Infinitely generated symbolic Rees rings of space monomial curves having negative curves

TL;DR

This paper investigates the finite generation of symbolic Rees rings of space monomial curves with negative curves, providing criteria, examples, and computational methods to determine when these rings are Noetherian.

Contribution

It offers new conditions and criteria for the finite generation of symbolic Rees rings of space monomial curves with negative curves, including examples of infinite generation.

Findings

Determined minimal degree elements satisfying Huneke's criterion for Noetherian rings.

Provided necessary and sufficient conditions for finite generation of symbolic Rees rings.

Constructed an example of an infinitely generated symbolic Rees ring with a negative curve.

Abstract

In this paper, we shall study finite generation of symbolic Rees rings of the defining ideal ${\frak p}$ of the space monomial curve $(t^a, t^b, t^c)$ for pairwise coprime integers $a$, $b$, $c$. Suppose that the base field is of characteristic $0$ and the above ideal ${\frak p}$ is minimally generated by three polynomials. Under the assumption that the homogeneous element $\xi$ of the minimal degree in ${\frak p}$ is the negative curve, we determine the minimal degree of an element $\eta$ such that the pair $\{ \xi, \eta \}$ satisfies Huneke's criterion in the case where the symbolic Rees ring is Noetherian. By this result, we can decide whether the symbolic Rees ring ${\cal R}_s({\frak p})$ is Notherian using computers. We give a necessary and sufficient conditions for finite generation of the symbolic Rees ring of ${\frak p}$ under some assumptions. We give an example of an infinitely generated symbolic Rees ring of ${\frak p}$ in which the homogeneous element of the minimal degree in ${\frak p}^{(2)}$ is the negative curve. We give a simple proof to (generalized) Huneke's criterion.

Equations

$${\varphi }_K:S⟶K[t]$$

$$\begin{array}[]{lll}s=\min\{{\mathbb{N}}a\cap({\mathbb{N}}_{0}b+{\mathbb{N}}_{0}c)\},\\ t=\min\{{\mathbb{N}}b\cap({\mathbb{N}}_{0}a+{\mathbb{N}}_{0}c)\},\\ u=\min\{{\mathbb{N}}c\cap({\mathbb{N}}_{0}a+{\mathbb{N}}_{0}b)\},\end{array}$$

$$x^s-y^{t_1}{z}^{u_1},y^t-x^{s_2}{z}^{u_2},z^u-x^{s_3}{y}^{t_3},$$

$${\mathcal{R}}_s(P)={\oplus }_{n\ge 0}{P}^{(n)}{T}^n\subset S[T],$$

$${\ell }_S(S/(x,z^u-x^{s_3}{y}^{t_3},\eta ))=u\cdot {\ell }_S(S/(x)+\mathfrak{p})$$

$$I^{(n)}=\bigcap _{\mathfrak{p}\in {\mathrm{Assh}}_{S}S/I}(I_{\mathfrak{p}}^n\cap S),$$

$${\mathcal{R}}_s(I)$$

$${\mathcal{R}}_s^{\prime }(I)$$

$${\mathcal{G}}_s(I)$$

$${\mathfrak{a}}^{(n)}=\bigcap _{P\in {\mathrm{Assh}}_{R}R/\mathfrak{a}}({\mathfrak{a}}_P^n\cap R).$$

$${\ell }_R(R/(x,\xi ,\eta )R)\ge kl\cdot {\ell }_S(S/(x)+I),$$

$${\ell }_{S_{\mathfrak{p}}}(S_{\mathfrak{p}}/(\xi ,\eta )S_{\mathfrak{p}})=k\ell \cdot {\ell }_{S_{\mathfrak{p}}}(S_{\mathfrak{p}}/I_{\mathfrak{p}})$$

$${\ell }_A(A/(\xi ,\eta ))\ge k\ell \cdot {\ell }_A(A/Q),$$

$$k\ell \cdot {\ell }_A(A/(\xi ,\eta ))={\ell }_A(A/J)={\mathrm{e}}_J(A),$$

$${\mathrm{e}}_J(A)\ge {\mathrm{e}}_{Q^{k\ell }}(A)=(k\ell {)}^2\cdot {\mathrm{e}}_Q(A)=(k\ell {)}^2\cdot {\ell }_A(A/Q).$$

$$(QT{)}^{k+\ell -1}\subseteq \xi T^k\cdot Q^{\ell -1}{T}^{\ell -1}+\eta T^{\ell }\cdot Q^{k-1}{T}^{k-1}\subseteq (\xi T^k,\eta T^{\ell })\mathcal{R}(Q),$$

$${\ell }_R(R/(x,\xi ,\eta )R)={\mathrm{e}}_{xR}(R/(\xi ,\eta )R).$$

$${\mathrm{e}}_{xR}(R/(\xi ,\eta )R))$$

$${\ell }_R(R/(x,\xi ,\eta )R)=k\ell \cdot {\ell }_S(S/(x)+I).$$

$${\ell }_S(S/(x,{\xi }^{\prime },{\eta }^{\prime }))={\ell }_R(R/(x,\xi ,\eta )R).$$

$${\ell }_R(R/(x,\xi ,\eta )R)={\ell }_R(R/(x,y^i,\eta )R)=i\cdot {\ell }_R(R/(x,y,\eta )R).$$

$$\eta \equiv {\alpha }_j{z}^j+{\alpha }_{j+1}{z}^{j+1}+\cdots \text{mbox}mod(x,y)S,$$

$${\ell }_R(R/(x,y,\eta )R)={\ell }_R(R/(x,y,z^j)R)=j.$$

$${\ell }_R(R/(x,\xi ,\eta )R)=ij.$$

$${\ell }_S(S/(x,{\xi }^{\prime },{\eta }^{\prime }))={\ell }_S(S/(x,y^i,{\eta }^{\prime }))=i\cdot {\ell }_S(S/(x,y,z^j))=ij.$$

$$2=0p{t}_R\mathfrak{b}\le \lambda (\mathfrak{b})\le 3-\underset{n>0}{\inf }\{0ptR/{\mathfrak{b}}^n\}=2,$$

$$I_{\mathfrak{p}}^n={\mathfrak{b}}_{\mathfrak{p}R}^r{I}_{\mathfrak{p}}^{n-mr}=(\xi {\mathfrak{b}}_{\mathfrak{p}R}^{r-i}+\eta {\mathfrak{b}}_{\mathfrak{p}R}^{r-j})I_{\mathfrak{p}}^{n-rm}=\xi I_{\mathfrak{p}}^{n-mi}+\eta I_{\mathfrak{p}}^{n-mj},$$

$$0⟶{\mathfrak{c}}^r/\mathfrak{b}{\mathfrak{c}}^r⟶R/\mathfrak{b}{\mathfrak{c}}^r⟶R/{\mathfrak{c}}^r⟶0$$

$${\mathrm{Ass}}_R{\mathfrak{c}}^r/\mathfrak{b}{\mathfrak{c}}^r={\mathrm{Ass}}_{R}R/\mathfrak{b}={\mathrm{Assh}}_{R}R/\mathfrak{a}.$$

$${\mathrm{Ass}}_{R}R/{\mathfrak{c}}^r={\mathrm{Assh}}_{R}R/\mathfrak{c}={\mathrm{Assh}}_{R}R/\mathfrak{a}$$

$${\mathrm{Ass}}_{R}R/\mathfrak{b}{\mathfrak{c}}^r={\mathrm{Assh}}_{R}R/\mathfrak{a}.$$

Full text

1. Introduction

Let ${\mathfrak{p}}_{K}(a,b,c)$ be the defining ideal of the space monomial curve $(t^{a},t^{b},t^{c})$ in $K^{3}$, where $K$ is a field. The ideal ${\mathfrak{p}}_{K}(a,b,c)$ is generated by at most three binomials in $K[x,y,z]$ (Herzog [11]). The symbolic Rees rings of space monomial primes are deeply studied by many authors. Huneke [12] and Cutkosky [2] developed criterions for finite generation of such rings. In 1994, Goto-Nishida-Watanabe [7] first found examples of infinitely generated symbolic Rees rings of space monomial primes. Recently, using toric geometry, Gonzáles-Karu [5] found some sufficient conditions for the symbolic Rees rings of space monomial primes to be infinitely generated.

Cutkosky [2] found the geometric meaning of the symbolic Rees rings of space monomial primes. Let ${\mathbb{P}}_{K}(a,b,c)$ be the weighted projective surface with degree $a$, $b$, $c$. Let $X_{K}(a,b,c)$ be the blow-up at a point in the open orbit of the toric variety ${\mathbb{P}}_{K}(a,b,c)$. Then, the Cox ring of $X_{K}(a,b,c)$ is isomorphic to the extended symbolic Rees ring of the space monomial prime ${\mathfrak{p}}_{K}(a,b,c)$. Therefore, the symbolic Rees ring of the space monomial prime ${\mathfrak{p}}_{K}(a,b,c)$ is finitely generated if and only if the Cox ring of $X_{K}(a,b,c)$ is finitely generated, that is, $X_{K}(a,b,c)$ is a Mori dream space. A curve $C$ on $X_{K}(a,b,c)$ is called the negative curve if $C^{2}<0$ and $C$ is different from the exceptional curve $E$. Here suppose $\sqrt{abc}\not\in{\mathbb{Q}}$. Cutkosky [2] proved that the symbolic Rees ring of the space monomial prime ${\mathfrak{p}}_{K}(a,b,c)$ is finitely generated if and only if the following two conditions are satisfied:

• (1)

There exists a negative curve $C$.

• (2)

There exists a curve $D$ on $X_{K}(a,b,c)$ such that $C\cap D=\emptyset$.

All known examples ([8], [5]) of the infinitely generated symbolic Rees rings of ${\mathfrak{p}}_{K}(a,b,c)$ satisfy the following conditions:

• (I)

there exists a negative curve $C$ such that $C.E=1$.

• (II)

the characteristic of $K$ is [math]

In this paper, we give an example of an infinitely generated symbolic Rees ring such that there exists the negative curve $C$ with $C.E=2$. Furthermore, in the case where both (I) and (II) as above are satisfied, we determine the minimal value of the degree of the curve $D$ which satisfies the condition (2) as above.

The existence of negative curves is a very difficult problem, that is deeply related to Nagata conjecture (Proposition 5.2 in Cutkosky-Kurano [3]).

In the rest of this section, we state the results of this paper precisely.

Let $a$, $b$, $c$ be pairwise coprime integers. We regard the polynomial ring $S=K[x,y,z]$ as a ${\mathbb{Z}}$-graded ring by $\deg(x)=a$, $\deg(y)=b$ and $\deg(z)=c$. Let ${\mathfrak{p}}_{K}(a,b,c)$ be the kernel of the $K$-algebra homomorphism

[TABLE]

given by $\phi_{K}(x)=t^{a}$, $\phi_{K}(y)=t^{b}$, $\phi_{K}(z)=t^{c}$. If no confusion is possible, we simply denote ${\mathfrak{p}}_{K}(a,b,c)$ by ${\mathfrak{p}}$.

By a result of Herzog [11], we know that ${\mathfrak{p}}_{K}(a,b,c)$ is generated by at most three binomials. We define $s$, $t$, $u$ to be

[TABLE]

where ${\mathbb{N}}$ (resp. ${\mathbb{N}}_{0}$) denotes the set of positive integers (resp. non-negative integers). Let $t_{1}$, $u_{1}$, $s_{2}$, $u_{2}$, $s_{3}$, $t_{3}$ be non-negative integers such that $sa=t_{1}b+u_{1}c$, $tb=s_{2}a+u_{2}c$, $uc=s_{3}a+t_{3}b$. Then, ${\mathfrak{p}}_{K}(a,b,c)$ is minimally generated by three elements if and only if $s$, $t$, $u\geq 2$. When this is the case, ${\mathfrak{p}}_{K}(a,b,c)$ is minimally generated by three elements

[TABLE]

and $t_{1}$, $u_{1}$, $s_{2}$, $u_{2}$, $s_{3}$, $t_{3}$ must be positive integers satisfying $s=s_{2}+s_{3}$, $t=t_{1}+t_{3}$, $u=u_{1}+u_{2}$.

For a prime ideal $P$ of $S$, we define the symbolic Rees ring of $P$ to be

[TABLE]

where $P^{(n)}=P^{n}S_{P}\cap S$ is the $n$th symbolic power of $P$ and $T$ is an indeterminate. Here, $\mathcal{R}_{s}(P)$ is a Noetherian ring if and only if $\mathcal{R}_{s}(P)$ is finitely generated over $S$ as a ring.

In Section 2, we give a simple proof to Huneke’s criterion [12]. We slightly generalize Huneke’s criterion here. Furthermore, we develop the method of mod $p$ reduction introduced in Goto-Nishida-Watanabe [7].

In Section 3, we give a proof to the following theorem:

Theorem 1.1**.**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the following three conditions:

• (i)

$K$* is a field of characteristic [math],*

• (ii)

${\mathfrak{p}}_{K}(a,b,c)$* is minimally generated by the three elements as in (2),*

• (iii)

$uc<\sqrt{abc}$.

Then, $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is a Noetherian ring if and only if there exists $\eta$ in $[{\mathfrak{p}}^{(u)}]_{ab}$ such that $z^{u}-x^{s_{3}}y^{t_{3}}$ and $\eta$ satisfy Huneke’s condition [12] (see Theorem 2.5), that is,

[TABLE]

holds.

