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Article

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

Department of Mathematics, Kyoto University of Education, 1 Fujinomori, Fukakusa, Fushimi-ku, Kyoto 612-8522, Japan
AppliedMath 2023, 3(2), 305-315; https://doi.org/10.3390/appliedmath3020017
Submission received: 21 February 2023 / Revised: 21 March 2023 / Accepted: 22 March 2023 / Published: 3 April 2023
(This article belongs to the Special Issue Feature Papers in AppliedMath)

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.
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 0 , h 1 , , h s ) , h s 0 , of a Cohen–Macaulay standard graded algebra is flawless if h i h s i for 0 i s / 2 and h i 1 h i for 1 i s / 2 . Niesi and Robbiano [2] 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 length is at least 9 has non-flawless h-vectors ([5], Theorem 5.2) by proving the conjecture of Hibi and Tsuchiya ([3], Conjecture 1).
However, these examples have the slightest flaws, i.e., there exists i with 0 i s / 2 and h i = h s i + 1 . In this paper, we construct a sequence of standard graded Cohen–Macaulay domains that have h-vectors with exponentially deep flaws, i.e., we show the following.
Theorem 1. 
Let K be a field and ℓ an integer with 2 . Then, there exists a standard graded Cohen–Macaulay domain A over K such that dim A = 8 3 , a ( A ) = 4 , and an h-vector ( h 0 , h 1 , , h s ) , h s 0 , with h s / 2 = h s s / 2 + 2 2 3 . In particular, A 2 , A 3 , …is a sequence of Cohen–Macaulay standard graded domains over K who have exponentially deep flaws.
This theorem is proved at the end of this paper.

