arXiv:1708.06090v1 [] 21 Aug 2017

THE STRONG REES PROPERTY OF POWERS OF THE
MAXIMAL IDEAL AND TAKAHASHI-DAO’S QUESTION
TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA
Dedicated to Craig Huneke on the occasion of his 65th Birthday.
Abstract. In this paper, we introduce the notion of the strong Rees property
(SRP) for m-primary ideals of a Noetherian local ring and prove that any power
of the maximal ideal m has its property if the associated graded ring G of m
satisfies depth G ≥ 2. As its application, we characterize two-dimensional
excellent normal local domains so that m is a pg -ideal.
Finally we ask what m-primary ideals have SRP and state a conjecture
which characterizes the case when mn are the only ideals which have SRP.

1. Introduction
Let (A, m) be a Noetherian local ring with d = dim A ≥ 1 and I an m-primary
ideal of A. The notion of m-full ideals was introduced by D. Rees and J. Watanabe
([22]) and they proved the “Rees property” for m-full ideals, namely, if I is m-full
ideal and J is an ideal containing I, then µ(J) ≤ µ(I), where µ(I) = ℓA (I/mI) is
the minimal number of generators of I. Also, they proved that integrally closed
ideals are m-full if A is normal.
fn , the Ratliff-Rush closure of mn , is m-full ([1,
Suppose depth A > 0. Then m
Proposition 2.2]). Thus mn is m-full for sufficiently large n.
Sometimes we need stronger property for µ(I) and we will call it “Strong Rees
property” (SRP for short).
Definition (Strong Rees Property). Let I be an m-primary ideal of A. Then
we say that I satisfies the strong Rees property if for every ideal J ) I, we have
µ(J) < µ(I).
So it will be natural to ask the following questions.
Question 1.1. Let (A, m) be a normal local ring. Then
(1) Does mn has the strong Rees property for every n ≥ 1?
(2) If I has the strong Rees property, then is I = mn for some n ≥ 1?
Date: August 22, 2017.
2000 Mathematics Subject Classification. Primary 13A30; Secondary 13H15, 13B22, 14B05.
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TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

Actually, both questions are not true in general. But we can show that (1) in
the Question 1.1 holds under suitable mild condition. Also, we will give an example
of two-dimensional normal local rings where m2 does not satisfy the strong Rees
property.
As for (2) of Question 1.1, we will discuss it in Section 6.
Assume that I is an m-primary ideal. Then the multiplicity (resp. the minimal
number of system of generators) of I is denoted by e(I) (resp. µ(I)). The Loewy
length ℓℓ(I) is defined by
ℓℓ(I) = min{r ∈ Z≥0 | mr ⊂ I}
Notice that the notion of Loewy length of an Artinian ring measures the nilpotency
of the maximal ideal. It is natural to ask if e(I) is bounded by the product of µ(I)
and ℓℓ(I) after adjusting some error terms.
The origin of this work was a discussion of second and third authors with Hailong
Dao. He presented an inequality between µ(I), the multiplicity e(I) of I and Loewy
length ℓℓ(I).
Question 5.1 ([2]). Let (A, m) be a d-dimensional Cohen-Macaulay local ring.
Let I ⊂ A be an m-primary (integrally closed) ideal. When does the inequality
(d − 1)!(µ(I) − d + 1) · ℓℓ(I) ≥ e(I)
hold true?
Dao and Smirnov [2] proved that the Question 5.1 holds true if A is a twodimensional analytically unramified and the maximal ideal m is a pg -ideal (or,
equivalently, the Rees algebra R(m) is normal and Cohen-Macaulay).
We are interested in the converse of the Question 5.1 in the case of d = 2. Here,
since e(I) does not change after taking integral closure, it is natural to assume that
I is integrally closed. Namely, our Question is
Question 1.2. Assume that (A, m) is a two-dimensional excellent normal local
domain. If an inequality
(µ(I) − 1) · ℓℓ(I) ≥ e(I)
holds for any m-primary integrally closed ideal I, then is m a pg -ideal?
It turns out that this question is related to the “strong Rees property” of powers
of the maximal ideal. The main result in this paper is the following theorem.
Theorem 3.2. Let (A, m) be a Noetherian local ring. Assume that depth A ≥ 2
1
and HM
(G) has finite length, where G = G(m) = ⊕n≥0 mn /mn+1 . If mℓ is RatliffRush closed, then mℓ has the strong Rees property.
By Remark 2.5, any power mℓ is Ratliff-Rush closed whenever depth G ≥ 1.
Hence we have the following corollary.

THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

3

Corollary 3.5. If depth G ≥ 2, then mℓ has the strong Rees property for every
ℓ ≥ 1.
In general, we cannot relax the assumption that depth G ≥ 2 even if A is normal.
See Section 4 for more details.
In Section 5, as an application of the theorem above, we prove the above question
has an affirmative answer.
Theorem 5.2.
Let (A, m) be a two-dimensional excellent normal local domain containing an
algebraically closed field. Then the following conditions are equivalent:
(1) m is a pg -ideal.
(2) For every m-primary integrally closed ideal I,
(µ(I) − 1) · ℓℓA (I) ≥ e(I)

(∗)

holds true.
(3) For any power I = mℓ of m, the inequality (∗) holds true.
After proving this theorem, Dao and Smirnov informed us that they proved the
same theorem independently.
2. Preliminaries
Throughout this paper, let (A, m) be a Noetherian local ring with d = dim A ≥ 1,
and let I, J ⊂ A be ideals of positive height. We recall the notion which we will
need later.
2.1. m-full ideals.
Definition 2.1 (m-full, Rees property[22]). An ideal I is called m-full if there
exists an element x ∈ m so that mI : x = I.
An ideal I is said to have the Rees property if µ(J) ≤ µ(I) for any ideal J
containing I.
Proposition 2.2 (See [22, Theorem 3]). Any m-primary m-full ideal has the Rees
property.
Proof. See the proof of [3, Lemma 2.2].



