Normal Reduction Numbers for Normal Surface Singularities with Application to Elliptic Singularities of Brieskorn Type

Abstract

In this paper, we give a formula for normal reduction number of an integrally closed $\mathfrak m$-primary ideal of a two-dimensional normal local ring $(A,\mathfrak m)$ in terms of the geometric genus p_g(A) of A. Also, we compute the normal reduction number of the maximal ideal of Brieskorn hypersurfaces. As an application, we give a short proof of a classification of Brieskorn hypersurfaces having elliptic singularities.

1 Introduction

For a Noetherian local ring $(A,\mathfrak m)$ and an $\mathfrak m$-primary ideal I, let $\overline {I}$ denote the integral closure, that is, $z \in \overline {I}$ if and only if zⁿ + c₁z^n− 1 + ⋯ + c_n = 0 for some n ≥ 1 and c_i ∈ Iⁱ (i = 1,…,n).

For a given Noetherian local ring $(A,\mathfrak m)$ and an integrally closed $\mathfrak m$-primary ideal I (i.e., $\overline {I}=I)$ with minimal reduction Q, we are interested in the question:

Question 1

What is the minimal number r such that $\overline {I^{r}} \subset Q$ for every $\mathfrak m$-primary ideal I of A and its minimal reduction Q?

One example of this direction is the Briançon-Skoda Theorem saying; If $(A,\mathfrak m)$ is a d-dimensional rational singularity (characteristic 0) or an F-rational ring (characteristic p > 0), then $\overline {I^{d}} \subset Q$ and d is the minimal possible number in this case (cf. [3, 9]).

The aim of our paper is to answer this question in the case of normal two-dimensional local rings using resolution of singularities.

In what follows, we always assume that $(A,\mathfrak {m})$ is an excellent two-dimensional normal local domain. For any $\mathfrak m$-primary integrally closed ideal I ⊂ A (e.g., the maximal ideal $\mathfrak m$) and its minimal reduction Q of I, we define two normal reduction numbers as follows:

$$\begin{array}{@{}rcl@{}} \text{nr}(I) &=& \min\{n \in \mathbb Z_{\ge 0} | \overline{I^{n + 1}}=Q\overline{I^{n}}\}, \\ \bar{\text{r}}(I) &=& \min\{n \in \mathbb Z_{\ge 0} | \overline{I^{N + 1}}=Q\overline{I^{N}} \text{ for every \(N \ge n\)}\}. \end{array} $$

These are analogues of the reduction number r_Q(I) of an ideal I ⊂ A. But in general, r_Q(I) is not independent of the choice of a minimal reduction Q. On the other hand, $\text {nr}(I)=\bar {\text {r}}(I)$ is not known in general.

Also, we define the following:

$$\begin{array}{@{}rcl@{}} \text{nr}(A)=\max\{\text{nr}(I) | I \text{is an} \mathfrak m\text{-primary integrally closed ideal of} A\}, \\ \bar{\text{r}}(A)=\max\{\bar{\text{r}}(I) | I \text{is an} \mathfrak m\text{-primary integrally closed ideal of} A\}. \end{array} $$

These invariants of A characterize “good” singularities.

Example 1.1 (See [8] for (1), [12] for (2))

Suppose that A is not regular.

• (1) A is a rational singularity (p_g(A) = 0) if and only if $\text {nr}(A)= \bar {\text {r}}(A) = 1$.

• (2) If A is an elliptic singularity, then $\bar {\text {r}}(A) = 2$, where we say that A is an elliptic singularity if the arithmetic genus of the fundamental cycle on any resolution of A is 1.

One of the main aims is to compare these invariants with geometric invariants (e.g., geometric genus p_g(A)). In [13], we have shown that nr(A) ≤ p_g(A) + 1. But actually, it turns out that we have a much better bound (see Theorem 2.9).

Theorem 1.2

If $(A,\mathfrak m)$ is a normal two-dimensional local ring, then $p_{g}(A) \ge \binom {\text {nr}(A)}{2}$ .

On the other hand, sometimes we have $\text {nr}(A) = \text {nr}(\mathfrak m)$. For example, if A = K[[x,y,z]]/(f), where f is a homogeneous polynomial of degree d ≥ 2 with isolated singularity, it is easy to see $\text {nr}(\mathfrak m) = d-1$. If d ≤ 4, we can see by Theorem 1.2 that $\text {nr}(A) = \text {nr}(\mathfrak m)=d-1$. We do not have an answer yet if d = 5.

Question 1.3

If A is a homogeneous hypersurface singularity of degree d, then nr(A) = d − 1?

To have examples for this theory, we compute $\text {nr}(\mathfrak m)$ of Brieskorn hypersurface singularities, that is, two-dimensional normal local domains

$$A=K[[x,y,z]]/(x^{a}+y^{b}+z^{c}), $$

where K is an algebraically closed field of any characteristic and 2 ≤ a ≤ b ≤ c.

Note that our approach in this paper will be extended to the case of Brieskorn complete intersection singularity (see [11]).

We can get an explicit value of $\text {nr}(\mathfrak m)$ in this case.

Theorem 3.1 1

Let A be a Brieskorn hypersurface singularity as above.Put$\mathfrak m=(x,y,z)A$andQ = (y,z)A.Then

$$\text{nr}(\mathfrak m)=\bar{\mathrm{r}}(\mathfrak m)= \left\lfloor \frac{(a-1)b}{a} \right\rfloor. $$

Moreover, if we put $n_{k}=\lfloor \frac {kb}{a}\rfloor$ for each k ≥ 0, then

$$\overline{\mathfrak m^{n}}=Q^{n} + xQ^{n-n_{1}}+x^{2}Q^{n-n_{2}}+\cdots +x^{a-1}Q^{n-n_{a-1}}. $$

As an application of the theorem, we can show that the Rees algebra $\mathcal {R}(\mathfrak m)$ is normal if and only if $\bar {\text {r}}(\mathfrak m)=a-1$ (see Corollary 3.7). Moreover, we can determine $\ell _{A}(\mathfrak m^{n + 1}/Q\mathfrak m^{n})$ for every n ≥ 0 and $q(\mathfrak m)=\ell _{A}(H^{1}(X,\mathcal {O}_{X}(-M)))$, where X →SpecA denotes the resolution of singularity of SpecA and M denotes the maximal ideal cycle on X.

