arXiv:1404.5026v1 [] 20 Apr 2014

THE SECOND HILBERT COEFFICIENTS AND
THE SECTIONAL GENERA OF IDEALS
KAZUHO OZEKI

1. Introduction
The purpose of our paper is to study the relationship between the second Hilbert
coefficients and the sectional genera of ideals.
Let A be a Cohen-Macaulay local ring with maximal ideal m and d = dim A > 0.
For simplicity, throughout this paper, we assume that the residue class field A/m of A
is infinite. Let I be a fixed m-primary ideal in A and Q = (a1 , a2 , · · · , ad ) a parameter
ideal of A which forms a reduction of I, that is the equality I n+1 = QI n holds true for
some n > 0. Let ℓA (N) denote, for an A-module N, the length of N. Then there exist
integers {ei (I)}0≤i≤d such that




n+d−1
n+d
n+1
+ · · · + (−1)d ed (I)
− e1 (I)
ℓA (A/I ) = e0 (I)
d−1
d
for all n ≫ 0. We call ei (I) the i-th Hilbert coefficient of I and especially call the
leading coefficient e0 (I) = e0 (Q) = ℓA (A/Q) (> 0) the multiplicity of I. Let
R′ (I) = A[It, t−1 ] ⊆ A[t, t−1 ] and gr(I) = R′ (I)/t−1 R′ (I)
where t is an indeterminate over A.
As a classical result of Northcott [No], the inequality e1 (I) ≥ e0 (I) − ℓA (A/I) (≥ 0)
holds true for every m-primary ideals I in A, and Huneke [H] and Ooishi [O] showed
that the equality e1 (I) = e0 (I) − ℓA (A/I) holds true if and only if I 2 = QI.
this
L When
n
is the case, the associated graded ring gr(I) and the fiber cone F(I) = n≥0 I /mI n
of I are both Cohen-Macaulay, and the Rees algebra R(I) = A[It] ⊆ A[t] of I is also
a Cohen-Macaulay ring, provided d ≥
 2. We also notice that, Kirby and Mehran [KM]
e0 (I)
were able to show that e1 (I) ≤ 2 .
The purpose of this paper is to study the second Hilbert coefficients e2 (I) of I. As
is well known, Narita [Na] showed that the inequality e2 (I) ≥ 0 holds true for any
m-primary ideal I in A. In [KM], it was proved that for m-primary ideals I in A, an
upper bound


e1 (I) + 1
e2 (I) ≤
2
of the second Hilbert coefficient e2 (I) in terms of the first Hilbert coefficient e1 (I).
Key words and phrases: Hilbert function, second Hilbert coefficient, associated graded rings
AMS 2010 Mathematics Subject Classification: 13D40, 13A30, 13H15, 13H10.
1

Our first purpose is to give an upper bound of the second Hilbert coefficients e2 (I)
in terms of the sectional genera of I. Let
gs (I) = ℓA (A/I) − e0 (I) + e1 (I)
denotes the sectional genus of I. When d = dim A ≥ 2, we shall show that the inequality


gs (I) + 1
e2 (I) ≤
2
holds true for every m-primary ideal I in A (Proposition 3.2). The following theorem
which is the first main result of this paper shows that the upper bound e2 (I) ≤ gs (I)+1
2

is sharp, clarifying when the equality e2 (I) = gs (I)+1
holds
true.
2

Theorem 1.1. Suppose that d ≥ 2 and let g = gs (I). Then the following three conditions are equivalent:


g+1
,
(1) e2 (I) =
2
(2) ℓA (I 2 /QI) ≤ 1,
(3) ℓA (I k+1 /QI k ) = 1 for all 1 ≤ k ≤ g and I g+2 = QI g+1 .

When this is the case we have the following.


g+1
for 3 ≤ i ≤ d,
(i) ei (I) =
i
(ii) depth gr(I) ≥ d − 1, and
(iii) gr(I) is a Cohen-Macaulay ring if and only if Q ∩ I 2 = QI and I 3 = QI 2 .