The following result is due to Rees in the case of normal integral domains.
Proposition 2.3 (See [22, Theorem 5]). Any integrally closed ideal of positive
height is m-full.
Proof. See the proof of Theorem [3, Theorem 2.4].



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TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

Definition 2.4 (Ratliff-Rush closure [17]). The Ratliff-Rush closure of I is
defined by
[
Ie =
I n+1 : I n .
n≥0

The ideal I is called Ratliff-Rush closed if Ie = I.

Remark 2.5. Note that I ⊂ Ie ⊂ I, where I is the integrally closure of I. If we
L
put G = G(m) = n≥0 mn /mn+1 , then
M
n+1 ∩ mn )/mn+1 .
]
H 0 (G) =
(m
M

n≥0

Lemma 2.6. Assume that for some z ∈ mn−1 \ mn we have zm ⊂ mn−1 . Then,
putting I = (x) + mn , µ(I) > µ(mn ), hence mn does not have Rees property.
Proof. Let {y1 , . . . , yµ } be a minimal set of generators of mn . Then we can see that
{z, y1, . . . , yµ } is a minimal set of generators of I. Hence µ(I) = µ(mn ) + 1.

Any m-full ideal has the Rees property. However, in order to prove our theorem,
we need stronger inequalities.
Recall the definition of the strong Rees property.
Definition 2.7. An m-primary ideal I is said to have the strong Rees property
(SRP for short) if µ(J) < µ(I) holds true for every ideal J with I ( J.
We can show the existence of m-primary ideals with SRP by the following
Lemma.
Lemma 2.8. Let (A, m) be a Noetherian local ring. Fix an m-primary ideal I with
µ(I) = n.
(1) Any maximal element of the set of m-primary ideals
I = {J ⊂ A | J ⊃ I and µ(J) ≥ n}
has the strong Rees property.
(2) If, moreover, I is m-full, then there is an ideal I ′ ⊃ I with the strong Rees
property and µ(I ′ ) = µ(I).
Proof. (1) is obvious by the definition of I.
(2) If I ′ ⊃ I has SRP and µ(I ′ ) ≥ µ(I), we should have µ(I ′ ) = µ(I) since I is
m-full.

The next example gives a motivation for us to study the strong Rees property
of powers of the maximal ideal. See also [22, Theorem 5] and Section 6.
Example 2.9. Assume that A is a two-dimensional regular local ring. Then for
any m-primary ideal I, the following conditions are equivalent.
(1) I has the strong Rees property.

THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

5

(2) I = mn for some integer n ≥ 1.
Indeed, assume that I has the strong Rees property. If we put n = ord(I) =
max{n ∈ Z | I ⊂ mn }, then we can take an element f ∈ I so that ord(f ) =
n. Then since I/(f ) is an m/(f )-primary ideal of A/(f ), we have µ(I/(f )) ≤
n = e(m/(f )) because A/(f ) is a one-dimensional Cohen-Macaulay local ring. In
particular, µ(I) ≤ n + 1 = µ(mn ). If I 6= mn , then n + 1 = µ(mn ) < µ(I) by the
assumption (1). But this is a contradiction.
Conversely, if I ) mn , then ord(I) < n and µ(I) ≤ ord(I) + 1 < n + 1 = µ(mn ).

2.2. pg -ideals. In what follows, let (A, m) be a two-dimensional excellent normal
local domain containing an algebraically closed field k = A/m. Let f : X → Spec A
be a resolution of singularities. Then pg (A) = dimk H 1 (OX ) is called the geometric
genus of A. Note that it does not depend on the choice of resolution of singularities.
Let I ⊂ A be an m-primary integrally closed ideal. Let f : X → Spec A be a
resolution of singularities on which I is represented, that is, IOX is invertible and
IOX = OX (−Z) for some anti-nef cycle Z on X.
Okuma and the last two authors [13] proved that dimk H 0 (OX (−Z)) ≤ pg (A)
holds true if H 0 (OX (−Z)) has no fixed component.
Definition 2.10 (See [13]). An anti-nef cycle Z is a pg -cycle if OX (−Z) is generated
and dimk H 0 (OX (−Z)) = pg (A). An m-primary integrally closed ideal I is called
a pg -ideal if I is represented as I = H 0 (OX (−Z)) by a pg -cycle Z.
The following theorem gives a characterization of pg -ideals in terms of Rees
algebras.
Proposition 2.11 (See [14]). Let (A, m) be a two-dimensional excellent normal
local domain containing an algebraically closed field. Let I ⊂ A be an m-primary
ideal. Then the following conditions are equivalent:
(1) I is a pg -ideal.
(2) I n = I n for every n ≥ 1 and I 2 = QI for some minimal reduction Q ⊂ I.
(3) The Rees algebra R(I) = A[It](⊂ A[t]) is a Cohen-Macaulay normal domain, where t is an indeterminate over A.
For instance, any integrally closed m-primary ideal in a two-dimensional rational
singularity (i.e. pg (A) = 0) is a pg -ideal. On the other hand, any two-dimensional
excellent normal local domain containing algebraically closed field has a pg -ideal;
see [13, 14].