In the last section, we discuss Brieskorn hypersurfaces with elliptic singularities. In fact, the first author proved that if A is an elliptic singularity then nr(A) = 2. In particular, if A is an elliptic singularity then $\text {nr}(\mathfrak m) \le 2$. If, in addition, A is a Brieskorn hypersurface singularity A = K[[x,y,z]]/(x^a + y^b + z^c), then our theorem shows that ⌊(a − 1)b/a⌋≤ 2. Using this fact, we can classify all Brieskorn hypersurfaces with elliptic singularity (see Theorem 4.4).

We are interested to know if nr(A) characterizes elliptic singularities or not. Namely, the question is equivalent to say, if A is not rational or elliptic, then does there exist I such that nr(I) ≥ 3? We can find such an ideal for all non-elliptic Brieskorn hypersurface singularity except (a,b,c) = (3,4,6) or (3,4,7).

2 Normal Reduction Numbers and Geometric Genus

Throughout this paper, let $(A,\mathfrak m)$ be a two-dimensional excellent normal local domain. In another word, A is a local domain with a resolution of singularities f : X →Spec(A). For a coherent $\mathcal {O}_{X}$-Module $\mathcal {F}$, we denote by $h^{i}(\mathcal {F})$ the length $\ell _{A}(H^{i}(\mathcal {F}))$.

We define the geometric genus of A by the following:

$$p_{g}(A)=h^{1}(\mathcal{O}_{X}), $$

which is independent of the choice of resolution of singularities. When p_g(A) = 0, A is called a rational singularity.

Let I ⊂ A be an $\mathfrak m$-primary integrally closed ideal. Then, there exist a resolution of singularity X →SpecA and an anti-nef cycle Z on X so that $I\mathcal {O}_{X}=\mathcal {O}_{X}(-Z)$ and $I=H^{0}(\mathcal {O}_{X}(-Z))$. Then, we say that I is represented by Z on X and write I = I_Z. Then, $I_{nZ}=\overline {I^{n}}$ for every integer n ≥ 1.

In what follows, let A, X, I = I_Z be as above.

The authors have studied p_g-ideals in [13,14,15]. So, we first recall the notion of p_g-ideals in terms of q(kI).

Definition 2.1

Put $q(0 I)=h^{1}(\mathcal {O_{X}})$, $q(I):= h^{1}(\mathcal {O}_{X}(-Z))$ and $q(nI)= q(\overline {I^{n}})$ for every integer n ≥ 1.

Theorem 2.2

[13] The following statements hold.

• (1) 0 ≤ q(I) ≤ p_g(A).

• (2)q(kI) ≥ q((k + 1)I) forevery integerk ≥ 1.

• (3)q(nI) = q((n + 1)I) = q((n + 2)I) = ⋯ forsome integern ≥ 0.

Definition 2.3

[13] The ideal I is called the p_g-ideal if q(I) = p_g(A).

Example 2.4

Any two-dimensional excellent normal local domain over an algebraically closed field admits a p_g-ideal. Moreover, if A is a rational singularity, then every $\mathfrak m$-primary integrally closed ideal is a p_g-ideal.

2.1 Upper Bound on Normal Reduction Numbers

Let Q be a minimal reduction of I. Then, there exists a nonnegative integer r such that $\overline {I^{r + 1}}=Q\overline {I^{r}}$. This is independent of the choice of a minimal reduction Q of I (see, e.g., [5, Theorem 4.5]). So we can define the following notion.

Definition 2.5 (Normal reduction number)

Put

$$\begin{array}{@{}rcl@{}} \text{nr}(I) &=& \min\{n \in \mathbb Z_{\ge 0} | \overline{I^{n + 1}}=Q\overline{I^{n}}\},\\ \bar{\text{r}}(I) &=& \min\{n \in \mathbb Z_{\ge 0} | \overline{I^{N + 1}}=Q\overline{I^{N}} \text{ for every } N \ge n \}. \end{array} $$

We call them the normal reduction numbers of I. We also define

$$\begin{array}{@{}rcl@{}} \text{nr}(A)&=&\max\{\text{nr}(I)| I \text{is a} \mathfrak m\text{-primary integrally closed ideal of} A\}, \\ \bar{\text{r}}(A)&=&\max\{\bar{\text{r}}(I)| I \text{is a} \mathfrak m\text{-primary integrally closed ideal of} A\}, \end{array} $$

which are called the normal reduction numbers of A.

Our study on normal reduction numbers is motivated by the following observation: For an $\mathfrak m$-primary ideal I in a two-dimensional excellent normal local domain A, I is a p_g-ideal if and only if $\bar {\text {r}}(I)= 1$.

By definition, $\text {nr}(I) \le \bar {\text {r}}(I)$ holds in general. In the next section, we show that $\text {nr}(\mathfrak m)=\bar {\text {r}}(\mathfrak m)$ holds true for any Brieskorn hypersurface A = K[[x,y,z]]/(x^a + y^b + z^c). But it seems to be open whether equality always holds for other integrally closed $\mathfrak m$-primary ideals.

Question 2.6

When does $\text {nr}(I)=\bar {\text {r}}(I)$ hold?

In order to state the main result in this section, we recall the following lemma, which gives a relationship between nr(I) and q(kI).