Thus the second Hilbert coefficients e2 (I) bounded above by gs (I)+1
. It seems now
2

natural to ask what happens on the m-primary ideals I with e2 (I) < gs (I)+1
. The
2
second main result of this paper answers the question and is stated as follows (Theorem
3.4).
Theorem 1.2. Suppose that d ≥ 2. Let g = gs (I) and assume that g ≥ 2. Then the
following three conditions are equivalent:
 
g
+ 1,
(1) e2 (I) =
2


 
g+1
g
,
+ 1 ≤ e2 (I) <
(2)
2
2
(3) ℓA (I 2 /QI) = 2, ℓA (I k+1 /QI k ) = 1 for all 2 ≤ k ≤ g − 1, and I g+1 = QI g .
When this is the case, we have the following:

(i) ei (I) = gi for 3 ≤ i ≤ d,
(ii) depth gr(I) ≥ d − 1, and
(iii) gr(I) is a Cohen-Macaulay ring if and only if Q ∩ I 2 = QI, Q ∩ I 3 = QI 2 , and
I 4 = QI 3 .
2



As a direct consequence of Theorem 1.2, we get e2 (I) = gs2(I) +1 or e2 (I) = gs (I)+1
,
2

gs (I)
if e2 (I) ≥ 2 + 1 (Corollary 4.2).
We now briefly explain how this paper is organized. In Section 2 we will summarize,
for the later use in this paper, some auxiliary results on the Hilbert coefficients and
the sectional genera of m-primary ideals. We shall prove Theorem 1.1 in Section 3
(Theorem 3.4). Theorem 1.2 will be proven in Section 4 (Theorem 4.1). We will show
in Section 5 examples of m-primary ideals satisfying the equality in Theorem 1.1 (1)
and Theorem 1.2 (1), respectively.
In what follows, unless otherwise specified, let A be a Cohen-Macaulay local ring
with maximal ideal m and d = dim A > 0. We throughout assume that the field A/m
is infinite. Let I be an m-primary ideal in A and Q = (a1 , a2 , · · · , ad ) a parameter ideal
of A which forms a reduction of I. Let rQ (I) = sup{n ≥ 0 | I n+1 = QI n } denotes the
reduction number of I with respect to Q. For each m-primary ideal I in A we set
R′ = R′ (I) = A[It, t−1 ] ⊆ A[t, t−1 ] and gr(I) = R′ /t−1 R′
where t is an indeterminate over A.
2. Preliminary steps
In this section we summarize some basic properties and known results of the Hilbert
coefficients, and the sectional genera of ideals, which we need throughout this paper.
The following result is, more or less, known (c.f. [CPP, Corollary 1.3], [RV2, Section
2.2]). Let us indicate a brief proof for the sake of completeness, because it plays a key
role in the proofs of our main theorems. Set H(gr(I), t) denotes the Hilbert series of
the associated graded ring gr(I) of I.
Lemma 2.1. Suppose that d = 1. Then we have
Pr−1
{ℓA (I k /QI k−1 ) − ℓA (I k+1/QI k )}tk + ℓA (I r /QI r−1 )tr
ℓA (A/I) + k=1
H(gr(I), t) =
1−t
where r = rQ (I).
Proof. Let

P

hk tk
1−t
with hk ∈ Z for k ≥ 0. Then we have h0 = ℓA ([gr(I)]0 ) = ℓA (A/I) and
H(gr(I), t) =

k≥0

hk = ℓA ([gr(I)]k ) − ℓA ([gr(I)]k−1) = ℓA (I k /I k+1) − ℓA (I k−1 /I k )
for all k ≥ 1. Thus, by the exact sequences
0 → I k+1 /QI k → I k /QI k → I k /I k+1 → 0
and
a

0 → I k−1 /I k →1 I k /QI k → I k /QI k−1 → 0
3

of A-modules, we get
hk = ℓA (I k /I k+1 ) − ℓA (I k−1 /I k )
= {ℓA (I k /QI k ) − ℓA (I k+1 /QI k )} − {ℓA (I k /QI k ) − ℓA (I k /QI k−1)}
= ℓA (I k /QI k−1 ) − ℓA (I k+1 /QI k )
for all k ≥ 1 as required.



We also need the following lemma.
Lemma 2.2. Suppose that d ≥ 2 and let a ∈ I be a superficial element for I. Then we
have gs (I) = gs (I/(a)).
Proof. We set A = A/(a) and I = I/(a). Then we have gs (I) = ℓA (A/I)−e0 (I)+e1 (I) =
ℓA (A/I) − e0 (I) + e1 (I) = gs (I) as ei (I) = ei (I) for all i = 0, 1, and ℓA (A/I) =
ℓA (A/I).

