Z by 1 z 2 B, h (z) = k z 2 C, 2k + 2 z = ¥. (1) + It is easily verified that h 2 tU and `h (¥) h (C) = 2` k. Further, we see k k the following. Lemma 3. It holds that a( A) = 4. (1) Proof. Since for any h 2 tU , h(¥)  h (K ) + 1  #K + 1 = 4 and h (¥) = 4, 0,1 0,1 1 (1) we see that a( A) = minfh(¥) j h 2 tU g = 4. n n+h Consider the graded A-homomorphism, j : A ! w (4), T 7! T , of degree 0. Then Imj is a submodule of w (4) generated by T . Further, we have the following. Lemma 4. It holds that Imj(4) = KT . (1) n2tU ,`n(¥)n (C)2`1 Further, Imj is a rank-1-free A-module with basis T . Proof. This lemma is proved almost identically to Lemma 4.2 in [5]. Set D = KT (1) h2tU ,`h(¥)h (C)=2`k for 2  k  2` 1. Then, the following holds. (0) h Lemma 5. D is a rank-1-free A -module with basis T for 2  k  2` 1. Proof. This lemma is proved almost identically to Lemma 4.5 in [5]. h`i Now, we prove Theorem 1. First, note that dim A = #V + 1 = 8` 3. Let h`i (h , h , . . . , h ), h 6= 0, be the h-vector of A . Then, 0 1 s s ` ` h`i h`i s = dim A + a( A ) = 8` 7 and bs /2c = 4` 4. ` ` By the second proof of Theorem 4.1 in [9], we see that h + h l + + h l s s 1 0 ` ` H(w (4), l) = , h`i A 8`3 (1 l) where H( M, l) denotes the Hilbert series of a graded module M. Since w = KT h`i (1) h2tU (G ) 0 1 0 1 B C 2`1B C M M M B C B C h h = B KT C B KT C, @ A @ A (1) (1) k=2 h2tU (G ) h2tU (G ) ` ` + + `h(¥)h (C )2`1 `h(¥)h (C )=2`k ` ` AppliedMath 2023, 3 314 and there is an exact sequence h`i 0 ! A ! w (4) ! Cokj ! 0, h`i we see by Lemmas 4 and 5 that 2`1 h`i h`i (0) 2k2 H(w (4), l) = H( A , l) + H(( A ) , l)l , h`i k=2 since deg T = h (¥) = 2k + 2 for 1  k  2` 1. Therefore, (h h ) + (h h )l + + (h h )l s 0 s 1 1 0 s ` ` 8`3 (1 l) 2`1 2 2`3 2k2 (1 + 4l + l ) l 8`4 (1 l) k=2 2`1 2 2`3 2k2 2k1 (1 + 4l + l ) (l l ) 8`3 (1 l) k=2 2 2`3 2 3 4`4 4`3 (1 + 4l + l ) (l l + + l l ) = . 8`3 (1 l) 4`3 By comparing the coefficient of l in the numerators, we see that h h 4`4 4`3 the sum of the coefficients of the odd powers of l of 2 2`3 (1 + 4l + l ) the sum of the coefficients of the even powers of l of 2 2`3 (1 + 4l + l ) 2`3 = (1 + 4 1) 2`3 = 2 . Since bs /2c = 4` 4 and s bs /2c = 4` 3, we see that ` ` ` 2`3 h = h + 2 . bs /2c s bs /2c ` ` ` Funding: This research was funded by JSPS KAKENHI JP20K03556. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Conflicts of Interest: The author declares no conflict of interest. References 1. Hibi, T. Flawless O-sequences and Hilbert functions of Cohen-Macaulay integral domains. J. Pure Appl. Algebra 1989, 60, 245–251. [CrossRef] 2. Niesi, G.; Robbiano, L.; Disproving Hibi’s Conjecture with CoCoA or Projective Curves with bad Hilbert Functions. In Computational Algebraic Geometry; Eyssette, F., Galligo, A., Eds.; Birkhäuser: Boston, MA, USA, 1993; pp. 195–201. 3. Hibi, T.; Tsuchiya, A. Odd Cycles and Hilbert Functions of Their Toric Rings. Mathematics 2020, 8, 22. [CrossRef] 4. Bruns, W.; Ichim, B.; Römer, T.; Sieg, R.; Söger, C. Normaliz, Algorithms for Rational Cones and Affine Monoids. Available online: https://www.normaliz.uni-osnabrueck.de (accessed on 2 March 2023). 5. Miyazaki, M. Non-Gorenstein Locus and Almost Gorenstein Property of the Ehrhart Ring of the Stable Set Polytope of a Cycle Graph. Taiwan. J. Math. 2022, 1–19. Available online: https://projecteuclid.org/journals/taiwanese-journal-of-mathematics/ advance-publication/Non-Gorenstein-Locus-and-Almost-Gorenstein-Property-of-the-Ehrhart/10.11650/tjm/221104.full (accessed on 2 March 2023). AppliedMath 2023, 3 315 6. Bruns, W.; Herzog, J. Cohen–Macaulay Rings; Cambridge University Press: Cambridge, UK, 1998; No. 39 7. Diestel, R. Graph Theory, 5th ed.; Springer: Berlin, Germany, 2017; GTM, Volume 173. 