Lemma 2.7

For any integern ≥ 1,we have

$$2 \cdot q(nI) + \ell_{A}(\overline{I^{n + 1}}/Q\overline{I^{n}}) =q((n + 1)I)+q((n-1)I). $$

Proof

Assume Q = (a,b) and consider the exact sequence as follows:

$$0\to \mathcal O_{X}((n-1)Z) \to \mathcal O_{X}(-Z)(-nZ)^{\oplus 2}\to \mathcal O_{X}(-(n + 1)Z)\to 0, $$

where the map $\mathcal O_{X}(-nZ)^{\oplus 2}\to \mathcal O_{X}(-(n + 1)Z)$ is defined by (x,y)↦ax + by as in Lemma 4.3 of [15]. By taking the cohomology long exact sequence, we have the following exact sequence:

$$\begin{array}{lll} & \to H^{0}(\mathcal{O}_{X}(-nZ))^{\oplus 2} & \stackrel{\varphi}{\to} H^{0}(\mathcal{O}_{X}(-(n + 1)Z)) \\ \to H^{1}(\mathcal{O}_{X}(-(n-1)Z)) & \to H^{1}(\mathcal{O}_{X}(-nZ))^{\oplus 2} & \to H^{1}(\mathcal{O}_{X}(-(n + 1)Z)) \to 0. \end{array} $$

Since $\text {Coker}(\varphi )\cong \overline {I^{n + 1}}/Q\overline {I^{n}}$, we obtain the required assertion. □

The lemma gives another description of nr(I) in terms of q(kI):

$$\text{nr}(I)=\min\{n\in \mathbb Z_{\ge 1} | q((n-1)I), q(nI), q((n + 1)I) \text{forms an arithmetic sequence} \}. $$

In particular,

$$\text{nr}(I) \le \min\{n \in \mathbb Z_{\ge 0}| q((n-1)I)=q(nI)=q((n + 1)I)={\cdots} \}=\bar{\text{r}}(I). $$

If the following question has an affirmative answer for I, then $\text {nr}(I)=\bar {\text {r}}(I)$ holds true.

Question 2.8

When is $\ell _{A}(\overline {I^{n + 1}}/Q\overline {I^{n}})$ a non-increasing function of n?

The main result in this section is the following theorem, which refines an inequality nr(I) ≤ p_g(A) + 1 (see [14, Lemma 3.1]).

Theorem 2.9

For any$\mathfrak m$-primaryintegrally closed idealI ⊂ A,we have

$$p_{g}(A) \ge \binom{r}{2} + q(rI), $$

where r = nr(I). In particular, $p_{g}(A) \ge \binom {\text {nr}(A)}{2}$.

Proof

Suppose nr(I) = r. Then, since $\overline {I^{k + 1}} \ne Q\ \overline {I^{k}}$ for every k = 1,2,…,r − 1 and $\overline {I^{r + 1}} = Q\ \overline {I^{r}}$, we have

$$\begin{array}{@{}rcl@{}} q((r-1)I)-q(rI) &=& q(rI)-q((r + 1)I), \\ q((r-2)I)-q((r-1)I)& \ge & q((r-1)I)-q(rI)+ 1, \\ &{\vdots} & \\ p_{g}(A)-q(I) &\ge & q(I) -q(2 I)+ 1. \end{array} $$

Thus, if we put a_k = q((r − k)I) for k = 0,1,…,r, then we get

$$a_{k}-a_{k-1} \ge a_{k-1}-a_{k-2}+ 1 \ge \cdots \ge \big\{a_{1}-a_{0} \big\} +(k-1) \ge k-1. $$

Hence,

$$p_{g}(A)=a_{r}={\sum}_{k = 1}^{r} (a_{k}-a_{k-1})+a_{0} \ge {\sum}_{k = 1}^{r} (k-1) + a_{0} =\frac{r(r-1)}{2}+q(rI), $$

as required.

The last assertion immediately follows from the definition of nr(A). □

The above theorem gives a best possible bound (see also the next section).

Example 2.10

If $p_{g}(A) < \binom {\text {nr}(J)+ 1}{2}$ for some $\mathfrak m$-primary integrally closed ideal J ⊂ A, then nr(A) = nr(J).

Proof

Suppose nr(A)≠nr(J). Then, nr(A) ≥nr(J) + 1. By assumption and the theorem, we have

$$\binom{\text{nr}(A)}{2} \le p_{g}(A) < \binom{\text{nr}(J)+ 1}{2} \le \binom{\text{nr}(A)}{2}. $$

This is a contradiction. Therefore, nr(A) = nr(J). □

3 Normal Reduction Numbers of the Maximal Ideal of Brieskorn Hypersurfaces

Let K be a field of any characteristic, and let a,b,c be integers with 2 ≤ a ≤ b ≤ c. Then, a hypersurface singularity

$$A=K[[x,y,z]]/(x^{a}+y^{b}+z^{c}),\quad \mathfrak m=(x,y,z)A $$

is called a Brieskorn hypersurface singularity if A is normal.

3.1 Normal Reduction Number of the Maximal Ideal

The main purpose in this section to give a formula for the reduction number of the maximal ideal $\mathfrak m$ in a hypersurface of Brieskorn type: A = K[[x,y,z]]/(x^a + y^b + z^c). Namely, we prove the following theorem.

Theorem 3.1

LetA = K[[x,y,z]]/(x^a + y^b + z^c) bea Brieskorn hypersurface singularity. If we putQ = (y,z)Aand$n_{k}=\lfloor \frac {kb}{a} \rfloor$fork = 1,2,…,a − 1,then,$\mathfrak m=\overline {Q}$andwe have

• (1)$\overline {\mathfrak m^{n}}=Q^{n}+xQ^{n-n_{1}}+x^{2}Q^{n-n_{2}}+ {\cdots } +x^{a-1}Q^{n-n_{a-1}}$foreveryn ≥ 1.

• (2)$\bar {\mathrm {r}}(\mathfrak m)=\text {nr}(\mathfrak m)=n_{a-1}$.In particular, if$\bar {\mathrm {r}}(\mathfrak m)\le 2$,then,$\lfloor \frac {(a-1)b}{a} \rfloor \le 2$.

• (3)$\overline {\mathcal {R}^{\prime }(\mathfrak m)}$and$\overline {G}(\mathfrak m)$areCohen-Macaulay.