2. Preliminaries

In this section, we establish notation and terminology. For unexplained terminology of commutative algebra and graphy theory we consult [6] and [7], respectively.
In this paper, all rings and algebras are assumed to be commutative with an identity element. Further, all graphs are assumed to be finite, simple and without loops. We denote the set of non-negative integers, the set of integers, the set of rational numbers, the set of real numbers and the set of non-negative real numbers by N , Z , Q , R and R 0 , respectively.
For a set X, the cardinality of X is denoted by # X . For sets X and Y, we define X Y : = { x X x Y } . For non-empty sets X and Y, we denote the set of maps from X to Y by Y X . If Y 1 is a subset of Y 2 , then we treat Y 1 X as a subset of Y 2 X . If X is a finite set, we identify R X with the Euclidean space R # X . For f, f 1 , f 2 R X and a R , we define maps f 1 ± f 2 and a f by ( f 1 ± f 2 ) ( x ) = f 1 ( x ) ± f 2 ( x ) and ( a f ) ( x ) = a ( f ( x ) ) , for x X . Let A be a subset of X. We define the characteristic function χ A R X of A by χ A ( x ) = 1 for x A and χ A ( x ) = 0 for x X A . We denote the zero map, i.e., a map which sends all elements of X to 0, by 0. Further, if X 1 is a subset of X, then we treat R X 1 as a coordinate subspace of R X , i.e., we identify R X 1 with { f R X f ( x ) = 0 for any x X X 1 } . For a non-empty subset X of R X , the convex hull (resp. affine span) of X is denoted by conv X (resp. aff X ).
Definition 1. 
Let X be a finite set and ξ R X . For B X , we set ξ + ( B ) : = b B ξ ( b ) .
For a field K , the polynomial ring with n variables over K is denoted by K [ n ] . Let R = n N R n be an N -graded ring. We say that R is a standard graded K -algebra if R 0 = K and R is generated by R 1 as a K -algebra. Let R = n N R n and S = n N S n be N -graded rings with R 0 = S 0 = K . We denote the Segre product n N R n K S n of R and S by R # S .
Let Y be a finite set. Suppose that there is a family { T y } y Y of indeterminates indexed by Y. For f Z Y , the Laurent monomial, y Y T y f ( y ) , is denoted by T f . A convex polyhedral cone in R Y is a set C of the form C = R 0 a 1 + + R 0 a r , where a 1 , …, a r R Y . If one can take a 1 , …, a r Q Y , we say that C is rational.
Let C be a rational convex polyhedral cone. For a field K , we define K [ Z Y C ] by K [ Z Y C ] : = f Z Y C K T f . By Gordon’s lemma, we see that K [ Z Y C ] is a finitely generated K -algebra. In particular, K [ Z Y C ] is Noetherian. Further, by the result of Hochster [8], we see that K [ Z Y C ] is normal and Cohen–Macaulay.
A subspace W of R Y is rational if there is a basis of W contained in Q Y . Let W 1 and W 2 be rational subspaces of R Y with W 1 W 2 = { 0 } and C i be a rational convex polyhedral cone in W i for i = 1 , 2 . Then, C 1 + C 2 is a rational convex polyhedral cone in R Y that is isomorphic to the Cartesian product C 1 × C 2 and K [ Z Y ( C 1 + C 2 ) ] K [ Z Y C 1 ] K [ Z Y C 2 ] .
Let X be a finite set and let P be a rational convex polytope in R X , i.e., a convex polytope in R X whose vertices are in Q X . In addition, let be a new element that is not contained in X. We set X : = X { } . Further, we set C ( P ) : = R 0 { f R X f ( ) = 1 , f | X P } . Then, C ( P ) is a rational convex polyhedral cone in R X . We define the Ehrhart ring E K [ P ] of P over a field K by E K [ P ] : = K [ Z X C ( P ) ] . We define deg T = 1 and deg T x = 0 for x X . Then, E K [ P ] is an N -graded K -algebra.
Note that if W 1 and W 2 are rational subspaces of R X with W 1 W 2 = { 0 } and P i is a rational convex polytope in W i for i = 1 , 2, then P 1 + P 2 is a rational convex polytope in R X that is isomorphic to the Cartesian product P 1 × P 2 and E K [ P 1 + P 2 ] = E K [ P 1 ] # E K [ P 2 ] .
It is known that dim E K [ P ] = dim P + 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
f Z X relint ( C ( P ) ) K T f
of E K [ P ] is the canonical module of E K [ P ] , where relint ( C ( P ) ) denotes the interior of C ( P ) in the topological space aff ( C ( P ) ) .
The ideal of the above lemma is denoted by ω E K [ P ] and is called the canonical ideal of E K [ P ] . Note that the a-invariant (cf. ([10], Definition 3.1.4)), a ( E K [ P ] ) , of E K [ P ] is min { f ( ) f Z X relint ( C ( P ) ) } .
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
conv { χ S R V S   i s   a   s t a b l e   s e t   o f   G } .
Note that χ { v } STAB ( G ) for any v V and χ STAB ( G ) . In particular, dim STAB ( G ) = # V .
We set
TSTAB ( G ) : = { f R V | 0 f ( x ) 1   for   any   x V , f + ( e ) 1   for   any   e E and   f + ( C ) # C 1 2   for   any   cycle   C } .
Then, TSTAB ( G ) is a rational convex polytope in R V 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 Z . Set K : = { K V K is a clique and # K 3 } . We define t U ( n ) ( G ) by
t U ( n ) ( G ) : = { μ Z V | μ ( z ) n   for   any z V , μ + ( K ) + n μ ( )   for any   maximal   element   of   K   and   μ + ( C ) + n μ ( ) . # C 1 2 for   any   odd   cycle   C   without   chord   and   length   at least 5 } .
We abbreviate t U ( n ) ( G ) as t U ( n ) if it is clear from the context.
By the definition of E K [ TSTAB ( G ) ] , we see that
E K [ TSTAB ( G ) ] = μ t U ( 0 ) K T μ .
Further, for μ Z V , μ relint ( C ( E K [ TSTAB ( G ) ] ) ) if and only if μ ( z ) > 0 , μ + ( K ) < μ ( ) and μ + ( C ) < μ ( ) · # C 1 2 , where z 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 μ ( z ) 1 , μ + ( K ) + 1 μ ( ) and μ + ( C ) + 1 μ ( ) · # C 1 2 , respectively. Therefore, by Lemma 1, we see that
ω E K [ TSTAB ( G ) ] = μ t U ( 1 ) K T μ .