8. Hochster, M. Rings of invariants of tori, Cohen-Macaulay rings generated by monomials and polytopes. Ann. Math. 1972, 96, 318–337. [CrossRef] 9. Stanley, R.P. Hilbert Functions of Graded Algebras. Adv. Math. 1978, 28, 57–83. [CrossRef] 10. Goto, S.; Watanabe, K. On graded rings, I. J. Math. Soc. Jpn. 1978, 30, 179–213. [CrossRef] 11. Fischer, I.; Kubitzke, M. Spectra and eigenvectors of the Segre transformation. Adv. Appl. Math. 2014, 56, 1–19. [CrossRef] Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png AppliedMath Multidisciplinary Digital Publishing Institute

A Sequence of Cohen&ndash;Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially Deep Flaws

Miyazaki, Mitsuhiro
AppliedMath , Volume 3 (2) – Apr 3, 2023
Download PDF
11 pages

Loading next page...
 
/lp/multidisciplinary-digital-publishing-institute/a-sequence-of-cohen-ndash-macaulay-standard-graded-domains-whose-h-TOVAWw6nlL
Publisher
Multidisciplinary Digital Publishing Institute
Copyright
© 1996-2023 MDPI (Basel, Switzerland) unless otherwise stated Disclaimer Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. Terms and Conditions Privacy Policy
ISSN
2673-9909
DOI
10.3390/appliedmath3020017
Publisher site
See Article on Publisher Site

Abstract

Article A Sequence of Cohen–Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially Deep Flaws Mitsuhiro Miyazaki Department of Mathematics, Kyoto University of Education, 1 Fujinomori, Fukakusa, Fushimi-ku, Kyoto 612-8522, Japan; g53448@kyokyo-u.ac.jp Abstract: Let K be a field. In this paper, we construct a sequence of Cohen–Macaulay standard graded K-domains whose h-vectors are non-flawless and have exponentially deep flaws. Keywords: flawless O-sequence; stable-set polytope; t-perfect; Ehrhart ring MSC: 13H10; 52B20; 05E40; 05C17 1. Introduction In 1989, Hibi [1] made several conjectures on the h-vectors of Cohen–Macaulay standard graded algebras over a field. In particular, he conjectured that the h-vector of a standard graded Cohen–Macaulay domain is flawless ([1], Conjecture 1.4). The h- vector (h , h , . . . , h ), h 6= 0, of a Cohen–Macaulay standard graded algebra is flawless 0 1 s s if h  h for 0  i  bs/2c and h  h for 1  i  bs/2c. Niesi and Robbiano [2] i si i1 i disproved this conjecture by constructing a Cohen–Macaulay standard graded domain whose h-vector is (1, 3, 5, 4, 4, 1). Further, Hibi and Tsuchiya [3] showed that the Ehrhart rings of the stable-set polytopes of cycle graphs of length 9 and 11 have non-flawless h-vectors by computation using the software Normaliz [4]. Moreover, the present author showed that the Ehrhart ring of the stable-set polytope of any odd cycle graph whose Citation: Miyazaki, M. A Sequence of length is at least 9 has non-flawless h-vectors ([5], Theorem 5.2) by proving the conjecture Cohen–Macaulay Standard Graded of Hibi and Tsuchiya ([3], Conjecture 1). Domains Whose h-Vectors Have However, these examples have the slightest flaws, i.e., there exists i with 0  i Exponentially Deep Flaws. bs/2c and h = h + 1. In this paper, we construct a sequence of standard graded i si AppliedMath 2023, 3, 305–315. Cohen–Macaulay domains that have h-vectors with exponentially deep flaws, i.e., we show https://doi.org/10.3390/ the following. appliedmath3020017 