Remark 3.2

Note 0 := n₀ ≤ n₁ < n₂ < ⋯ < n_a− 1. In particular, n_k ≥ k for each k = 0,1,…, a − 1.

In the following, we use the notation in this theorem and prove it.

Lemma 3.3

For integers k, n withn ≥ 1 and1 ≤ n ≤ a − 1,we have that$x^{k} \in \overline {Q^{n}}$ifand only ifn ≤ n_k.

Proof

Suppose n ≤ n_k. Then,

$$(x^{k})^{a}=(x^{a})^{k}=(-1)^{k} (y^{b}+z^{c})^{k} \in Q^{bk} \subset Q^{an_{k}} =(Q^{n_{k}})^{a}. $$

Hence, $x^{k} \in \overline {Q^{n_{k}}} \subset \overline {Q^{n}}$.

Next, we prove the converse. Suppose $x^{k} \in \overline {Q^{n}}$. Then, there exists a nonzero element c ∈ A such that c(x^k)^ℓ ∈ Q^nℓ for all large integers ℓ. By Artin-Rees’ lemma [10, Theorem 8.5], we can choose an integer ℓ₀ ≥ 1 such that $Q^{\ell } \cap cA =cQ^{\ell -\ell _{0}}$ for every ℓ ≥ ℓ₀.

Now suppose that n ≥ n_k + 1. Since $\frac {kb}{a}+\frac {1}{a} \le n_{k}+ 1 \le n$, we get

$$(y^{b}+z^{c})^{k\ell}=(-1)^{k}x^{ka\ell} \in Q^{na\ell} \colon c \subset Q^{na\ell -\ell_{0}} \subset Q^{(n_{k}+ 1)a\ell-\ell_{0}} \subset Q^{(bk+ 1)\ell-n_{0}} $$

for sufficiently large ℓ. This implies that y^bkℓ ∈ (y^bkℓ+ 1,z) and this is a contradiction because y, z forms a regular sequence. Therefore, n ≤ n_k, as required. □

Corollary 3.4

For an integern ≥ 1, if weput

$$L_{n}=Q^{n}+xQ^{n-n_{1}}+x^{2}Q^{n-n_{2}}+ {\cdots} + x^{a-1}Q^{n-n_{a-1}}, $$

then $Q^{n} \subset L_{n} \subset \overline {Q^{n}} = \overline {\mathfrak m^{n}}$.

Proof

It is enough to prove $x^{k}y^{i}z^{j} \in \overline {Q^{n}}$ if and only if i + j ≥ n − n_k. In fact, since Q = (y,z) is a parameter ideal in A, [6, Corollary 6.8.13] and Lemma 3.3 imply

$$\begin{array}{@{}rcl@{}} x^{k}y^{i}z^{j} \in \overline{Q^{n}} & \Longleftrightarrow & x^{k}y^{i-1}z^{j} \in \overline{Q^{n-1}} \\ & \Longleftrightarrow & \cdots{\cdots} \\ & \Longleftrightarrow & x^{k} \in \overline{Q^{n-i-j}} \\ & \Longleftrightarrow & n-n_{k} \le i+j . \end{array} $$

Hence, $L_{n} \subset \overline {Q^{n}}$. □

Put $d=\gcd (a,b)$, $a^{\prime }=\frac {a}{d}$ and $b^{\prime }=\frac {b}{d}$. If we put

$$I_{n}=(x^{k}y^{i}z^{j} | kb^{\prime}+ia^{\prime}+ja^{\prime} \ge n)A $$

for every n ≥ 1, then $\{I_{n}\}_{n = 1,2,\dots }$ is a filtration of A.

Lemma 3.5

G({I_n}) isalways reduced. In particular,$\mathcal {R}^{\prime }(\{I_{n}\})$isa Gorenstein normal domain.

Proof

One can easily see

$$ G(\{I_{n}\}) \cong \left\{ \begin{array}{ll} K[X,Y,Z]/(X^{a}+Y^{b}+Z^{c}) & \text{if } b=c \\ K[X,Y,Z]/(X^{a}+Y^{b}) & \text{if } b<c. \end{array} \right. $$

(3.1)

By assumption, K[X,Y,Z]/(X^a + Y^b + Z^c) is a normal domain. If charK = 0, then K[X,Y,Z]/(X^a + Y^b) is reduced. Otherwise, we put p = charK > 0. Since A is normal, we have that p does not divide $\gcd (a,b)=d$. Hence K[X,Y ]/(X^a + Y^b) is reduced.

As A is normal, $R=\mathcal {R}^{\prime }(\{I_{n}\})$ is a Gorenstein normal domain because G({I_n})≅R/t^− 1R. □

Lemma 3.6

$L_{n}=I_{na^{\prime }}$foreveryn ≥ 1.

Proof

Since L_n and $I_{na^{\prime }}$ are monomial ideals, it suffices to show that x^kyⁱz^j ∈ L_n if and only if $x^{k}y^{i}z^{j} \in I_{na^{\prime }}$. But this is clear from the definition. □

We are now ready to prove the theorem.

Proof of Theorem 3.1

(1) Since $\mathcal {R}^{\prime }(\{I_{n}\})$ is normal by Lemma 3.5, we have that every I_n is integrally closed. In particular, $L_{n} = I_{na^{\prime }}$ is also integrally closed by Lemma 3.6. Therefore, $L_{n}=\overline {Q^{n}}=\overline {\mathfrak m^{n}}$ by Corollary 3.4.

(2) One can easily see that L_n+ 1 = QL_n if and only if n ≥ n_a− 1. Hence, (2) is immediately follows from (1).

(3) $\overline {\mathcal {R}^{\prime }(\mathfrak m)}$ is Cohen-Macaulay since it is a Veronese subring of a Cohen-Macaulay ring $\mathcal {R}^{\prime }(\{I_{n}\})$. Then $\overline {G}(\mathfrak m)=\overline {\mathcal {R}^{\prime }(\mathfrak m)}/t^{-1}\overline {\mathcal {R}^{\prime }(\mathfrak m)}$ is also Cohen-Macaulay by [14, Theorem 4.1]. □

Corollary 3.7

Let $(A,\mathfrak m)$ be a Brieskorn hypersurface as in Theorem 3.1. Then,

• (1)$\mathcal {R}(\mathfrak m)$isnormal if and only if$\bar {\mathrm {r}}(\mathfrak m) = a-1$.