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TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

3. Strong Rees Property of powers of the maximal ideal
In what follows, let (A, m) be a Noetherian local ring and set R = R(m) and
G = G(m) and M = mR + R+ . The main purpose of this section is to consider the
following question.
Question 3.1. Assume that I is Ratliff-Rush closed. When does I have the strong
Rees property?
As an answer, we show that some powers of the maximal ideal m have the strong
Rees property if depth G ≥ 2; see Corollary 3.5. More generally, we can show the
following theorem.
1
Theorem 3.2 (Strong Rees Property). Assume that depth A ≥ 2 and HM
(G)
has finite length. If mℓ is Ratliff-Rush closed, then mℓ has the strong Rees property.

We first need to prove the following lemma.
Lemma 3.3. Suppose that for some x ∈ m, mℓ+1 : x = mℓ . Then for every ideal
J with J ⊃ mℓ , µ(J) ≤ µ(mℓ ) is always satisfied, and the following conditions are
equivalent:
(1) µ(J) = µ(mℓ ).
(2) mJ = xJ + mℓ+1 .
When this is the case, if we put C = JR/mℓ R, C/(xt)C = (C/(xt)C)0 = J/mℓ .
In particular, dim C ≤ 1.
Proof. Consider the following two short exact sequences:
0 →

mℓ ∩ mJ
mℓ
mℓ
→
→
→ 0.
mℓ+1
mℓ+1
mℓ ∩ mJ

J
J
mℓ + mJ
→
→ ℓ
→ 0.
mJ
mJ
m + mJ
Since mℓ /(mℓ ∩ mJ) ∼
= (mℓ + mJ)/mJ, combining two exact sequence implies
0 →

(3.1)

0 →

mℓ ∩ mJ
mℓ
J
J
→ ℓ+1 →
→ ℓ
→ 0.
ℓ+1
m
m
mJ
m + mJ

It follows that

ℓA (mℓ ∩ mJ/mℓ+1 ) = µ(mℓ ) − µ(J) + ℓA (J/mℓ + mJ).

Furthermore, since mJ/(mℓ ∩ mJ) ∼
= (mℓ + mJ)/mℓ , we get

ℓA (mJ/mℓ+1 ) = ℓA (J/mℓ ) + µ(mℓ ) − µ(J)

We now consider an R-module C = JR/mℓ R. The assumption that mℓ+1 : x =
m implies that the multiplication map
ℓ

·x : C0 = J/mℓ → C1 = mJ/mℓ+1

THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

7

is injective. Hence µ(mℓ ) − µ(J) = ℓA (C1 ) − ℓA (C0 ) ≥ 0. Moreover, equality holds
true if and only if the multiplication map by x is isomorphism, which means that
mJ = xJ + mℓ+1 .
When this is the case, we have
mn+1 J = xmn J + mn+ℓ+1
for every n ≥ 1. The last assertion immediately follows from this.



Lemma 3.4. Let J ⊂ A be an ideal with J ) mℓ , where ℓ ≥ 1. Put C = JR/mℓ R.
1
Assume that depth A ≥ 2, and HM
(G) has finite length. If mℓ is Ratliff-Rush
closed, then
i
(1) HM
(C) has finite length for i = 0, 1.
0
(2) [HM
(C)]0 = 0.

Proof. First we define an R-module L(−1) as follows:
0 → R → A[t] → L(−1) → 0 (ex),
that is, L(−1) =

L

n≥0 (A/m

n

)tn .

i
(L(−1)) has finite length for i = 0, 1.
Claim 1. HM

By [15, Proposition 4.7] we have
0
HM
(L(−1)) =

Mm
fn
n≥0

mn

0
= HM
(G),

1
which has finite length and then it is proved in [16, Theorem 6.2] that HM
(L(−1))
1
has finite length, if and only if HM (G) has finite length. For instance, depth G ≥ 2,
i
then HM
(L(−1)) = 0 for each i = 0, 1.
L
Secondly, we define D = L(−1)≥ℓ (ℓ) = n≥0 (A/mℓ+n )tn .

0
1
Claim 2. [HM
(D)]0 = 0 and HM
(D) has finite length.

By definition, we have
0 → D(−ℓ) → L(−1) → W → 0,
where W is an R-module of finite length. Then since in the exact sequence
0
1
1
W = HM
(W ) → HM
(D(−ℓ)) → HM
(L(−1))
1
the modules of the both sides have finite length, so does HM
(D(−ℓ)). Moreover,
0
0
0
as HM (D(−ℓ)) ⊂ HM (L(−1)), HM (D(−ℓ)) is also of finite length.
0
fℓ /mℓ and our assumption.
The first assertion follows from the fact [HM
(D)]0 ⊂ m
L
Thirdly, we define an R-module V = n≥0 (A/mn J)tn as follows:

0 → JR → A[t] → V → 0 (ex).

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TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

By definition of V , we have
0
0
1
1
0 = HM
(A[t]) → HM
(V ) → HM
(JR) → HM
(A[t]) = 0,

where two vanishing follows from the fact that depth A ≥ 2. It follows that
1
0
0
(JR)]n = 0 for large enough n. On the other hand, as HM
(V ) ⊂
[HM
(V )]n ∼
= [HM
0
0
V , [HM (V )]n = 0 for each n ≤ −1. Thus HM (V ) has finite length.
0
1
Claim 3. [HM
(C)]0 = 0 and HM
(C) has finite length.