The condition (iii) as above implies that $z^{u}-x^{s_{3}}y^{u_{3}}$ is the negative curve, that is, there exists the negative curve $C$ such that $C.E=1$. Theorem 1.1 says that there exists a curve $D$ such that $D\cap C=\emptyset$ and $D\sim abA-uE$ if and only if $\mathcal{R}_{s}({\mathfrak{p}}_{\mathbb{Q}}(a,b,c))$ is a Noetherian ring, where $A$ is an Weil divisor on $X$ satisfying ${\mathcal{O}}_{X}(A)=\pi^{*}{\mathcal{O}}_{\mathbb{P}}(1)$.

We emphasis that it is possible to verify weather there exists $\eta$ in $[{\mathfrak{p}}^{(u)}]_{ab}$ satisfying (3) as above using computers. We shall prove this theorem using the mod $p$ reduction as in Goto-Nishida-Watanabe [8], Fujita’s vanishing theorem (Theorem 1.4.35 in [13]) and Cutkosky’s method [2] in characteristic $p>0$. The most important point is that the negative curve is isomorphic to ${\mathbb{P}}^{1}_{K}$ in this case.

In Section 4, we introduce the condition EU. In Ebina [4] and Uchisawa [15], the condition EU was defined and they proved that the condition EU is a sufficient condition for finite generation under the assumptions (i), (ii), (iii) in Theorem 1.1. For the convenience of the reader, we shall give a proof of it in this paper. Furthermore, in the case where $u\leq 6$, we show that the condition EU is a necessary and sufficient condition for the finite generation of the symbolic Rees ring of ${\mathfrak{p}}$ in Propositin 4.10.

In Section 5 we give an example of infinitely generated symbolic Rees ring of ${\mathfrak{p}}$ where the homogeneous element of the minimal degree in ${\mathfrak{p}}^{(2)}$ is the negative curve in Example 5.6. We emphasis that one of the minimal generators of ${\mathfrak{p}}$ is the negative curve in all known examples of infinitely generated $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$, except for this example.

2. Huneke’s condition and mod $p$ reduction

Let $S=K[x,y,z]$, where $K$ is a field and $x,y,z$ are indeterminates. We regard $S$ as a $\mathbb{Z}$-graded ring putting suitable weights on $x,y$ and $z$. We set $\mathfrak{m}=(x,y,z)S$ and $R=S_{\mathfrak{m}}$. Let $I$ be a homogeneous proper ideal of $S$ satisfying the following conditions;

• •

$(x)+I$ is $\mathfrak{m}$-primary,

• •

$\mathop{\rm Ass}\nolimits_{S}S/I=\mathop{\rm Assh}\nolimits_{S}S/I:=\{\mathfrak{p}\in\mathop{\rm Ass}\nolimits_{S}S/I\mid\dim S/\mathfrak{p}=\dim S/I\}$, and

• •

$I_{\mathfrak{p}}$ is generated by $2$ elements for any $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$.

Then, $S/I$ is a $\mathbb{Z}$-graded Cohen-Macaulay ring of $\dim S/I=1$. If we replace $x$ in the first assumption stated above with $y$ or $z$, it can play the same role as $x$ in the arguments of this section. So, homogeneous prime ideals of height $2$ are typical examples of $I$. For any $n\in\mathbb{Z}$, we set

[TABLE]

where $I_{\mathfrak{p}}^{n}$ denotes the ideal $(I^{n})_{\mathfrak{p}}=(I_{\mathfrak{p}})^{n}$ of $S_{\mathfrak{p}}$. Then, we have $\mathop{\rm Ass}\nolimits_{S}S/I^{(n)}=\mathop{\rm Assh}\nolimits_{S}S/I$ if $n>0$, and the equality $I^{(n)}=I^{n}:_{S}x^{i}$ holds for $i\gg 0$, which means that $I^{(n)}$ is a homogeneous ideal of $S$ and $(I^{(n)})_{x}=I_{x}^{n}$, where $I_{x}^{n}$ denotes the ideal $(I^{n})_{x}=(I_{x})^{n}$ of $S_{x}$. Moreover, we set

[TABLE]

where $T$ is an indeterminate and $I^{(n)}=S$ for $n\leq 0$. Let us call $\mathcal{R}_{s}(I)$ the symbolic Rees ring of $I$.

We set $\mathfrak{a}=I_{\mathfrak{m}}=IR$. It is easy to see that $R/\mathfrak{a}$ is a Cohen-Macaulay local ring of $\dim R/\mathfrak{a}=1$ and $\mathop{\rm Ass}\nolimits_{R}R/\mathfrak{a}=\mathop{\rm Assh}\nolimits_{R}R/\mathfrak{a}=\{\mathfrak{p}R\mid\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I\}$. Moreover, for $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$, we have $\mathfrak{a}_{\mathfrak{p}R}=I_{\mathfrak{p}}$, which becomes a parameter ideal of $R_{\mathfrak{p}R}=S_{\mathfrak{p}}$. For any $n\in\mathbb{Z}$, we set

[TABLE]

Then, we have $\mathfrak{a}^{(n)}=I^{(n)}R$ and $\mathop{\rm Ass}\nolimits_{R}R/\mathfrak{a}^{(n)}=\mathop{\rm Assh}\nolimits_{R}R/\mathfrak{a}$ if $n>0$. As $\mathfrak{a}^{(n)}=\mathfrak{a}^{n}:_{R}x^{i}$ holds for $i\gg 0$, we have $(\mathfrak{a}^{(n)})_{x}=\mathfrak{a}_{x}^{n}$. The $R$-algebras $\mathcal{R}_{s}(\mathfrak{a})$ and $\mathcal{G}_{s}(\mathfrak{a})$ are derived from $\mathcal{R}_{s}(I)$ and $\mathcal{G}_{s}(I)$ respectively applying $R\otimes_{S}\ast$ . If $\mathcal{R}_{s}(\mathfrak{a})$ is finitely generated, then there exists $0<m\in\mathbb{Z}$ such that $\mathfrak{a}^{(mn)}=(\mathfrak{a}^{(m)})^{n}$ for any $n\in\mathbb{Z}$. This equality implies $I^{(mn)}=(I^{(m)})^{n}$ since $I^{(mn)}\supseteq(I^{(m)})^{n}$ and $(I^{(m)})^{n}$ is a homogeneous ideal. Thus we see that $\mathcal{R}_{s}(I)$ is finitely generated if so is $\mathcal{R}_{s}(\mathfrak{a})$. The converse of this assertion holds obviously.

For a proper ideal $J$ of $S$ such that $S/J$ is Artinian, we have $\ell_{S}(\,S/J\,)\geq\ell_{R}(\,R/JR\,)$, and the equality holds if and only if $J$ is $\mathfrak{m}$-primary, which holds if $J$ is homogeneous.

The purpose of this section is to review the condition on $I$ for its symbolic Rees ring to be finitely generated, which was originally given by Huneke [12] in the case where $I$ is a prime ideal of a $3$-dimensional regular local ring. Furthermore, using mod $p$ reduction technique for prime numbers $p\gg 0$, we give a condition on $I$ for $\mathcal{R}_{s}(I)$ to be infinitely generated, which is a modification to the method introduced in [8].

Let us begin with the following

Proposition 2.1**.**

Let $0<k,\ell\in\mathbb{Z}$, $\xi\in I^{(k)}$ and $\eta\in I^{(\ell)}$. Then we have

[TABLE]

where the equality holds if and only if $\mathfrak{a}\subseteq\sqrt{(\xi,\eta)R}$ and

[TABLE]

for all $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$.

In order to prove Proposition 2.1, let us recall the following fact.

Lemma 2.2**.**

Let $A$ be a $2$-dimensional Cohen-Macaulay local ring and $Q$ a parameter ideal of $A$. Let $0<k,\ell\in\mathbb{Z}$, $\xi\in Q^{k}$ and $\eta\in Q^{\ell}$. We assume that $\xi,\eta$ is an sop for $A$. Then, we have

[TABLE]

and the equality holds if and only if one of the following conditions, which are equivalent to each other, is satisfied;

• (1)

$QT\subseteq\sqrt{(\xi T^{k},\eta T^{\ell})\mathcal{R}(Q)}$, where $\mathcal{R}(Q)=\sum_{n\geq 0}Q^{n}T^{n}\subset A[T]$,

• (2)

$\xi T^{k},\eta T^{\ell}$* is an sop for $\mathcal{G}(Q)=\mathcal{R}(Q)/Q\mathcal{R}(Q)$,*

• (3)

$Q^{k+\ell-1}=\xi Q^{\ell-1}+\eta Q^{k-1}$,

• (4)

$Q^{n}\cap(\xi,\eta)A=\xi Q^{n-k}+\eta Q^{n-\ell}$* for any $n\in\mathbb{Z}$.*

Proof. We set $J=(\xi^{\ell},\eta^{k})A$. Then we have

[TABLE]

where ${\mathrm{e}}_{J}(A)$ denotes the multiplicity of $A$ with respect to $J$. Because $J\subseteq Q^{k\ell}$, it follows that

[TABLE]

Hence we get the required inequality. Moreover, we see that the equality holds if and only if $J$ is a reduction of $Q^{k\ell}$, which is a condition equivalent to (1). The equivalence of the conditions (1) and (2) is obvious. Let us notice that $\mathcal{G}(Q)$ is isomorphic to a polynomial ring with $2$ variables over $A/Q$, so its homogeneous sop is always a regular sequence, which implies the equivalence of the conditions (2) and (4). Moreover, if the condition (2) is satisfied, it follows that $\mathcal{G}(Q)/(\xi T^{k},\eta T^{\ell})\mathcal{G}(Q)$ is an Artinian $\mathbb{Z}$-graded ring whose a-invariant is $k+\ell-2$ (cf. [10]), so the equality of the condition (3) holds. Finally, if the condition (3) is satisfied, we have

[TABLE]

and hence the condition (1) is satisfied. ∎

Proof of Proposition 2.1. We may assume that $(x,\xi,\eta)R$ is $\mathfrak{m}R$-primary. Then, as $R/(\xi,\eta)R$ is a Cohen-Macaulay $R$-module, for which $x$ is an sop, we have

[TABLE]

Here we notice that $\mathfrak{p}R\in\mathop{\rm Assh}\nolimits_{R}R/(\xi,\eta)R$ for any $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$. Hence, using additive formula of multiplicity and Lemma 2.2, we get

[TABLE]

Thus we get the required inequality. Moreover, we see that the equality holds if and only if $\mathop{\rm Assh}\nolimits_{R}R/(\xi,\eta)R=\mathop{\rm Assh}\nolimits_{R}R/\mathfrak{a}$ and $\ell_{S_{\mathfrak{p}}}(\,S_{\mathfrak{p}}/(\xi,\eta)S_{\mathfrak{p}}\,)=k\ell\cdot\ell_{S_{\mathfrak{p}}}(\,S_{\mathfrak{p}}/I_{\mathfrak{p}}\,)$ for all $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$. Since $\mathfrak{a}\subseteq\sqrt{(\xi,\eta)R}$ holds if and only if $\mathop{\rm Assh}\nolimits_{R}R/(\xi,\eta)R=\mathop{\rm Assh}\nolimits_{R}R/\mathfrak{a}$, the proof is complete. ∎

Definition 2.3**.**

Let $0<k,\ell\in\mathbb{Z},\xi\in I^{(k)}$ and $\eta\in I^{(\ell)}$. We say that $\xi$ and $\eta$ satisfy Huneke’s condition on $I$ (with respect to $x$) if

[TABLE]

When this is the case, for any $0<i,j\in\mathbb{Z}$, $\xi^{i}\in I^{(ki)}$ and $\eta^{j}\in I^{(\ell j)}$ also satisfy Huneke’s condition on $I$.

Even if there exist elements satisfying Huneke’s condition, those elements may not be homogeneous. Although the existance of homogeneous elements satisfying Huneke’s condition is not clear, but it can be verified elementary in special cases. For example, the following remark implies that if $\xi$ and $\eta$ satisfy Huneke’s condition and $\xi\equiv y^{i}$ mod $xS$ for some $0<i\in\mathbb{Z}$, then we can choose homogeneous parts of $\xi$ and $\eta$ so that they also satisfy Huneke’s condition.

Lemma 2.4**.**

Suppose $\xi\in S$ and $\xi\equiv y^{i}$ mod $xS$, where $0<i\in\mathbb{Z}$. Let $\xi^{\prime}$ be the homogeneous part of $\xi$ containing $y^{i}$ as a term. Then, the following assertions hold.

• (1)

$(x,\xi)S=(x,\xi^{\prime})S=(x,y^{i})S$.

• (2)

For any $\eta\in S$, we can choose its homogeneous part $\eta^{\prime}$ so that

[TABLE]

Proof. The assertion (1) holds obviously. Let us verify the assertion (2). We may assume $\eta\not\in(x,y)S$. Then, as $x,y,\eta$ is an $R$-regular sequence, we have

[TABLE]

We write

[TABLE]

where $0<j\in\mathbb{Z}$ and $\alpha_{j},\alpha_{j+1},\dots$ are elements of $K$ with $\alpha_{j}\neq 0$. Since $\alpha_{j}+\alpha_{j+1}z+\cdots$ is a unit of $R$, we have

[TABLE]

Thus we get

[TABLE]

Let $\eta^{\prime}$ be the homogeneous part of $\eta$ containing $\alpha_{j}z^{j}$ as a term. Then, as $\eta^{\prime}\equiv\alpha_{j}z^{j}$ mod $(x,y)S$, it follows that

[TABLE]

Thus we get the required equality. ∎

Theorem 2.5**.**

The symbolic Rees algebra $\mathcal{R}_{s}(I)$ is finitely generated over $R$ if and only if there exist elements in $I^{(k)}$ and $I^{(\ell)}$ satisfying Huneke’s condition on $I$ for some $0<k,\ell\in\mathbb{Z}$.