• (2)$\overline {\mathcal {R}(\mathfrak m)}$isCohen-Macaulay if and only if$\bar {\mathrm {r}}(\mathfrak m) = 1$.

• (3)$\mathfrak m$isap_g-idealif and only ifa = 2 and$\bar {\mathrm {r}}(\mathfrak m)= 1$.

Proof

(1) Suppose $\bar {\text {r}}(\mathfrak m)=a-1$. Then, n_a− 1 = a − 1 by (1) and this implies that n_k = k for each k = 1,2,…,a − 1. Then, one can easily see that $\overline {\mathfrak m^{n}} = (Q,x)^{n}=\mathfrak m^{n}$ for every n ≥ 1. Hence, $\mathcal {R}(\mathfrak m)$ is normal.

Conversely, if $\mathcal {R}(\mathfrak m)$ is normal, then, $\overline {\mathfrak m^{n}} = \mathfrak m^{n}=(Q,x)^{n}$. Then, n_a− 1 = a − 1.

(2) Since $F=\{\overline {\mathfrak m^{n}}\}$ is a good $\mathfrak m$-adic filtration, $\overline {\mathcal {R}(\mathfrak m)}=\mathcal {R}(F)$ is Cohen-Macaulay if and only if G(F) is Cohen-Macaulay and $\bar {\text {r}}(\mathfrak m)-2=a(G(F))<0$ by [2, Part 2, Corollary 1.2] and [4, Theorem 3.8].

(3) $\mathfrak m$ is a p_g-ideal if and only if $R(\mathfrak m)$ is normal and Cohen-Macaulay. Hence, the assertion follows from (1), (2). □

3.2 $q(\mathfrak m)$ and $\ell _{A}(\overline {\mathfrak m^{n + 1}}/Q\overline {\mathfrak m^{n}})$

In the proof of Theorem 3.1, we gave a formula of the integral closure of $\mathfrak m^{n}$. As an application, we give a formula of $q(\mathfrak m)$ for Brieskorn hypersurface singularities.

Proposition 3.8

LetA = K[[x,y,z]]/(x^a + y^b + z^c) bea Brieskorn hypersurface singularity. Under the same notation as in Theorem 3.1, wehave

• (1)$\ell _{A}(\overline {\mathfrak m^{n + 1}}/Q\overline {\mathfrak m^{n}}) = \max \left (a-\lceil \frac {a(n + 1)}{b}\rceil ,0\right )$.

• (2)$q(\mathfrak m)=p_{g}(A)-\displaystyle {{\sum }_{k = 1}^{a-1}} (n_{k}-n_{k-1})(a-k)$.

Proof

Suppose n_k ≤ n < n_k+ 1 for some 0 ≤ k ≤ a − 2. Then $\overline {\mathfrak m^{n}}=Q^{n}+xQ^{n-n_{1}} +{\cdots } + x^{k}Q^{n-n_{k}}+(x^{k + 1})$ and x^k+ 1, x^k+ 2,…, x^a− 1 forms a K-basis of $\overline {\mathfrak m^{n + 1}}/Q\overline {\mathfrak m^{n}}$ and thus $\ell _{A}(\overline {\mathfrak m^{n + 1}}/Q\overline {\mathfrak m^{n}})=a-1-k$. Hence,

$$\ell_{A}(\overline{\mathfrak m^{n + 1}}/Q\overline{\mathfrak m^{n}}) = \left\{ \begin{array}{ll} a-1-k & \text{if } n_{k} \le n < n_{k + 1}, k = 0,1,\ldots,a-2; \\ 0 & \text{if } n \ge n_{a-1}. \end{array} \right. $$

Moreover, one can easily see $k=a-\lceil \frac {a(n + 1)}{b}\rceil -1$.

(2) Put $a_{n}=p_{g}(A)-q(n\mathfrak m)$ and $v_{n}=\ell _{A}(\overline {\mathfrak m^{n + 1}}/Q\overline {\mathfrak m^{n}})$ for every n ≥ 0. Then, a₀ = 0 and {a_n} is an increasing sequence and a_n+ 1 = a_n for sufficiently large n. By Lemma 2.7, we have

$$0=a_{n + 1}-a_{n}=a_{n}-a_{n-1}-v_{n} = {\cdots} = a_{1} - a_{0} - {\sum}_{k = 1}^{n} v_{k} $$

for sufficiently large n ≥ 1. Hence, (1) yields

$$p_{g}(A)-q(\mathfrak m)=a_{1}= {\sum}_{k = 1}^{n} v_{k} ={\sum}_{k = 1}^{a-1} (n_{k}-n_{k-1})(a-k), $$

as required. □

When a = 2, one can obtain the following.

Example 3.9

Let A = K[[x,y,z]]/(x² + y^b + z^c) be a Brieskorn hypersurface singularity and put $r=\lfloor \frac {b}{2} \rfloor$. Then, (1) $q(i \mathfrak m)=\left \{\begin {array}{ll}p_{g}(A)-i(r-1) + \binom {i}{2} & \text {if } 1 \le i \le r-1; \\ p_{g}(A)-\binom {r}{2} & \text {if } i \ge r. \end {array} \right .$ (2) The normal Hilbert coefficients of $\mathfrak m$ are given as follows:

$$\overline{e}_{0}(\mathfrak m)= 2, \quad \overline{e}_{1}(\mathfrak m)=r, \quad \overline{e}_{2}(\mathfrak m)=\binom{r}{2}, $$

where

$$\ell_{A}(A/\overline{I^{n + 1}}) = \overline{e}_{0}(I) \binom{n + 2}{2} - \overline{e}_{1}(I)\binom{n + 1}{1}+\overline{e}_{2}(I) $$

for sufficiently large n.