Thanks to Lemma 2.1, the following result holds true.
Lemma 2.3. Suppose that d = 2 and let a ∈ I be a superficial element for I. Then we
have the following, where A = A/(a), I = I/(a), Q = Q/(a), r ′ = rQ (I), and m be a
positive integer such that I n+1 :A a = I n for all n ≥ m.
′ −1
rX
k+1
k
(1) gs (I) =
ℓA (I /Q I ),
(2) e2 (I) =

k=1
′ −1
rX

k·ℓA (I

k+1

k

/Q I ) −

k=1

m−1
X

ℓA ([I k+1 :A a]/I k ).

k=1

k+1

k

/Q I ) for all k ≥ 0. Then, by Lemma 2.1, we have
P ′ −1
′
{vk−1 − vk }tk + vr′ −1 tr
ℓA (A/I) + rk=1
H(gr(I), t) =
1−t
Pr′ −1
Pr′ −1
so that e1 (I) = k=0 vk and e2 (I) = k=1 k·vk . Therefore, we get

Proof. We set vk = ℓA (I

gs (I) = gs (I) = ℓA (A/I) − e0 (I) + e1 (I) =

′ −1
rX

vk

k=1

by Lemma 2.2, because v0 = ℓA (I/Q) = e0 (I) − ℓA (A/I). We also get
e2 (I) = e2 (I) −

m−1
X

ℓA ([I

k+1

k

:A a]/I ) =

k=1

′ −1
rX

k=1

k·vk −

m−1
X

ℓA ([I k+1 :A a]/I k )

k=1

as required (c.f. [RV2, Proposition 1.2]).



The following result is due to Huckaba and Marley [HM].
4

Theorem 2.4. ([HM, Theorem 4.7]) Suppose that d ≥ 1. Then we have the inequality
X
e1 (I) ≤
ℓA (I k+1 /QI k )
k≥0

and the following two conditions are equivalent:
P
(1) e1 (I) = k≥0 ℓA (I k+1/QI k ),
(2) depth gr(I) ≥ d − 1.
When this is the case we have

X  k 
ℓA (I k+1 /QI k )
ei (I) =
i
−
1
k≥i−1

for all 2 ≤ i ≤ d.
Proof. See, for example, [RV2, Theorem 2.5].



The following result was proved by Wang [W] and, at the same time by Rossi and
Valla [RV1].
Theorem 2.5. ([RV1, W]) Suppose that d ≥ 1 and assume that ℓA (I 2 /QI) ≤ 1. Then
we have the following.
(1) depth gr(I) ≥ d − 1,
(2) ei (I) = gs (I)+1
for all 2 ≤ i ≤ d.
i
Proof. (1) See [S1, W] ([RV2, Theorem 4.4] also).
(2) We have ℓA (I k+1 /QI k ) ≤ 1 for all k ≥ 1 by [CPP, Lemma 3.8]P(see [S1, W] also).
Since depth gr(I) ≥ d−1, we have gs (I) = ℓA (A/I)−e0 (I)+e1 (I) = k≥1 ℓA (I k+1 /QI k )
k+1
k
g+2
by Theorem 2.4, so that we
= QI g+1 .
 have ℓA (I /QI ) = 1 for all 1 ≤ k ≤ g and I
g+1
Thus, we get ei (I) = i for all 2 ≤ i ≤ d by Theorem 2.4.


Thus the proofs of the implication (2) ⇒ (1) and the last assertion (i) in Theorem 1.1
are given by Theorem 2.5.
3. An upper bound for the second Hilbert coefficient
The purpose of this section is to estimate the second Hilbert coefficients of m-primary
ideals in terms of the sectional genera.
Let us begin with the following lemma.

Lemma 3.1. Let ℓ ≥ 0 be an integer. Suppose that {vk }k≥1 is the set of integers such
that
(i) P
vk ≥ 0 for all k ≥ 1,
(ii) k≥1 vk ≤ ℓ, and
(iii) vj = 0 for all j ≥ k once vk = 0 for some k ≥ 1.
5

Then we have



ℓ+1
k·vk ≤
2
k≥1

X

and the following two conditions are equivalent:

P
(1) k≥1 k·vk = ℓ+1
,
2
(2) vk = 1 for all 1 ≤ k ≤ ℓ, and vk = 0 for all k ≥ ℓ + 1.
Proof. We proceed by induction on ℓ. When ℓ ≤ 1, we have nothing to do. Assume
that ℓ ≥ 2 and that our assertion holds true for ℓ − 1. We may assume that v1 ≥ 1. We
set wk = vk+1 for all k ≥ 1. Then, by the hypothesis of induction on ℓ, we have
  

X
X
ℓ
ℓ − v1 + 1
≤
(k − 1)·vk =
k·wk ≤
2
2
k≥2
k≥1
P
P
because k≥1 wk = k≥1 vk − v1 ≤ ℓ − v1 ≤ ℓ − 1. Therefore we get the required
inequality


 
X
X
X
ℓ+1
ℓ
.
+ℓ=
k·vk =
(k − 1)·vk +
vk ≤
2
2
k≥1
k≥2
k≥1
In the rest of
3.1, we have only to show the implication (1) ⇒ (2).