Proof. First, let us assume that $\mathcal{R}_{s}(I)$ is finitely generated. Then, there exists a positive integer $m$ such that $I^{(mn)}=(I^{(m)})^{n}$ for any $n\in\mathbb{Z}$. We set $\mathfrak{b}=\mathfrak{a}^{(m)}$. Then, for any $0<n\in\mathbb{Z}$, we have $\mathfrak{a}^{(mn)}=\mathfrak{b}^{n}$, which means $0pt_{R}R/\mathfrak{b}^{n}=1$. Hence, by Burch’s theorem (cf. [1]), we see

[TABLE]

where $\lambda(\mathfrak{b})$ denotes the Krull dimension of $R/\mathfrak{m}\otimes\mathcal{G}(\mathfrak{b})$, which is called the analytic spread of $\mathfrak{b}$. Thus we get $\lambda(\mathfrak{b})=2$. Hence, we can choose $0<i,j\in\mathbb{Z}$, $\xi\in I^{(mi)}$ and $\eta\in I^{(mj)}$ such that $\xi T^{i},\eta T^{j}$ is an sop for $R/\mathfrak{m}\otimes\mathcal{G}(\mathfrak{b})$. (Here, we notice that we don’t have to assume that $K$ is infinite since we don’t require $i=j=1$.) Let us take $r\gg 0$. Then, we have $\mathfrak{b}^{r}=\xi\mathfrak{b}^{r-i}+\eta\mathfrak{b}^{r-j}$, which means $\mathfrak{a}^{mr}\subseteq(\xi,\eta)R$, and so $\mathfrak{a}\subseteq\sqrt{(\xi,\eta)R}$. Moreover, if $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$ and $mr<n$, we have

[TABLE]

which means that $\xi T^{mi},\eta T^{mj}$ is an sop for $\mathcal{G}(I_{\mathfrak{p}})$. Therefore, by Proposition 2.1 and Lemma 2.2, it follows that $\xi$ and $\eta$ satisfy Huneke’s condition on $I$.

Next, we assume that there exist $0<k,\ell\in\mathbb{Z}$, $\xi\in I^{(k)}$ and $\eta\in I^{(\ell)}$ such that $\xi$ and $\eta$ satisfy Huneke’s condition on $I$. We set $m=k\ell$, $\mathfrak{b}=\mathfrak{a}^{(m)}$ and $\mathfrak{c}=(\xi^{\ell},\eta^{k})R\subseteq\mathfrak{b}$. Let us look at the exact sequence

[TABLE]

of $R$-modules, where $r$ is any non-negative integer. Since $\xi^{\ell},\eta^{k}$ is an $R$-regular sequence, $\mathfrak{c}^{r}/\mathfrak{c}^{r+1}$ is $R/\mathfrak{c}$-free, so $\mathfrak{c}^{r}/\mathfrak{b}\mathfrak{c}^{r}\cong R/\mathfrak{b}\otimes_{R}\mathfrak{c}^{r}/\mathfrak{c}^{r+1}$ is $R/\mathfrak{b}$-free, which means

[TABLE]

On the other hand, by Proposition 2.1 we have

[TABLE]

since $\xi^{\ell}$ and $\eta^{k}$ also satisfy Huneke’s condition on $I$. Thus we see

[TABLE]

Now we take any $P\in\mathop{\rm Assh}\nolimits_{R}R/\mathfrak{a}$, and write $P=\mathfrak{p}R$, where $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$. Then, by Proposition 2.1 and Lemma 2.2 we have $I_{\mathfrak{p}}^{2m-1}=\xi^{\ell}I_{\mathfrak{p}}^{m-1}+\eta^{k}I_{\mathfrak{p}}^{m-1}$, which means $\mathfrak{b}_{P}^{2}=(\mathfrak{b}\mathfrak{c})_{P}$, and so $\mathfrak{b}_{P}^{r+1}=(\mathfrak{b}\mathfrak{c}^{r})_{P}$. Hence, we get

[TABLE]

and hence $\mathfrak{a}^{(mr+m)}=\mathfrak{b}\mathfrak{c}^{r}=\mathfrak{b}^{r+1}$. Thus we see that the $m$-th veronese subring of $\mathcal{R}_{s}(\mathfrak{a})$ is generated in degree one. Therefore $\mathcal{R}_{s}(\mathfrak{a})$ is Noetherian by [6, Lemma (2.4)]. Then $\mathcal{R}_{s}(I)$ itself must be Noetherian. ∎

Lemma 2.6**.**

Let $0<k,\ell\in\mathbb{Z}$, $\xi\in I^{(k)}$ and $\eta\in I^{(\ell)}$. Suppose that $\xi$ and $\eta$ satisfy Huneke’s condition on $I$. Then the following assertions hold.

• (1)

$\mathcal{R}_{s}(\mathfrak{a})_{+}=\sqrt{(\xi T^{k},\eta T^{\ell})\mathcal{R}_{s}(\mathfrak{a})}$, and hence $\mathcal{G}_{s}(\mathfrak{a})_{+}\subseteq\sqrt{(\xi T^{k},\eta T^{\ell})\mathcal{G}_{s}(\mathfrak{a})}$.

• (2)

$\mathfrak{a}^{(k+\ell-1)}\subseteq(\xi,\eta)R$.

• (3)

$\mathfrak{a}_{x}^{n}\cap(\xi,\eta)R_{x}=\xi\mathfrak{a}_{x}^{n-k}+\eta\mathfrak{a}_{x}^{n-\ell}$* for any $n\in\mathbb{Z}$.*

• (4)

$\mathfrak{a}^{(n)}\cap(\xi,\eta)R=\xi\mathfrak{a}^{(n-k)}+\eta\mathfrak{a}^{(n-\ell)}$* if $n\leq k+\ell$.*

• (5)

If $k=1$ or $2$, then we have

[TABLE]

for any $n\in\mathbb{Z}$, which means that $\xi T^{k},\eta T^{\ell}$ is a regular sequence on $\mathcal{G}_{s}(\mathfrak{a})$, and hence $\mathop{\rm grade}\nolimits\mathcal{G}_{s}(\mathfrak{a})_{+}=2$.

Proof. (1) We set $m=k\ell,\mathfrak{b}=\mathfrak{a}^{(m)}$ and $\mathfrak{c}=(\xi^{\ell},\eta^{k})R$. Then, as is stated in the proof of Theorem 2.5, we have $\mathfrak{a}^{(mr+m)}=\mathfrak{b}^{r+1}=\mathfrak{b}\mathfrak{c}^{r}$ for any $0\leq r\in\mathbb{Z}$. Let us take any $0<n\in\mathbb{Z}$ and $\rho\in\mathfrak{a}^{(n)}$. Then we have $\rho^{2m}\in\mathfrak{a}^{(m(2n-1)+m)}=\mathfrak{b}^{2n}=\mathfrak{b}\mathfrak{c}^{2n-1}\subseteq\mathfrak{c}\mathfrak{b}^{2n-1}=\xi^{\ell}\mathfrak{b}^{2n-1}+\eta^{k}\mathfrak{b}^{2n-1}\subseteq\xi\mathfrak{a}^{(2mn-k)}+\eta\mathfrak{a}^{(2mn-\ell)}$, so

[TABLE]

Hence we get the assertion (1)

(2) Let us take any $P\in\mathop{\rm Assh}\nolimits_{R}R/\mathfrak{a}$ and write $P=\mathfrak{p}R$, where $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$. Then, as $R_{P}=S_{\mathfrak{p}}$ and $\mathfrak{a}_{P}=I_{\mathfrak{p}}$, by Proposition 2.1 and Lemma 2.2, we have $\mathfrak{a}_{P}^{k+\ell-1}=\xi\mathfrak{a}_{P}^{\ell-1}+\eta\mathfrak{a}_{P}^{k-1}\subseteq(\xi,\eta)R_{P}$. Therefore we get

[TABLE]

(3) Since $\xi\in I_{x}^{k}$ and $\eta\in I_{x}^{\ell}$, the inclusion $\mathfrak{a}_{x}^{n}\supseteq\xi\mathfrak{a}_{x}^{n-k}+\eta\mathfrak{a}_{x}^{n-\ell}$ holds obviously. So, it is enough to show

[TABLE]

for any $P\in\mathop{\rm Spec}\nolimits R$ satisfying $\xi\mathfrak{a}^{n-k}+\eta\mathfrak{a}^{n-\ell}\subseteq P$ and $x\not\in P$. Such $P$ must contains $\mathfrak{a}$ since $\mathfrak{a}\subseteq\sqrt{(\xi,\eta)R}$, so there exists $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$ such that $P=\mathfrak{p}R$. Then, by Proposition 2.1 and Lemma 2.2, we get the required equality as $R_{P}=S_{\mathfrak{p}}$ and $\mathfrak{a}_{P}=I_{\mathfrak{p}}$.

(4) Let $n\leq k+\ell$ and $\varphi\in\mathfrak{a}^{(n)}\cap(\xi,\eta)R$. We write $\varphi=\xi u+\eta v$, where $u,v\in R$. Since $\varphi\in\mathfrak{a}_{x}^{n}\cap(\xi,\eta)R_{x}=\xi\mathfrak{a}_{x}^{n-k}+\eta\mathfrak{a}_{x}^{n-\ell}$ by (3), there exist $\alpha\in\mathfrak{a}_{x}^{n-k}$ and $\beta\in\mathfrak{a}_{x}^{n-\ell}$ such that $\varphi=\xi\alpha+\eta\beta$. Here, we take $i\gg 0$ so that $x^{i}\alpha\in\mathfrak{a}^{n-k}$ and $x^{i}\beta\in\mathfrak{a}^{n-\ell}$. Then we have $x^{i}(\xi u+\eta v)=x^{i}\varphi=x^{i}(\xi\alpha+\eta\beta)$, so $\xi(x^{i}u-x^{i}\alpha)=\eta(x^{i}\beta-x^{i}v)$. Since $\xi,\eta$ is an $R$-regular sequence, it follows that $x^{i}u-x^{i}\alpha\in\eta R\subseteq\mathfrak{a}^{(\ell)}\subseteq\mathfrak{a}^{(n-k)}$ and $x^{i}\beta-x^{i}v\in\xi R\subseteq\mathfrak{a}^{(k)}\subseteq\mathfrak{a}^{(n-\ell)}$. Hence $x^{i}u\in\mathfrak{a}^{(n-k)}$ and $x^{i}v\in\mathfrak{a}^{(n-\ell)}$, which means $u\in\mathfrak{a}^{(n-k)}$ and $v\in\mathfrak{a}^{(n-\ell)}$. Thus we get $\varphi\in\xi\mathfrak{a}^{(n-k)}+\eta\mathfrak{a}^{(n-\ell)}$.

(5) Let $k=1$ or $2$. By (2) and (4), it is enough to show

[TABLE]

assuming $n>k+\ell$. We take positive integers $m$ and $r$ such that $n-\ell=km-r$ and $0\leq r<k$. Then, $m\geq 2$ and $r$ is [math] or $1$. Since $\xi^{m}\in I^{(km)}$ and $\eta\in I^{(\ell)}$ also satisfy Huneke’s condition on $I$ and $km+\ell-1\leq km+\ell-r=n\leq km+\ell$, we have $\mathfrak{a}^{(n)}\subseteq(\xi^{m},\eta)R$ by (2) and $\mathfrak{a}^{(n)}\cap(\xi^{m},\eta)R=\xi^{m}\mathfrak{a}^{(\ell-r)}+\eta\mathfrak{a}^{(n-\ell)}$ by (4). Let us notice that $\xi^{m-1}\mathfrak{a}^{(\ell-r)}\subseteq\mathfrak{a}^{(n-k)}$ as $k(m-1)+(\ell-r)=n-k$. Thus we see $\mathfrak{a}^{(n)}\subseteq\xi\mathfrak{a}^{(n-k)}+\eta\mathfrak{a}^{(n-\ell)}$. Since the converse inclusion is obvious, we get the required equality. ∎

Definition 2.7**.**

Let $0<k\in\mathbb{Z}$ and $\xi\in I^{(k)}$. We denote by $\mathrm{HC}(I;k,\xi)$ the set of positive integers $\ell$ for which there exists $\eta\in I^{(\ell)}$ such that $\xi$ and $\eta$ satisfy Huneke’s condition on $I$.

Remark 2.8**.**

Let $k$ and $\xi$ be as in Definition 2.7. If $\xi\equiv y^{i}$ mod $xS$, where $0<i\in\mathbb{Z}$, and $\xi^{\prime}$ is the homogeneous part of $\xi$ containing $y^{i}$ as a term, we have $\mathrm{HC}(I;k,\xi)=\mathrm{HC}(I;k,\xi^{\prime})$ by Lemma 2.4 (1).

Proposition 2.9**.**

Let $k=1$ or $2$, and let $\xi\in I^{(k)}$. Suppose that $\xi\equiv y^{i}\,\,{\rm mod}\,\,xS$ for some $0<i\in\mathbb{Z}$ and $\mathrm{HC}(I;k,\xi)\neq\phi$. We set $m=\min\mathrm{HC}(I;k,\xi)$. Then the following assertions hold.

• (1)

$\mathrm{HC}(I;k,\xi)=\{m,2m,3m,\cdots\}$.

• (2)

$S[\{I^{(n)}T^{n}\mid 1\leq n\leq m-1\}]\subsetneq\mathcal{R}_{s}(I)$.

• (3)

If there exist elements in $I^{(k^{\prime})}$ and $I^{(\ell^{\prime})}$ satisfying Huneke’s condition on $I$ for $0<k^{\prime},\ell^{\prime}\in\mathbb{Z}$, we have

[TABLE]

In particular,

[TABLE]

Proof. By Remark 2.8, we may assume that $\xi$ is homogeneous. Then, by Lemma 2.4 (2), we can choose a homogeneous element $\eta\in I^{(m)}$ such that $\xi$ and $\eta$ satisfy Huneke’s condition on $I$.