3.3 Geometric Genus

In this subsection, let us consider a graded ring

$$B=K[x,y,z]/(x^{a}+y^{b}+z^{c}) $$

with $\deg x=q_{0}=bc$, $\deg y=q_{1}=ac$ and $\deg z=q_{2}=ab$. Put $\mathfrak m=(x,y,z)A$ and D = abc. In particular, the a-invariant of B is given by a(B) = D − q₀ − q₁ − q₂. Also, we have that $A=\widehat {B_{\mathfrak m}}$ is the completion of the local ring $B_{\mathfrak m}$. Then, we can calculate p_g(A) using this formula.

Lemma 3.10

Under the above notation, we have

$$p_{g}(A)=\sum\limits_{i = 0}^{a(B)} \dim_{K} B_{i} =\sharp\{(t_{0}, t_{1},t_{2}) \in \mathbb Z_{\ge 0}^{\oplus 3} | D - q_{0}-q_{1}-q_{2} \ge q_{0}t_{0}+q_{1}t_{1}+q_{2}t_{2}\}. $$

We can find many examples of Brieskorn hypersurfaces with p_g(A) = p for a given p ≥ 1 if $\text {nr}(\mathfrak m)= 1,2$.

Example 3.11

Let p ≥ 1 be an integer.

• (1) If $A=\mathbb {C}[[x,y,z]]/(x^{2}+y^{3}+z^{6p + 1})$, then p_g(A) = p and $\text {nr}(\mathfrak m)=\bar {\text {r}}(\mathfrak m)= 1$.

• (2) If $A=\mathbb {C}[[x,y,z]]/(x^{2}+y^{4}+z^{4p + 1})$, then p_g(A) = p and $\text {nr}(\mathfrak m)=\bar {\text {r}}(\mathfrak m)= 2$.

Example 3.12

Let k ≥ 1 be an integer.(1) Put $A=\mathbb {C}[[x,y,z]]/(x^{2}+y^{6}+z^{10k+i})$ for i = 0,1,…,9. Then, $\text {nr}(\mathfrak m)=\bar {\text {r}}(\mathfrak m)= 3$ and

$${}p_{g}(A)=\left\{ \begin{array}{ll} 6k, & \text{(if \(i = 0,1,2\))}; \\ 6k + 1, & \text{(if \(i = 3,4,5\))}; \end{array} \right. \qquad p_{g}(A)=\left\{ \begin{array}{ll} 6k + 3, & \text{(if \(i = 6,7,8\))}; \\ 6k + 4, & \text{(if \(i = 9,10,11\))}. \end{array} \right. $$

(2) Put $A=\mathbb {C}[[x,y,z]]/(x^{2}+y^{7}+z^{14k+i})$ for i = 0,1,…,13. Then, $\text {nr}(\mathfrak m)=\bar {\text {r}}(\mathfrak m)= 3$ and

$${}p_{g}(A)=\left\{ \begin{array}{ll} 9k, & \text{(if \(i = 0,1,2\))}; \\ 9k + 1, & \text{(if \(i = 3,4\))}; \\ 9k + 2, & \text{(if \(i = 5\))}; \\ 9k + 3, & \text{(if \(i = 6,7,8\))}; \end{array} \right. \qquad p_{g}(A)=\left\{ \begin{array}{ll} 9k + 4, & \text{(if \(i = 9\))}; \\ 9k + 5, & \text{(if \(i = 10,11\))}; \\ 9k + 6, & \text{(if \(i = 12,13\))}. \end{array} \right. $$

We discuss when $p_{g}(A)=\binom {\text {nr}(\mathfrak m)}{2}$ holds.

Proposition 3.13

Let$A=\mathbb {C}[[x,y,z]]/(x^{a}+y^{b}+z^{c})$with2 ≤ a ≤ b ≤ c.Then,$p_{g}(A)=\binom {\text {nr}(\mathfrak m)}{2}$ifand only if one of the following cases: ∙ (a,b,c) = (2,2,n) (n ≥ 1).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)= 1$andp_g(A) = 0.∙ (a,b,c) = (2,3,3),(2,3,4),(2,3,5).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)= 1$andp_g(A) = 0.∙ (a,b,c) = (2,4,4),(2,4,5),(2,4,6),(2,4,7).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)= 2$andp_g(A) = 1.∙ (a,b,c) = (2,2r,2r),(2,2r,2r + 1),(2,2r,2r + 2) (r ≥ 3).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)=r$and$p_{g}(A)=\binom {r}{2} \ge 3$.∙ (a,b,c) = (2,2r + 1,2r + 1),(2,2r + 1,2r + 2) (r ≥ 2).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)=r$and$p_{g}(A)=\binom {r}{2}$.∙ (a,b,c) = (3,3,3),(3,3,4),(3,3,5).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)= 2$andp_g(A) = 1.∙ (a,b,c) = (3,3s + 1,3s + 1).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)= 2s$and$p_{g}(A)=\binom {2s}{2}$.∙ (a,b,c) = (3,3s + 2,3s + 2),(3,3s + 2,3s + 3).In this case,$\text {nr}(A)=\text {nr}(\mathfrak m)= 2s + 1$and$p_{g}(A)=\binom {2s + 1}{2}$.