3. Construction

Let K be a field. In this section, for each integer 2 , we construct a standard graded 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
I , i : = { 1 , 2 , 3 } 0 i 2 4 { 1 } 2 3 i 2 , C : = { c 0 , c 1 , , c 2 } , B : = { b i k 0 i 2 , k I , i } , V : = C B , E : = { { c i , c j } j i 1 ( mod 2 + 1 ) } { { c i , b i k } k I , i } { { c j , b i k } j i 1 ( mod 2 + 1 ) , k I , i } and G : = ( V , E ) .
The cases where = 3 and 4 are as follows.
Appliedmath 03 00017 i001
In addition, set
A : = E K [ TSTAB ( G ) ] and R : = E K [ TSTAB ( G ( C ) ) ] ,
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 , C , B , A , R and I , i as just G, V, E, C, B, A, R and I i , respectively. Further, we consider the subscripts of c i , I i and the first subscript of b i , k modulo 2 + 1 . For example, c 2 + 1 = c 0 , I 2 = I 2 1 and b 3 , 1 = b 2 2 , 1 .
We set
e i : = { c i , c i + 1 } and K i , k : = { c i , c i + 1 , b i , k }
for 0 i 2 and k I i . We also consider the subscript of e i and the first subscript of K i , k modulo 2 + 1 .
We define μ i J Z V for 0 i 2 and J I i 2 by
μ i J ( z ) = 1 z = c j for   some   j   with   j i 0 , 2 , , 2 2 ( mod 2 + 1 ) , z = b i 2 , k with k J   or z = , 0 otherwise .
It is easily verified that μ i J t U ( 0 ) . We also consider the subscript of μ i J modulo 2 + 1 . Note that ( μ i J ) + ( C ) = and ( μ i J ) + ( K j , k ) = 1 if j i 2 ( mod 2 + 1 ) or j i 2 ( mod 2 + 1 ) and k J . Otherwise, ( μ i J ) + ( K j , k ) = 0 .
First we show the following.
Proposition 1. 
The ring A is a standard graded K -algebra.
Proof. 
Since
A = μ t U ( 0 ) K T μ ,
it is enough to show that for any μ t U ( 0 ) with μ ( ) = n > 0 there are μ 1 , …, μ n t U ( 0 ) with μ i ( ) = 1 for 1 i n and μ = μ 1 + + μ n (i.e., TSTAB ( G ) has the integer 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 μ ( c ) > 0 for any c C . Since
i = 0 2 μ + ( e i ) = 2 μ + ( C ) 2 n < ( 2 + 1 ) n ,
we see that there exists j with μ + ( e j ) < n . Set J = { k μ ( b j , k ) > 0 } . Then, we claim that μ μ j + 2 J t U ( 0 ) .
First, since μ ( c ) > 0 for any c C by assumption and μ ( b j , k ) > 0 for any k J , we see that ( μ μ j + 2 J ) ( z ) 0 for any z V .
Next let i be an integer with 0 i 2 and k I i . If i j ( mod 2 + 1 ) or i j ( mod 2 + 1 ) and k J , then ( μ j + 2 J ) + ( K i , k ) = 1 . Thus, ( μ μ j + 2 J ) + ( K i , k ) μ ( ) 1 = ( μ μ j + 2 J ) ( ) . If i j ( mod 2 + 1 ) and k J , then μ + ( e i ) < n and μ ( b i , k ) = 0 . Therefore, ( μ μ j + 2 J ) + ( K i , k ) = μ + ( e i ) n 1 = ( μ μ j + 2 J ) ( ) .