(1) We obviously have $\mathrm{HC}(I;k,\xi)\supseteq\{m,2m,3m,\cdots\}$. In order to show the converse inclusion, We suppose that there exists $\ell\in\mathrm{HC}(I;k,\xi)$ which is not a multiple of $m$. Let us choose such $\ell$ as small as possible. Then, there exists a homogeneous element $\rho\in I^{(\ell)}$ such that $\xi$ and $\rho$ satisfy Huneke’s condition on $I$. Since $m<\ell$, by Lemma 2.6 (2) and (5), we have $\mathfrak{a}^{(\ell)}=\xi\mathfrak{a}^{(\ell-k)}+\eta\mathfrak{a}^{(\ell-m)}$, which implies

[TABLE]

as $\xi$ and $\eta$ are homogeneous. Hence, there exists a homogeneous element $\rho^{\prime}\in I^{(\ell-m)}$ such that

[TABLE]

Then $\rho\in(\xi,\rho^{\prime})S$, and hence we get

[TABLE]

as $\mathfrak{a}\subseteq\sqrt{(\xi,\rho)R}$ by Proposition 2.1. Now we take any $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$ and $n\gg 0$. Then, by Proposition 2.1 and Lemma 2.2, we have

[TABLE]

so we get

[TABLE]

Therefore, $\xi$ and $\rho^{\prime}$ satisfy Huneke’s condition on $I$, so $\ell-m\in\mathrm{HC}(I;k,\xi)$, which contradicts to the minimality of $\ell$ as $\ell-m$ is not a multiple of $m$. Consequently, we see that any $\ell\in\mathrm{HC}(I;k,\xi)$ is a multiple of $m$.

(2) The assertion holds obviously if $m=1$, so let us consider the case where $m\geq 2$. Suppose

[TABLE]

Then we have

[TABLE]

We set $\overline{S}=S/(x,y)\cong K[z]$. Since any homogeneous ideal of $\overline{S}$ is a power of $z\overline{S}$,

[TABLE]

holds for some $\beta=1,2,\dots,m-1$. Moreover, we can choose homogeneous elements $\rho\in I^{(\beta)}$ and $\rho^{\prime}\in I^{(m-\beta)}$ such that $\eta$ and $\rho\rho^{\prime}$ have the same class in $\overline{S}$, which is equivalent to

[TABLE]

Then, by Proposition 2.1, we have

[TABLE]

Since $(x,y,\eta)$, $(x,y,\rho)$ and $(x,y,\rho^{\prime})$ are all homogeneous $\mathfrak{m}$-primary ideals, we have

[TABLE]

Consequently, it follows that

[TABLE]

Hence we get $\beta,m-\beta\in\mathrm{HC}(I;k,\xi)$, which contradicts to the minimality of $m$. Thus we see

[TABLE]

(3) Let $0<k^{\prime},\ell^{\prime}\in\mathbb{Z}$, $\xi^{\prime}\in I^{(k^{\prime})}$ and $\eta^{\prime}\in I^{(\ell^{\prime})}$. Suppose that $\xi^{\prime}$ and $\eta^{\prime}$ satisfy Huneke’s condition on $I$. Then, by Lemma 2.6 (1), we have

[TABLE]

On the other hand, from the existance of $\xi$ and $\eta$, we see $\mathop{\rm grade}\nolimits\mathcal{G}_{s}(\mathfrak{a})_{+}=2$ by Lemma 2.6 (5). Hence, it follows that $\xi^{\prime}T^{k^{\prime}},\eta^{\prime}T^{\ell^{\prime}}$ is a regular sequence on $\mathcal{G}_{s}(\mathfrak{a})$. If $k^{\prime}+\ell^{\prime}-1\leq n$, we have $\mathfrak{a}^{(n)}\subseteq(\xi^{\prime},\eta^{\prime})R$ by Lemma 2.6 (2), so

[TABLE]

Thus we see

[TABLE]

which means that the first assertion of (3) holds. We get the last assertion taking $k$ and $m$ as $k^{\prime}$ and $\ell^{\prime}$, respectively. ∎

In the rest of this section, let $S_{\mathbb{Z}}=\mathbb{Z}[x,y,z]$. Moreover, for a field $K$, we denote $K[x,y,z]$ by $S_{K}$ instead of $S$ in order to emphasize that the coefficient field is $K$. Putting suitable weights on $x,y$ and $z$, we regard $S_{\mathbb{Z}}$ and $S_{K}$ as $\mathbb{Z}$-graded rings. We set $\mathfrak{m}_{\mathbb{Z}}=(x,y,z)S_{\mathbb{Z}}$, $\mathfrak{m}_{K}=(x,y,z)S_{K}$ and $R_{K}=(S_{K})_{m_{K}}$. When we denote an ideal of $S_{\mathbb{Z}}$ by $J_{\mathbb{Z}}$, the ideal $J_{\mathbb{Z}}S_{K}$ is denoted by $J_{K}$. Similarly, when we denote an element of $S_{\mathbb{Z}}$ by $\xi_{\mathbb{Z}}$, its image in $S_{K}$ is denoted by $\xi_{K}$. For a prime number $p$, we set ${\mathbb{F}_{p}}=\mathbb{Z}/p\mathbb{Z}$. Of course, $S_{\mathbb{Q}}=(\mathbb{Z}\setminus\{0\})^{-1}S_{\mathbb{Z}}$ and $S_{\mathbb{F}_{p}}=S_{\mathbb{Z}}/pS_{\mathbb{Z}}$.

Lemma 2.10**.**

Let $J_{\mathbb{Z}}$ be an ideal of $S_{\mathbb{Z}}$. Then, we have

[TABLE]

for any prime number $p\gg 0$. If $J_{\mathbb{Z}}$ is homogeneous, we may replace $R_{\mathbb{Q}},(J_{\mathbb{Q}})_{\mathfrak{m}_{\mathbb{Q}}},R_{\mathbb{F}_{p}}$ and $(J_{\mathbb{F}_{p}})_{\mathfrak{m}_{\mathbb{F}_{p}}}$ with $S_{\mathbb{Q}},J_{\mathbb{Q}},S_{\mathbb{F}_{p}}$ and $J_{\mathbb{F}_{p}}$, respectively.

Proof. First, let us consider the case where $R_{\mathbb{Q}}/(J_{\mathbb{Q}})_{\mathfrak{m}_{\mathbb{Q}}}$ is Artinian. We prove the required equality by induction on $\ell_{R_{\mathbb{Q}}}(\,R_{\mathbb{Q}}/(J_{\mathbb{Q}})_{\mathfrak{m}_{\mathbb{Q}}}\,)$.

If $\ell_{R_{\mathbb{Q}}}(\,R_{\mathbb{Q}}/(J_{\mathbb{Q}})_{\mathfrak{m}_{\mathbb{Q}}}\,)=0$, then $J_{\mathbb{Q}}$ contains an element which does not belong to $\mathfrak{m}_{\mathbb{Q}}$, so there exists $\xi_{\mathbb{Z}}\in J_{\mathbb{Z}}\setminus\mathfrak{m}_{\mathbb{Z}}$. Let us take a prime number $p\gg 0$ so that the constant term of $\xi_{\mathbb{Z}}$, which is non-zero, is not a multiple of $p$. Then, $\xi_{{\mathbb{F}_{p}}}\in J_{\mathbb{F}_{p}}\setminus\mathfrak{m}_{{\mathbb{F}_{p}}}$. Hence $(J_{\mathbb{F}_{p}})_{\mathfrak{m}_{\mathbb{F}_{p}}}=R_{\mathbb{F}_{p}}$, so $\ell_{R_{\mathbb{F}_{p}}}(\,R_{\mathbb{F}_{p}}/(J_{\mathbb{F}_{p}})_{\mathfrak{m}_{\mathbb{F}_{p}}}\,)=0$.

Now we suppose $\ell_{R_{\mathbb{Q}}}(\,R_{\mathbb{Q}}/(J_{\mathbb{Q}})_{\mathfrak{m}_{\mathbb{Q}}}\,)>0$. Then, as $\mathfrak{m}_{\mathbb{Z}}\in\mathop{\rm Ass}\nolimits_{S_{\mathbb{Z}}}S_{\mathbb{Z}}/J_{\mathbb{Z}}$, there exists $\eta_{\mathbb{Z}}\in S_{\mathbb{Z}}$ such that $J_{\mathbb{Z}}:\eta_{\mathbb{Z}}=\mathfrak{m}_{\mathbb{Z}}$. We set $L_{\mathbb{Z}}=J_{\mathbb{Z}}+(\eta_{\mathbb{Z}})$. Let us notice $L_{\mathbb{Q}}/J_{\mathbb{Q}}\cong S_{\mathbb{Q}}/\mathfrak{m}_{\mathbb{Q}}$. Hence we have $\ell_{R_{\mathbb{Q}}}(\,(L_{\mathbb{Q}}/J_{\mathbb{Q}})_{\mathfrak{m}_{\mathbb{Q}}}\,)=1$, so

[TABLE]

Here, we take a prime number $p\gg 0$. Then the hypothesis of induction implies

[TABLE]

Moreover, by taking larger $p$ if necessary, we may assume that $p$ is regular on $S_{\mathbb{Z}}/L_{\mathbb{Z}}$. If $\eta_{\mathbb{F}_{p}}\in J_{\mathbb{F}_{p}}$, we have $\eta_{\mathbb{Z}}\in J_{\mathbb{Z}}+pS_{\mathbb{Z}}$, so there exists $\rho_{\mathbb{Z}}\in S_{\mathbb{Z}}$ such that $\eta_{\mathbb{Z}}\equiv p\cdot\rho_{\mathbb{Z}}$ mod $J_{\mathbb{Z}}$. Since $p$ is regular on $S_{\mathbb{Z}}/L_{\mathbb{Z}}$, we have $\rho_{\mathbb{Z}}\in L_{\mathbb{Z}}$, so there exists $\sigma_{\mathbb{Z}}\in S_{\mathbb{Z}}$ such that $\rho_{\mathbb{Z}}\equiv\eta_{\mathbb{Z}}\sigma_{\mathbb{Z}}$ mod $J_{\mathbb{Z}}$. Then, we have $\eta_{\mathbb{Z}}\equiv p\cdot\eta_{\mathbb{Z}}\sigma_{\mathbb{Z}}$ mod $J_{\mathbb{Z}}$, and hence $1-p\cdot\sigma_{\mathbb{Z}}\in J_{\mathbb{Z}}:\eta_{\mathbb{Z}}=\mathfrak{m}_{\mathbb{Z}}$, which is impossible. Thus we see $\eta_{\mathbb{F}_{p}}\not\in J_{\mathbb{F}_{p}}$. Hence we have $\mathfrak{m}_{\mathbb{F}_{p}}=J_{\mathbb{F}_{p}}:\eta_{\mathbb{F}_{p}}$ since $\eta_{\mathbb{F}_{p}}\cdot\mathfrak{m}_{\mathbb{F}_{p}}\subseteq J_{\mathbb{F}_{p}}$ holds obviously. Then, we get $L_{\mathbb{F}_{p}}/J_{\mathbb{F}_{p}}\cong R_{\mathbb{F}_{p}}/\mathfrak{m}_{\mathbb{F}_{p}}$, so $\ell_{R_{\mathbb{F}_{p}}}(\,L_{\mathbb{F}_{p}}/J_{\mathbb{F}_{p}}\,)=1$. Consequently,

[TABLE]

Therefore, the required equality follows.

Next, we assume $\dim R_{\mathbb{Q}}/(J_{\mathbb{Q}})_{\mathfrak{m}_{\mathbb{Q}}}>0$, and aim to prove $\dim R_{\mathbb{F}_{p}}/(J_{\mathbb{F}_{p}})_{\mathfrak{m}_{\mathbb{F}_{p}}}>0$ for $p\gg 0$. In this case, there exists $P_{\mathbb{Z}}\in\mathop{\rm Spec}\nolimits S_{\mathbb{Z}}$ such that $J_{\mathbb{Z}}\subseteq P_{\mathbb{Z}}\subsetneq\mathfrak{m}_{\mathbb{Z}}$. Let us take any $\tau_{\mathbb{Z}}\in\mathfrak{m}_{\mathbb{Z}}\setminus P_{\mathbb{Z}}$ and choose a prime number $p\gg 0$ so that $p$ is regular on $S_{\mathbb{Z}}/(\tau_{\mathbb{Z}})+P_{\mathbb{Z}}$. Then, as $p,\tau_{\mathbb{Z}}$ is a regular sequence on $(S_{\mathbb{Z}}/P_{\mathbb{Z}})_{(pS_{\mathbb{Z}}+\mathfrak{m}_{\mathbb{Z}})}$, it follows that $\tau_{\mathbb{Z}}$ is regular on $(S_{\mathbb{Z}}/pS_{\mathbb{Z}}+P_{\mathbb{Z}})_{(pS_{\mathbb{Z}}+\mathfrak{m}_{\mathbb{Z}})}\cong R_{\mathbb{F}_{p}}/(P_{\mathbb{F}_{p}})_{\mathfrak{m}_{\mathbb{F}_{p}}}$. Hence, we have $\dim R_{\mathbb{F}_{p}}/(P_{\mathbb{F}_{p}})_{\mathfrak{m}_{\mathbb{F}_{p}}}>0$, and so $\dim R_{\mathbb{F}_{p}}/(J_{\mathbb{F}_{p}})_{\mathfrak{m}_{\mathbb{F}_{p}}}>0$ as $J_{\mathbb{F}_{p}}\subseteq P_{\mathbb{F}_{p}}$. ∎

In the rest of this section, let $I_{\mathbb{Z}}$ be a homogeneous ideal of $S_{\mathbb{Z}}$ contained in $\mathfrak{m}_{\mathbb{Z}}$. We assume that the following conditions are satisfied for any field $K$;

• •

$(x)+I_{K}$ is $\mathfrak{m}_{K}$-primary,

• •

$\mathop{\rm Ass}\nolimits_{S_{K}}S_{K}/I_{K}=\mathop{\rm Assh}\nolimits_{S_{K}}S_{K}/I_{K}$, and

• •

$(I_{K})_{\mathfrak{p}}$ is generated by $2$ elements for any $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S_{K}}S_{K}/I_{K}$.

