arXiv:1707.09843v1 [] 31 Jul 2017

RESULTS ON THE HILBERT COEFFICIENTS AND REDUCTION
NUMBERS
AMIR MAFI AND DLER NADERI
Abstract. Let (R, m) be a d-dimensional Cohen-Macaulay local ring, I an m-primary
ideal and J a minimal reduction of I. In this paper we study the independence of
reduction ideals and the behavior of the higher Hilbert coefficients. In addition, we give
some examples in this regards.

Introduction
Throughout this paper, we assume that (R, m) is a Cohen-Macaulay local ring of dimension d and the residue class field R/m is infinite. For an R-module M , let λ(M ) denote
the length of M . Let I be an m-primary ideal of R. The Hilbert-Samuel function HI (n)
of I is defined as HI (n) = λ(R/I n ). There exists a polynomial PI (n) of the form
PI (n) = e0

!

!

n+d−1
n+d−2
− e1
+ ... + (−1)d ed
d
d−1

such that PI (n) = HI (n) for all large n, where ei = ei (I) ∈ Z is called the Hilbert
coefficients of R with respect to I.
An ideal J ⊆ I is called a reduction ideal of I if I r+1 = JI r for some nonnegative
integer r (see [22]). The least such r is called the reduction number of I with respect
to J and denoted by rJ (I). A reduction ideal J is called a minimal reduction if it does
not properly contain a reduction ideal of I, under our assumption it is generated by a
regular sequence. The reduction number of I is defined as r(I) = min{rJ (I) : J is a
minimal reduction ideal of I}. The reduction number r(I) is said to be independent if
r(I) = rJ (I) for all minimal reduction J of I. Sally in [28] raised the following question: If
(R, m) is a d-dimensional Cohen-Macaulay local ring having an infinite residue field, then
is r(m) independent?
A natural extension of this question is to replace r(m) with r(I).
L
Let G(I) = n≥0 I n /I n+1 be the associated graded ring of I. Huckaba in [11] and Trung
in [30] independently proved that if depth G(I) ≥ d − 1, then r(I) is independent (see also
[19], [9], [10], [29], [17] and [18]). Moreover, Wu in [31] with some conditions proved that
if depth G(I) ≥ d − 2, then r(I) is independent. However if d ≥ 2 and depth G(I) ≤ d − 2,
then r(I) is not independent in general. Counter-examples have been obtained in [11], [19]
and [18].
It is known that e0 , e1 and e2 are nonnegative integers. Unfortunately, the well-behavior
of the Hilbert coefficients stops with e2 . Indeed, Narita in [21] showed that it is possible for
e3 to be negative (see also [3]). However, Itoh in [15] proved that if I is normal ideal then
e3 is a nonnegative integer (see also [3]). Puthenpurakal in [25] characterized remarkable
results about negativity of e3 .
2010 Mathematics Subject Classification. 13A30, 13D40, 13H10.
Key words and phrases. Hilbert coefficient, Minimal reduction, Associated graded ring.
1

The main purpose of this paper is to study the independence of reduction ideals and also
the behavior of the higher Hilbert coefficients. In the last section we collect some examples
which are disprove a question one can make about the behavior of Hilbert coefficients.
1. Main results
We begin this section by recalling some known notations which is studied in [13] and
[12]. An element x ∈ I \ I 2 is said to be superficial for I if there is an integer c such that
(I n+1 : x) ∩ I c = I n for all n ≥ c. If grade(I) ≥ 1 and x is a superficial element, then x is
a regular element of R and so by Artin-Rees Theorem I n+1 : x = I n for all n ≫ 0. If R/m
is infinite, then a superficial element for I always exists. A sequence x1 , ..., xs is called a
superficial sequence for I if x1 is superficial for I and xi is superficial for I/(x1 , ..., xi−1 )
for 2 ≤ i ≤ s. If I is an m-primary ideal and J is a minimal reduction of I, then there is
a superficial sequence x1 , ..., xd in I such that J = (x1 , ..., xd ). For any element x ∈ I we
let x∗ denote the image of x in I/I 2 . We note that if x∗ is a regular element of G(I), then
x is a regular element of R and G(I/(x)) = G(I)/(x∗ ).
Huckaba and Marley in [13] constructed the Koszul complex C. (x1 , ..., xd , n) which has
the following form
.

