On the Ratio of the Multiplicity and the Embedding
Dimension of Hibi Rings
Takahiro CHIBA, Kazunori MATSUDA and Ken-ichi YOSHIDA
（Accepted November 14, 2012）
ON THE RATIO OF THE MULTIPLICITY AND THE EMBEDDING DIMENSION OF HIBI RINGS
TAKAHIRO CHIBA, KAZUNORI MATSUDA, AND KEN-ICHI YOSHIDA Abstract. We investigate the ratio e/v of the multiplicity e and the embedding dimension v of a Hibi ring R, and study posets whose ratio attains to the lower bound or the upper bound. Moreover, we show that for any given positive rational number α, there exists a Gorenstein Hibi ring R whose ratio e/v is equal to α.
1. Introduction
Let R be a commutative Noetherian local ring with unit element 1Rand the unique
maximal ideal m. Set d = dim R, the Krull dimension. Set v = v(R), the embedding
dimension, that is, the dimension of the vector space m/m2 over k = R/m. Let e(R)
denote the multiplicity of R, that is, e(R) is a positive integer defined by
lR(R/mn+1) =
e(R) d! n
d+ (lower terms) for n
≫ 0.
These three invariants are basic and important (in the theory of Cohen-Macaulay rings). The starting point of our research is the following theorem due to Sally, which gives a fundamental inequality with respect to these invariants:
Theorem 1.1 (Sally [Sal]). If (R, m) is a Cohen-Macaulay local ring of dimension
d, then
v ≤ e + d − 1.
We can rewrite this theorem in terms of ratio as follows (see Proposition 2.2): 1
d ≤ e v .
So it is natural to ask the following question.
Question 1.2. Does there exist an upper bound on e/v for all Cohen-Macaulay local rings R of dimension d?
Such an upper bound does not exist in general. For instance, there exists a Stanley-Reisner ring k[∆] with v(R) = v and e(R) = (vd) for any integers v ≥ d ≥ 2 (see Example 2.4). How about normal domains?
Recently, Huneke and Watanabe proved the following theorem:
Theorem 1.3. ([HuW]) Let (R, m) be a Noetherian local ring of d = dim R, e =
e(R) and v = v(R). Then:
(1) If R is F -pure, then e≤(vd
) .
(2) If R is F -rational(in char R = p > 0) or rational singularity (in char R = 0), then e≤(vd−1−1
) .
See [HoR, HoHu1, HoHu2] for definitions.
It seems to be difficult to find examples of F -rational rings which satisfies e =(vd−1−1) in higher dimensional case. This suggests the possibility of the existence of upper bounds.
On the other hand, Example 2.5 shows that if R = Sm(n), where S = k[X1, . . . , Xd]
and m is the unique graded maximal ideal, then one can easily see that lim
n→∞
e(R)
v(R) = (d− 1)!.
Furthermore, many examples of F-rational local rings R of dimension d≥ 3 satisfies an inequality
e(R)
v(R) ≤ (d − 1)!.
So we believe that “good ” F-rational rings satisfies this inequality. See also Question 2.6.
Our main result in this article is the following theorem, which gives an answer to the question above in the case of Hibi rings, which is very good examples of F-rational rings. See Section 3 for the definition of Hibi rings.
Theorem 1.4 (Main Theorem). Let P be a poset, and let R(P ) be the Hibi ring obtained from P . Set d = dim R(P ), e = e(R(P )), and v = v(R(P )). Then the following inequalities hold:
1 d ≤ e v ≤ (d− 1)! 2d−1 .
(1) The following conditions are equivalent: (a) Equality holds in the left inequality. (b) P is a totally ordered set.
(c) R(P ) is a polynomial ring over K. (2) The following conditions are equivalent:
(a) Equality holds in the right inequality.
(b) P is an anti-chain (i.e. any distinct two elements of P are incomparable). (c) R(P ) is a Gorenstein ring with a(R(P )) =−2, where
a(R) = max{n ∈ Z | [Hmdim R(R)]n̸= 0}
denotes the a-invariant of a graded ring R (see [GW]).