Finally, ( μ μ j + 2 J ) + ( C ) = μ + ( C ) ( μ j + 2 J ) + ( C ) n = ( μ μ j + 2 J ) ( ) . Therefore, μ μ j + 2 J t U ( 0 ) .
Next, suppose that μ ( c i ) = 0 for some i. Take i 0 with μ ( c i 0 ) = 0 . We define μ Z V by the following way.
First, we define μ ( c i 0 + j ) ( 0 j 2 ) by induction on j. We define μ ( c i 0 + 0 ) = 0 . Suppose that 1 j 2 and for any j with 0 j j 1 , μ ( c i 0 + j ) is defined so that μ ( c i 0 + j ) { 0 , 1 } , μ ( c i 0 + j ) μ ( c i 0 + j ) for 0 j j 1 , and μ ( c i 0 + j ) = 1 implies μ ( c i 0 + j + 1 ) = 0 for 0 j j 2 (these assumptions are trivially satisfied when j = 1 ). We set
μ ( c i 0 + j ) = 1 if   μ ( c i 0 + j 1 ) = 0   and   μ ( c i 0 + j ) > 0 , 0 if   μ ( c i 0 + j 1 ) = 1   or   μ ( c i 0 + j ) = 0 .
Then, μ ( c i 0 + j ) { 0 , 1 } , μ ( c i 0 + j ) μ ( c i 0 + j ) and μ ( c i 0 + j 1 ) = 1 implies μ ( c i 0 + j ) = 0 . Thus, we can continue the induction procedure up to j = 2 . We also set
μ ( b i , k ) = 1 if   μ ( c i ) = μ ( c i + 1 ) = 0   and   μ ( b i , k ) > 0 , 0 otherwise , μ ( ) = 1
and we define μ Z V . Note that Im μ { 0 , 1 } . Note also that μ ( c i ) = 1 implies μ ( c i + 1 ) = 0 and μ ( c i ) = 0 implies μ ( c i 1 ) = 1 or μ ( c i ) = 0 , for any i Z .
Next we prove that μ t U ( 0 ) .
First since Im μ { 0 , 1 } , we see that μ ( z ) 0 for any z V .
Next we show that μ ( K i , k ) 1 , for any i Z and k I i . First consider the case where μ ( c i ) = 1 . Then, μ ( c i + 1 ) = 0 . Further, μ ( b i , k ) = 0 by the definition of μ . Therefore, we see that ( μ ) + ( K i , k ) = 1 . Next, consider the case where μ ( c i ) = 0 . Since μ ( b i , k ) 1 and μ ( c i + 1 ) = 1 implies that μ ( b i , k ) = 0 , we see that ( μ ) + ( K i , k ) 1 .
Finally, since Im μ { 0 , 1 } and μ ( c i ) = 1 implies μ ( c i + 1 ) = 0 for any i Z , we see that ( μ ) + ( C ) = μ ( ) . Thus, we see that μ t U ( 0 ) .
Next, we prove that μ μ t U ( 0 ) .
First, by the definition of μ , we see that μ ( z ) = 0 implies μ ( z ) = 0 for any z V . Since μ ( z ) { 0 , 1 } for any z V , we see that ( μ μ ) ( z ) 0 for any z V .
