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 pg(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.
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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 + c1zn− 1 + ⋯ + cn = 0 for some n ≥ 1 and ci ∈ 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:
These are analogues of the reduction number rQ(I) of an ideal I ⊂ A. But in general, rQ(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:
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 (pg(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 pg(A)). In [13], we have shown that nr(A) ≤ pg(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
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
Moreover, if we put \(n_{k}=\lfloor \frac {kb}{a}\rfloor \) for each k ≥ 0, then
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]]/(xa + yb + zc), 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:
which is independent of the choice of resolution of singularities. When pg(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 = IZ. Then, \(I_{nZ}=\overline {I^{n}}\) for every integer n ≥ 1.
In what follows, let A, X, I = IZ be as above.
The authors have studied pg-ideals in [13,14,15]. So, we first recall the notion of pg-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) ≤ pg(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 pg-ideal if q(I) = pg(A).
Example 2.4
Any two-dimensional excellent normal local domain over an algebraically closed field admits a pg-ideal. Moreover, if A is a rational singularity, then every \(\mathfrak m\)-primary integrally closed ideal is a pg-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
We call them the normal reduction numbers of I. We also define
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 pg-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]]/(xa + yb + zc). 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
Proof
Assume Q = (a,b) and consider the exact sequence as follows:
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:
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):
In particular,
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) ≤ pg(A) + 1 (see [14, Lemma 3.1]).
Theorem 2.9
For any\(\mathfrak m\)-primaryintegrally closed idealI ⊂ A,we have
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
Thus, if we put ak = q((r − k)I) for k = 0,1,…,r, then we get
Hence,
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
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
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]]/(xa + yb + zc). Namely, we prove the following theorem.
Theorem 3.1
LetA = K[[x,y,z]]/(xa + yb + zc) 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 < ⋯ < na− 1. In particular, nk ≥ 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 ≤ nk.
Proof
Suppose n ≤ nk. Then,
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(xk)ℓ ∈ Qnℓ 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 ≥ nk + 1. Since \(\frac {kb}{a}+\frac {1}{a} \le n_{k}+ 1 \le n\), we get
for sufficiently large ℓ. This implies that ybkℓ ∈ (ybkℓ+ 1,z) and this is a contradiction because y, z forms a regular sequence. Therefore, n ≤ nk, as required. □
Corollary 3.4
For an integern ≥ 1, if weput
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 − nk. In fact, since Q = (y,z) is a parameter ideal in A, [6, Corollary 6.8.13] and Lemma 3.3 imply
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
for every n ≥ 1, then \(\{I_{n}\}_{n = 1,2,\dots }\) is a filtration of A.
Lemma 3.5
G({In}) isalways reduced. In particular,\(\mathcal {R}^{\prime }(\{I_{n}\})\)isa Gorenstein normal domain.
Proof
One can easily see
By assumption, K[X,Y,Z]/(Xa + Yb + Zc) is a normal domain. If charK = 0, then K[X,Y,Z]/(Xa + Yb) 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 ]/(Xa + Yb) is reduced.
As A is normal, \(R=\mathcal {R}^{\prime }(\{I_{n}\})\) is a Gorenstein normal domain because G({In})≅R/t− 1R. □
Lemma 3.6
\(L_{n}=I_{na^{\prime }}\)foreveryn ≥ 1.
Proof
Since Ln and \(I_{na^{\prime }}\) are monomial ideals, it suffices to show that xkyizj ∈ Ln 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 In 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 Ln+ 1 = QLn if and only if n ≥ na− 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)\(\mathfrak m\)isapg-idealif and only ifa = 2 and\(\bar {\mathrm {r}}(\mathfrak m)= 1\).
Proof
(1) Suppose \(\bar {\text {r}}(\mathfrak m)=a-1\). Then, na− 1 = a − 1 by (1) and this implies that nk = 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, na− 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 pg-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]]/(xa + yb + zc) 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 nk ≤ n < nk+ 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 xk+ 1, xk+ 2,…, xa− 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,
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 {an} is an increasing sequence and an+ 1 = an for sufficiently large n. By Lemma 2.7, we have
for sufficiently large n ≥ 1. Hence, (1) yields
as required. □
When a = 2, one can obtain the following.
Example 3.9
Let A = K[[x,y,z]]/(x2 + yb + zc) 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:
where
for sufficiently large n.
3.3 Geometric Genus
In this subsection, let us consider a graded ring
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 pg(A) using this formula.
Lemma 3.10
Under the above notation, we have
We can find many examples of Brieskorn hypersurfaces with pg(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 pg(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 pg(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
(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
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,\(\text {nr}(A)=\text {nr}(\mathfrak m)= 1\)andpg(A) = 0.∙ (a,b,c) = (2,3,3),(2,3,4),(2,3,5).In this case,\(\text {nr}(A)=\text {nr}(\mathfrak m)= 1\)andpg(A) = 0.∙ (a,b,c) = (2,4,4),(2,4,5),(2,4,6),(2,4,7).In this case,\(\text {nr}(A)=\text {nr}(\mathfrak m)= 2\)andpg(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,\(\text {nr}(A)=\text {nr}(\mathfrak m)= 2\)andpg(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
Then,
Suppose λ0 = 0. Then,
and thus \(\lambda _{1} < r-1+\frac {\varepsilon }{~a~}\). By Lemma 3.10 and assumption, we have
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:
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]]/(xa + yb + zc). 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 ai,ℓi,αi, λi, \(\widehat {g_{i}}\) (i = 1,2,3), \(\widehat {g}\), and ℓ as follows:
We put \(\widehat {g}=\frac {abc}{\text {lcm}(a,b,c)}\) and ℓ = lcm(a,b,c), and define integers βi by the following condition:
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:
where \(E_{w,j}^{2}=-c_{w,j}\) and
is a Hirzebruch-Jung continued fraction if αw ≥ 2, we regard Bw empty if αw = 1. Moreover, we have
For instance, if (a,b,c) = (3,4,7), then we have the following:
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.
See [11, 4.4] for more details.
4 Brieskorn Hypersurfaces with Elliptic Singularities
We use the notation of Section 3.4. Let ZE denote the fundamental cycle.
We call pf(A) := pa(ZE) the fundamental genus of A. The singularity A is said to be elliptic if pf(A) = 1. We have the following.
Theorem 4.1
[12] Ifpf(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 pf(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 pf(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
We are now ready to state our result in the case of pf(A) = 1.
Theorem 4.4
\((A,\mathfrak m)\)is elliptic(i.e.,pf(A) = 1) if andonly if (a,b,c) isone of the following.
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(1) (2,3,c),c ≥ 6.
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(2) (2,4,c),c ≥ 4.
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(3) (2,5,c),5 ≤ c ≤ 9.
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(4) (3,3,c),c ≥ 3.
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(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:
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 ifpf(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 pf(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 + zc)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 + zc)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 pg(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) pg(A) = 3, pf(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 ZE. 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.
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This work was partially supported by JSPS Grant-in-Aid for Scientific Research (C) Grant Nos., 25400050, 26400053, 17K05216.
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Okuma, T., Watanabe, Ki. & Yoshida, Ki. Normal Reduction Numbers for Normal Surface Singularities with Application to Elliptic Singularities of Brieskorn Type. Acta Math Vietnam 44, 87–100 (2019). https://doi.org/10.1007/s40306-018-00311-4
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DOI: https://doi.org/10.1007/s40306-018-00311-4