Theorem 1. Let K be a field and ` an integer with `  2. Then, there exists a standard graded Academic Editor: Takayuki Hibi h`i h`i h`i Cohen–Macaulay domain A over K such that dim A = 8` 3, a( A ) = 4, and an Received: 21 February 2023 2`3 h2i h3i h-vector (h , h , . . . , h ), h 6= 0, with h = h + 2 . In particular, A , A , 0 1 s s bs /2c s bs /2c ` ` ` ` ` Revised: 21 March 2023 . . . is a sequence of Cohen–Macaulay standard graded domains over K who have exponentially deep Accepted: 22 March 2023 flaws. Published: 3 April 2023 This theorem is proved at the end of this paper. 2. Preliminaries Copyright: © 2023 by the author. In this section, we establish notation and terminology. For unexplained terminology Licensee MDPI, Basel, Switzerland. of commutative algebra and graphy theory we consult [6] and [7], respectively. This article is an open access article In this paper, all rings and algebras are assumed to be commutative with an identity distributed under the terms and element. Further, all graphs are assumed to be finite, simple and without loops. We denote conditions of the Creative Commons the set of non-negative integers, the set of integers, the set of rational numbers, the set of Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ real numbers and the set of non-negative real numbers by N, Z, Q, R and R , respectively. 4.0/). AppliedMath 2023, 3, 305–315. https://doi.org/10.3390/appliedmath3020017 https://www.mdpi.com/journal/appliedmath AppliedMath 2023, 3 306 For a set X, the cardinality of X is denoted by #X. For sets X and Y, we define Xn Y := fx 2 X j x 62 Yg. For non-empty sets X and Y, we denote the set of maps from X X X X to Y by Y . If Y is a subset of Y , then we treat Y as a subset of Y . If X is a finite set, 1 2 1 2 X #X X we identify R with the Euclidean space R . For f , f , f 2 R and a 2 R, we define 1 2 maps f  f and a f by ( f  f )(x) = f (x) f (x) and (a f )(x) = a( f (x)), for x 2 X. Let 2 2 2 1 1 1 A be a subset of X. We define the characteristic function c 2 R of A by c (x) = 1 for A A x 2 A and c (x) = 0 for x 2 Xn A. We denote the zero map, i.e., a map which sends all elements of X to 0, by 0. Further, if X is a subset of X, then we treat R as a coordinate X X X subspace of R , i.e., we identify R with f f 2 R j f (x) = 0 for any x 2 Xn X g. For a non-empty subset X of R , the convex hull (resp. affine span) of X is denoted by convX (resp. affX ). X + Definition 1. Let X be a finite set and x 2 R . For B  X, we set x (B) := x(b). b2B [n] For a field K, the polynomial ring with n variables over K is denoted by K . Let R = R be an N-graded ring. We say that R is a standard graded K-algebra if R = K n 0 n2N L L and R is generated by R as a K-algebra. Let R = R and S = S be N-graded 1 n2N n n2N n rings with R = S = K. We denote the Segre product R S of R and S by R#S. 0 0 n n n2N K Let Y be a finite set. Suppose that there is a family fT g of indeterminates indexed y y2Y f (y) Y f by Y. For f 2 Z , the Laurent monomial, Õ T , is denoted by T . A convex polyhe- y2Y Y Y dral cone in R is a set C of the form C = R a + + R a , where a , . . . , a 2 R . If 0 1 0 r 1 r one can take a , . . . , a 2 Q , we say that C is rational. 