.



.

0 → R I n−d → (R I n−d+1 )d → (R I n−d+2 )
′

d
2



→

···

→ R/I n → 0.

Let C. (n) = C. (x1 , x2 , ..., xd , n) and C. (n) = C. (x1 , x2 , ..., xd−1 , n). For any n there is an
exact sequence of complexes
0 −→ C. ′ (n) −→ C. (n) −→ C. ′ (n − 1)[−1] −→ 0.
Thus, we have the corresponding long exact sequence on homology:
x

d
· · · −→ Hi (C. ′ (n)) −→ Hi (C. (n)) −→ Hi−1 (C. ′ (n − 1)) −→
Hi−1 (C. ′ (n)) −→ · · · .

For i ≥ 1, we define
∞
X

hi :=

λ(Hi (C. (n)))

n=1

and
ki :=

∞
X

(n − 1)λ(Hi (C. (n))).

n=2

By [13, §4], we have

∆d [PI (n) − HI (n)] = λ(I n /I n ∩ J) −

d
X

(−1)i λ(Hi (C. (n))

i=1

= λ(I n /JI n−1 ) −

d
X

(−1)i λ(Hi (C. (n))

i=2

and
ei (I) =

∞ 
X

n=i

n−1
i−1



∆d [PI (n) − HI (n)].
2

Hence by combining two previous formula, we have
e1 (I) =

∞
X

λ(I n /JI n−1 ) −

n=1

d
X

(−1)i hi =

∞
X

λ(I n /I n ∩ J) +

n=1

i=2

d
X

(−1)i−1 hi .

i=1

and
e2 (I) =

∞
X

(n − 1)λ(I n /JI n−1 ) −

n=2

d
X

(−1)i ki .

i=2

Lemma 1.1. Let I be an m-primary integrally closed ideal and J be a minimal reduction
of I. If e2 (I) = λ(I 2 /JI) + 1, then e2 (I) = e1 − e0 + λ(R/I), rJ (I) ≤ 3 and r(I) is
independent for any minimal reduction J of I.
Proof. By [14, Theorem 12] we have e1 − e0 + λ(R/I) ≤ e2 (I) = λ(I 2 /JI) + 1. Since I
is integrally closed, we have I 2 ∩ J = IJ. Suppose e1 − e0 + λ(R/I) ≤ λ(I 2 /JI). Thus
λ(I 2 /JI) +

d
P

λ(I n /I n ∩ J) +

n=3

have

d
P

λ(I n /I n

d
P

(−1)i−1 hi ≤ λ(I 2 /JI) and so by [13, Theorem 3.7] we

i=1
d
P

∩ J) = 0 and

n=3

and by [18, Lemma 3.2]

(−1)i−1 hi = 0. Therefore G(I) is Cohen-Macaulay

i=1

d
P

(−1)i−1 ki = 0. Since

d
P

(n − 1)λ(I n /I n ∩ J) = 0, we have

n=3

i=1

e2 (I) = λ(I 2 /JI) which is contradiction by our hypothesis. Therefore e1 − e0 + λ(R/I) =
λ(I 2 /JI)+1 and by [23, Theorem 3.4] I 4 = JI 3 and so r(I) is independent for any minimal
reduction J of I.

Corso, Polini and Rossi in [3, Remark 3.7] observed that if I is integrally closed ideal
and e2 (I) = 0, 1, 2, then G(I) is Cohen-Macaulay (see also [4], [5], [6] and [14]). Also they
observed that assumption on the ideal I being integrally closed cannot be weakened, see
[3, Example 3.8]. In the following proposition we prove that if I is an m-primary integrally
closed ideal and e2 (I) = 3 then depth G(I) ≥ d − 2 and r(I) is independent.
Proposition 1.2. Let I be an m-primary integrally closed ideal and J be a minimal
reduction of I. If e2 (I) = 3, then depth G(I) ≥ d − 2 and r(I) is independent.
Proof. Let e2 (I) = 3. By [14, Theorem 12] we have e1 − e0 + λ(R/I) ≤ e2 (I) = 3.
If e1 − e0 + λ(R/I) ≤ 2, then by [18, Lemma 3.15] ,depth G(I) ≥ d − 1 and r(I) is
independent. Hence we assume e1 − e0 + λ(R/I) = 3. In this case by [23, Corllary 4.7] we
have depth G(I) ≥ d − 2 and so by [18, Proposition 3.16] r(I) is independent.