One can easily obtain the following exact sequence:
0 → C → D → V → 0 (ex).
Hence
0

0
0
→ HM
(C) → HM
(D)

0
→ HM
(V )

1
1
→ HM
(C) → HM
(D)

→

···

0
0
As [HM
(D)]0 = 0 by Claim 2, we get [HM
(C)]0 = 0. Moreover, in the sequence
0
0
1
1
HM
(D) → HM
(V ) → HM
(C) → HM
(D),
1
1
the both sides of HM
(C) have finite length. Hence so does HM
(C).



Proof of Theorem 3.2. Choose an m-superficial element x ∈ m \ m2 . By assumption, we have
fℓ = mℓ .
mℓ ⊂ mℓ+1 : x ⊂ m

Then mℓ+1 : x = mℓ In particular, this means mℓ is m-full.
Let J be an ideal with J ) mℓ . By Lemma 3.3, µ(J) ≤ µ(mℓ ). We want to show
that this inequality is strict. Now suppose that equality holds true: µ(J) = µ(mℓ ).
Put C = JR/mℓ R. In Lemma 3.3, we showed that C/(xt)C has finite length.
Hence we have dim C ≤ 1.
1
0
(C) has
On the other hand, by Lemma 3.4, we have [HM
(C)]0 = 0 and HM
ℓ
0
finite length. If dim C = 0, then 0 6= J/m = [C]0 = [HM (C)]0 = 0. This is a
1
contradiction. Hence dim C = 1. Then HM
(C) is not finitely generated. This
1
contradicts the fact that HM (C) has finite length. Therefore we conclude that
µ(J) < µ(mℓ ), as required.

In the case where depth G ≥ 2, then all powers of the maximal ideal have the
strong Rees property.
Corollary 3.5. If depth G ≥ 2, then mℓ has the strong Rees property for every
ℓ ≥ 1.
1
(G) = 0 and
Proof. Assume Theorem 3.2 holds true. If depth G ≥ 2, then HM
depth A ≥ 2. Hence the ring A satisfies the assumption on Theorem 3.2.


THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

9

Corollary 3.6. Let (A, m) be a Cohen-Macaulay local ring with d = dim A ≥ 2. If
G = G(m) is Cohen-Macaulay, then mℓ has the strong Rees property for each ℓ ≥ 1.
Corollary 3.7. If depth A ≥ 2 and m is a normal ideal (i.e., mn is integrally closed
for all n ≥ 1) then mℓ has the strong Rees property for every ℓ ≥ 1.
Proof. By [8, Theorem 3.1] we get depth G(mn ) ≥ 2 for all n ≫ 0. Here G(mn )
1
denotes the associated graded ring of mn . It follows that HM
(G(m)) has finite
r
length. Also as m is integrally closed for all r ≥ 1 we get that it is Ratliff-Rush
closed. It follows that mr has SRP for all r ≥ 1 by Theorem 3.2.

Example 3.8. Let (A, m) be a two-dimensional excellent normal local domain.
If m is a pg -ideal (e.g. A is a rational singularity), then mℓ has the strong Rees
property for every ℓ ≥ 1.
On the other hand, “depth G ≥ 1” can be characterized by the Rees property.
Proposition 3.9. Put G = G(m), where d = dim G ≥ 1. Then the following
conditions are equivalent:
(1) depth G ≥ 1.
(2) mℓ has the Rees property for every ℓ ≥ 1.
fℓ = mℓ and mℓ is m-full for every ℓ ≥ 1.
Proof. (1) =⇒ (2) : If depth G ≥ 1, then m
ℓ
In particular, m has the Rees property.
0
(2) =⇒ (1) : Now suppose depth G = 0. Then HM
(G) 6= 0. There exist an
ℓ
ℓ+1
integer ℓ ≥ 1 and an element z ∈ m \ m
such that 0 6= z ∗ = z + mℓ+1 ∈
0
0
ℓ+2
[HM (G)]ℓ ∩ Soc(HM (G)). Then mz ⊂ m . If we put J = (z) + mℓ+1 , then
mJ = m(z) + mℓ+2 = mℓ+2 . Hence

J/mJ = ((z) + mℓ+1 )/mℓ+2 ) mℓ+1 /mℓ+2
and so µ(J) = ℓA (J/mJ) > ℓA (mℓ+1 /mℓ+2 ) = µ(mℓ+1 ). This contradicts the
assumption that mℓ+1 has the Rees property.

We now consider the case of depth G = 1. We need the following lemma.
Lemma 3.10. Suppose that depth G = 1. Assume that J ) mℓ so that µ(J) =
µ(mℓ ). Then one can find elements x ∈ m \ m2 and y ∈ J \ (mℓ + (x)) such that x∗
0
is G-regular and 0 6= z ∗ ∈ Soc(HM
(G)), where z = y ∈ A/xA and z ∗ denotes the
∗
initial form of z in G = G/x G.
Proof. Take an element x ∈ m \ m2 so that x∗ is G-regular and J 6⊂ mℓ + (x);
see the remark below for the exitence of x. Choose y ∈ J \ (mℓ + (x)). Then
y ∈ (mk + (x)) \ (mk+1 + (x)) for some k with 0 ≤ k ≤ ℓ − 1. By assumption and
Lemma 3.3, we have that my ⊂ mJ = xJ + mℓ+1 . Let denote the image of the
surjection A → A/xA and put z = y. Then mz ⊂ mℓ+1 ⊂ mk+2 , which means that
0
0 6= z ∗ ∈ [Soc(HM
(G/x∗ G))]k .