Pour proof of Lemma
ℓ+1
Assume that k≥1 k·ak = 2 . Then we have
  

X
ℓ
ℓ − v1 + 1
=
k·wk =
2
2
k≥1
by the above argument, and hence v1 = 1. Thus, by the hypothesis of induction on
ℓ, we get vk = wk−1 = 0 for all 2 ≤ k ≤ ℓ and vk = wk−1 = 0 for all k ≥ ℓ + 1 as
required.

We notice that the set of integers {ℓA (I k+1 /QI k )}k≥1, for an m-primary ideal I and
a reduction Q of I, satisfies all conditions (i), (ii), and (iii) in Lemma 3.1.
The following result gives an upper bound for e2 (I) in terms of the sectional genus
gs (I) of I, and also shows that the implication (1) ⇒ (3) and the assertion (ii) in
Theorem 1.1 (Theorem 3.4) hold true.
Proposition 3.2. Suppose that d ≥ 2 and let g = gs (I). Then we have the inequality


g+1
.
e2 (I) ≤
2

If e2 (I) = g+1
then the following two assertions hold true.
2
(1) depth gr(I) ≥ d − 1,
(2) ℓA (I k+1 /QI k ) = 1 for all 1 ≤ k ≤ g and I g+2 = QI g+1 .
Thanks to Proposition 3.2, we get the following inequality which is given by Kirby
and Mehran [KM].
6

Corollary 3.3. Suppose that d ≥ 2. Then we have the inequality


e1 (I) + 1
.
e2 (I) ≤
2

We furthermore have I = Q, once the equality e2 (I) = e1 (I)+1
holds true.
2

Proof. The inequality e2 (I) ≤ e1 (I)+1
holds true by Proposition 3.2. Assume that
2



e1 (I)+1
e2 (I) =
. Then, because e2 (I) = e1 (I)+1
= gs (I)+1
, we have e1 (I) = gs (I), so
2
2
2
that ℓA (I/Q) = e0 (I)−ℓA (A/I) = e1 (I)−gs (I) = 0. Thus we get I = Q as required. 
Proof of Proposition 3.2. We proceed by induction on d. Since the residue class field
A/m of A is infinite, we may choose an element a ∈ Q\mQ is superficial for I. We set
A = A/(a), I = I/(a), Q = Q/(a), and r ′ = rQ (I).
P ′ −1
k+1
k
Suppose that d = 2. Let vk = ℓA (I /Q I ) for all k ≥ 1. Then we have g = rk=1
vk
by Lemma 2.3. Therefore, by Lemma 2.3 and Lemma 3.1, we get
′ −1
′ −1


rX
m−1
rX
X
g+1
k+1
k
,
e2 (I) =
k·vk −
ℓA ([I
:A a]/I ) ≤
k·vk ≤
2
k=1
k=1
k=1

where m ≥ 1 be an integer such that I k+1 :A a = I k for all k ≥ m. Assume e2 (I) = g+1
2

P ′ −1
k+1
k
k·vk = g+1
. Therefore, we get ℓA (I /Q I ) = vk = 1 for all
then we have rk=1
2
k+1

k

1 ≤ k ≤ g and ℓA (I /Q I ) = vk = 0 for all k ≥ g + 1 by Lemma 3.1, and hence
g+2
g+1
we have I
= Q I . We also have depth gr(I) ≥ 1 because I k+1 :A a = I k for all
k ≥ 1. Then, since at forms a gr(I)-regular element, we get ℓA (I k+1/QI k ) = 1 for all
1 ≤ k ≤ g and I g+2 = QI g+1 as required.
Assume that d ≥ 3 and that our assertion holds true for d−1. Then by the hypothesis
of induction on d, we get the required inequality


 
gs (I) + 1
g+1
=
e2 (I) = e2 (I) ≤
2
2

then we have e2 (I) =
because gs (I) = g by Lemma 2.2. Assume that e2 (I) = g+1
2

gs (I)+1
. The hypothesis of induction on d says that depth gr(I) ≥ (d−1)−1 = d−2 > 0,
2
k+1

k

g+2

g+1

= Q I . Then, thanks to Sally’s
ℓA (I /Q I ) = 1 for all 1 ≤ k ≤ g, and I
technique ([S1], [HM, Lemma 2.2]), at forms a gr(I)-regular element. Thus we get
depth gr(I) ≥ d − 1, ℓA (I k+1 /QI k ) = 1 for all 1 ≤ k ≤ g, and I g+2 = QI g+1 as required.
This completes the proof of Proposition 3.2.