Furthermore, for any $n\in\mathbb{Z}$, we set $(I^{(n)})_{\mathbb{Z}}=\bigcup_{i>0}((I_{\mathbb{Z}})^{n}:_{S_{\mathbb{Z}}}x^{i})$, which is a homogeneous ideal of $S_{\mathbb{Z}}$. Let us denote $(I^{(n)})_{\mathbb{Z}}S_{K}$ by $(I^{(n)})_{K}$ for any field $K$.

Lemma 2.11**.**

The following assertions hold for any $n\in\mathbb{Z}$.

• (1)

$(I_{\mathbb{Q}})^{(n)}=(I^{(n)})_{\mathbb{Q}}$.

• (2)

$(I_{\mathbb{F}_{p}})^{(n)}=(I^{(n)})_{\mathbb{F}_{p}}$* for any prime number $p\gg 0$.*

Proof. First, let us notice that, for any field $K$, we have $(I_{K})^{(n)}=\bigcup_{i>0}{((I_{K})^{n}:_{S_{K}}x^{i})}$, and hence $(I_{K})^{(n)}\supseteq(I^{(n)})_{K}$ holds. The converse inclusion holds obviously if $K=\mathbb{Q}$. So, we have to prove $(I_{\mathbb{F}_{p}})^{(n)}\subseteq(I^{(n)})_{\mathbb{F}_{p}}$ for $p\gg 0$.

Let us take a prime number $p\gg 0$ so that $p$ is regular on $S_{\mathbb{Z}}/(x)+(I^{(n)})_{\mathbb{Z}}$. Moreover, we take any $\xi_{\mathbb{Z}}\in S_{\mathbb{Z}}$ satisfying $\xi_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(n)}$. Then, there exists $0<i\in\mathbb{Z}$ such that $x^{i}\xi_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{n}$, which means $x^{i}\xi_{\mathbb{Z}}\in pS_{\mathbb{Z}}+(I_{\mathbb{Z}})^{n}\subseteq pS_{\mathbb{Z}}+(I^{(n)})_{\mathbb{Z}}$. Hence, there exists $\eta_{\mathbb{Z}}\in S_{\mathbb{Z}}$ such that $x^{i}\xi_{\mathbb{Z}}\equiv p\eta_{\mathbb{Z}}$ mod $(I^{(n)})_{\mathbb{Z}}$. Since $x,p$ is a regular sequence on $S_{\mathbb{Z}}/(I^{(n)})_{\mathbb{Z}}$, so is $x^{i},p$. Hence $\eta_{\mathbb{Z}}\in(x^{i})+(I^{(n)})_{\mathbb{Z}}$, so there exists $\rho_{\mathbb{Z}}\in S_{\mathbb{Z}}$ such that $\eta_{\mathbb{Z}}\equiv x^{i}\rho_{\mathbb{Z}}$ mod $(I^{(n)})_{\mathbb{Z}}$. Then we have $x^{i}\xi_{\mathbb{Z}}\equiv px^{i}\rho_{\mathbb{Z}}$ mod $(I^{(n)})_{\mathbb{Z}}$, which means $\xi_{\mathbb{Z}}-p\rho_{\mathbb{Z}}\in(I^{(n)})_{\mathbb{Z}}$. Thus we get $\xi_{\mathbb{F}_{p}}\in(I^{(n)})_{\mathbb{F}_{p}}$. ∎

Proposition 2.12**.**

Let $0<k,\ell\in\mathbb{Z},\xi_{\mathbb{Z}}\in(I^{(k)})_{\mathbb{Z}}$ and $\eta_{\mathbb{Z}}\in(I^{(\ell)})_{\mathbb{Z}}$. Suppose that $\xi_{\mathbb{Q}}\in(I_{\mathbb{Q}})^{(k)}$ and $\eta_{\mathbb{Q}}\in(I_{\mathbb{Q}})^{(\ell)}$ satisfy Huneke’s condition on $I_{\mathbb{Q}}$. Then, for any prime number $p\gg 0$, $\xi_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(k)}$, $\eta_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(\ell)}$, and these elements satisfy Huneke’s condition on $I_{\mathbb{F}_{p}}$.

Proof. Let $p\gg 0$. Then, by Lemma 2.11, we have $\xi_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(k)}$ and $\eta_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(\ell)}$. Moreover, by Lemma 2.10, we have

[TABLE]

Thus we get the required assertion. ∎

Theorem 2.13**.**

Let $k=1$ or $2$. Let $\xi_{\mathbb{Z}}\in(I^{(k)})_{\mathbb{Z}}$ and $\xi_{\mathbb{Z}}\equiv y^{i}\,\,{\rm mod}\,\,xS_{\mathbb{Z}}$ for some $0<i\in\mathbb{Z}$. Suppose that there exists a positive integer $r$ such that, for any prime number $p\gg 0$, we have $rp^{e_{p}}\in\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})$ for some $0<e_{p}\in\mathbb{Z}$. Then the following conditions are equivalent:

• (1)

$\mathcal{R}_{s}(I_{\mathbb{Q}})$* is finitely generated.*

• (2)

$\mathrm{HC}(I_{\mathbb{Q}};k,\xi_{\mathbb{Q}})\neq\emptyset$.

• (3)

$r\in\mathrm{HC}(I_{\mathbb{Q}};k,\xi_{\mathbb{Q}})$.

• (4)

$r\in\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})$* for any prime number $p\gg 0$.*

Proof.  Let $\xi^{\prime}_{\mathbb{Z}}$ be the homogeneous part of $\xi_{\mathbb{Z}}$ containing $y^{i}$ as a term. Then, as $\xi^{\prime}_{\mathbb{Z}}\in(I^{(k)})_{\mathbb{Z}}$, we have $\xi^{\prime}_{\mathbb{Q}}\in(I_{\mathbb{Q}})^{(k)}$ and $\xi^{\prime}_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(k)}$ for $p\gg 0$ by Lemma 2.11. Moreover, by Remark 2.8, we have $\mathrm{HC}(I_{\mathbb{Q}};k,\xi_{\mathbb{Q}})=\mathrm{HC}(I_{\mathbb{Q}};k,\xi^{\prime}_{\mathbb{Q}})$ and $\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})=\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi^{\prime}_{\mathbb{F}_{p}})$ for $p\gg 0$. Hence, by replacing $\xi_{\mathbb{Z}}$ with $\xi^{\prime}_{\mathbb{Z}}$, we may assume that $\xi_{\mathbb{Z}}$ is homogeneous from the beginning. It is easy to see $(3)\Rightarrow(2)\Rightarrow(1)$.

Now, we start to prove $(1)\Rightarrow(4)$. By Theorem 2.5 and Lemma 2.11 (1), there exist $0<k^{\prime},\ell^{\prime}\in\mathbb{Z},\zeta_{\mathbb{Z}}\in(I^{(k^{\prime})})_{\mathbb{Z}}$ and $\rho_{\mathbb{Z}}\in(I^{(\ell^{\prime})})_{\mathbb{Z}}$ such that $\zeta_{\mathbb{Q}}\in(I_{\mathbb{Q}})^{(k^{\prime})}$ and $\rho_{\mathbb{Q}}\in(I_{\mathbb{Q}})^{(\ell^{\prime})}$ satisfy Huneke’s condition on $I_{\mathbb{Q}}$. Here, we take a prime number $p\gg 0$ such that $\zeta_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(k^{\prime})}$, $\rho_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(\ell^{\prime})}$, and these elements satisfy Huneke’s condition on $I_{\mathbb{F}_{p}}$, which is possible by Proposition 2.12. By taking larger $p$ if necessary, we may assume $p>\max\{k^{\prime},\ell^{\prime},k^{\prime}+\ell^{\prime}-2\}$ and our assumption on $\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})$ is satisfied. Then, as $\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})\neq\phi$, we have

[TABLE]

by Proposition 2.9 (3). We set $m=\min\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})$ and take $0<e_{p}\in\mathbb{Z}$ such that $rp^{e_{p}}\in\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})$. Then, by Proposition 2.9 (1), there exists $m^{\prime}\in\mathbb{Z}$ such that $rp^{e_{p}}=mm^{\prime}$. Since Proposition 2.9 (2) implies $m<p$, $m$ is not a multiple of $p$, so $m^{\prime}$ is a multiple of $p^{e_{p}}$. Hence $r$ is a multiple of $m$, which means $r\in\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})$.

Next, we shall prove $(4)\Rightarrow(3)$. Let us take a prime number $p\gg 0$ such that $r\in\mathrm{HC}(I_{\mathbb{F}_{p}};k,\xi_{\mathbb{F}_{p}})$, $\ell_{S_{\mathbb{Q}}}(\,S_{\mathbb{Q}}/(x)+I_{\mathbb{Q}}\,)=\ell_{S_{\mathbb{F}_{p}}}(\,S_{\mathbb{F}_{p}}/(x)+I_{\mathbb{F}_{p}}\,)$ and $(I_{\mathbb{F}_{p}})^{(r)}=(I^{(r)})_{\mathbb{F}_{p}}$, which is possible by Lemma 2.10 and Lemma 2.11. Then, by Lemma 2.4 (2) and Lemma 2.11, there exists a homogeneous element $\eta_{\mathbb{Z}}\in(I^{(r)})_{\mathbb{Z}}$ such that $\xi_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(k)}$ and $\eta_{\mathbb{F}_{p}}\in(I_{\mathbb{F}_{p}})^{(r)}$ satisfy Huneke’s condition on $I_{\mathbb{F}_{p}}$. We write

[TABLE]

where $j$ is a positive integer and $\alpha$ is an integer which is not a multiple of $p$. Let $K=\mathbb{Q}$ or ${\mathbb{F}_{p}}$. Then, as the image of $\alpha$ in $K$ is not vanished, we have $(x,y,\eta_{K})S_{K}=(x,y,z^{j})S_{K}$. Hence we get

[TABLE]

Consequently, we have

[TABLE]

which means $r\in\mathrm{HC}(I_{\mathbb{Q}};k,\xi_{\mathbb{Q}})$. ∎

3. Proof of Theorem 1.1

We shall prove Theorem 1.1 in this section.

Let $K$ be a field and $a$, $b$, $c$ be pairwise coprime positive integers. We regard the polynomial ring $S=K[x,y,z]$ as a ${\mathbb{Z}}$-graded ring by $\deg(x)=a$, $\deg(y)=b$ and $\deg(z)=c$.

We denote by ${\mathbb{P}}_{K}(a,b,c)$ the weighted projective space $\mathop{\rm Proj}\nolimits S$. Let

[TABLE]

be the blow-up at the point corresponding to ${\mathfrak{p}}_{K}(a,b,c)$. We remark that ${\mathbb{P}}_{K}(a,b,c)$ is non-singular at this point (e.g., Lemma 9 in [2]). If no confusion is possible, we denote ${\mathfrak{p}}_{K}(a,b,c)$ (resp. $X_{K}(a,b,c)$, ${\mathbb{P}}_{K}(a,b,c)$) simply by ${\mathfrak{p}}$ (resp. $X$, ${\mathbb{P}}$). Let $E$ be the exceptional divisor of $\pi$. Let $A$ be a Weil divisor on $X$ which satisfies ${\mathcal{O}}_{X}(A)=\pi^{*}{\mathcal{O}}_{\mathbb{P}}(1)$. Since $a$, $b$, $c$ are pairwise coprime, we have ${\mathcal{O}}_{X}(nA)=\pi^{*}{\mathcal{O}}_{\mathbb{P}}(n)$ for any $n\in{\mathbb{Z}}$ (e.g., [14]). Then,

[TABLE]

with the intersection pairing

[TABLE]

Definition 3.1**.**

A curve $C$ on $X_{K}(a,b,c)$ is called a negative curve on $X_{K}(a,b,c)$ if $C^{2}<0$ and $C\neq E$.

An irreducible homogeneous polynomial $\xi$ in $[{\mathfrak{p}}_{K}(a,b,c)^{(r)}]_{d}$ is called a negative curve in ${\mathfrak{p}}_{K}(a,b,c)^{(r)}$ if $d/r<\sqrt{abc}$.

If a negative curve $C$ on $X_{K}(a,b,c)$ exists, then it is unique. If a negative curve $\xi$ in $[{\mathfrak{p}}_{K}(a,b,c)^{(r)}]_{d}$ exists, then $r$ and $d$ are uniquely determined, and $\xi$ is also unique up to multiplication by an element in $K^{\times}$. The negative curve $C$ on $X_{K}(a,b,c)$ is the proper transform of $V_{+}(\xi)$.

Lemma 3.2**.**

Let $K$ be a field and $a$, $b$, $c$ be pairwise coprime positive integers. We assume that ${\mathfrak{p}}_{K}(a,b,c)$ is minimally generated by the three elements in (2).

Then the curve $V_{+}(z^{u}-x^{s_{3}}y^{t_{3}})$ in ${\mathbb{P}}_{K}(a,b,c)$ is isomorphic to ${\mathbb{P}}^{1}_{K}$. The proper transform $C$ (in $X$) of this curve is also isomorphic to ${\mathbb{P}}^{1}_{K}$.