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TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

Remark 3.11. Suppose depth G = 1. For any ideal J ) mℓ , one can find an
element x ∈ m \ m2 so that x∗ is G-regular and J is not contained in mℓ + (x). In
order to prove this, it suffices to consider the case ℓ(J/mℓ ) = 1, that is, J = (f )+mℓ,
where f 6∈ mℓ and mf ⊂ mℓ .
If J ⊂ mℓ + (x) for such an element x as above, then f − ax ∈ mℓ for some
a ∈ A. As x∗ is G-regular (and thus a∗ x∗ 6= 0), we obtain the relation a∗ x∗ = f ∗
in G. Let G = k[X]/a and let F be the inverse image of f ∗ in the polynomial
ring k[X]. As k is an infinite field and dim k[X]/(a + (F )) ≥ 1, one can find a
homogeneous element X of degree one which does not vanish on V (a + (F )). The
required assertion follows from here.
Hence we have the following.
Proposition 3.12. Suppose that depth G = 1. Let x∗ ∈ G1 be a nonzero divisor
0
(G)]ℓ = 0 for all ℓ < n, then mℓ has the
of G and put G = G/x∗ G. If [Soc(HM
strong Rees property for all ℓ ≤ n.
0
Proof. First we note that Soc(HM
(G/x∗ G)) is independent of the choice of x. In
fact, the short exact sequence
x∗

0 −→ G(−1) −→ G −→ G := G/x∗ G −→ 0
yields a short exact seuence
x∗

0
0
1
1
HM
(G) = 0 −→ HM
(G) −→ HM
(G)(−1) −→ HM
(G).

Taking a socle, we get
x∗

1
1
0
(G))(−1) −→ Soc(HM
(G)).
0 −→ Soc(HM
(G)) −→ Soc(HM
0
1
Since the last map is a zero map, Soc(HM
(G)) ∼
(G))(−1) is independent
= Soc(HM
of the choice of x.
Now suppose that mℓ does not have SRP. Then we can find an ideal J ) mℓ so
that µ(J) = µ(mℓ ). By Lemma 3.10, there exist an element x ∈ m \ m2 such that
0
x∗ is G-regular and 0 6= [Soc(HM
(G/x∗ G)]k for some 0 ≤ k ≤ ℓ − 1 ≤ n − 1. This
contradicts the assumption.


Proposition 3.13. When depth G = 1, there exists an n ∈ N ∪ {∞} such that mℓ
has the strong Rees property if and only if 1 ≤ ℓ ≤ n.
In order to prove the proposition, it suffices to show the following lemma.
Lemma 3.14. Suppose that depth G = 1. If mℓ does not have the strong Rees
property, then neither does mℓ+1 .
Proof. Since depth G = 1, there exists an element x ∈ m \ m2 so that x∗ = x + m2 is
G-regular. In particular, mℓ and mℓ+1 are m-full. By assumption and Lemma 3.3,

THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

11

we can take an ideal I ) mℓ so that mI = xI + mℓ+1 . Put J = xI + mℓ+1 . Then
mJ = m(xI + mℓ+1 ) = m(xI) + mℓ+2 .
Moreover, we suppose that J = mℓ+1 . Then xI ⊂ mℓ+1 and thus I ⊂ mℓ+1 : x = mℓ .
This contradicts the choice of I. Hence J ) mℓ+1 and µ(J) = µ(mℓ+1 ). This implies
that mℓ+1 does not have SRP.

Example 3.15. For any n ≥ 1, there exists a triple (a, b, c) ∈ N3 such that A =
k[[s, ta , tb , tc ]] is a two-dimensional Cohen-Macaulay local domain such that mℓ has
the strong Rees property if and only if 1 ≤ ℓ ≤ n.
Proof. For a given n ≥ 1, we can choose an integer a = 10N > 2n. Set b = a + 1
and c = (a − 1)(a + 1) − an. Then s, ta , tb , tc is a minimal system of generators
because ab − a − b = (a − 1)(a + 1) − a ≥ c(> b > a).
Since
A = k[[s, x, y, z]]/(yz − xa+1−n , xn z − y a−1 , z 2 − xa+1−2n y a−2 ),
we get
G = G(m) ∼
= k[S, X, Y, Z]/(Y Z, X n Z, Z 2 , Y a ).
0
(G/SG)) is generated by X n−1 Z ∈ [G/SG]n .
Then S is an G-regular and Soc(HM
ℓ
Hence m has SRP if 1 ≤ ℓ ≤ n.
If we put I = (xn−1 z) + mn+1 ) mn+1 , then x · xn−1 z = y a−1 ∈ ma−1 ⊆ mn+2 .
Similarly, we have that y · xn−1 z ∈ ma ⊆ mn+2 and z · xn−1 z ∈ m2a−n−2 ⊂ mn+2 .
Hence mI = (s · xn−1 z) + mn+2 , and this implies that mn+1 does not have SRP. 