Proof

We give a proof of only if part. Put $r=\text {nr}(\mathfrak m)$. By Theorem 3.1, we have $\lfloor \frac {(a-1)b}{a}\rfloor$. So, we can write (a − 1)b = ra + ε, where ε is an integer with 0 ≤ ε ≤ a − 1. Now suppose

$$abc-bc-ca-ab \ge bc \lambda_{0} + ca \lambda_{1} + ab \lambda_{2}. $$

Then,

$$ (ra+\varepsilon)c-ca-ab \ge bc \lambda_{0} + ca \lambda_{1} + ab \lambda_{2}. $$

(eq.pg)

Suppose λ₀ = 0. Then,

$$\left( r-1+\frac{\varepsilon}{~a~}-\lambda_{1} \right)\frac{~c~}{b} \ge \lambda_{2}+ 1 $$

and thus $\lambda _{1} < r-1+\frac {\varepsilon }{~a~}$. By Lemma 3.10 and assumption, we have

$$\begin{array}{@{}rcl@{}} \binom{r}{2} = p_{g}(A) &\ge & \sum\limits_{k = 0}^{r-1} \left\lfloor\left( r-1+\frac{\varepsilon}{~a~}-k \right)\frac{~c~}{b} \right\rfloor \\ &\ge & \sum\limits_{k = 0}^{r-2} \left\lfloor\left( r-1-k \right)\frac{~c~}{b} \right\rfloor + \left\lfloor\frac{\varepsilon}{~a~} \cdot \frac{~c~}{b} \right\rfloor \\ &\ge & \sum\limits_{k = 0}^{r-2} \left( r-1-k \right) + \left\lfloor\frac{\varepsilon}{~a~} \cdot \frac{~c~}{b} \right\rfloor \ge \binom{r}{2}. \end{array} $$

Hence, $\left (r-1+\frac {\varepsilon }{~a~}-k \right )\frac {~c~}{b} < r-k$ for each k = 0,1,…,r − 1. Moreover, if λ₀ ≥ 1, then, since λ₁ = λ₂ = 0 does not satisfy the condition (eq.pg) by Theorem 2.9, we get the following:

$$abc-bc-ca-ab < bc, \quad \text{that is,}\quad \frac{2}{~a~} + \frac{~1~}{b}+\frac{~1~}{c} > 1. $$

This implies a = 2,3. If a = 2, then ε = 0,1. If a = 3, then ε = 0,1,2.

Now suppose a = 3 and ε = 2. Then, as 2b = 3r + 2, we can write r = 2s, b = 3s + 1, where s ≥ 1. Moreover, the condition holds true if and only if $(2s+\frac {2}{3}-1)\frac {c}{3s + 1}< 2s$. This means $c < 3s + 1 + \frac {3s + 1}{6s-1}$. Hence, c = 3s + 1 because c ≥ b = 3s + 1. Similarly, easy calculation yields the required assertion.

□

3.4 Weighted Dual Graph

In this subsection, let us explain how to construct the weighted dual graph of the minimal good resolution of singularity X →SpecA for a Brieskorn hypersurface singularity A = K[[x,y,z]]/(x^a + y^b + z^c). Though it is obtained in [7], we use the notation of [11] in which the first author studies complete intersection singularities of Brieskorn type. Let E be the exceptional set of X →SpecA and E₀ the central curve with genus g and ${E_{0}^{2}}=-c_{0}$. We define positive integers a_i,ℓ_i,α_i, λ_i, $\widehat {g_{i}}$ (i = 1,2,3), $\widehat {g}$, and ℓ as follows:

$$\begin{array}{lll} a_{1} =a, & a_{2}=b, & a_{3}=c, \\ \ell_{1} = \text{lcm}(b,c), & \ell_{2} = \text{lcm}(a,c), & \ell_{3} = \text{lcm}(a,b), \\ \alpha_{1} = \frac{a_{1}}{(a_{1},\ell_{1})}, & \alpha_{2} = \frac{a_{2}}{(a_{2},\ell_{2})}, & \alpha_{3} = \frac{a_{3}}{(a_{3},\ell_{3})}, \\ \lambda_{1} = \frac{\ell_{1}}{(a_{1},\ell_{1})}, & \lambda_{2} = \frac{\ell_{2}}{(a_{2},\ell_{2})}, & \lambda_{3} = \frac{\ell_{3}}{(a_{3},\ell_{3})}, \\ \hat g_{1}=(b,c), & \hat g_{2}=(a,c), & \hat g_{3}=(a,b). \end{array} $$

We put $\widehat {g}=\frac {abc}{\text {lcm}(a,b,c)}$ and ℓ = lcm(a,b,c), and define integers β_i by the following condition:

$$\lambda_{i}\beta_{i}+ 1 \equiv 0 \pmod{\alpha_{i}},\quad 0 \le \beta_{i} < \alpha_{i}. $$

Then, E₀ has $\hat g_{1}+\hat g_{2} +\hat g_{3}$ branches. For each w = 1,2,3, we have $\hat g_{w}$ branches as follows:

$$B_{w} : \quad E_{w,1} - E_{w,2} - {\cdots} - E_{w,s_{w}}, $$

where $E_{w,j}^{2}=-c_{w,j}$ and

$$\frac{\alpha_{w}}{\beta_{w}} = [[c_{w,1},c_{w,2},\ldots,c_{w,s_{w}}]] $$

is a Hirzebruch-Jung continued fraction if α_w ≥ 2, we regard B_w empty if α_w = 1. Moreover, we have

$$2g-2=\hat g- \sum\limits_{w = 1}^{3} \hat g_{w}, \qquad c_{0}= \sum\limits_{w = 1}^{3} \frac{\hat g_{w} \beta_{w}}{\alpha_{w}} + \frac{\hat g}{\ell}. $$

For instance, if (a,b,c) = (3,4,7), then we have the following:

$$\begin{array}{llllll} a_{1} = 3, & a_{2}= 4, & a_{3}= 7, & \ell_{1} = 28, & \ell_{2} = 21, & \ell_{3} = 12, \\ \alpha_{1} = 3, & \alpha_{2} = 4, & \alpha_{3} = 7, & \lambda_{1} = 28, & \lambda_{2} = 21, & \lambda_{3} = 12, \\ \beta_{1} = 2, & \beta_{2} = 3, & \beta_{3} = 4, & \hat g_{1}= 1, & \hat g_{2}= 1, & \hat g_{3}= 1, \\ \hat g = 1, & \ell= 84. & & & & \end{array} $$

Thus, g = 0 and c₀ = 2. Therefore, each irreducible component of E is a rational curve, and the weighted dual graph of E is represented as in Fig. 1, where the vertex $\blacksquare$ has weight − 4 and other vertices ∙ have weight − 2.