Proof.

First of all, we remark that $z^{u}-x^{s_{3}}y^{t_{3}}$ is an irreducible polynomial by definition of $u$ (see (1)). We put $v=x^{s_{2}}z^{u_{2}}/y^{t}$ and $w=x^{s_{3}}y^{t_{3}}/z^{u}$.

Since ${\mathfrak{p}}_{K}(a,b,c)$ is generated by the three elements as in (2), we have

[TABLE]

Then, we have

[TABLE]

Taking the degree [math] component of

[TABLE]

we obtain

[TABLE]

Let $\phi:K[v^{\pm 1},w^{\pm 1}]\rightarrow K[v^{\pm 1}]$ be the map given by $\phi(w)=1$. The kernel of the map $\phi$ is $(w-1)K[v^{\pm 1},w^{\pm 1}]$. By (4), we have $\phi(S[y^{-1}]_{0})=K[v]$. Hence, we have

[TABLE]

In the same way, we know that

[TABLE]

is also isomorphic to a polynomial ring over $K$ with one variable. Hence, the curve $V_{+}(z^{u}-x^{s_{3}}y^{t_{3}})$ in ${\mathbb{P}}_{K}(a,b,c)$ is isomorphic to ${\mathbb{P}}^{1}_{K}$.

Since the map $C\rightarrow V_{+}(z^{u}-x^{s_{3}}y^{t_{3}})$ is a finite birational map, $C$ is also isomorphic to ${\mathbb{P}}^{1}_{K}$. ∎

Lemma 3.3**.**

Let $K$ be a field of prime characteristic $p$. Let $a$, $b$, $c$ be pairwise coprime positive integers. We assume the conditions (ii) and (iii) in Theorem 1.1.

Then, there exist integers $q_{1}$ and $q_{2}$ such that

[TABLE]

for $i>0$, $n\geq q_{1}u$ and $m\geq(ab/u)n+q_{2}$.

Proof.

First of all, remember that ${\mathcal{O}}_{X}(mA-nE)$ is invertible if $abc$ divides $m$ (e.g., Lemma 1.3 in [14]). By the condition (iii), $z^{u}-x^{s_{3}}y^{t_{3}}$ is the negative curve in ${\mathfrak{p}}$. Letting $C$ be the proper transform of $V_{+}(z^{u}-x^{s_{3}}y^{t_{3}})$, $C$ is the negative curve that is linearly equivalent to $cuA-E$. Therefore, $mA-nE$ is a nef Cartier divisor if $m\geq(ab/u)n\geq 0$ and $abc\mid m$.

Let $m$ and $n$ be integers such that $m\geq(ab/u)n\geq 0$. Then, there exists a nef Cartier divisor $m_{1}A-n_{1}E$ such that $0\leq m-m_{1}<abc$ and $0\leq n-n_{1}<cu$.

We use Fujita’s vanishing theorem (Theorem 1.4.35 in [13]) for finitely many coherent shaves

[TABLE]

Then, there exists an ample Cartier divisor $q_{2}A+q_{1}(abA-uE)$ such that

[TABLE]

for $i>0$, $0\leq m_{2}<abc$, $0\leq n_{2}<cu$ and any nef Cartier divisor $m_{1}A-n_{1}E$. Then, $q_{1}$ and $q_{2}$ satisfy the requirement in Lemma 3.3. ∎

Lemma 3.4**.**

Let $K$ be a field of prime characteristic $p$. Let $a$, $b$, $c$ be pairwise coprime positive integers. We assume the conditions (ii) and (iii) in Theorem 1.1.

Then, there exist $e>0$ and $\eta\in[{\mathfrak{p}}^{(p^{e}u)}]_{p^{e}ab}$ such that $z^{u}-x^{s_{3}}y^{t_{3}}$ and $\eta$ satisfy Huneke’s condition on ${\mathfrak{p}}$, that is,

[TABLE]

holds. (The above integer $e$ depends on $a$, $b$, $c$ and $p$.)

Proof.

Let $C$ be the proper transform of $V_{+}(z^{u}-x^{s_{3}}y^{t_{3}})$. Then, $C$ is the negative curve on $X$ that is linearly equivalent to $ucA-E$.

Consider the reflexive sheaf ${\mathcal{O}}_{\mathbb{P}}(ab)$. Since $S_{ab}$ contains both $x^{b}$ and $y^{a}$, ${\mathcal{O}}_{\mathbb{P}}(ab)$ is invertible away from the point $V_{+}(x,y)$. Therefore ${\mathcal{O}}_{X}(abA)$ is invertible away from the point $\pi^{-1}(V_{+}(x,y))$. Since $C$ does not contain the point $\pi^{-1}(V_{+}(x,y))$, ${\mathcal{O}}_{X}(abA-uE)\otimes{\mathcal{O}}_{nC}$ is an invertible sheaf on $nC$ for any $n>0$.

We choose integers $q_{1}$ and $q_{2}$ that satisfy Lemma 3.3. By the condition (iii) in Theorem 1.1, we obtain

[TABLE]

Let $q$ be an integer which satisfies

[TABLE]

Consider the invertible sheaf ${\mathcal{O}}_{X}(abA-uE)\otimes{\mathcal{O}}_{C}$. Since $(abcA-cuE).C=0$, the degree of ${\mathcal{O}}_{X}(abcA-cuE)\otimes{\mathcal{O}}_{C}$ is [math]. Since $C$ is isomorphic to ${\mathbb{P}}^{1}_{K}$, ${\mathcal{O}}_{X}(abcA-cuE)\otimes{\mathcal{O}}_{C}\simeq{\mathcal{O}}_{C}$. Therefore,

[TABLE]

since ${\rm Pic}({\mathbb{P}}^{1}_{K})\simeq{\mathbb{Z}}$.

We have the exact sequences

[TABLE]

for $\ell=1,2,\ldots,qu-1$. They induce the exact sequences 111Suppose that $I$ is an ideal of a ring $A$ with $I^{2}=(0)$. Then, consider the map $I\rightarrow A^{\times}$ defined by $a\mapsto 1+a$. It induces the exact sequence $0\rightarrow I\rightarrow A^{\times}\rightarrow(A/I)^{\times}\rightarrow 1$.

[TABLE]

and

[TABLE]

for $\ell=1,2,\ldots,qu-1$. Therefore we know that the order of an element in the kernel of the map ${\rm Pic}((\ell+1)C)\rightarrow{\rm Pic}(\ell C)$ is $1$ or $p$. Hence, the order of ${\mathcal{O}}_{X}(abA-uE)\otimes{\mathcal{O}}_{quC}$ (in ${\rm Pic}(quC)$) is a power of $p$ by (6). Thus, there exists $e>0$ such that

[TABLE]

and

[TABLE]

Since

[TABLE]

is exact, we have the following exact sequence:

[TABLE]

Since $C\sim cuA-E$, we have

[TABLE]

By (7), we have $p^{e}-q>q_{1}$. By (5), we have

[TABLE]

Then, by Lemma 3.3, we have

[TABLE]

Then, we have

[TABLE]

Therefore, the natural map

[TABLE]

induces the surjection

[TABLE]

Thus, there exists an effective Weil divisor $D$ such that $D\sim p^{e}(abA-uE)$ and the support of $D$ does not intersect with $C$. Let $\eta$ be the equation of $\pi(D)$. The degree of $\eta$ is $p^{e}ab$. Since $D\cap C=\emptyset$, $V_{+}(z^{u}-x^{s_{3}}y^{t_{3}})\cap V_{+}(\eta)\subset V_{+}({\mathfrak{p}})$ as a set. Therefore, ${\mathfrak{p}}$ is the only one minimal prime ideal of $(z^{u}-x^{s_{3}}y^{t_{3}},\eta)$. Hence $x$, $z^{u}-x^{s_{3}}y^{t_{3}}$, $\eta$ form a regular sequence of $S$. We obtain

[TABLE]

which is the required equality. ∎

Proof of Theorem 1.1.  By Lemma 3.4, we know that

[TABLE]

Then, by Theorem 2.13, we know that $\mathcal{R}_{s}({\mathfrak{p}}_{\mathbb{Q}}(a,b,c))$ is Noetherian if and only if

[TABLE]

Thus, We have completed the proof of Theorem 1.1. ∎

Remark 3.5**.**

Let $a$, $b$, $c$ be pairwise coprime positive integers.

Let $\xi\in[{\mathfrak{p}}_{K}(a,b,c)^{(k)}]_{d}$ be a negative curve with $d/k<\sqrt{abc}$. Then, $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is Noetherian if and only if $\mathrm{HC}({\mathfrak{p}}_{K}(a,b,c);k,\xi)\neq\emptyset$.

Let $\xi$ be a homogeneous element in $S_{\mathbb{Z}}$. Then, $\xi_{\mathbb{Q}}\in[{\mathfrak{p}}_{\mathbb{Q}}(a,b,c)^{(k)}]_{d}$ is a negative curve with $d/k<\sqrt{abc}$ if and only if, for $p\gg 0$, $\xi_{\mathbb{F}_{p}}\in[{\mathfrak{p}}_{\mathbb{F}_{p}}(a,b,c)^{(k)}]_{d}$ is a negative curve with $d/k<\sqrt{abc}$.

Let $\xi$ be a homogeneous element in $S_{\mathbb{Z}}$. Assume that $\xi_{\mathbb{Q}}\in[{\mathfrak{p}}_{\mathbb{Q}}(a,b,c)^{(k)}]_{d}$ is a negative curve with $d/k<\sqrt{abc}$, and $\mathcal{R}_{s}({\mathfrak{p}}_{\mathbb{Q}}(a,b,c))$ is Noetherian. Then, $\mathrm{HC}({\mathfrak{p}}_{\mathbb{Q}}(a,b,c);k,\xi_{\mathbb{Q}})=\mathrm{HC}({\mathfrak{p}}_{{\mathbb{F}_{p}}}(a,b,c);k,\xi_{\mathbb{F}_{p}})$ for $p\gg 0$.

Let $\xi\in[{\mathfrak{p}}_{K}(a,b,c)^{(k)}]_{d}$ be a negative curve with $d/k<\sqrt{abc}$. Assume that $k=1$ or $2$. Suppose that

[TABLE]

for some $i$. Furthermore, assume that $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is Noetherian. Then, there exists a positive integer $m$ such that

[TABLE]

by Proposition 2.9 (1). **

4. The condition EU

In this section, we introduce a sufficient condition (which is called as “the condition EU” below) for finite generation of $\mathcal{R}_{s}({\mathfrak{p}})$ under the assumption in Theorem 1.1. The condition EU was defined in Ebina [4] and Uchisawa [15]. We shall prove that, if $u\leq 6$, the condition EU is a necessary and sufficient condition for finite generation of $\mathcal{R}_{s}({\mathfrak{p}})$ in Proposition 4.10.

Let us remember the method introduced in Gonzáles-Karu [5]. Let $a$, $b$, $c$ be pairwise coprime positive integers and $K$ be a field. Let $S=K[x,y,z]$ be a ${\mathbb{Z}}$-graded ring with $\deg(x)=a$, $\deg(y)=b$ and $\deg(z)=c$. Suppose that the prime ideal ${\mathfrak{p}}_{K}(a,b,c)$ is minimally generated by the three elements in (2).

We put

[TABLE]

Since ${\mathfrak{p}}_{K}(a,b,c)$ is generated by the three elements in (2), we have

[TABLE]

Therefore, for each non-negative integer $e$, we have

[TABLE]

Let $\Delta_{u}$ be the domain (with boundary) surrounded by the following three lines

[TABLE]

Let $(0,0)$, $(u,u_{2})$, $(\delta_{1},\delta_{2})$ be the vertices of $\Delta_{u}$. Here, $\delta_{1}$ and $\delta_{2}$ may not be integers.

[TABLE]

For a non-negative integer $e$, we put

[TABLE]

Then, it is easy to see that the euality (8) induces

[TABLE]

Since

[TABLE]

we have

[TABLE]

and

[TABLE]

for any $n>0$. Therefore,

[TABLE]

Remark 4.1**.**

Let $K$ be a field of characteristic [math]. Let $\varphi(v,w)$ be an element in $K[v^{\pm 1},w^{\pm 1}]$. Then, $\varphi(v,w)\in(v-1,w-1)^{n}K[v^{\pm 1},w^{\pm 1}]$ if and only if

[TABLE]

for $k+\ell<n$. **

Remark 4.2**.**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the conditions (ii) and (iii) in Theorem 1.1. Then, by Huneke’s condition on ${\mathfrak{p}}$, $\mathcal{R}_{s}({\mathfrak{p}})$ is Noetherian if and only if $[{\mathfrak{p}}^{(eu)}]_{eab}$ contains an element whose coefficient of $y^{ea}$ is not [math] for a sufficiently divisible $e$.

Assume the conditions (i), (ii) and (iii) in Theorem 1.1. Then, by Theorem 1.1, $\mathcal{R}_{s}({\mathfrak{p}})$ is Noetherian if and only if $[{\mathfrak{p}}^{(u)}]_{ab}$ contains an element whose coefficient of $y^{a}$ is not [math]. By (9), it is equivalent to that

[TABLE]

contains an element whose constant term is not [math]. It is not so difficult to check whether it is satisfied or not using computers. **

Now, we introduce the condition EU which is defined by Ebina [4] and Uchisawa [15].

Definition 4.3** (Ebina [4], Uchisawa [15]).**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Suppose that the prime ideal ${\mathfrak{p}}$ is minimally generated by the three elements in (2). For $i=1,2,\ldots,u$, we put

[TABLE]

We sort the sequence $\ell_{1}$, $\ell_{2}$, …, $\ell_{u}$ into assending order

[TABLE]

We say that the condition EU is satisfied for $(a,b,c)$ if

[TABLE]

for $i=1,2,\ldots,u$. **

Example 4.4**.**

• (I)

Assume $(a,b,c)=(8,19,9)$. Then,

[TABLE]

and the conditions (ii) and (iii) in Theorem 1.1 is satisfied.