Lemma 1.3. Let (R, m) be a Cohen-Macaulay local ring of dimension 2. Let I be an
m-primary ideal and J be a minimal reduction of I. If rJ (I) ≤ 2 and I˜ = I, then
depth G(I) ≥ 1.
Proof. If rJ (I) ≤ 2, then we have I n ∩ J = JI n−1 for all n ≥ 3. Since I˜ = I, we have
I 2 : x = I for any superficial element x in I. Hence by [16, Proposition 2.1], we have
Ifn = I n for any n ≥ 1 and so depth G(I) ≥ 1.


An ideal I is said to be asymptotically normal if there exists an integer k ≥ 1 such
that I n is integrally closed for all n ≥ k. In [3, Remark 4.3] there is interesting examples
of asymptotically normal ideals that are not normal. The following result improves [14,
Proposition 16].
3

Theorem 1.4. Let (R, m) be a Cohen-Macaulay local ring of dimension 3. Let I be an
m-primary ideal and J be a minimal reduction of I. Assume that I is an asymptotically
normal ideal and I˜ = I. Then PI (n) = HI (n) for n = 1, 2 if and only if rJ (I) ≤ 2.
Proof. Let x be superficial element of I and set A = R/(x), B = IA and C = JA. Then
dim A = 2. Since I is an asymptotically normal ideal by [24, Corollary 7.11] B̃ = B. If
rJ (I) ≤ 2, then rC (B) ≤ 2. Therefore by Lemma 1.3 depth G(B) ≥ 1 and so by [13,
Lemma 2.2] depth G(I) ≥ 2. Hence by [20, Theorem 2] we have rJ (I) = n(I) + 3. Thus
n(I) ≤ −1 and PI (n) = HI (n) for all n ≥ 0. Conversely, If PI (n) = HI (n) for n = 1, 2,
then by [31, Lemma 2.3] we have PB (n) = HB (n) for n = 1, 2. So by [14, Proposition 16]
rC (B) ≤ 2 and by Lemma 1.3 depth G(B) ≥ 1. Therefore depth G(I) ≥ 2 and rJ (I) ≤ 2.

Let R(I) be the Rees-algebra of I and E be an R(I)-module. Then in the following
theorem we set H i (E) to be the ith- local cohomology module of E with respect to the
maximal homogeneous ideal of R(I).
Theorem 1.5. Let (R, m) be a Cohen-Macaulay local ring of dimension 4. Let I be an
m-primary ideal and J be a minimal reduction of I. If I is an asymptotically normal ideal
and rJ (I) ≤ 3, then e4 (I) ≤ 0.
Proof. By [9, Lemma 2.4] for n ≫ 0 we have
H i (G(I n ))j = 0 f or j ≥ 1 and i = 0, 1, 2, 3, 4
and since rJ (I) ≤ 3 we get by [30, Proposition 3.2] and [9, Lemma 2.4]
H 4 (G(I n ))0 = 0.
Let q be the integer such that I q is normal. By [24, Theorem 7.3], depth G(I n ) ≥ 2 for
n ≫ 0 and so H i (G(I n )) = 0 for n = 0, 1. Thus
a2 (I n ) < a3 (I n ) = 0
and therefore a2 (I n ) ≤ −1 and H 2 (G(I n ))0 = 0.
Set hi = λ(H i (G(I n ))0 ) for i = 0, 1, 2, 3, 4. Then
h0 = h1 = h2 = h4 = 0.
Set K := I n for n ≫ 0 and let
PK (z) = c0



z+3
3



− c1



z+2
2

.





+ c2



z+1
1



− c3

be the Hilbert polynomial of G(K) i.e.