Fig. 1

The weighted dual graph of k[[x,y,z]]/(x³ + y⁴ + z⁷)

See [11, 4.4] for more details.

4 Brieskorn Hypersurfaces with Elliptic Singularities

We use the notation of Section 3.4. Let Z_E denote the fundamental cycle.

We call p_f(A) := p_a(Z_E) the fundamental genus of A. The singularity A is said to be elliptic if p_f(A) = 1. We have the following.

Theorem 4.1

[12] Ifp_f(A) = 1,then$\bar {\mathrm {r}}(A)= 2$.

Remark 4.2

By [1], nr(A) = 1 if and only if A is rational. Therefore, $\text {nr}(A)=\bar {\text {r}}(A)$ if $\bar {\text {r}}(A)= 2$.

It is natural to ask whether the converse of Theorem 4.1 holds or not. In the following, we classify Brieskorn hypersurface singularities with p_f(A) = 1 or $\bar {\text {r}}(A)= 2$ as an application of results in Section 3. Before doing that, we need the following formula of p_f(A) in the case of Brieskorn hypersurfaces. Put α = α₁α₂α₃.

Lemma 4.3 (16, [7, Theorem 1.7], [11, 5.4])

Ifλ₃ ≤ α,then$-{Z_{E}^{2}} =\hat g_{3}\left \lceil \lambda _{3}/\alpha _{3}\right \rceil$and

$$\begin{array}{@{}rcl@{}} p_{f}(A)&=&\frac{1}{2}\lambda_{3} \left\{\hat g -\frac{(2\left\lceil\lambda_{3}/\alpha_{3}\right\rceil-1)\hat g_{3}}{\lambda_{3}} -\frac{\hat g_{1}}{\alpha_{1}}-\frac{\hat g_{2}}{\alpha_{2}}\right\}+ 1 \\ &=&\frac{1}{2}\left( ab-a-b-(2\left\lceil\lambda_{3}/\alpha_{3}\right\rceil-1)(a,b)\right)+ 1. \end{array} $$

We are now ready to state our result in the case of p_f(A) = 1.

Theorem 4.4

$(A,\mathfrak m)$is elliptic(i.e.,p_f(A) = 1) if andonly if (a,b,c) isone of the following.

• (1) (2,3,c),c ≥ 6.

• (2) (2,4,c),c ≥ 4.

• (3) (2,5,c),5 ≤ c ≤ 9.

• (4) (3,3,c),c ≥ 3.

• (5) (3,4,c),4 ≤ c ≤ 5.

Proof

If A is elliptic, then by Theorem 4.1 and Theorem 3.1, we have $\lfloor \frac {(a-1)b}{a} \rfloor \le 2$. Thus possible pairs (a,b) are as follows:

$$(2,2), (2,3), (2,4), (2,5), (3,3), (3,4). $$

We know that A is rational if (a,b,c) = (2,2,c) with 2 ≤ c or (2,3,c) with 3 ≤ c ≤ 5. We obtain the assertion by Lemma 4.3, for example, $p_{f}(A)= 3-\left \lceil 10/c\right \rceil$ for (a,b,c) = (2,5,c), and $p_{f}(A)= 4-\left \lceil 12/c\right \rceil$ for (a,b,c) = (3,4,c). □

We can classify Brieskorn hypersurface singularities with $\bar {\text {r}}(A)= 2$ except (a,b,c) = (3,4,6) or (3,4,7).

Proposition 4.5

$\bar {\mathrm {r}}(A)= 2$ifand only ifp_f(A) = 1,except (a,b,c) = (3,4,6),or (3,4,7).

Proof

It suffices to check whether nr(A) ≥ 3 for singularities with $\bar {\text {r}}(\mathfrak m)= 2$ and p_f(A) ≥ 2.

Suppose (a,b,c) = (2,5,c), c ≥ 10. Let Q = (y,z²) and $J=\overline {Q}$. Then, xz∉Q and (xz)² = (y⁵ + z^c)z² ∈ Q⁶ = (Q³)². Hence, nr(J) ≥ 3.

Next suppose that (a,b,c) = (3,4,c), c ≥ 8. Let Q = (y,z²) and $J=\overline {Q}$, again. Then, x²z∉Q and (x²z)³ = (y⁴ + z^c)²z³ ∈ Q⁹ = (Q³)³. Hence, nr(J) ≥ 3. □

Applying the result of [11], we can show that the formula for $\bar {\text {r}}(\mathfrak m)$ for Brieskorn complete intersection singularities. Thus, the statement above can be extended to those singularities.

Remark 4.6

Suppose that p_g(A) = 3. It follows from Theorem 2.9 and its proof that nr(I) = 3 if and only if q(I) = 1 and q(nI) = 0 for n ≥ 2. In particular, q(nI) = q(I) for n ≥ 2 if q(I) ≥ 2.

Remark 4.7

If (a,b,c) = (3,4,6) or (3,4,7), we have the following:

(1) p_g(A) = 3, p_f(A) = 2, $h^{1}(\mathcal O_{X}(-Z_{E}))= 1$.

(2) There exists a point p ∈ E such that $\mathfrak m\mathcal O_{X}=\mathcal I_{p}\mathcal O_{X}(-Z_{E})$, where $\mathcal I_{p}\subset \mathcal O_{X}$ is the ideal sheaf of the point p; so $\mathfrak m=H^{0}(\mathcal O_{X}(-Z_{E}))$, but $\mathfrak m$ is not represented by Z_E. Note that $H^{0}(\mathcal O_{X}(-nZ_{E}))\ne \overline {\mathfrak m^{n}}$ for n ≥ 2. On the other hand, $\mathcal O_{X}(-2Z_{E})=\mathcal O_{X}(K_{X})$ is generated by global sections. By the vanishing theorem, $h^{1}(\mathcal O_{X}(-nZ_{E}))= 0$ for n ≥ 2.