1 r Let C be a rational convex polyhedral cone. For a field K, we define K[Z \ C] by Y f Y K[Z \ C] := Y KT . By Gordon’s lemma, we see that K[Z \ C] is a finitely f2Z \C generated K-algebra. In particular, K[Z \ C] is Noetherian. Further, by the result of Hochster [8], we see that K[Z \ C] is normal and Cohen–Macaulay. Y Y A subspace W of R is rational if there is a basis of W contained in Q . Let W and W 1 2 be rational subspaces of R with W \ W = f0g and C be a rational convex polyhedral 1 2 i cone in W for i = 1, 2. Then, C + C is a rational convex polyhedral cone in R that i 1 2 Y Y is isomorphic to the Cartesian product C  C and K[Z \ (C + C )] K[Z \ C ] 1 2 1 2 1 K[Z \ C ]. Let X be a finite set and let P be a rational convex polytope in R , i.e., a convex X X polytope in R whose vertices are in Q . In addition, let ¥ be a new element that is not contained in X. We set X := X[f¥g. Further, we set C(P) := R f f 2 R j f (¥) = 1, fj 2 Pg. Then, C(P) is a rational convex polyhedral cone in R . We define the Ehrhart ring E [P] of P over a field K by E [P] := K[Z \ C(P)]. We K K define deg T = 1 and deg T = 0 for x 2 X. Then, E [P] is an N-graded K-algebra. ¥ x K Note that if W and W are rational subspaces of R with W \ W = f0g andP is a ra- 1 2 1 2 tional convex polytope in W for i = 1, 2, then P +P is a rational convex polytope in R i 1 2 that is isomorphic to the Cartesian product P P and E [P +P ] = E [P ]#E [P ]. 2 2 2 1 K 1 K 1 K It is known that dim E [P] = dimP + 1. Moreover, by the description of the canonical module of a normal affine semigroup ring by Stanley ([9], p. 82), we have the following. Lemma 1. The ideal KT f2Z \relint(C(P)) of E [P] is the canonical module of E [P], where relint(C(P)) denotes the interior of C(P) in K K the topological space aff(C(P)). The ideal of the above lemma is denoted by w and is called the canonical ideal E [P] of E [P]. Note that the a-invariant (cf. ([10], Definition 3.1.4)) , a(E [P]), of E [P] is K K K minf f (¥) j f 2 Z \ relint(C(P))g. AppliedMath 2023, 3 307 A stable set of a graph G = (V, E) is a subset S of V whose no two elements are adjacent. We treat the empty set as a stable set. Definition 2. The stable-set polytope STAB(G) of a graph G = (V, E) is convfc 2 R j S is a stable set of Gg. Note that c 2 STAB(G) for any v 2 V and c 2 STAB(G). In particular, fvg dim STAB(G) = #V. We set 0  f (x)  1 for any x 2 V, f (e)  1 for any e 2 E TSTAB(G) := f 2 R . + #C1 and f (C)  for any odd cycle C Then, TSTAB(G) is a rational convex polytope in R with TSTAB(G)  STAB(G). If TSTAB(G) = STAB(G), we say that G is t-perfect. Let G = (V, E) be an arbitrary graph and n 2 Z. Set K := fK  V j K is a clique and (n) #K  3g. We define tU (G) by 8 9 m(z)  n for any z 2 V, m (K) + n  m(¥) for > > < = any maximal element of K and m (C) + n  m(¥) (n) V tU (G) := m 2 Z . #C1 > > for any odd cycle C without chord and length at : ; least 5 (n) (n) We abbreviate tU (G) as tU if it is clear from the context. By the definition of E [TSTAB(G)], we see that E [TSTAB(G)] = KT . (0) m2tU V + Further, for m 2 Z , m 2 relint(C(E [TSTAB(G)])) if and only if m(z) > 0, m (K) < + #C1 m(¥) and m (C) < m(¥) , where z 2 V, K is a maximal element of K and C is an odd cycle without chords. However, since the values appearing in these inequalities are integers, these inequalities are equivalent to m(z)  1, m (K) + 1  m(¥) and + #C1 m (C) + 1  m(¥) , respectively. Therefore, by Lemma 1, we see that w = KT . E [TSTAB(G)] (1) m2tU 3. Construction Let K be a field. In this section, for each integer `  2, we construct a standard graded h`i Cohen–Macaulay K-algebra, A , which has a non-flawless h-vector. The flaw of the h-vector is computed in the next section. Let ` be an integer with `  2. We define a graph G = (V , E ) by the following way. ` ` ` Set AppliedMath 2023, 3 308 f1, 2, 3g 0  i  2` 4 I := , `,i f1g 2` 3  i  2` C := fc , c , . . . , c g, ` 0 1 2` B := fb j 0  i  2`, k 2 I g, ` ik `,i V := C [ B , ` ` ` E := ffc , c g j j i  1 (mod 2` + 1)g[ffc , b g j k 2 I g ` i j i ik `,i [ffc , b g j j i  1 (mod 2` + 1), k 2 I g ik `,i and G := (V , E ). ` ` ` The cases where ` = 3 and 4 are as follows. 