PK (i) = λ K i K i+1 f or i ≫ 0.
By Grothendieck-Serre formula we get

HK (i) − PK (i) =

4
X

(−1)s λ(H s (G(K))i ).

s=0

Set i = 0 we get
λ (R/I n ) − [c0 − c1 + c2 − c3 ] = h0 − h1 + h2 − h3 + h4 = −h3 .
4

Let ϕI (n) be the Hilbert-Samuel polynomial of I i.e.,
ϕI (z) = e0 (I)



z+3
4

Write
ϕK (z) = c0





− e1 (I)

z+3
4





z+2
3


− c1



+ e2 (I)


z+2
3

+ c2





z+1
2

z+1
2





− e3 (I)

− c3



z
1




z
1



+ e4 (I).

+ c4 .

Clearly ϕK (z) = ϕI (nz). In particular c4 = e4 (I). Also notice that
ϕK (1) = c0 − c1 + c2 − c3 + c4 = ϕI (n) = λ (R/I n ) .
So we get c4 = −h3 . Thus e4 = −h3 . Thus e4 ≤ 0.
For any ideal I the set of ideals (I n+1 : I n ) forms an ascending chain. Let Ie denote the
union of these ideals. Ratliff and Rush [26] showed that Ie is the largest ideal containing I
which has the same Hilbert polynomial as I. In the following proposition we use notations:
L
L
I n+1 :x
n+1 .
B I (x, R) = ∞
and LI (R) = ∞
n=0 I n
n=0 R/I
Proposition 1.6. Let (R, m) be a Cohen-Macaulay local ring of dimension 3. Let I be
an m-primary ideal and J be a minimal reduction of I. If e2 (I) ≤ 1,then e3 (I) ≤ 0.

Proof. By [25, Proposition 6.4] we can assume that e2 (I) = 1. Let x be a superficial
element of I. Set A = R/(x) , B = IA and C = JA, then dim A = 2, ei (I) = ei (B) for
i = 0, 1, 2 and e2 (B) = 1. By [13, Corollary 4.13] we have
e2 (B) = e^
2 (B) =

X

n≥1

Therefore we have

X

n≥2

.

]
n+1 C g
nλ(B
B n ) = 1.

.

]
n+1 C g
λ(B
B n ) = 0.

Thus by [8, Proposition 1.9] we have
e^
3 (B) =

X n 

n≥2

2

.

]
n+1 C g
λ(B
B n ) = 0.

By using [24, 6.3] we have the following exact sequence

0 → B I (x, R) → H 0 (LI (R))(−1) → H 0 (LI (R)) → H 0 (LI (A)).
Therefore we obtain
b := λ(B I (x, R)) ≤ λ(H 0 (LI (A))) =: r.
By using [25, 1.5] we get that e3 (B) = e^
3 (B) − r and by [27, Proposition 1.2] e3 (I) =
e3 (B) + b then e3 (I) ≤ 0.

The computation of examples is performed by using Maxaulay2 [7] and CoCoA [1].
As an application of Proposition 1.6 we give the following example.
Example 1.7. Let R = k[x, y, z](x,y,z) , where k is a field, and let I = (x3 , y 3 , x2 y +
z 3 , xz 2 , y 2 z + x2 z). Then depth G(I) = 0 and we have the following Hilbert series
PI (t) =

16 + 5t + 5t2 − 5t3 + 6t4 + 10t5 − 13t6 + 2t7 + t8
.
(1 − t)3
5

and Hilbert polynomial
PI (n) = 27



n+2
3



− 18



n+1
2



+





n
1

+ 15.

Hence e2 (I) = 1 and e3 (I) ≤ 0.
Theorem 1.8. Let (R, m) be a Cohen-Macaulay local ring of dimension 3, I an m-primary
integrally closed ideal and J a minimal reduction of I. If e1 (I) − e0 (I) + λ(R/I) = e2 (I),
then e3 (I) ≤ 0.
Proof. By [14, Lemma 11] there exists a superficial element x of I so that I/(x) is an
integrally closed ideal of A = R/(x). Set B = IA and C = JA. Then dim A = 2,
ei (I) = ei (B) for i = 0, 1, 2 and e1 (B) − e0 (B) + λ(A/B) = e2 (B). By [13, Corollary 4.13]
we have
^
e1 (B) − e0 (B) + λ(A/B) = e^
1 (B) − e0 (B) + λ(A/B) =

X

n≥1

and
e2 (B) = e^
2 (B) =

X

n≥1

Therefore

X

n≥2

and so
e^
3 (B) =

.