Thus the
 second Hilbert coefficients e2 (I) of m-primary ideals I are bounded above
gs (I)+1
by
. It seems now natural to ask what happens on the m-primary ideals I of A,
2

once the equality e2 (I) = gs (I)+1
is attained. The main result of this section answers
2
the question and is stated as follows (Theorem 1.1).
7

Theorem 3.4. Suppose that d ≥ 2 and let g = gs (I). Then the following three conditions are equivalent:


g+1
,
(1) e2 (I) =
2
(2) ℓA (I 2 /QI) ≤ 1,
(3) ℓA (I k+1 /QI k ) = 1 for all 1 ≤ k ≤ g and I g+2 = QI g+1 .
When this is the case we have the following.


g+1
for 3 ≤ i ≤ d,
(i) ei (I) =
i
(ii) depth gr(I) ≥ d − 1, and
(iii) gr(I) is a Cohen-Macaulay ring if and only if Q ∩ I 2 = QI and I 3 = QI 2 .
Proof. (1) ⇒ (3) and (ii) By Proposition 3.2.
(3) ⇒ (2) It is clear.
(2) ⇒ (1) and (i) By Theorem 2.5.
Now we have only to show that the assertion (iii) holds true. Assume that gr(I) is a
Cohen-Macaulay ring. Then we have Q ∩ I n+1 = QI n for all n ≥ 1 by Valabrega-Valla
criterion ([VV]). Since ℓA (I 2 /QI) ≤ 1, we have I 3 ⊆ QI, so that I 3 = Q ∩ I 3 = QI 2 .
The converse also holds true by Valabrega-Valla criterion ([VV]). This completes the
proof of Theorem 3.4.

From now on, we introduce two consequences of Theorem 3.4.
The following result gives a characterization of ideals I with ℓA (I/Q) ≤ 1 in terms
of e1 (I) and e2 (I).
Corollary 3.5. Suppose that d ≥ 2. Then the following two conditions are equivalent:


e1 (I)
,
(1) e2 (I) =
2
(2) e0 (I) ≤ ℓA (A/I) + 1, that is ℓA (I/Q) ≤ 1.
When this is the case, we have the following.


e1 (I)
for 3 ≤ i ≤ d,
(i) ei (I) =
i
(ii) depth gr(I) ≥ d − 1, and
(iii) gr(I) is a Cohen-Macaulay ring if and only if I 2 = QI.
Proof. (1) ⇒ (2) We may assume that ℓA (I/Q) ≥ 1, hence we have gs (I) = ℓA (A/I) −
e0 (I) + e1 (I) = e1 (I) − ℓA (I/Q) ≤ e1 (I) − 1. Therefore the inequalities

 

e1 (I)
gs (I) + 1
= e2 (I)
≤
e2 (I) ≤
2
2
follow by Proposition 3.2. Then, since e1 (I) = gs (I) + 1, we get ℓA (I/Q) = e0 (I) −
ℓA (A/I) = e1 (I) − gs (I) = 1.
(2) ⇒ (1) We may assume that ℓA (I/Q) = 1. Let I = Q + (x) with x ∈ I. Then
I 2 = QI + (x2 ), so that ℓA (I 2 /QI) ≤ 1 because mI ⊆ Q. Therefore, because gs (I) =
8


ℓA (A/I) − e0 (I) + e1 (I) = e1 (I) − ℓA (I/Q) = e1 (I) − 1, the equality e2 (I) = gs (I)+1
=
2

e1 (I)
holds true by Theorem 2.5.
2
(i) and (ii) By Theorem 1.1.
(iii) Assume that gr(I) is a Cohen-Macaulay ring. Then, because ℓA (I/Q) ≤ 1, we
have I 2 ⊆ Q, so that I 2 = Q ∩ I 2 = QI by Theorem 3.4 (iii). The converse also follows
by Theorem 3.4 (iii).

In the end of this section we review a result of Sally [S1] in order to see how our
Theorem 3.4 works to prove or improve them.
Corollary 3.6. Suppose that d ≥ 2. Assume that gs (I) = 1 and e2 (I) 6= 0. Then the
following assertions hold true.
(1) e2 (I) = 1 and ei (I) = 0 for all 3 ≤ i ≤ d,
(2) ℓA (I 2 /QI) = 1 and I 3 = QI 2 , and
(3) depth gr(I) ≥ d − 1.
4. The second border
The second Hilbert coefficients e2 (I) of m-primary ideals I are bounded above by
. In this section we explore the second border of e2 (I) in terms of the sectional
genus gs (I). The main result of this section is stated as follows.
gs (I)+1
2

Theorem 4.1. Suppose that d ≥ 2. Let g = gs (I, M) and assume that g ≥ 2. Then
the following three conditions are equivalent:
 
g
+ 1,
(1) e2 (I) =
2


 
g+1
g
,
+ 1 ≤ e2 (I) <
(2)
2
2
(3) ℓA (I 2 /QI) = 2, ℓA (I k+1 /QI k ) = 1 for all 2 ≤ k ≤ g − 1, and I g+1 = QI g .
When this is the case, we have the following.