Next, we show that ( μ μ ) + ( K i , k ) ( μ μ ) ( ) for any i and k I i . If ( μ ) + ( K i , k ) = 1 , then ( μ μ ) + ( K i , k ) = μ + ( K i , k ) 1 μ ( ) 1 = ( μ μ ) ( ) . Assume that ( μ ) + ( K i , k ) = 0 . Then, μ ( c i ) = μ ( c i + 1 ) = μ ( b i , k ) = 0 . Thus, we see that μ ( b i , k ) = 0 by the definition of μ . Since μ ( c ı ) = 0 implies μ ( c ı 1 ) = 1 or μ ( c ı ) = 0 , for any ı Z , we see that μ ( c i + 1 ) = 0 . If μ ( c i ) = 0 , then ( μ μ ) + ( K i , k ) = 0 ( μ μ ) ( ) . Suppose that μ ( c i ) > 0 . Then, μ ( c i 1 ) = 1 by the property of μ noted above. Therefore, μ ( c i 1 ) > 0 . Since μ ( c i 1 ) + μ ( c i ) μ + ( K i 1 , 1 ) μ ( ) , we see that μ ( c i ) μ ( ) 1 . Therefore, ( μ μ ) + ( K i , k ) = μ ( c i ) ( μ μ ) ( ) .
Finally we show that ( μ μ ) + ( C ) ( μ μ ) ( ) . Since μ ( c i 0 ) = μ ( c i 0 ) = 0 , we see that
( μ μ ) + ( C ) = ( μ μ ) ( c i 0 ) + j = 1 ( μ μ ) + ( e i 0 + 2 j 1 ) = j = 1 ( μ μ ) + ( e i 0 + 2 j 1 ) j = 1 ( μ μ ) + ( K i 0 + 2 j 1 , 1 ) j = 1 ( μ μ ) ( ) = ( μ μ ) ( ) .
Remark 1. 
The functions μ j + 2 J and μ in the proof of Proposition 1 are the characteristic function 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
W : = { f R V f + ( C ) = f ( ) } .
Then, W is a codimension 1 vector subspace of R V with W R B . Further, we set
t U 0 ( 0 ) = t U 0 ( 0 ) ( G ) : = { μ t U ( 0 ) ( G ) μ + ( C ) = μ ( ) } ,
t U 0 ( 0 ) ( G ( C ) ) : = { μ t U ( 0 ) ( G ( C ) ) μ + ( C ) = μ ( ) }
and
A ( 0 ) : = μ t U 0 ( 0 ) K T μ .
Then, A ( 0 ) is a K -subalgebra of A (we denote this ring by ( A ) ( 0 ) when it is necessary to express ). Further, since
μ t U 0 ( 0 ) ( G ) μ | C t U 0 ( 0 ) ( G ( C ) ) ,
for μ t U ( 0 ) , we see that
R A ( 0 ) = μ t U 0 ( 0 ) ( G ( C ) ) K T μ .
We denote this ring by R ( 0 ) . Note that μ i J t U 0 ( 0 ) for any i Z and J I i 2 . By ([5], Lemma 4.3) and the argument following the proof of it, we see the following.
Theorem 2. 
The elements μ 0 , μ 1 , …, μ 2 of R V are linearly independent and
R ( 0 ) = K [ T μ 0 , T μ 1 , , T μ 2 ] .
Further, we see the following.
Lemma 2. 
It holds that
A ( 0 ) = K [ T μ i J 0 i 2 , J I i 2 ] = K [ Z V i = 0 2 J I i 2 R 0 μ i J ] .
Proof. 
It is clear that
K [ T μ i J 0 i 2 , J I i 2 ] K [ Z V i = 0 2 J I i 2 R 0 μ i J ] A ( 0 ) .
In order to prove the inclusion A ( 0 ) K [ T μ i J 0 i 2 , J I i 2 ] , it is enough to show that for any μ t U 0 ( 0 ) , T μ K [ T μ i J 0 i 2 , J I i 2 ] . We prove this fact by induction on μ ( ) .
The case where μ ( ) = 0 is trivial. Let μ be an arbitrary element of t U 0 ( 0 ) with μ ( ) > 0 . By the proof of Lemma 4.3 in [5], we see that there is i with ( μ μ i ) ( c ) 0 for any c C and ( μ μ i ) + ( e j ) ( μ μ i ) ( ) for any j. Set J = { k μ ( b i 2 , k ) > 0 } . Then, it holds that μ μ i J t U 0 ( 0 ) .