Example 3.16. Let A = k[[s, t4 , t5 , t11 ]] and m = (s, t4 , t5 , t11 ). Then m2 does
not have strong Rees property. In fact, m2 = (s2 , st4 , st5 , st11 , t8 , t9 , t10 ) and thus
µ(m2 ) = 7. If we put I = (s2 , st4 , st5 , t8 , t9 , t10 , t11 ) = m2 + (t11 ), then I ) m2 and
µ(I) = 7 = µ(m2 ).
On the other hand, since G ∼
= k[S, X, Y, Z]/(XZ, Y Z, Z 2 , Y 4 ) (see e.g. [20,
f2 is m-full. Moreover, since
Section 2]), we have depth G = 1 and thus m2 = m
2
2
11
2
t ∈ m \ m , m is not integrally closed.
We can find an example of two-dimensional excellent normal local domains (A, m)
for which m2 does not satisfy the strong Rees property and A/sA ∼
= k[[t4 , t5 , t11 ]]
for some nonzero divisor s of A. See the next section.
4. Point divisor on a smooth curve – An example mn does not have
strong Rees property
In this section we treat a class of normal graded rings of dimension 2 and discuss
whether mn has the strong Rees property in such rings.

12

TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

Definition 4.1. Let k be an algebraically closed field and C be a smooth connected
projective curve of genus g over k. We take a point P ∈ C and define
H = HC,P = {n ∈ Z | h0 (C, OC (nP )) > h0 (C, OC ((n − 1)P ))},
where hi (C, F ) = dimk H i (C, F ). It is easy to see that HC,P is an additive semigroup and N \ HC,P has just g elements.
We define
R = RH = RC,P = ⊕n≥0 H 0 (C, OC (nP )T n ,
as a subring of k(C)[T ], where H 0 (C, OC (nP ) = {f ∈ k(C) | divC (f ) + nP ≥ 0}.
Namely, f ∈ H 0 (C, OC (nP ), if and only iff has pole of order at most n at P and
no other poles.
Then R = RC,P is a normal graded ring of dimension 2 as treated in [5], Chapter
5, §2. In the following, we fix H = HC,P and write A = k[H] and R = RH so that
R/T R ∼
= A, where T = 1.T ∈ R1 .
Pe
We write H = hn1 , . . . , ne i if H = { i=1 ai ni | ai ∈ Z≥0 (i = 1, . . . , e)}. In
this case, we say that H is generated by e elements. We denote by H+ the set of
positive elements of H and denote n ∈ rH+ if n = h1 + . . . + hr with hi ∈ H+
(i = 1, . . . , r).
Remark 4.2. Given a semigroup H, sometimes there does not exist the pair (C, P )
such that HC,P = H. But at least we know the existence of (C, P ) such that
H = HC,P in the following cases (cf. [10], [11]) ;
(1)
(2)
(3)
(4)

k[H] is a complete intersection.
H is generated by 3 elements.
H is generated by 4 elements and H is symmetric or pseudo-symmetric.
g(H) ≤ 9, where g(H) is the number of positive integers not in H.

We summarize some property of R = RC,P . We put m = R+ .
Proposition 4.3. Let R = RC,P . An element of Rn is denoted by f T n , where
f ∈ k(C). We denote by v(f ) the order of the pole of f at P . For non zero
elements f, g ∈ k(C), v(f g) = v(f ) + v(g).
(1) f T n ∈ Rn if and only if v(f ) ≤ n and f has no other poles on C.
(2) Hence if v(f ) < n, then f T n ∈ T n−v(f ) R, because f T v(f ) ∈ Rv(f ) .
(3) If HC,P = hn1 , . . . , ne i, which are minimal generating system, then there
are elements f1 , . . . , fe ∈ k(C) with v(fi ) = ni (i = 1, . . . , e) such that
R = k[T, f1 T n1 , . . . , fe T ne ].
(4) If f T n ∈ Rn and v(f ) = n, then f T n ∈ mr if and only if n ∈ rH+ .
(5) T ∈ R1 is a super regular element of R. Namely, if T x ∈ mr for some
x ∈ R, then x ∈ mr−1 .

THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

13

Theorem 4.4. Let R = RC,P and H = HC,P = hn1 , . . . , ne i. If for some n ∈
rH+ , n 6∈ (r − 1)H+ , n + ni ∈ (r + 2)H+ for i = 1, . . . , e, then mn+1 does not have
the strong Rees property.
Proof. By the assumption, there is some n ∈ (r − 1)H+ , n 6∈ rH+ such that n +
H+ ⊂ (r + 1)H+ . Then we can take f ∈ k(C) so that v(f ) = n and f T n ∈
Rn . Since f T n 6∈ T R, f T n ∈ mr−1 and f T n 6∈ mr+1 . We put I = (mr , f T n )
and show that µ(I) = µ(mr ). Now, let homogeneous minimal generators of m
be {T, g1T n1 , . . . , ge T ne }. Then among the homogeneous minimal generators of
m(f T n), (f T n )(gi T ni ) ∈ mr+1 by our assumption. Hence we can obtain minimal
generating system of I from that of mr , interchanging f T n and f T n+1, obtaining
µ(I) = µ(mr ).