2. Ratio of multiplicity and embedding dimension
Throughout this section, let (R, m, k) be a Cohen-Macaulay local ring of dimension
d ≥ 1. Let v = v(R) (resp. e = e(R)) be the embedding dimension (resp. the
multiplicity with respect to m) of R. Then the following theorem is classical but very important in commutative ring theory. A Similar result holds for standard graded algebras over an algebraically closed field ([Ab]).
Theorem 2.1 (Sally [Sal]). Let (R, m) be a Cohen-Macaulay local ring. Then we have
v ≤ e + d − 1.
The ring R is said to be a Cohen-Macaulay local ring of minimal multiplicity (or of
maximal embedding dimension) if equality holds.
We can rewrite this theorem in terms of ratio.
Proposition 2.2. Let (R, m) be a Cohen-Macaulay local ring of dimension d ≥ 2.
Then
1
d ≤ e v
and equality holds if and only if R is regular.
Proof. Sally’s theorem (Theorem 2.1) implies that e v − 1 d = de− v dv ≥ de− e − d + 1 dv = (d− 1)(e − 1) dv ≥ 0.
Note that equality holds if and only if v = e + d− 1 and e = 1. This condition is
equivalent to saying that R is regular. □
Remark 2.3. The same inequality holds true in the case of d = 1. In this case, equality holds if and only if R has a minimal multiplicity.
How about upper bounds of the ratio? Namely, does there exist a function f (or polynomial) of d such that if R is any Cohen-Macaulay local ring of dimension d, then
e
v ≤ f(d)?
However, in general, we cannot find an upper bound on e/v as the following example shows.
Example 2.4. Let v ≥ d ≥ 2 be integers, and put V = {1, 2, . . . , v}. If we set ∆ = {F | F ⊆ V, ♯(F ) ≤ d}, then ∆ is an abstract simplicial complex on V , where ♯(X) denotes the cardinality of a set X. The Stanley-Reisner ring k[∆] is defined to
be k[X1, . . . , Xv]/I∆, where
I∆ = (Xi1· · · Xip| 1 < i1 <· · · < ip ≤ v, {i1, . . . , ip} /∈ ∆)
is a Cohen-Macaulay ring of dimension d (see e.g. [BH2]). If we put R = k[∆](x1,...,xv),
then we have v(R) = v, e(R) = ( v d ) .
Thus if we fix d and tend v to ∞, then
e(R) v(R) =
(v− 1) · · · (v − d + 1)
d! → ∞.
On the other hand, the following observation yields an open question below. 3
Example 2.5. Let S = k[x1, . . . , xv] be a standard graded ring of dimension d, that
is, S = ⊕n≥0Sn is an N-graded ring, S0 = k and Sn = S1n for every n ≥ 1. Let R = Sm(n)be the localization of the nth Veronese subring of S, where m is the unique
graded maximal ideal of S(n). Then we have
lim
n→∞
e(R)
v(R) = (d− 1)!,
Let S = k[X1, . . . , Xd] be a polynomial ring over a field k. For any integer n≥ 2,
let R = Sm(n) be as above. Then we have
v(R) = ( n + d− 1 n ) , e(R) = nd−1.
Hence e(S(n))/v(S(n))↑ (d − 1)! if n tends to ∞.
So we pose the following question.
Question 2.6. When does the inequality e/v ≤ (d − 1)! hold true?
In the next section, we give an affirmative answer to this question in the case of Hibi rings.
3. Hibi rings
First, we set up the notions of posets and define the Hibi ring. Let P ={p1, p2, . . . , pN}
be a finite partially ordered set (poset for short). A subset I ⊆ P is called a poset ideal of P if x ∈ I, y ∈ P and y ≤ x then y ∈ I. Let J(P ) be the set of all poset
ideals of P . Then D = J(P ) is a distributive lattice. Conversely, by structure the-orem of Birkhoff (see [Bir]), for any nonempty distributive lattice D, there exists a poset P such that D ∼= J(P ) ordered by inclusion. In this article, we use the Hasse diagram (see [Bir]) to represent posets.
A chain X of P is a totally ordered subset of P . The length of a chain X of P is ♯(X)− 1. The rank of P , denoted by rank P , is the maximum of the lengths of chains in P . A poset is called pure if its all maximal chains have the same length. Definition 3.1 ([Hib]). Let the notation be as above. We consider the following map: φ : D(= J(P )) −→ k[T, X1, . . . , XN] ∈ ∈ I �−→ T ∏ pi∈I Xi
Then we define the Hibi ring Rk[D] as follows:
Rk[D] = k[φ(I)| I ∈ J(P )].