[TABLE]

Then, $u=3$ and

[TABLE]

Therefore

[TABLE]

The condition EU is satisfied in this case.

• (II)

Assume $(a,b,c)=(25,29,72)$. Then,

[TABLE]

and the conditions (ii) and (iii) in Theorem 1.1 is satisfied.

[TABLE]

Then, $u=3$ and

[TABLE]

Therefore

[TABLE]

The condition EU is not satisfied in this case.

• (III)

Assume $(a,b,c)=(17,503,169)$. Then,

[TABLE]

and the conditions (ii) and (iii) in Theorem 1.1 is satisfied.

[TABLE]

Then, $u=7$ and

[TABLE]

Therefore

[TABLE]

The condition EU is not satisfied in this case.

In order to show that the condition EU is a sufficient condition for finite generation of $\mathcal{R}_{s}({\mathfrak{p}})$ under some assumptions, we need the following lemma ([4], [15]). For the convenience of the reader, we give a proof of it here.

Lemma 4.5** (Ebina [4], Uchisawa [15]).**

Let $K$ be a field of characteristic [math] and $v$, $w$ be variables.

Let $u$ be a positive integer and $\alpha_{1}$, $\alpha_{2}$, …, $\alpha_{u}$ be mutually distinct integers. For $i=1,2,\ldots,u$, consider the integers $\beta_{i1}$, $\beta_{i2}$, …, $\beta_{ii}$ satisfying

[TABLE]

Put

[TABLE]

Then, we have

[TABLE]

Proof.

We shall prove it by induction on $u$.

If $u=1$, then ${}^{\#}T=1$. It is easily verified in this case.

Assume $u\geq 2$. Take

[TABLE]

Considering $v^{-\alpha_{u}}\varphi(v,w)$, we may assume $\alpha_{u}=0$. Then, $\frac{\partial\varphi}{\partial v}$ satisfies all the assumptions with $u-1$. Here, recall $\frac{\partial\varphi}{\partial v}\in(v-1,w-1)^{u-1}K[v^{\pm 1},w^{\pm 1}]$ by Remark 4.1. By induction, we obtain $\frac{\partial\varphi}{\partial v}=0$. Therefore, we may suppose

[TABLE]

where $C_{1},\ldots,C_{u}\in K$. Since $\varphi(v,w)\in(v-1,w-1)^{u}K[v^{\pm 1},w^{\pm 1}]$, we have

[TABLE]

for $k=0,1,\ldots,u-1$. Then, we have

[TABLE]

for $k=0,1,\ldots,u-1$. It is easy to see $C_{1}=C_{2}=\cdots=C_{u}=0$. ∎

By this lemma, we can prove that the condition EU is a sufficient condition for finite generation of $\mathcal{R}_{s}({\mathfrak{p}})$ under some assumptions.

Proposition 4.6** (Ebina [4], Uchisawa [15]).**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the conditions (i), (ii), (iii) in Theorem 1.1.

If the condition EU is satisfied, then $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is Noetherian.

Proof.

By the condition EU, we can choose a set $T$ as in Lemma 4.5 which satisfies

[TABLE]

By (9), we obtain

[TABLE]

By this equality, we know that $[{\mathfrak{p}}^{(u)}]_{ab}$ is defined by $\frac{u(u+1)}{2}$ linear equations in $y^{a}\left(\oplus_{(\alpha,\beta)\in\Delta_{u}\cap{\mathbb{Z}}^{2}}Kv^{\alpha}w^{\beta}\right)$. We put $T^{\prime}=T\cup\{(0,0)\}$. Recall that $T^{\prime}\subset\Delta_{u}\cap{\mathbb{Z}}^{2}$ and ${}^{\#}T^{\prime}=\frac{u(u+1)}{2}+1$. Then, $[{\mathfrak{p}}^{(u)}]_{ab}$ contains a non-zero element in the form

[TABLE]

where $C_{(\alpha,\beta)}\in K$.

If $C_{(0,0)}=0$, we have

[TABLE]

by Lemma 4.5. It is a contradiction. Therefore, $C_{(0,0)}\neq 0$. Then,

[TABLE]

holds. Hence, $\mathcal{R}_{s}({\mathfrak{p}})$ is Noetherian by Huneke’s condition. ∎

The aim in the rest of this section is to prove the converse of Proposition 4.6 in the case $u\leq 6$.

Definition 4.7**.**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the condition (ii) in Theorem 1.1.

We define

[TABLE]

where $[\ ,\ ]$ is the closed interval.

We say that the condition GK is satisfied if one of the following two conditions is satisfied:

• (I)

${}^{\#}\left\{(n-1)[(u_{2}/u),(t/t_{3})]\cap{\mathbb{Z}}\right\}=n$* and $(u_{2}/u)n\not\in{\mathbb{Z}}$,*

• (II)

${}^{\#}\left\{(m-1)[-(s_{2}/s_{3}),(u_{2}/u)]\cap{\mathbb{Z}}\right\}=m$* and $(u_{1}/u)m\not\in{\mathbb{Z}}$.*

We remark that the above condition (I) is satisfied for $a$, $b$, $c$ if and only if the above condition (II) is satisfied for $b$, $a$, $c$.

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the conditions (i), (ii), (iii) in Theorem 1.1. If the condition GK is satisfied, then $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is not Noetherian by Theorem 1.2 in Gonzáles-Karu [5].

Proposition 4.8**.**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the conditions (i), (ii), (iii) in Theorem 1.1.

Then, the condition GK is satisfied if and only if one of the following five conditions is satisfied:

• (GK1)

$n=1$,

• (GK2)

$m=1$,

• (GK3)

$n=m=2<u$,

• (GK4)

$3\leq n<u$, $m=2$ and ${}^{\#}\left\{(n-1)[(u_{2}/u),(t/t_{3})]\cap{\mathbb{Z}}\right\}=n$,

• (GK5)

$n=2$, $3\leq m<u$ and ${}^{\#}\left\{(m-1)[-(s_{2}/s_{3}),(u_{2}/u)]\cap{\mathbb{Z}}\right\}=m$.

Proof.

Let $(\delta_{1},\delta_{2})$ be one of the vertices of $\Delta_{u}$ as in the beginning of this section. First, we remark that, if $0\leq i<i+1\leq\delta_{1}$, then $\ell_{i+1}\geq\ell_{i}+(n-1)$. In the same way, if $\delta_{1}\leq i<i+1\leq u$, then $\ell_{i}\geq\ell_{i+1}+(m-1)$. Thus, it is easy to see the following:

[TABLE]

[TABLE]

[TABLE]

Next, recall $s=s_{2}+s_{3}$, $t=t_{1}+t_{3}$ and $u=u_{1}+u_{2}$ by the condition (ii). Then, we have

[TABLE]

Since $a$ and $b$ are coprime, $u_{1}$, $u_{2}$ and $u$ are pairwise coprime. Therefore, $(u_{2}/u)n\not\in{\mathbb{Z}}$ if and only if $n/u\not\in{\mathbb{Z}}$, and $(u_{3}/u)m\not\in{\mathbb{Z}}$ if and only if $m/u\not\in{\mathbb{Z}}$.

It is easy to see that, if the condition (GK$i$) is satisfied for some $i$, then the condition GK is satisfied.

Conversely, assume that the condition GK is satisfied. Then $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is not Noetherian by Theorem 1.2 in Gonzáles-Karu [5]. By Theorem 4.6 and (10), either $n<3$ or $m<3$ is satisfied.

If $n=1$ (resp. $m=1$), then (GK1) (resp. (GK2)) holds.

Suppose $n=2$ and $m\geq 2$. Since the condition EU is not satisfied, we have $m<u$ by (11). If (I) of the condition GK is satisfied, then $n=m=2<u$, and therefore (GK3) is satisfied. If (II) of the condition GK is satisfied, then (GK3) or (GK5) is satisfied.

Suppose $n\geq 3$ and $m=2$. We know $n<u$ by (12). Then (II) is not satisfied. If (I) is satisfied, then (GK4) is satisfied. ∎

Lemma 4.9**.**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the conditions (i), (ii), (iii) in Theorem 1.1.

Assume $n=2$ and $3\leq m<u$. If either $u_{1}=1$ or $u_{2}=1$ is satisfied, then either the condition GK or EU is satisfied.

• 2)

If $n=2$ and $u>m\geq(u+1)/2$, then either the condition GK or EU is satisfied.

Proof.

First of all, remark that the condition EU is satisfied for $a$, $b$, $c$ if and only if so for $b$, $a$, $c$. Furthermore, the condition GK is satisfied for $a$, $b$, $c$ if and only if so for $b$, $a$, $c$.

First, we shall prove 1). Assume $n=2$, $3\leq m<u$ and $u_{2}=1$. If (GK5) is not satisfied, then

[TABLE]

and

[TABLE]

since $u_{2}=1$. Thus, the condition EU is satisfied.

Next, assume $n=2$, $3\leq m<u$ and $u_{1}=1$. Considering $b$, $a$, $c$, we may assume $3\leq n<u$, $m=2$ and $u_{2}=1$. Then, we have

[TABLE]

and

[TABLE]

In this case, the condition EU is always satisfied.

Here, we start to prove 2). Assume that (GK5) is not satisfied. Then, we have

[TABLE]

and

[TABLE]

Thus, the condition EU is satisfied. ∎

Proposition 4.10**.**

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the conditions (i), (ii), (iii) in Theorem 1.1.

If $u\leq 6$, then the condition EU is a necessary and sufficient condition for finite generation of $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$.

If $u\leq 6$, then the condition GK is a necessary and sufficient condition for infinite generation of $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$.

Proof.

We shall prove that either the condition GK or EU is satisfied if $u\leq 6$.

If $n=1$ or $m=1$, then (GK1) or (GK2) is satisfied.

If $n\geq 3$ and $m\geq 3$, then EU is satisfied as in (10).

If $u=n=m=2$, then EU is satisfied.

If $u>n=m=2$, then (GK3) is satisfied.

If $n=2$ and $m\geq u$, then then EU is satisfied by (11).

If $n\geq u$ and $m=2$, then then EU is satisfied by (12).

Now assume that $n=2$ and $3\leq m<u$. If $u>m\geq(u+1)/2$, then either the condition GK or EU is satisfied by Lemma 4.9 2). Assume $3\leq m<(u+1)/2$. If $u\leq 5$, then such $m$ does not exist. Suppose $u=6$ and $m=3$. Since $u$, $u_{1}$, $u_{2}$ are pairwise coprime, either $u_{1}$ or $u_{2}$ is $1$. Then, by Lemma 4.9 1), either the condition GK or EU is satisfied.

Assume $3\leq n<u$ and $m=2$. If $u\leq 6$, we can prove that either the condition GK or EU is satisfied in the same way as above. ∎

Example 4.11**.**

Let $K$ be a field of characteristic [math].

Remember the three examples in Example 4.4.

Assume $(a,b,c)=(8,19,9)$. In this case, $u=3$, and the conditions (i), (ii) and (iii) in Theorem 1.1 are satisfied. Since the conditin EU is satisfied, $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is Noetherian.

Assume $(a,b,c)=(25,29,72)$. In this case, $u=3$, and the conditions (i), (ii) and (iii) in Theorem 1.1 are satisfied. Since the conditin EU is not satisfied, $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is not Noetherian. Infinite generation of this ring was proved by Goto-Nishida-Watanabe [8].

Assume $(a,b,c)=(17,503,169)$. In this case, $u=7$, and the conditions (i), (ii) and (iii) in Theorem 1.1 are satisfied. In this case, neither GK nor EU is satisfied. Applying Theorem 1.1, we know that $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ is not Noetherian by calculation using computers (see Remark 4.2). **

Let $a$, $b$, $c$ be pairwise coprime positive integers. Assume the conditions (i), (ii), (iii) in Theorem 1.1. We do not know any example of finitely generated $\mathcal{R}_{s}({\mathfrak{p}}_{K}(a,b,c))$ such that the condition EU is not satisfied.

5. An example having negative curve in the second symbolic power

Let $S=K[x,y,z]$, where $K$ is a field and $x,y,z$ are indeterminates. We set $\mathfrak{m}=(x,y,z)S$ and $R=S_{\mathfrak{m}}$. In this section, we first take positive integers $s_{2},s_{3},t_{1},t_{3},u_{1},u_{2}$ arbitrarily, and set

[TABLE]

where $s=s_{2}+s_{3},t=t_{1}+t_{3},u=u_{1}+u_{2}$. Moreover, we set

[TABLE]

Let us regard $S$ as a $\mathbb{Z}$-graded ring by setting

[TABLE]

Then, we can check directly that $f,g,h$ are all homogeneous. We set $I=(f,g,h)S$ and $\mathfrak{a}=IR$.

Lemma 5.1**.**

We have the following relations;

• (1)

$y^{t_{3}}f+z^{u_{1}}g+x^{s_{2}}h=0$,

• (2)

$z^{u_{2}}f+x^{s_{3}}g+y^{t_{3}}h=0$.

Proof.  Since $f,g,h$ are the maximal minors of the matrix

[TABLE]

we get the relations stated above. ∎

Lemma 5.2**.**

The following assertions hold.

• (1)

$(x)+I$* is $\mathfrak{m}$-primary.*

• (2)

$\mathop{\rm Ass}\nolimits_{S}S/I=\mathop{\rm Assh}\nolimits_{S}S/I$.