12 b G : b 3 13 2 5 21 c 3 4 22 b b b 02 81 b 0 b 12 71 c c 1 8 c c 2 7 G : b 4 21 b b 22 61 c c 3 6 c c 4 5 b b 32 51 b b 41 b 43 In addition, set h`i h`i A := E [TSTAB(G )] and R := E [TSTAB(G(C ))], K ` K ` where G(C ) is the induced subgraph of G by C . ` ` ` In the following, up to the end of the proof of Lemma 5, we fix ` and write G , V , E , ` ` ` h`i h`i C , B , A , R and I as just G, V, E, C, B, A, R and I , respectively. Further, we consider ` ` `,i the subscripts of c , I and the first subscript of b modulo 2` + 1. For example, c = c , i i i,k 2`+1 0 I = I and b = b . 2 3,1 2`1 2`2,1 We set e := fc , c g and K := fc , c , b g i i i+1 i,k i i+1 i,k AppliedMath 2023, 3 309 for 0  i  2` and k 2 I . We also consider the subscript of e and the first subscript of K i i i,k modulo 2` + 1. We define m 2 Z for 0  i  2` and J  I by i2 1 z = c for some j with j i  0, 2, . . . , 2` 2 (mod 2` + m (z) = 1), z = b with k 2 J or z = ¥, i2,k 0 otherwise. J J (0) It is easily verified that m 2 tU . We also consider the subscript of m modulo i i J J + + 2` + 1. Note that (m ) (C) = ` and (m ) (K ) = 1 if j 6 i 2 (mod 2` + 1) or j  i 2 j,k i i (mod 2` + 1) and k 2 J. Otherwise, (m ) (K ) = 0. j,k First we show the following. Proposition 1. The ring A is a standard graded K-algebra. Proof. Since A = KT , (0) m2tU (0) (0) it is enough to show that for any m 2 tU with m(¥) = n > 0 there are m , . . . , m 2 tU 1 n with m (¥) = 1 for 1  i  n and m = m + + m (i.e., TSTAB(G) has the integer i 1 n decomposition property). We prove this fact by induction on n. The case where n = 1 is trivial. Suppose that n > 1. We first consider the case where m(c) > 0 for any c 2 C. Since 2` + + m (e ) = 2m (C)  2`n < (2` + 1)n, å i i=0 we see that there exists j with m (e ) < n. Set J = fk j m(b ) > 0g. Then, we claim that j j,k (0) m m 2 tU . j+2 First, since m(c) > 0 for any c 2 C by assumption and m(b ) > 0 for any k 2 J, we see j,k that (m m )(z)  0 for any z 2 V. j+2 Next let i be an integer with 0  i  2` and k 2 I . If i 6 j (mod 2` + 1) or i  j J J + + (mod 2` + 1) and k 2 J, then (m ) (K ) = 1. Thus, (m m ) (K )  m(¥) 1 = i,k i,k j+2 j+2 (m m )(¥). If i  j (mod 2` + 1) and k 62 J, then m (e ) < n and m(b ) = 0. i i,k j+2 J J + + Therefore, (m m ) (K ) = m (e )  n 1 = (m m )(¥). i,k j+2 j+2 J J J + + + Finally, (m m ) (C) = m (C) (m ) (C)  n` ` = `(m m )(¥). There- j+2 j+2 j+2 (0) fore, m m 2 tU . j+2 0 V Next, suppose that m(c ) = 0 for some i. Take i with m(c ) = 0. We define m 2 Z i 0 i by the following way. 0 0 First, we define m (c ) (0  j  2`) by induction on j. We define m (c ) = 0. i +j i +0 0 0 0 0 0 Suppose that 1  j  2` and for any â with 0  â  j 1, m (c 0) is defined so that i +â 0 0 0 0 m (c 0) 2 f0, 1g, m (c 0)  m(c 0) for 0  â  j 1, and m (c 0) = 1 implies i +â i +â i +â i +â 0 0 0 0 0 0 m (c 0 ) = 0 for 0  â  j 2 (these assumptions are trivially satisfied when j = 1). We i +â +1 set 1 if m (c ) = 0 and m(c ) > 0, i +j1 i +j 0 0 m (c ) = i +j 0 0 if m (c ) = 1 or m(c ) = 0. i +j1 i +j 0 0 AppliedMath 2023, 3 310 0 0 0 0 Then, m (c ) 2 f0, 1g, m (c )  m(c ) and m (c ) = 1 implies m (c ) = 0. i +j i +j i +j i +j1 i +j 0 0 0 0 0 Thus, we can continue the induction procedure up to j = 2`. We also set 0 0 1 if m (c ) = m (c ) = 0 and m(b ) > 0, 0 i i+1 i,k m (b ) = i,k 0 otherwise, m (¥) = 1 0 V 0 0 and we define m 2 Z . Note that Imm  f0, 1g. Note also that m (c ) = 1 implies 0 0 0 m (c ) = 0 and m (c ) = 0 implies m (c ) = 1 or m(c ) = 0, for any i 2 Z. i+1 i i1 i (0) Next we prove that m 2 tU . First since Imm  f0, 1g, we see that m(z)  0 for any z 2 V. Next we show that m (K )  1, for any i 2 Z and k 2 I . First consider the case where i,k i 0 0 0 0 m (c ) = 1. Then, m (c ) = 0. Further, m (b ) = 0 by the definition of m . Therefore, we i i+1 i,k 0 + 0 0 see that (m ) (K ) = 1. Next, consider the case where m (c ) = 0. Since m (b )  1 and i,k i i,k 0 0 0 + m (c ) = 1 implies that m (b ) = 0, we see that (m ) (K )  1. i+1 i,k i,k 0 0 0 Finally, since Imm  f0, 1g and m (c ) = 1 implies m (c ) = 0 for any i 2 Z, we see i i+1 0 + 0 0 (0) that (m ) (C)  ` = `m (¥). Thus, we see that m 2 tU . (0) Next, we prove that m m 2 tU . 0 0 First, by the definition of m , we see that m(z) = 0 implies m (z) = 0 for any z 2 V. 0 0 Since m (z) 2 f0, 1g for any z 2 V, we see that (m m )(z)  0 for any z 2 V. 0 + 0 0 + Next, we show that (m m ) (K )  (m m )(¥) for any i and k 2 I . If (m ) (K ) = i,k i i,k 0 + + 0 1, then (m m ) (K ) = m (K ) 1  m(¥) 1 = (m m )(¥). Assume that i,k i,k 0 + 0 0 0 (m ) (K ) = 0. Then, m (c ) = m (c ) = m (b ) = 0. Thus, we see that m(b ) = 0 by i,k i i+1 i,k i,k 0 0 0 0 the definition of m . Since m (c 0) = 0 implies m (c 0 ) = 1 or m(c 0) = 0, for any ı 2 Z, we ı ı 1 ı 0 + 0 see that m(c ) = 0. If m(c ) = 0, then (m m ) (K ) = 0  (m m )(¥). Suppose that i+1 i i,k 0 0 m(c ) > 0. Then, m (c ) = 1 by the property of m noted above. Therefore, m(c ) > 0. i i1 i1 Since m(c ) + m(c )  m (K )  m(¥), we see that m(c )  m(¥) 1. Therefore, i1 i i1,1 i 0 + 0 (m m ) (K ) = m(c )  (m m )(¥). i,k i 0 + 0 0 Finally we show that (m m ) (C)  `(m m )(¥). Since m(c ) = m (c ) = 0, we i i 0 0 see that 0 + 0 0 + (m m ) (C) = (m m )(c ) + (m m ) (e ) i å i +2j1 0 0 j=1 0 + = (m m ) (e ) å i +2j1 j=1 0 + (m m ) (K ) i +2j1,1 j=1 (m m )(¥) j=1 = `(m m )(¥). Remark 1. The functions m and m in the proof of Proposition 1 are the characteristic function j+2 of some stable set of G. Therefore, the above proof shows that G is a t-perfect graph. 4. Structure of the Canonical Module In this section, we study the generators and the structure of the canonical module of A. First, we set V + W := f f 2 R j f (C) = ` f (¥)g. AppliedMath 2023, 3 311 V B Then, W is a codimension 1 vector subspace of R with W  R . Further, we set (0) (0) (0) + tU = tU (G) := fm 2 tU (G) j m (C) = `m(¥)g, 0 0 (0) (0) + tU (G(C)) := fm 2 tU (G(C)) j m (C) = `m(¥)g and (0) m A := KT . (0) m2tU (0) h`i (0) Then, A is a K-subalgebra of A (we denote this ring by ( A ) when it is necessary to express `). Further, since (0) (0) m 2 tU (G) () mj 2 tU (G(C)), 0 0 (0) for m 2 tU , we see that (0) m R\ A = KT . (0) m2tU (G(C)) J (0) (0) We denote this ring by R . Note that m 2 tU for any i 2 Z and J  I . By ([5], i2 i 0 Lemma 4.3) and the argument following the proof of it, we see the following. Æ Æ Æ V Theorem 2. The elements m , m , . . . , m of R are linearly independent and 0 1 2` Æ Æ Æ (0) m m m 0 1 2` R = K[T , T , . . . , T ]. Further, we see the following. Lemma 2. It holds that 2` (0) m V J A = K[T j 0  i  2`, J  I ] = K[Z \ R m ]. i2 0 å å