In fact, ( μ μ i J ) ( z ) 0 for any z V by the choice of i and the definition of J. If j i 2 ( mod 2 + 1 ) or j i 2 ( mod 2 + 1 ) and k J , then ( μ i J ) + ( K j , k ) = 1 . Thus, ( μ μ i J ) + ( K j , k ) = μ + ( K j , k ) 1 μ ( ) 1 = ( μ μ i J ) ( ) . If j i 2 ( mod 2 + 1 ) and k J , then μ ( b i 2 , k ) = μ i J ( b i 2 , k ) = 0 by the definition of J. Therefore, by the choice of i, we see that ( μ μ i J ) + ( K j , k ) = ( μ μ i J ) + ( e i 2 ) = ( μ μ i ) + ( e i 2 ) ( μ μ i ) ( ) = ( μ μ i J ) ( ) . Finally, ( μ μ i J ) + ( C ) = μ + ( C ) ( μ i J ) + ( C ) = μ ( ) = ( μ μ i J ) ( C ) . Thus, we see that μ μ i J t U 0 ( 0 ) .
Since ( μ μ i J ) ( ) = μ ( ) 1 , we see, by the induction hypothesis, that
T μ μ i J K [ T μ i J 0 i 2 , J I i 2 ] .
Thus, we see that
T μ = T μ i J T μ μ i J K [ T μ i J 0 i 2 , J I i 2 ] .
Since
R V = R C R B a n d W R B ,
we see that
W = ( R C W ) R B .
Thus, R C W is a codimension 1 vector subspace of R C . Since dim R C = # C = 2 + 2 and μ i R C W for any 0 i 2 , we see, by Theorem 2, that μ 0 , μ 1 , …, μ 2 is a basis of R C W . Set
W i = k I i 2 R χ { b i 2 , k } and W i = R μ i W i
for 0 i 2 . Then,
R B = W 0 W 1 W 2 , W = W 0 W 1 W 2 a n d W i = J I i 2 R μ i J
for 0 i 2 . Set
C i = J I i 2 R 0 μ i J
for 0 i 2 . Then, by Lemma 2, we see that
A ( 0 ) = K [ Z V i = 0 2 C i ] K [ Z V C 0 ] K [ Z V C 2 ] .
It is easily verified that K [ Z V C i ] is isomorphic to the Ehrhart ring of the unit cube for 2 i 2 2 . Therefore,
K [ Z V C i ] K [ 2 ] # K [ 2 ] # K [ 2 ]
for 2 i 2 2 . Further, it is easily verified that
K [ Z V C i ] K [ 2 ]
for i = 0 , 1, 2 1 and 2 . Thus, we see that
A ( 0 ) ( K [ 2 ] # K [ 2 ] # K [ 2 ] ) 2 3 K [ 8 ] .
It is verified by a direct computation, or by Theorem 2.1 in [11], that the Hilbert series of K [ 2 ] # K [ 2 ] # K [ 2 ] is 1 + 4 λ + λ 2 ( 1 λ ) 4 . Therefore, the Hilbert series of A ( 0 ) is
( 1 + 4 λ + λ 2 ) 2 3 ( 1 λ ) 8 4 .
For each integer k with 1 k 2 1 , we define η k Z V by
η k ( z ) = 1 z B , k z C , 2 k + 2 z = .
It is easily verified that η k t U ( 1 ) and η k ( ) η k + ( C ) = 2 k . Further, we see the following.
Lemma 3. 
It holds that a ( A ) = 4 .
Proof. 
Since for any η t U ( 1 ) , η ( ) η + ( K 0 , 1 ) + 1 # K 0 , 1 + 1 = 4 and η 1 ( ) = 4 , we see that a ( A ) = min { η ( ) η t U ( 1 ) } = 4 . □
Consider the graded A-homomorphism, φ : A ω A ( 4 ) , T ν T ν + η 1 , of degree 0. Then Im φ is a submodule of ω A ( 4 ) generated by T η 1 . Further, we have the following.
Lemma 4. 
It holds that
Im φ ( 4 ) = ν t U ( 1 ) , ν ( ) ν + ( C ) 2 1 K T ν .