In this article, we use the notation R(P ) instead of Rk[J(P )] and call it the Hibi
ring obtained from P for simplicity.
We recall several basic properties of Hibi rings. 4
Proposition 3.2 (See [Hib], [BH1, Theorem 1.1], [CM]). A Hibi ring R(P ) is an affine toric ring and an algebra with straightening law (ASL), thus a normal Cohen-Macaulay domain.
(1) R(P ) is Gorenstein if and only if P is pure. (2) dim R(P ) = ♯(P ) + 1.
(3) v(R(P )) = ♯(J(P )), the numbers of all poset ideals in P . (4) −a(R) = rank P + 2.
The following proposition may be well-known (e.g. [Wa]). But we give a sketch of the proof for the sake of completeness.
Proposition 3.3. Let R(P ) be the Hibi ring obtained from a poset P . Then
e(R(P )) = ♯{C | C is a maximal chain in J(P )}
Proof. By [Hib]ɼthe Hibi ring R(P ) can be represented as a homomorphic image
of a polynomial ring: R(P ) ∼= S/I, where
S = k[za| a ∈ J(P )], I ={zazb− za∧bza∨b | a, b ∈ J(P )}.
For a suitable reverse lexicographic order <, S/in<(I) can be regarded as the
Stanley-Reisner ring with respect to the order complex ∆ of J(P ) ([BH1]). Hence
e(R(P )) = e(S/in<(I)) = ♯{F | F is a facet of ∆}
= ♯{C | C is a maximal chain in J(P )},
as required. □
Now let us recall typical examples of Hibi rings.
Example 3.4. Any polynomial ring k[X1, . . . , Xd] over a field k is isomorphic to
the Hibi ring obtained from a totally ordered set Td−1 of (d− 1) points.
Td−1 : J(Td−1) : p1 p2 pd−1 ∅ {p1} Td−1\ {pd−1} Td−1
Example 3.5. Let P be a poset consisting of two disjoint totally ordered sets Tm−1 and Tn−1. Then R(P ) is isomorphic to the Segre product
k[x1, . . . , xm] ♯ k[y1, . . . , yn].
p1 p2 pm−1 q1 q2 qn−1 Tm−1 Tn−1 P : D : ∅ {p1} {q1} Tm−1 Tn−1 P
4. Main Theorem
The main result in this article is the following.
Theorem 4.1. Let P be a poset, and let R(P ) be the Hibi ring obtained from P . Set d = dim R(P ), e = e(R(P )), and v = v(R(P )). Then the following inequalities hold: 1 d ≤ e v ≤ (d− 1)! 2d−1 .
(1) The following conditions are equivalent: (a) Equality holds in the left inequlaity. (b) P is a totally ordered set.
(c) R(P ) is a polynomial ring over K. (2) The following conditions are equivalent:
(a) Equality holds in the right inequality. (b) P is an anti-chain.
(c) R(P ) is a Gorenstein Hibi ring with a(R(P )) =−2.
In what follows, we prove this theorem. In order to do that, we first give a way to construct a new poset Pq from a pair of a given poset P and a maximal element
q of P .
Definition 4.2. Let P be a poset, and let q be a maximal element of P . Then we define a poset Pq as follows:
Pq = P \ {q} as a set
and for any p1, p2 ∈ Pq, we have
p1 < p2 in Pq
def
⇐⇒ p1 < p2 in P .
P : Pq :
q
The next lemma yields a relationship between the invariants of P and Pq.
Lemma 4.3. Let P be a poset. Set W ={q1, . . . , qs}, the set of maximal elements of P . Then we have: (1) dim R(Pqi) = dim R(P )− 1. (2) v(R(Pqi)) < v(R(P )) for every i. (3) e(R(P )) = s ∑ i=1 e(R(Pqi)).
Proof. (1) It follows from ♯(Pqi) = ♯(P )− 1.
(2) The set P is a poset ideal of P but not a poset ideal of Pqi.