(i) ei (I) = gi for 3 ≤ i ≤ d,
(ii) depth gr(I) ≥ d − 1, and
(iii) gr(I) is a Cohen-Macaulay ring if and only if Q ∩ I 2 = QI, Q ∩ I 3 = QI 2 , and
I 4 = QI 3 .
As a direct consequence of Theorem 4.1, we have the following.


Corollary 4.2. Suppose that d ≥ 2 and assume that e2 (I) ≥ gs2(I) + 1. Then we have


e2 (I) = gs2(I) + 1 or e2 (I) = gs (I)+1
, and depth gr(I) ≥ d − 1.
2
Before giving a proof of Theorem 4.1, let us begin with the following. We set
[
Ie =
[I n+1 :A I n ]
n≥0

denotes the Ratliff-Rush closure of I, which is the largest m-primary ideal in A such
e = ei (I) for all 0 ≤ i ≤ d (c.f. [RR]).
that I ⊆ Ie and ei (I)
9

Lemma 4.3. Suppose that d ≥ 2. We have Ie = I, if e2 (I) ≥
Proof. Assume that Ie ) I. Then
e = ℓA (A/I)
e − e0 (I)
e + e1 (I)
e
gs (I)

gs (I)
2



+ 1.

e
= {ℓA (A/I) − ℓA (I/I)}
− e0 (I) + e1 (I)
e
= gs (I) − ℓA (I/I)
≤ gs (I) − 1

e = ei (I) for all i = 0, 1. Therefore we get
because ei (I)
 
 e
e + 1 gs (I)
g
(I)
−
ℓ
(
I/I)
g
(
I)
+
1
s
A
s
e ≤
≤
=
e2 (I) = e2 (I)
2
2
2


e
e ≤ gs (I)+1
because e2 (I)
by Proposition 3.2. However, since e2 (I) ≥ gs2(I) + 1 by our
2
assumption, it is impossible, so that we can get a required contradiction. Thus Ie = I
as required.

We are now in a position to prove Theorem 4.1.
Proof of Theorem 4.1. (1) ⇒ (2) It is clear.
(2) ⇒ (3) and (ii). We proceed by induction on d. Since the residue class field
A/m of A is infinite, we may choose a ∈ Q\mQ is a superficial element for I. We set
A = A/(a), I = /(a), Q = /(a), and r ′ = rQ (I).
k+2

k+1

Suppose that d = 2. We set vk = ℓA (I /Q I ) for all k ≥ 0. Then, we have
P ′ −2
2
vk by Lemma 2.3. If v0 = ℓA (I /Q I) ≤ 1 then we have ℓA (I 2 /QI) ≤ 1,
g = rk=0

because Ie = I by Lemma 4.3. Then we have e2 (I) = g+1
by Theorem 1.2, but it is
2
Pr′ −2
2
impossible. Therefore v0 = ℓA (I /Q I) ≥ 2. Then because k=1 vk = g − v0 ≤ g − 2,
we have

 

X
g−1
g − v0 + 1
≤
k·vk ≤
2
2
k≥1
by Lemma 3.1. Then, by Lemma 2.3, we have
e2 (I) =

′ −2
rX

(k + 1)·vk −

m−1
X

ℓA ([I k+1 :A a]/I k )

k=1

k=0


 

g
g−1
+ 1 ≤ e2 (I),
+g =
k·vk +
vk ≤
≤
2
2
k=1
k=0
′ −2
rX

′ −2
rX

where m ≥ 1 be an integer such that I k+1 :A a = I k for all k ≥ m. Therefore, we have

 

X
g−1
g − v0 + 1
=
k·vk =
2
2
k≥1
2

k+2

k+1

so that v0 = ℓA (I /Q I) = 2, vk = ℓA (I /Q I ) = 1 for all 1 ≤ k ≤ g − 2, and
g+1
g
vg−1 = ℓA (I /Q I ) = 0 by Lemma 3.1. We also have I k+1 :A a = I k for all k ≥ 1,
10

so that depth gr(I) ≥ 1. Thus, we get ℓA (I 2 /QI) = 2, ℓA (I k+1 /QI k ) = 1 for all
1 ≤ k ≤ g − 1, and I g+1 = QI g since at forms a gr(I)-regular element.
Assume that d ≥ 3 and that our assertion holds true for d − 1. Since gs (I) = g ≥ 2
by Lemma 2.2, we have