.

]
n+1 C g
λ(B
B n)

.

]
n+1 C g
nλ(B
B n ).

]
n+1 C g
B n) = 0
λ(B

X n 

2

n≥2

.

]
n+1 C g
B n ) = 0.
λ(B

By using [24, 6.3] we have the following exact sequence

0 → B I (x, R) → H 0 (LI (R))(−1) → H 0 (LI (R)) → H 0 (LI (A)).
Hence
b := λ(B I (x, R)) ≤ λ(H 0 (LI (A))) =: r.
by using [25, 1.5] we get that e3 (B) = e^
3 (B) − r and so by [27, Proposition 1.2] e3 (I) =
e3 (B) + b. Thus e3 (I) ≤ 0.
As an application of Theorem 1.8 we give the following example.
Example 1.9. Let R = k[x, y, z, u, v, w](x,y,z,u,v,w) , where k is a field. Let Q = (z 2 , zu, zv, uv, u3 −
yz, v 3 − xz), m = (x, y, z, u, v, w) and S = R/Q. Then S is a 3-dimensional CohenMacaulay local ring and m is a maximal ideal (integrally closed m-primary ideal) of S.
Then depth G(I) = 1 and we have the following Hilbert series
PI (t) =

1 + 3t + 3t3 − t4
,
(1 − t)3

and Hilbert polynomial
PI (n) = 6



n+2
3



−8



n+1
2

Thus e1 (I) − e0 (I) + λ(R/I) = e2 (I) and e3 (I) ≤ 0.
6



+3



n
1



+ 1.

2. Examples
Marley in [20] proved that if (R, m) is a d-dimensional Cohen-Macaulay local ring
and I an m-primary ideal such that depth G(I) ≥ d − 1, then all Hilbert coefficients
e0 (I), e1 (I), ..., ed (I) are non-negative integers.
Thus it is natural to ask if d − 3 ≤ depth G(I) = t ≤ d − 2 whenever d ≥ 3, then is
ei (I) ≥ 0 for i = t, t + 1, ..., d? In the following examples we show that the question is
negative.
Example 2.1. Let R = k[x, y, z, u](x,y,z,u) , where k is a field, and
I = (x3 , y 3 , z 3 , u3 , xy 2 , yz 2 , zu2 , xyz, xyu). Then we have depth G(I) = 2, Hilbert series
PI (t) =

33 + 19t + 21t2 + 7t3 + 5t4 − 3t5 − t6
(1 − t)4

and Hilbert polynomial
PI (n) = 81



n+3
4



− 81





n+2
3

+ 27



n+1
2



+ 23



n
1



− 50.

Hence e0 (I) = 81, e1 (I) = 81, e2 (I) = 27, e3 (I) = −23 and e4 (I) = −50.
In the following example we show that if (R, m) is a 5-dimensional Cohen-Macaulay
local ring and I an m-primary ideal and depth G(I) = 3, then ei (I) < 0 for i = 3, 4, 5.
Example 2.2. Let R = k[x, y, z, u, v](x,y,z,u,v) , where k is a field and
I = (x3 , y 3 , z 3 , u2 , v, xy 2 , yz 2 , xyz, xyu). Then we have depth G(I) = 3, Hilbert series
PI (t) =

28 + 11t + 10t2 + 5t3 + t4 − t5
(1 − t)5

and Hilbert polynomial
PI (n) = 54



n+4
5



− 45



n+3
4



+ 21



n+2
3



+



n+1
2



−4



n
1



+ 1.