(3) We can compute e(R(P )) as the number of maximal chains in J(P ). Each maximal chain C in J(P ) is an increasing sequence of poset ideals as follows:
C :∅ = I0 ⊂ I1 ⊂ · · · ⊂ In−1 ⊂ In = P
such that ♯(Ii\ Ii−1) = 1. In particular, In−1 is equal to P \ {qi} for some unique i.
Then the sequence I0 =∅ ⊂ I1 ⊂ · · · ⊂ In−1 can be regarded as a maximal chain of
Pqi. Thus we obtain the required formula. □
Secondly, we define a poset P as an extension of a given poset P .
Definition 4.4. Let q′ be a point which is not contained in P and fix it. Then we
define a poset P as follows:
P = P ∪ {q′} as a set
and for any p1, p2 ∈ P , we have p1 < p2 in P
def
⇐⇒ p1 < p2 in P if p1, p2 ̸= q′
and q′ is incomparable to any element p∈ P .
P : P :¯
q′
We recall the definition of a direct sum of posets. Let P1, P2 be distinct two
posets. Then the direct sum P of P1 and P2 is a poset as follows: P = P1∪ P2 as a set
and for any elements p1, p2 ∈ P , we have p1 < p2 in P
def
⇐⇒ p1 < p2 if p1, p2 ∈ P1 or p1, p2 ∈ P2
and p1 and p2 are incomparable, otherwise.
P can be regarded as a direct sum of a poset P and one point. Thus P does
not depend on the choice of q′. In order to describe the embedding dimension, the
multiplicty of R(P ), we need the following lemma.
Lemma 4.5. Let P1 (resp. P2) be a poset consisting of m (resp. n) elements. Let P be the direct sum of two posets P1 and P2. Then we have:
(1) dim R(P ) = dim R(P1) + dim R(P2)− 1 = m + n + 1.
(2) v(R(P )) = v(R(P1))· v(R(P2)).
(3) e(R(P )) =(m+nm )e(R(P1))· e(R(P2)).
Proof. (1) is trivial.
(2) It follows from J(P ) ={I1⨿ I2 | I1 ∈ J(P1), I2 ∈ J(P2)}.
(3) We obtain(m+nm )maximal chains in P from a maximal chain in P1and a maximal
chain in P2. □
Corollary 4.6. Let P be a poset with n = ♯(P ). Let q be a maximal element of P . Then we have:
(1) dim R(P ) = dim R(P ) + 1 = n + 2. (2) v(R(P )) = 2· v(R(P )).
(3) e(R(P )) = (n + 1)· e(R(P )).
Remark 4.7. Note that R(P ) can be regarded as a Rees algebra of the graded maximal ideal m of R(P ).
Corollary 4.8. If P is an anti-chain poset with ♯(P ) = n, then
e(R(P )) = n!, v(R(P )) = 2n.
In what follows, we put e = e(R(P )), v = v(R(P )). Moreover, for a maximal element q of P , we put eq = e(R(Pq))), eq = e(R(Pq)), vq = v(R(Pq)) and vq =
v(R(Pq)). The following lemma is technical but it plays a key role in the proof of
the theorem.
Lemma 4.9. Let P be a poset with n = ♯(P ), and set W ={q1, . . . , qs}, the set of
maximal elements of P . If s≤ n
2 , then there exists a maximal element q such that
2· e ≤ eq.
Proof. By Lemma 4.3 and Corollary 4.6, we have e = s ∑ i=1 eqi = 1 n s ∑ i=1 eqi.
Suppose the contrary. Namely, we suppose that 2e > eqi for all i = 1, . . . , s. Then
we get n· e = s ∑ i=1 eqi < s ∑ i=1 2e = 2s· e ≤ n · e.
This is a contradiction. Hence we obtain the required assertion. □ If a poset P does not satify the assumption of Lemma 4.9, then it suffices to consider the poset of P instead of P . Let us recall the definition of the anti-poset.
Definition 4.10. Let P be a poset. Then the anti-poset of P , denoted by Po, is
the poset defined as follows:
Po = P as a set and for any elements p1, p2 ∈ Po, we have
p1 < p2 in Po ⇐⇒ p1 > p2 in P .
Lemma 4.11. If Po is the anti-poset of P , then R(P ) ∼= R(Po) as rings.