gs (I)
gs (I) + 1
+ 1 ≤ e2 (I) <
.
2
2
2

k+1

k

Thus, by the hypothesis of induction on d, we get ℓA (I /Q I) = 2, ℓA (I /Q I ) = 1
g+1
g
= Q I , and depth gr(I) ≥ d − 2 > 0. Then, thanks to
for all 2 ≤ k ≤ g − 1, I
Sally’s technique ([S1], [HM, Lemma 2.2]), the element at is gr(I)-regular. Therefore,
we get ℓA (I 2 /QI) = 2, ℓA (I k+1/QI k ) = 1 for all 2 ≤ k ≤ g − 1, I g+1 = QI g , and
depth gr(I) ≥ d − 1 as required.
P
(3) ⇒ (1) and (i) We have k≥1 ℓA (I k+1 /QI k ) = g by our assumption, so that the
P
k+1
equality e1 (I) = e0 (I) − ℓA (A/I) + g =
/QI k ) holds true. Therefore,
k≥0 ℓA (I


thanks to Theorem 2.4, we get e2 (I) = g2 + 1 and ei (I) = gi for all 3 ≤ i ≤ d.
(iii) Assume that gr(I) is a Cohen-Macaulay ring. Then we have Q ∩ I n+1 = QI n
for all n ≥ 1 ([VV]). Since ℓA (I 3 /QI 2 ) ≤ 1, we have I 4 ⊆ QI 2 , so that we get
I 4 = Q ∩ I 4 = QI 3 . The converse also holds true by Valabrega-Valla criterion ([VV]).
This completes the proof of Theorem 4.1.

In the end of this section we introduce one consequence of Theorem 4.1. The following
result is the main theorem in [S2], which is exactly the case where gs (I) = e2 (I) = 2 in
Theorem 4.1.
Corollary 4.4. ([S2, Theorem 4.4]) Suppose that d ≥ 2. Assume that gs (I) = e2 (I) =
2. Then the following assertions hold true.
(1) ei (I) = 0 for all 3 ≤ i ≤ d,
(2) ℓA (I 2 /QI) = 2 and I 3 = QI 2 , and
(3) depth gr(I) ≥ d − 1.
5. Examples
In this section, we shall construct several examples of m-primary ideals I satisfying
the conditions of our main theorems (Theorem 3.4 and Theorem 4.1).
Remark 5.1. To construct necessary examples we may assume that dim A = 1. In fact,
let d ≥ 2 be an integer and let J be an n-primary ideal in a certain 1-dimensional CohenMacaulay local ring (B, n) with q = aB a reduction of J. Let A = B[[X1 , X2 , · · · , Xd−1 ]]
be the formal power series ring. We set I = JA + (X1 , X2 , · · · , Xd−1 )A and Q =
qA + (X1 , X2 , · · · , Xd−1 )A. Then A is a Cohen-Macaulay local ring with dim A = d
and the maximal ideal m = nA + (X1 , X2 , · · · , Xd−1 )A. The ideal Q is a reduction
of I and because X1 , X2 , · · · , Xd−1 forms a super regular sequence for I (recall that
gr(I) = gr(J)[Y1 , Y2, · · · , Yd−1 ] is the polynomial ring, where Yi ’s are the initial forms
11

of Xi ’s), we have depth gr(I) = depth gr(J) + d − 1 ≥ d − 1, ei (I) = ei (J) (i = 0, 1, 2),
gs (I) = gs (J), and I k+1 /QI k ∼
= J k+1 /qJ k for all k ≥ 0. This observation allows us to
concentrate our attention on the case where dim A = 1.
We give
 the following example of m-primary ideals I satisfying the equality e2 (I) =
in Theorem 3.4 (Theorem 1.1).

gs (I)+1
2

Example 5.2. Let e ≥ 3 be an integer and let H = h2e, 2e+1, · · · , 4e−1i be the numerical semi-group generated by 2e, 2e + 1, · · · , 4e − 1. Let B = k[[u2e , u2e+1 , · · · , u4e−1 ]] ⊆
k[[u]], where k[[u]] denotes the formal power series ring with one indeterminate u over
an infinite field k. Then B is a 1-dimensional Cohen-Macaulay local ring with the maximal ideal n = (u2e , u2e+1 , · · · , u4e−1). Let q = (u2e ) and J = (u2e , u2e+2, u4e+1 ). Then
we have the following.
(i) gs (J) = e − 2,
(ii) ℓB (J k+1 /qI k ) = 1 for all 1 ≤ k ≤ e − 2, and I e = QI e−1 , and
(iii) depth gr(J) = 0.
Let d ≥ 2 be an integer and A = B[[X1 , X2 , · · · , Xd−1 ]] be a formal power series ring
over B, and I = JA + (X1 , X2 , · · · , Xd−1 )A and Q = q + (X1 , X2 , · · · , Xd−1 )A. Then
Q is a reduction of I and the following assertions hold true.
(1) A is a Cohen-Macaulay ring with dim A = d ≥ 2,
(2) gs (I) = e − 2,