Therefore e0 (I) = 54, e1 (I) = 45, e2 (I) = 21, e3 (I) = −1, e4 (I) = −4 and e5 (I) = −1.
In the following example we show that if (R, m) is a 4-dimensional Cohen-Macaulay
local ring and I an m-primary ideal, depth G(I) = 2 and e3 (I) > 0, then e4 (I) < 0.
Example 2.3. Let R = k[x, y, z, u](x,y,z,u) , where k is a field and
I = (x4 , y 4 , z 4 , u4 , x2 y 2 , y 2 z 2 , z 2 u2 , xyz, xyu). Then we have depth G(I) = 2, Hilbert series
PI (t) =

81 + 58t + 31t2 + 7t3 − t4
(1 − t)4

and Hilbert polynomial
PI (n) = 176



n+3
4



− 137



n+2
3



+ 46



n+1
2



−3



n
1



− 1.

So we have e0 (I) = 176, e1 (I) = 137, e2 (I) = 46, e3 (I) = 3 and e4 (I) = −1.
In the following example we show that if (R, m) is a 4-dimensional Cohen-Macaulay
local ring and I an m-primary ideal, depth G(I) = 1 then e4 (I) > 0 but e3 (I) < 0.
7

Example 2.4. Let R = k[x, y, z, u](x,y,z,u) , where k is a field and
I = (x3 , y 3 , z 3 , u3 , xy 2 , xz 2 , xu2 , xyz, xyu). Then we have depth G(I) = 1, Hilbert series
PI (t) =

37 + 14t + 17t2 + 15t3 + 6t4 − 12t5 + 4t6
(1 − t)4

and Hilbert polynomial
PI (n) = 81



n+3
4



− 81



n+2
3



+ 38



n+1
2





+

n
1



+ 6.

Thus e0 (I) = 81, e1 (I) = 81, e2 (I) = 38, e3 (I) = −1 and e4 (I) = 6.
In the following example (R, m) is a 5-dimensional Cohen-Macaulay local ring, I an
m-primary ideal, depth G(I) = 3 and e4 (I) < 0 but e5 (I) ≥ 0.
Example 2.5. Let R = k[x, y, z, u, v](x,y,z,u,v) , where k is a field and
I = (x4 , y 4 , z 4 , u4 , v, x2 y 2 , y 2 z 2 , z 2 u2 , xyz, xyu). Then depth G(I) = 3, Hilbert series
PI (t) =
and Hilbert polynomial
PI (n) = 176



n+4
5



− 137



81 + 58t + 31t2 + 7t3 − t4
(1 − t)5
n+3
4



+ 46



n+2
3



−3



n+1
2



+



n
1



.

Thus e0 (I) = 176, e1 (I) = 137, e2 (I) = 46, e3 (I) = 3, e4 (I) = −1 and e5 (I) = 0.
In the following example we prove that if (R, m) is a d-dimensional Cohen-Macaulay
local ring and I an m-primary ideal, then all Hilbert coefficients are positive however
depth G(I) ≤ d − 2.
Example 2.6. Let R = k[x, y, z](x,y,z) , where k is a field, and I = (x4 , y 4 , z 4 , x3 y, y 3 z, xyz).
Then we have depth G(I) = 1, Hilbert series
PI (t) =
and Hilbert polynomial
PI (n) = 48



n+2
3



31 + 9t + 7t2 + t3
(1 − t)3

− 26



n+1
2



+ 10



n
1



− 1.

Thus all Hilbert coefficients are positive.
In the following example we show that depth G(I) = 0 but all Hilbert coefficients are
non negative.
Example 2.7. Let R = k[x, y, z](x,y,z) , where k is a field and I = (x4 , y 4 , z 4 , x3 y, xy 3 , y 3 z, yz 3 , x3 y, xy 3 ).
Then depth G(I) = 0, Hilbert series
PI (t) =

30 + 12t + 22t2 + 8t3 − 2t4 − 12t5 + 6t6
(1 − t)3

and Hilbert polynomial
PI (n) = 64



n+2
3



− 48



Hence all Hilbert coefficients are no negative.
8

n+1
2



+4



n
1



.

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Amir Mafi, Department of Mathematics, University Of Kurdistan, P.O. Box: 416, Sanandaj, Iran.
E-mail address: A_Mafi@ipm.ir
Dler Naderi, Department of Mathematics, University of Kurdistan, P.O. Box: 416, Sanandaj, Iran.
E-mail address: dler.naderi65@gmail.com

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