• (3)

$I_{\mathfrak{p}}$* is generated by $2$ elements for any $\mathfrak{p}\in\mathop{\rm Assh}\nolimits_{S}S/I$.*

• (4)

$\ell_{S}(\,S/(x)+I^{(n)}\,)=(n(n+1)/2)\cdot a$* for any $0<n\in\mathbb{Z}$.*

• (5)

We have $I\subseteq\mathfrak{p}_{K}(a,b,c)$, and the equality holds if $\mathrm{GCD}(a,b,c)=1$.

Proof.  (1)  This holds as $(x)+I$ contains $x,y^{t}$ and $z^{u}$.

(2)  We get this assertion by Hilbert-Burch’s theorem.

(3)  Let us take any $\mathfrak{p}\in\mathop{\rm Ass}\nolimits_{S}S/I$. Then, as $x\not\in\mathfrak{p}$, we have $h\in(f,g)S_{\mathfrak{p}}$ by Lemma 5.1 (1), so $I_{\mathfrak{p}}=(f,g)S_{\mathfrak{p}}$.

(4)  Since $(x)+I=(x,y^{t},y^{t_{1}}z^{u_{1}},z^{u})$, we have ${\mathrm{e}}_{xR}(R/I)=\ell_{S}(\,S/(x)+I)\,)=a$. Let us take any $0<n\in\mathbb{Z}$. Then,

[TABLE]

For any $P\in\mathop{\rm Assh}\nolimits_{R}R/\mathfrak{a}$, $\mathcal{G}(\mathfrak{a}_{P})$ is isomorphic to a polynomial ring with $2$ variables over $R_{P}/\mathfrak{a}_{P}$, so

[TABLE]

Thus we get

[TABLE]

(5)  We set $\mathfrak{p}=\mathfrak{p}_{K}(a,b,c)$. Since $f,g,h$ are all homogeneous, we have $I\subseteq\mathfrak{p}$. Hence, we have

[TABLE]

Now, we assume $\mathrm{GCD}(a,b,c)=1$. Then, as is well known, we have ${\mathrm{e}}_{xR}(R/\mathfrak{p}R)=a$, so we see

[TABLE]

which means $(x)+I=(x)+\mathfrak{p}$. Then, we have

[TABLE]

from which the equality $\mathfrak{p}=I$ follows. ∎

Lemma 5.3**.**

Suppose $s_{2}>s_{3}$, $t_{1}=t_{3}=1$ and $u_{1}<u_{2}$. Then, the following assertions hold.

• (1)

There exists $\xi\in I^{(2)}$ such that

• (i)

$x^{s_{3}}\xi=z^{u_{2}-u_{1}}f^{2}-gh$,

• (ii)

$z^{u_{1}}\xi=x^{s_{2}-s_{3}}h^{2}-fg$, and

• (iii)

$\xi\equiv y^{3}$* mod $(x)$.*

• (2)

$(x)+I^{(2)}=(x,y^{3},y^{2}z^{2u_{1}},yz^{u+u_{1}},z^{2u})$.

Proof.  (1)  ¿From the relations (1) and (2) of Lemma 5.1, we get

[TABLE]

respectively. Hence, we have

[TABLE]

so we get

[TABLE]

Since $x^{s_{3}},z^{u_{1}}$ is a regular sequence on $S$, there exists $\xi\in S$ such that

[TABLE]

The first equality implies $x^{s_{3}}\xi\in I^{2}$, so $\xi\in I^{(2)}$. The second equality implies $z^{u_{1}}\xi\equiv-fg$ mod $(x)$, so $z^{u_{1}}\xi\equiv yz^{u_{1}}\cdot y^{2}$ mod $(x)$ as $f\equiv-yz^{u_{1}}$ mod $(x)$ and $g\equiv y^{2}$ mod $(x)$. Hence, we get $\xi\equiv y^{3}$ mod $(x)$ since $z^{u_{1}}$ is regular on $S/(x)$.

(2)  Since $(x)+I=(x,y^{2},yz^{u_{1}},z^{u})$ and $u_{1}<u_{2}$, we have

[TABLE]

We set $J=(\xi)+I^{2}\subseteq I^{(2)}$. Then, as

[TABLE]

we have

[TABLE]

by Lemma 5.2 (4). Hence we get the required assertion. ∎

Lemma 5.4**.**

Suppose $s_{2}>2s_{3}$, $t_{1}=t_{3}=1$ and $u_{1}<u_{2}<2u_{1}$. Then, the following assertions hold.

• (1)

There exists $\zeta\in I^{(3)}$ such that

• (i)

$x^{s_{3}}\zeta=f^{3}+z^{2u_{1}-u_{2}}h\xi$,

• (ii)

$z^{u_{2}-u_{1}}\zeta=f\xi+x^{s_{2}-2s_{3}}h^{3}$, and

• (iii)

$\zeta\equiv-y^{4}z^{2u_{1}-u_{2}}$* mod $(x)$.*

• (2)

$(x)+I^{(3)}=(x,y^{5},y^{4}z^{2u_{1}-u_{2}},y^{3}z^{u},y^{2}z^{u+2u_{1}},yz^{2u+u_{1}},z^{3u})$.

• (3)

$S[IT,I^{(2)}T^{2},I^{(3)}T^{3}]\subsetneq\mathcal{R}_{s}(I)$.

Proof.  (1)  ¿From the relations (i) and (ii) of Lemma 5.3, we get

[TABLE]

respectively. Hence, we have

[TABLE]

so we get

[TABLE]

Since $x^{s_{3}},z^{u_{2}-u_{1}}$ is a regular sequence on $S$, there exists $\zeta\in S$ such that

[TABLE]

The first equality implies $x^{s_{3}}\zeta\in II^{(2)}\subseteq I^{(3)}$, so $\zeta\in I^{(3)}$ as $x^{s_{3}}$ is regular on $S/I^{(3)}$. The second equality implies $z^{u_{2}-u_{1}}\zeta\equiv f\xi$ mod $(x)$, so $z^{u_{2}-u_{1}}\zeta\equiv-yz^{u_{1}}\cdot y^{3}$ mod $(x)$ as $f\equiv-yz^{u_{1}}$ mod $(x)$ and $\xi\equiv y^{3}$ mod $(x)$. Hence, we get $\zeta\equiv-y^{4}z^{2u_{1}-u_{2}}$ mod $(x)$ since $z^{u_{2}-u_{1}}$ is regular on $S/(x)$.

(2)  By Lemma 5.3 (2), we have

[TABLE]

as $u_{1}<u_{2}<2u_{1}$. We set $J=(\zeta)+II^{(2)}\subseteq I^{(3)}$. Then, as

[TABLE]

we have

[TABLE]

by Lemma 5.2 (4). Hence we get the required assertion.

(3)  It is enough to show $I^{(2)}I^{(3)}\subsetneq I^{(5)}$. (We have $(I^{(2)})^{2}+II^{(3)}=I^{(4)}$, which can be verified in the same way.) By Lemma 5.3 (2) and Lemma 5.4 (2), we have

[TABLE]

so we get

[TABLE]

On the other hand, by Lemma 5.2 (4), we have

[TABLE]

Since $\min\{29u_{1}+16u_{2},32u_{1}+14u_{2}\}-(30u_{1}+15u_{2})=\min\{u_{2}-u_{1},2u_{1}-u_{2}\}>0$, we see

[TABLE]

which means $I^{(2)}I^{(3)}\subsetneq I^{(5)}$. ∎

In the rest of this section, let us denote $S,I,\xi$ and $\zeta$ by $S_{K},I_{K},\xi_{K}$ and $\zeta_{K}$, respectively, in order to emphasize that the coefficient field is $K$.

Theorem 5.5**.**

Let us choose any rational numbers $\alpha$ and $\beta$ such that

[TABLE]

Moreover, we choose positive integers $s_{2},s_{3},u_{1}$ and $u_{2}$ such that

[TABLE]

Then, setting $t_{1}=t_{3}=1$, we get the following assertions.

• (1)

$s_{2}>2s_{3}$* and $u_{1}<u_{2}<2u_{1}$.*

• (2)

Let $0\ll q\in\mathbb{Z}$. We denote by $k$ the largest integer which is not bigger than $q/3$. Then we have $(x^{s_{3}},z^{2u_{1}-u_{2}})^{k}\subseteq(x^{q(s_{2}-2s_{3})+1},z^{q(u_{2}-u_{1})})$.

• (3)

Let $p$ be any prime number. Then $3p^{e_{p}}\in\mathrm{HC}(I_{\mathbb{F}_{p}};2,\xi_{\mathbb{F}_{p}})$ for any $e_{p}\gg 0$.

• (4)

$3\not\in\mathrm{HC}(I_{K};2,\xi_{K})$* for any field $K$.*

• (5)

$\mathcal{R}_{s}(I_{\mathbb{Q}})$* is infinitely generated.*

Proof.  (1)  These inequalities hold since $s_{2}/s_{3}>2$ and $1<u_{2}/u_{1}<2$.

(2)  Since $(x^{s_{3}},z^{2u_{1}-u_{2}})^{k}$ is generated my

[TABLE]

it is enough to show that

[TABLE]

holds for any $i=0,1,\dots,k$. So, we suppose $(k-i)s_{3}\leq q(s_{2}-2s_{3})$, where $i=0,1,\dots,k$. Then, dividing both sides of this inequality by $s_{3}$, we get

[TABLE]

Here, we write $q=3k+\ell$, where $\ell=0,1,2$. Then, as

[TABLE]

we have

[TABLE]

Hence, we get

[TABLE]

where $m=-\ell\{(\beta-2)(2-\alpha)+(\alpha-1)\}$. Now, we notice that our assumption $\alpha<5/4$ and $\beta<7/3-(\alpha-1)/(2-\alpha)$ implies

[TABLE]

so we see

[TABLE]

Since $q\gg 0$, we have $k\gg 0$ too, so it follows that

[TABLE]

as $m$ is a bounded number. Then, multiplying both sides of $i(2-\alpha)>q(\alpha-1)$ by $u_{1}$, we get

[TABLE]

(3)  By Lemma 5.4 (ii), we have a relation

[TABLE]

in $S_{\mathbb{F}_{p}}$. We take $0\ll e_{p}\in\mathbb{Z}$ and put $q=p^{e_{p}}$. Then, we have

[TABLE]

Here, we write $q=3k+\ell$, where $\ell=0,1,2$. Then,

[TABLE]

The relation (i) of Lemma 5.4 means $f^{3}\in(x^{s_{3}},z^{2u_{1}-u_{2}})I_{\mathbb{F}_{p}}^{(3)}$. Hence, by (2), we have

[TABLE]

Thus we see

[TABLE]

so there exist $\sigma_{\mathbb{F}_{p}},\tau_{\mathbb{F}_{p}}\in I_{\mathbb{F}_{p}}^{(3q)}$ such that

[TABLE]

Then, we have

[TABLE]

Since $x^{q(s_{2}-2s_{3})},z^{q(u_{2}-u_{1})}$ is a regular sequence on $S_{\mathbb{F}_{p}}$, there exists $\eta_{\mathbb{F}_{p}}\in S_{\mathbb{F}_{p}}$ such that

[TABLE]

The first equality implies $x^{q(s_{2}-2s_{3})}\eta_{\mathbb{F}_{p}}\in I_{\mathbb{F}_{p}}^{(3q)}$, so we have $\eta_{\mathbb{F}_{p}}\in I_{\mathbb{F}_{p}}^{(3q)}$ as $x^{q(s_{2}-s_{3})}$ is regular on $S/I_{\mathbb{F}_{p}}^{(3q)}$. The second equality implies $z^{q(u_{2}-u_{1})}\eta_{\mathbb{F}_{p}}\equiv(-1)^{q+1}h^{3q}$ mod $xS_{\mathbb{F}_{p}}$, so $z^{q(u_{2}-u_{1})}\eta_{\mathbb{F}_{p}}\equiv(-1)^{q+1}z^{3qu}$ mod $xS_{\mathbb{F}_{p}}$ as $h\equiv z^{u}$ mod $xS_{\mathbb{F}_{p}}$. Hence, we get $\eta_{\mathbb{F}_{p}}\equiv(-1)z^{2q(u+u_{1})}$ mod $xS_{\mathbb{F}_{p}}$ since $z^{q(u_{2}-u_{1})}$ is regular on $S_{\mathbb{F}_{p}}/xS_{\mathbb{F}_{p}}$. Then, we have

[TABLE]

and hence $3q\in\mathrm{HC}(I_{\mathbb{F}_{p}};2,\xi_{\mathbb{F}_{p}})$.

(4)  If $3\in\mathrm{HC}(I_{K};2,\xi_{K})$, by Proposition 2.9 (3), we have

[TABLE]

which contradicts to Lemma 5.4 (3).

(5)  Let us notice that $\xi_{\mathbb{Q}}\in\mathbb{Z}[x,y,z]$. Then, setting $k=2$ and $r=3$ in Theorem 2.13, we see that $\mathcal{R}_{s}(I_{\mathbb{Q}})$ is not finitely generated. ∎

Example 5.6**.**

Let $\alpha=6/5$ and $\beta=49/24$, which satisfy the assumptions on $\alpha$ and $\beta$ of Theorem 5.5. We set

[TABLE]

where $m,n$ are coprime positive integers such that $m$ is odd and $n$ is not a multiple of $97$. Then, we have

[TABLE]

Since $683$ and $97$ are prime numbers, we get $\mathrm{GCD}(a,b,c)=1$. Hence, by Lemma 5.2 (5) and Theorem 5.5 (5), we see that $\mathcal{R}_{s}(\mathfrak{p}_{\mathbb{Q}}(a,b,c))$ is infinitely generated. If $m=n=1$, then $a,b,c$ are pairwise coprime, and one can check directly that $\xi_{K}$ is the negative curve for any field $K$.