(3) e0 (I) = e1 (I) = e and ei (I) = e−1
for all 2 ≤ i ≤ d, and
i
(4) depth gr(I) = d − 1.
Proof. Thanks to Theorem 3.4 and Remark 5.1, we have only to show that the assertions
(i), (ii), and (iii) hold true.
P
It is routine to show that J k+1 = qk J + kℓ=1 qk−ℓ u2(ℓ+1)(e+1) = qJ k + u2(k+1)(e+1) B
for all k ≥ 1. Then, since u2k(e+1) ∈
/ qJ k for all 1 ≤ k ≤ e − 2 and mJ ⊆ q, we
k+1
k
get ℓB (J /qJ ) = 1 for all
≤ e − 2 and J e = qJ e−1 . Therefore gs (J) =
P 1 ≤ k k+1
ℓB (B/J) − e0 (I) + e1 (I) = k≥1 ℓB (J /qJ k ) = e − 2 by Lemma 2.1. Since J 2 ⊆ q
and J 2 6= qJ, the associated graded ring gr(J) is not Cohen-Macaulay.

In the end of this paper,
 we introduce the example of m-primary ideals I satisfying
gs (I)
the equality e2 (I) = 2 + 1 in Theorem 4.1 (Theorem 1.2).

Example 5.3. Let e ≥ 5 be an integer and let H = he, e + 1, · · · , 2e − 1i be the
numerical semi-group generated by e, e + 1, · · · , 2e − 1. Let d ≥ 2 be an integer and let
A = k[[ue , ue+1, · · · , u2e−1 ]] ⊆ k[[u]], where k[[u]] denotes the formal power series ring
with one indeterminate u over an infinite field k. Then A is a 1-dimensional CohenMacaulay local ring with the maximal ideal m = (ue , ue+1, · · · , u2e−1). Let Q = (ue )
and I = (ue , ue+1, u2e−2 ). We then have the following.
(i) gs (J) = e − 3,
(ii) ℓB (J 2 /qJ) = 2, ℓB (J k+1 /qI k ) = 1 for all 2 ≤ k ≤ e − 4, and J e−2 = qJ e−3 , and
12

(iii) depth gr(J) = 0.
Let d ≥ 2 be an integer and A = B[[X1 , X2 , · · · , Xd−1 ]] be a formal power series ring
over B, and I = JA + (X1 , X2 , · · · , Xd−1 )A and Q = qA + (X1 , X2 , · · · , Xd−1 )A. Then
Q is a reduction of I and the following assertions hold true.
(1) A is a Cohen-Macaulay local ring with dim A = d ≥ 2,
(2) gs (I) = e − 3,

(3) e0 (I) = e, e1 (I) = e − 1, and ei (I) = e−3
for all 2 ≤ i ≤ d, and
i
(4) depth gr(I) = d − 1.
Proof. It is routine to show that J 2 = qJ + (u2e+2 , u3e−1)B and J k+1 = qk J +
P
qk−1 (u2e+2, u3e−1 ) + kℓ=2 qk−ℓ u(ℓ+1)(e+1) = qJ k + u(k+1)(e+1) B for all k ≥ 2. Therefore, since u(k+1)(e+1) ∈
/ qJ k for all 2 ≤ k ≤ e − 4 and mJ ⊆ q, we get ℓB (J k+1 /qJ k ) = 1
2
for all 2 ≤ k ≤ e − 4. We also have
/qJ) = 2 and J e−2 = qJ e−3. Hence
P ℓB (J k+1
gs (J) = ℓB (B/J) − e0 (I) + e1 (I) = k≥1 ℓB (J /qJ k ) = e − 3 by Lemma 2.1. Since
J 2 ⊆ q and J 2 6= qJ, the associated graded ring gr(J) is not Cohen-Macaulay. Thus,
thanks to Remark 5.1 and Theorem 4.1, all assertions (1), (2), (3), and (4) hold true. 
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Department of Mathematical Science, Faculty of Science, Yamaguchi University,
1677-1 Yoshida, Yamaguchi 753-8512, Japan
E-mail address: ozeki@yamaguchi-u.ac.jp
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