Corollary 4.5. Let R = RC,P , H = HC,P and m = R+ . Then the following
conditions are equivalent:
(1) For all n ≥ 2, mn has the strong Rees property.
(2) The associated graded ring of k[H] with respect to k[H]+ is Cohen-Macaulay.
(3) The associated graded ring of R with respect to R+ is Cohen-Macaulay.
Example 4.6. Let H = h4, 5, 11i, C a smooth curve of genus 5 such that there is a
point P with HC,P = h4, 5, 11i. We put R = RC,P and m = R+ . Since 11+4 ∈ 3H+
and 11 + 5 ∈ 4H+ , we see that m2 does not have the strong Rees property. In this
example, we can easily see that mn is integrally closed for all n ≥ 2.
Remark 4.7. For 3 generated semigroup H = ha, b, ci, we know when the associated graded ring is Cohen-Macaulay (cf, [7],[12]).
5. Takahashi-Dao’s question
Dao and Takahashi [21] gave two upper bounds of the dimension of the singularity
category dim Dsg (A):
dim Dsg (A) ≤

(µ(I) − dim A + 1)ℓℓ(I) − 1

dim Dsg (A) ≤

e(I) − 1

for any m-primary ideal I contained in the sum N A of the Noether differents of A.
They posed the following question.
Question 5.1 (Takahashi-Dao). Let (A, m) be a d-dimensional Cohen-Macaulay
local ring. Let I ⊂ A be an m-primary (integrally closed) ideal. When does an
inequality
(d − 1)!(µ(I) − d + 1) · ℓℓA (I) ≥ e(I)
hold true (cf. [2])?

14

TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

The following theorem is motivated by the question as above. In fact, (1) ⇒ (2)
is due to Dao and Smirnov, and (2) ⇒ (1) is also proved by them independently.
Theorem 5.2. Let (A, m) be a two-dimensional excellent normal local domain
containing k = k ∼
= A/m. Then the following conditions are equivalent:
(1) m is a pg -ideal.
(2) For every m-primary integrally closed ideal I,
(µ(I) − 1) · ℓℓA (I) ≥ e(I)

(∗)

holds true.
(3) For any power I = mℓ of m, the inequality (∗) holds true.
Proof. (1) =⇒ (2) : We give a sketch of proof here for the sake of the completeness.
Assume that m is a pg -ideal. Let I ⊂ A be an m-primary integrally closed ideal.
Then there exists a resolution of singularities f : X → Spec A so that IOX =
OX (−Z) for some anti-nef cycle Z on X. By [13, Theorem 6.1], we have
µ(I) = −M Z + 1,
where M is an anti-nef cycle on X so that mOX = OX (−M ).
Put r = ℓℓA (I), that is, mr ⊂ I and mr−1 6⊂ I. Then mr ⊂ I = I. Thus
rM ≥ Z. Since e(I) = −Z 2 , we have
(µ(I) − 1)ℓℓA (I) − e(I) = (−M Z)r + Z 2 = −(rM − Z)Z ≥ 0,
as required.
(2) =⇒ (3) : Trivial.
(3) =⇒ (1) : Now assume that inequalities
(µ(mℓ ) − 1) · ℓℓA (mℓ ) ≥ e(mℓ )
hold true for all integers ℓ ≥ 1. This shows that
(µ(mℓ ) − 1) · ℓ ≥ e(mℓ ) = e(mℓ ) = ℓ2 · e,
where e = e(m). Hence µ(mℓ ) ≥ ℓe + 1.
In the case of ℓ = 1, we have that µ(m) − 1 ≥ e(m). On the other hand,
Abhyankar’s inequality implies that µ(m) − 1 ≤ e(m), and thus equality holds true.
That is, A has maximal embedding dimension in the sense of Sally ([19]). Then
µ(mℓ ) = ℓe + 1. Moreover, since G = G(m) is Cohen-Macaulay ([18]), we obtain
fℓ is m-full for every ℓ ≥ 1. Then µ(mℓ ) ≤ µ(mℓ ) = ℓe + 1 by Rees
that mℓ = m
property of m-full ideals. Hence µ(mℓ ) = µ(mℓ ). Then Corollary 3.6 yields that
mℓ = mℓ because G(m) is Cohen-Macaulay in our case (Sally [18]). Therefore m is
a pg -ideal by [14].


THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

15

Remark 5.3. We can show that I = mℓ satisfies the above inequality if (A, m) has
maximal embedding dimension. Note that mℓ does not necessarily have the strong
Rees property; see Sections 3 and 4.
It is known that the maximal ideal m of any two-dimensional rational singularity
is a pg -ideal. So it is natural to ask the following question. Which ring the maximal
ideal of which is a pg -ideal? By a similar argument as in [6, Corollary 11.4], we can
show the following. Notice that this gives a slight generalization of the fact that
any two-dimensional rational singularity is an almost Gorenstein local ring.
Proposition 5.4. Let (A, m) be a two-dimensional excellent normal local domain
containing an algebraically closed field. Let KA denote the canonical module of A.
If m is a pg -ideal, then A is an almost Gorenstein local ring in the sense of [6].
That is, there exists a short exact sequence of A-modules:
0 → A → KA → C → 0
such that mC = xC for some regular element x over C.
Example 5.5. In the notation of Section 4, let P ∈ C be such that HC,P =
{0, g + 1, g + 2, . . .}. Then the maximal ideal m of RC,P is a pg -ideal.
Proof. Take f ∈ k(C) with f T g+1 ∈ R and v(f ) = g + 1. Then putting Q =
(T, f T g+1 ), we see that m2 = Qm.
Introduce a valuation w on R such that


v(f )
m
w(f T ) = (m − v(f )) +
g+1
To show that mn is integrally closed it suffices to show that w(gT m ) ≥ n if and
only if gT m ∈ mn . If w(gT m ) ≥ n and m − v(g) = r, since v(g) ≥ (g + 1)(n − r),
gT v(g) ∈ mn−r and then gT m ∈ mn .
Hence mn is integrally closed for all n ≥ 1 and m is a pg -ideal by Proposition
2.11.