i=0 J I i2 Proof. It is clear that 2` m V (0) K[T j 0  i  2`, J  I ]  K[Z \ R m ]  A . i2 å å i=0 J I i2 (0) m In order to prove the inclusion A  K[T j 0  i  2`, J  I ], it is enough to i2 (0) m m show that for any m 2 tU , T 2 K[T j 0  i  2`, J  I ]. We prove this fact by i2 induction on m(¥). (0) The case where m(¥) = 0 is trivial. Let m be an arbitrary element of tU with m(¥) > 0. By the proof of Lemma 4.3 in [5], we see that there is i with (m m )(c)  0 Æ + Æ for any c 2 C and (m m ) (e )  (m m )(¥) for any j. Set J = fk j m(b ) > 0g. j i2,k i i (0) Then, it holds that m m 2 tU . i 0 In fact, (m m )(z)  0 for any z 2 V by the choice of i and the definition of J. If j 6 i 2 (mod 2` + 1) or j  i 2 (mod 2` + 1) and k 2 J, then (m ) (K ) = 1. Thus, j,k J J + + (m m ) (K ) = m (K ) 1  m(¥) 1 = (m m )(¥). If j  i 2 (mod 2` + 1) j,k j,k i i and k 62 J, then m(b ) = m (b ) = 0 by the definition of J. Therefore, by the choice of i2,k i2,k J J + + Æ + Æ i, we see that (m m ) (K ) = (m m ) (e ) = (m m ) (e )  (m m )(¥) = j,k i2 i2 i i i i J J J J + + + (m m )(¥). Finally, (m m ) (C) = m (C) (m ) (C) = `m(¥) ` = `(m m )(C). i i i i J (0) Thus, we see that m m 2 tU . i 0 AppliedMath 2023, 3 312 Since (m m )(¥) = m(¥) 1, we see, by the induction hypothesis, that J J mm m 0 i i T 2 K[T j 0  i  2`, J  I ]. i2 Thus, we see that J J J m m mm m 0 i i i T = T T 2 K[T j 0  i  2`, J  I ]. i2 Since V C B B R = R  R and W  R , we see that C B W = (R \ W) R . C C C Thus, R \ W is a codimension 1 vector subspace of R . Since dimR = #C = Æ C Æ Æ Æ 2` + 2 and m 2 R \ W for any 0  i  2`, we see, by Theorem 2, that m , m , . . . , m is i 1 2` a basis of R \ W. Set 0 Æ 0 W = Rc and W = Rm  W å fb g i i i i i2,k k2 I i2 for 0  i  2`. Then, B 0 0 0 R = W  W  W , W = W  W  W and W = Rm 0 1 1 2` i å 2` i J I i2 for 0  i  2`. Set C = R m i å 0 J I i2 for 0  i  2`. Then, by Lemma 2, we see that 2` (0) V V V A = K[Z \ C ] = K[Z \ C ] K[Z \ C ]. å i 0 2` i=0 It is easily verified that K[Z \ C ] is isomorphic to the Ehrhart ring of the unit cube for 2  i  2` 2. Therefore, V [2] [2] [2] K[Z \ C ] K #K #K for 2  i  2` 2. Further, it is easily verified that V [2] K[Z \ C ] K for i = 0, 1, 2` 1 and 2`. Thus, we see that (0) [2] [2] [2] 2`3 [8] A (K #K #K ) K . It is verified by a direct computation, or by Theorem 2.1 in [11], that the Hilbert series [2] [2] [2] 1+4l+l (0) of K #K #K is . Therefore, the Hilbert series of A is (1l) 2 2`3 (1 + 4l + l ) 8`4 (1 l) AppliedMath 2023, 3 313 For each integer k with 1  k  2` 1, we define h 2 Z by 1 z 2 B, h (z) = k z 2 C, 2k + 2 z = ¥. (1) + It is easily verified that h 2 tU and `h (¥) h (C) = 2` k. Further, we see k k the following. Lemma 3. It holds that a( A) = 4. (1) Proof. Since for any h 2 tU , h(¥)  h (K ) + 1  #K + 1 = 4 and h (¥) = 4, 0,1 0,1 1 (1) we see that a( A) = minfh(¥) j h 2 tU g = 4. n n+h Consider the graded A-homomorphism, j : A ! w (4), T 7! T , of degree 0. Then Imj is a submodule of w (4) generated by T . Further, we have the following. Lemma 4. It holds that Imj(4) = KT . (1) n2tU ,`n(¥)n (C)2`1 Further, Imj is a rank-1-free A-module with basis T . Proof. This lemma is proved almost identically to Lemma 4.2 in [5]. Set D = KT (1) h2tU ,`h(¥)h (C)=2`k for 2  k  2` 1. Then, the following holds. (0) h Lemma 5. D is a rank-1-free A -module with basis T for 2  k  2` 1. Proof. This lemma is proved almost identically to Lemma 4.5 in [5]. h`i Now, we prove Theorem 1. First, note that dim A = #V + 1 = 8` 3. Let h`i (h , h , . . . , h ), h 6= 0, be the h-vector of A . Then, 0 1 s s ` ` h`i h`i s = dim A + a( A ) = 8` 7 and bs /2c = 4` 4. ` ` By the second proof of Theorem 4.1 in [9], we see that h + h l + + h l s s 1 0 ` ` H(w (4), l) = , h`i A 8`3 (1 l) where H( M, l) denotes the Hilbert series of a graded module M. Since w = KT h`i (1) h2tU (G ) 0 1 0 1 B C 2`1B C M M M B C B C h h = B KT C B KT C, @ A @ A (1) (1) k=2 h2tU (G ) h2tU (G ) ` ` + + `h(¥)h (C )2`1 `h(¥)h (C )=2`k ` ` AppliedMath 2023, 3 314 and there is an exact sequence h`i 0 ! A ! w (4) ! Cokj ! 0, h`i we see by Lemmas 4 and 5 that 2`1 h`i h`i (0) 2k2 H(w (4), l) = H( A , l) + H(( A ) , l)l , h`i k=2 since deg T = h (¥) = 2k + 2 for 1  k  2` 1. Therefore, (h h ) + (h h )l + + (h h )l s 0 s 1 1 0 s ` ` 8`3 (1 l) 2`1 2 2`3 2k2 (1 + 4l + l ) l 8`4 (1 l) k=2 2`1 2 2`3 2k2 2k1 (1 + 4l + l ) (l l ) 8`3 (1 l) k=2 2 2`3 2 3 4`4 4`3 (1 + 4l + l ) (l l + + l l ) = . 8`3 (1 l) 4`3 By comparing the coefficient of l in the numerators, we see that h h 4`4 4`3 the sum of the coefficients of the odd powers of l of 2 2`3 (1 + 4l + l ) the sum of the coefficients of the even powers of l of 2 2`3 (1 + 4l + l ) 2`3 = (1 + 4 1) 2`3 = 2 . Since bs /2c = 4` 4 and s bs /2c = 4` 3, we see that ` ` ` 2`3 h = h + 2 . bs /2c s bs /2c ` ` ` Funding: This research was funded by JSPS KAKENHI JP20K03556. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Conflicts of Interest: The author declares no conflict of interest. References 1. Hibi, T. Flawless O-sequences and Hilbert functions of Cohen-Macaulay integral domains. J. Pure Appl. Algebra 1989, 60, 245–251. [CrossRef] 2. Niesi, G.; Robbiano, L.; Disproving Hibi’s Conjecture with CoCoA or Projective Curves with bad Hilbert Functions. In Computational Algebraic Geometry; Eyssette, F., Galligo, A., Eds.; Birkhäuser: Boston, MA, USA, 1993; pp. 195–201. 3. Hibi, T.; Tsuchiya, A. Odd Cycles and Hilbert Functions of Their Toric Rings. Mathematics 2020, 8, 22. [CrossRef] 4. Bruns, W.; Ichim, B.; Römer, T.; Sieg, R.; Söger, C. Normaliz, Algorithms for Rational Cones and Affine Monoids. Available online: https://www.normaliz.uni-osnabrueck.de (accessed on 2 March 2023). 5. Miyazaki, M. Non-Gorenstein Locus and Almost Gorenstein Property of the Ehrhart Ring of the Stable Set Polytope of a Cycle Graph. Taiwan. J. Math. 2022, 1–19. Available online: https://projecteuclid.org/journals/taiwanese-journal-of-mathematics/ advance-publication/Non-Gorenstein-Locus-and-Almost-Gorenstein-Property-of-the-Ehrhart/10.11650/tjm/221104.full (accessed on 2 March 2023). AppliedMath 2023, 3 315 6. Bruns, W.; Herzog, J. Cohen–Macaulay Rings; Cambridge University Press: Cambridge, UK, 1998; No. 39 7. Diestel, R. Graph Theory, 5th ed.; Springer: Berlin, Germany, 2017; GTM, Volume 173. 8. Hochster, M. Rings of invariants of tori, Cohen-Macaulay rings generated by monomials and polytopes. Ann. Math. 1972, 96, 318–337. [CrossRef] 9. Stanley, R.P. Hilbert Functions of Graded Algebras. Adv. Math. 1978, 28, 57–83. [CrossRef] 10. Goto, S.; Watanabe, K. On graded rings, I. J. Math. Soc. Jpn. 1978, 30, 179–213. [CrossRef] 11. Fischer, I.; Kubitzke, M. Spectra and eigenvectors of the Segre transformation. Adv. Appl. Math. 2014, 56, 1–19. [CrossRef] Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

Journal

AppliedMath – Multidisciplinary Digital Publishing Institute

Published: Apr 3, 2023

Keywords: flawless O-sequence; stable-set polytope; t-perfect; Ehrhart ring

There are no references for this article.