Further, Im φ is a rank-1-free A-module with basis T η 1 .
Proof. 
This lemma is proved almost identically to Lemma 4.2 in [5]. □
Set
D k = η t U ( 1 ) , η ( ) η + ( C ) = 2 k K T η
for 2 k 2 1 . Then, the following holds.
Lemma 5. 
D k is a rank-1-free A ( 0 ) -module with basis T η k for 2 k 2 1 .
Proof. 
This lemma is proved almost identically to Lemma 4.5 in [5]. □
Now, we prove Theorem 1. First, note that dim A = # V + 1 = 8 3 . Let ( h 0 , h 1 , , h s ) , h s 0 , be the h-vector of A . Then,
s = dim A + a ( A ) = 8 7 and s / 2 = 4 4 .
By the second proof of Theorem 4.1 in [9], we see that
H ( ω A ( 4 ) , λ ) = h s + h s 1 λ + + h 0 λ s ( 1 λ ) 8 3 ,
where H ( M , λ ) denotes the Hilbert series of a graded module M. Since
ω A = η t U ( 1 ) ( G ) K T η = η t U ( 1 ) ( G ) η ( ) η + ( C ) 2 1 K T η k = 2 2 1 η t U ( 1 ) ( G ) η ( ) η + ( C ) = 2 k K T η ,
and there is an exact sequence
0 A φ ω A ( 4 ) Cok φ 0 ,
we see by Lemmas 4 and 5 that
H ( ω A ( 4 ) , λ ) = H ( A , λ ) + k = 2 2 1 H ( ( A ) ( 0 ) , λ ) λ 2 k 2 ,
since deg T η k = η k ( ) = 2 k + 2 for 1 k 2 1 . Therefore,
( h s h 0 ) + ( h s 1 h 1 ) λ + + ( h 0 h s ) λ s ( 1 λ ) 8 3 = k = 2 2 1 ( 1 + 4 λ + λ 2 ) 2 3 λ 2 k 2 ( 1 λ ) 8 4 = k = 2 2 1 ( 1 + 4 λ + λ 2 ) 2 3 ( λ 2 k 2 λ 2 k 1 ) ( 1 λ ) 8 3 = ( 1 + 4 λ + λ 2 ) 2 3 ( λ 2 λ 3 + + λ 4 4 λ 4 3 ) ( 1 λ ) 8 3 .
By comparing the coefficient of λ 4 3 in the numerators, we see that
h 4 4 h 4 3 = ( the   sum   of   the   coefficients   of   the   odd   powers   of   λ   of ( 1 + 4 λ + λ 2 ) 2 3 ) ( the   sum   of   the   coefficients   of   the   even   powers   of   λ   of ( 1 + 4 λ + λ 2 ) 2 3 ) = ( 1 + 4 1 ) 2 3 = 2 2 3
Since s / 2 = 4 4 and s s / 2 = 4 3 , we see that
h s / 2 = h s s / 2 + 2 2 3 .

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

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Miyazaki, M. A Sequence of Cohen–Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially Deep Flaws. AppliedMath 2023, 3, 305-315. https://doi.org/10.3390/appliedmath3020017

AMA Style

Miyazaki M. A Sequence of Cohen–Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially Deep Flaws. AppliedMath. 2023; 3(2):305-315. https://doi.org/10.3390/appliedmath3020017

Chicago/Turabian Style

Miyazaki, Mitsuhiro. 2023. "A Sequence of Cohen–Macaulay Standard Graded Domains Whose h-Vectors Have Exponentially Deep Flaws" AppliedMath 3, no. 2: 305-315. https://doi.org/10.3390/appliedmath3020017

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