In dimension 2, although Takahashi-Dao’s inequality does not hold for general
normal ring A, we have an inequality adding a constant depending on A. Also we
have converse inequality for e(I) changing ℓℓ(A) to ord(I).
Proposition 5.6. Let (A, m) be an excellent normal local ring as in Theorem 5.2
and I be an integrally closed ideal in A. Then we have the following inequalities;
(1) If we put c = pg (A) − ℓA (H 1 (X, OX (−M ))), then we have an inequality
(µ(I) − 1 + c) · ℓℓ(I) ≥ e(I).
Note that c is an invariant depending only on A.
(2) If we change ℓℓ(I) to ord(I), we have the converse inequality;
e(I) ≥ (µ(I) − 1) · ord(I).

16

TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

Proof. We modify our argument in the proof of Theorem 5.2.
(1) In the situation of the proof of Theorem 5.2, by [13, Theorem 6.1], we have
µ(I) ≥ −M Z + 1 − c and then the argument is the same as Theorem 5.2.
(2) Since I ⊂ mord(I) , we have Z ≥ ord(I)M and hence −Z 2 ≥ − ord(I)M Z and
−M Z ≥ µ(I) − 1.

Example 5.7. Put A = k[[X, Y, Z]]/(f ), where f is homogeneous of degree n ≥ 3

and we assume A is normal. Then c = pg (A) − ℓA (H 1 (X, OX (−M ))) = n−1
in
2
s
this case. In fact, if we take I = m with s ≥ n, then we have ℓℓ(I) = s, µ(I) =

sn − (n − 3)n/2 and we see that c = n−1
is best possible.
2
In dimension ≥ 3, we can take A so that there is no constant c for which the
inequality
(d − 1)!(µ(I) − d + 1 + c) · ℓℓ(I) ≥ e(I)
hold for all integrally closed ideal I. See the next example.
Example 5.8. Let A = k[[x, y, z, w]]/(f ), where f is a homogeneous polynomial
of degree n and we assume A is normal. Then we can see if n ≥ 5, putting I = ms ,
to have inequality (d − 1)!(µ(I) − d + 1 + c) · ℓℓ(I) ≥ e(I) holds we have to take c
arbitrary large when s tends to infinity.
6. What Ideals have SRP ?
We have shown under certain condition, mn has SRP and also in regular local
rings of dimension 2, mn (n ≥ 1) are only ideals with SRP in Example 2.9. In
this section, we ask if that this property characterize regular local rings and some
Veronese subrings in dimension 2. We get a partial result for rational singularities.
Proposition 6.1. Let (A, m) be a two-dimensional rational singularity and assume
that A/m is algebraically closed. Then we have the following results concerning the
strong Rees property for integral closed ideals.
(1) If an integrally closed ideal I has the strong Rees property, then I is a good
ideal in the sense of [4].
(2) If the minimal resolution of Spec(A) has more than two exceptional curves,
then there are integrally closed ideals I 6= mn with the property such that
µ(I ′ ) < µ(I) for every integrally closed ideal I ′ strictly containing I.
Proof. If A is a rational singularity, and if I = IZ as in the proof of Theorem 5.2,
we have shown µ(I) = −M Z + 1.
(1) If I is not good and I ′ is the minimal good ideal containing I ′ , we have
µ(I) = µ(I ′ ). Hence I does not have SRP. (2) If I = IZ has SRP, then Z is defined
on the minimal resolution of Spec(A) and the property “µ(I ′ ) < µ(I) for every
integrally closed ideal I ′ strictly containing I” is equivalent to say that “for every
cycle Z ′ < Z, −M Z ′ < −M Z or M (Z ′ − Z) > 0. If the minimal resolution of

THE STRONG REES PROPERTY OF POWERS OF THE MAXIMAL IDEAL

17

Spec(A) contains more than two curves, we can construct such Z 6= nM with that
property.

Conjecture 6.2. We believe that if (A, m) is a local ring of dimension d ≥ 2 and
if mn are only ideals of A which has the strong Rees property, then dim A = 2 and
either A is a regular local ring or Â ∼
= k[[X r , X r−1 Y, . . . , Y r ]] for some r.
Acknowledgment 1. The authors would like to thank Hailong Dao for giving
us a motivation of this work. Moreover, they would like to thank Shiro Goto,
Jürgen Herzog, Ryo Takahashi, Junzo Watanabe and Santiago Zarzuela for giving
us several valuable comments.
The second author is supported by Grant-in-Aid for Scientific Research (C)
26400053. The third author is supported by Grant-in-Aid for Scientific Research
(C) 16K05110.
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TONY J.PUTHENPURAKAL, KEI-ICHI WATANABE, KEN-ICHI YOSHIDA

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Tony J.Puthenpurakal, Department of Mathematics, IIT Bombay, Powai, Mumbai
400076, India, email: tputhen@gmail.com
Kei-ichi Watanabe, Department of Mathematics, College of Humanity and Sciences,
Nihon University, Setagaya-ku, Tokyo, 156-8550, Japan, email: watanabe@math.chs.nihonu.ac.jp
Ken-ichi Yoshida, Department of Mathematics, College of Humanities and Sciences,
Nihon University, Setagaya-ku, Tokyo, 156-8550, Japan, email: yoshida@math.chs.nihonu.ac.jp