We are now ready to prove the main theorem. 8
Proof of Theorem 4.1. The left inequality follows from Proposition 2.2. So it suffices
to show the right inequality and that equality holds if and only if P is an anti-chain poset. We use an induction on n = d− 1. First suppose that P is a direct sum of two nonempty posets P1, P2. By induction hypothesis, we have
e1 v1 ≤ n1! 2n1 and e2 v2 ≤ n2! 2n2 , where ni = ♯(Pi) for i = 1, 2.
Since d− 1 = dim R(P ) − 1 = n1 + n2, we get e v = ( n1+ n2 n1 ) e1e2 v1v2 ≤ ( n1+ n2 n1 ) n1! 2n1 n2! 2n2 = (n1+ n2)! 2n1+n2 = (d− 1)! 2d−1 .
In particular, if P is an anti-chain, then d = n + 1, v = 2n and e = n! and thus
e v =
(d− 1)! 2d−1 .
Conversely, if equality holds for P then equality holds for P1 and P2, respectively.
Hence P1 and P2 are anti-chains, and thus P is also an anti-chain.
In the following, we may assume that ♯(P ) ≥ 2 and that P cannot be written as a direct sum of two posets. In particular, P has no isolated points. If necessary, by considering Po, we may assume that ♯{q ∈ P | q is a maximal element} ≤ n
2 . Then
one can take a maximal element q of P such that 2e≤ eq := e(R(Pq)). On the other
hand, we have that vq = 2vq< 2v. Therefore
eq· v − vq· e ≥ e(2v − vq) > 0.
Since ♯(Pq) < ♯(P ) = n and dim R(Pq) = dim R(P ) = d, the induction hypothesis
implies that e v < eq vq ≤ (d− 1)! 2d−1 . □
5. Examples
We first give the complete list of all posets P consisting of three points and the ratios e/v of R(P ). Note that we remove P from the list when P ̸= Po and Po
appears in it. P e 1 2 3 6 v 4 5 6 8 e/v 1/4 2/5 1/2 3/4 −a(R) 4 3 3 2
Secondly, we give the complete list of all posets P consisting of four points and the ratios e/v of R = R(P ). Note that “reg” (resp. “CI”, “Gor”) means that R(P ) is regular (resp. complete intersection, Gorenstein) in the table below.
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P e 1 2 2 3 4 4 5 6 6 8 12 24 v 5 6 6 7 8 7 8 9 9 10 12 16 e/v 1/5 1/3 1/3 3/7 1/2 4/7 5/8 2/3 2/3 4/5 1 3/2 −a(R) 5 4 4 4 4 3 3 3 3 3 3 2
R(P ) reg CI CI CI Gor Gor Gor Gor
ɹ
Remark 5.1. In the table above, although R(P8) and R(P9) are not isomorphic as
Hibi rings, the ratios of them are equal.
In the final of this section, we consider the following question.
Question 5.2. For any given α ∈ Q>0, does there exist Hibi ring R = R(P ) such
that e(R)/v(R) = α?
As an answer to this question, we prove the following theorem.
Theorem 5.3. For any α ∈ Q>0, there exists a Gorenstein Hibi ring R = R(P )
such that e(R)/v(R) = α.
Proof. For a given α = a/b∈ Q>0, where a and b are positive integer, we set
t = α−1· (4a)! − 24a.
Then t is a nonnegative integer. Moreover, we consider the following pure poset Pα:
. . . q1 q2 q3 q4a pt p2 p1 Pα:
One can easily see that
e = e(R(Pα)) = (4a)!, v = v(R(Pα)) = 24a+ t = α(4a)!
It follows that e/v = a/b = α. □
Remark 5.4. Adding a totally ordered set at the bottom yields a polynomial ex-tension of a Hibi ring.
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(Takahiro Chiba) Graduate School of Mathematics, Nagoya University, Nagoya 464–8602, Japan
E-mail address: [email protected]
(Kazunori Matsuda) Graduate School of Mathematics, Nagoya University, Nagoya 464–8602, Japan
E-mail address: [email protected]
(Ken-ichi Yoshida) Department of Mathematics, College of Humanities and Sci-ences, Nihon University, 3-25-40 Sakurajosui, Setagaya-ku, Tokyo 156–8550, Japan
E-mail address: [email protected]
