The Structure of Chains of Ulrich Ideals in Cohen-Macaulay Local Rings of Dimension One

Abstract

This paper studies Ulrich ideals in one-dimensional Cohen-Macaulay local rings. A correspondence between Ulrich ideals and overrings is given. Using the correspondence, chains of Ulrich ideals are closely explored. The specific cases where the rings are of minimal multiplicity and GGL rings are analyzed.

Introduction

The purpose of this paper is to investigate the behavior of chains of Ulrich ideals, in a one-dimensional Cohen-Macaulay local ring, in connection with the structure of birational finite extensions of the base ring.

The notion of Ulrich ideals is a generalization of stable maximal ideals, which dates back to 1971, when the monumental paper [9] of J. Lipman was published. The modern treatment of Ulrich ideals was started by [4, 5] in 2014, and has been explored in connection with the representation theory of rings. In [4], the basic properties of Ulrich ideals are summarized; whereas in [5], Ulrich ideals in two-dimensional rational singularities are closely studied with a concrete classification. However, in contrast to the existing research on Ulrich ideals, the theory pertaining to the one-dimensional case does not seem capable of growth. Some part of the theory, including research on the ubiquity as well as the structure of the chains of Ulrich ideals, seems to have been left unchallenged. In the current paper, we focus our attention on the one-dimensional case, clarifying the relationship between Ulrich ideals and the birational finite extensions of the base ring. The main objective is to understand the behavior of chains of Ulrich ideals in one-dimensional Cohen-Macaulay local rings.

To explain our objective as well as our main results, let us begin with the definition of Ulrich ideals. Although we shall focus our attention on the one-dimensional case, we would like to state the general definition, in the case of any arbitrary dimension. Let \((R,\mathfrak m)\) be a Cohen-Macaulay local ring with \(d= \dim R \ge 0\).

Definition 1.1 ([4])

Let I be an \(\mathfrak m\)-primary ideal of R and assume that I contains a parameter ideal Q = (a1, a2,…, ad) of R as a reduction. We say that I is an Ulrichideal of R, if the following conditions are satisfied.

  1. (1)

    IQ,

  2. (2)

    I2 = QI, and

  3. (3)

    I/I2 is a free R/I-module.

We notice that Condition (2) together with Condition (1) is equivalent to saying that the associated graded ring \(\text {gr}_{I}(R) = \bigoplus _{n \ge 0}I^{n}/I^{+ 1}\) of I is a Cohen-Macaulay ring and a(grI(R)) = 1 − d, where a(grI(R)) denotes the a-invariant of grI(R) ([6, Remark 3.10], [8, Remark 3.1.6]). Therefore, these two conditions are independent of the choice of reductions Q of I. In addition, assuming Condition (2) is satisfied, Condition (3) is equivalent to saying that I/Q is a free R/I-module ([4, Lemma 2.3]). We also notice that Condition (3) is automatically satisfied if \(I = \mathfrak m\). Therefore, when the residue class field \(R/\mathfrak m\) of R is infinite, the maximal ideal \(\mathfrak m\) is an Ulrich ideal of R if and only if R is not a regular local ring, possessing minimal multiplicity ([10]). From this perspective, Ulrich ideals are a kind of generalization of stable maximal ideals, which Lipman [9] started to analyze in 1971.

Here, let us briefly summarize some basic properties of Ulrich ideals, as seen in [4, 7]. Although we need only a part of them, let us also include some superfluity in order to show what specific properties Ulrich ideals enjoy. Throughout this paper, let r(R) denote the Cohen-Macaulay type of R, and let \(\text {Syz}_{R}^{i}(M)\) denote, for each integer i ≥ 0 and for each finitely generated R-module M, the i-th syzygy module of M in its minimal free resolution.

Theorem 1.2

([4, 7]) Let I be an Ulrich ideal of a Cohen-Macaulay local ringR of dimensiond ≥ 0 andsett = nd (> 0),where n denotes the number of elements in a minimal system of generators of I.Let

$${\cdots} \to F_{i} \overset{\partial_{i}}{\to} F_{i-1} \to {\cdots} \to F_{1} \overset{\partial_{1}}{\to} F_{0} = R \to R/I \to 0 $$

be a minimal free resolution ofR/I. Then r(R) = t⋅r(R/I) and the following assertions hold true.

  1. (1)

    I(i) = I for i ≥ 1.

  2. (2)

    For i ≥ 0, \(\beta _{i} = \left \{\begin {array}{lll} t^{i-d}{\cdot }(t + 1)^{d} & (i \ge d),\\ \binom {d}{i}+ t{\cdot }\beta _{i-1} & (1 \le i \le d),\\ 1 & (i = 0). \end {array}\right .\)

  3. (3)

    \(\text {Syz}_{R}^{i + 1}(R/I) \cong [\text {Syz}_{R}^{i}(R/I)]^{\oplus t}\)forid.

  4. (4)

    Fori, \(\text {Ext}_{R}^{i}(R/I, R)\cong \left \{\begin {array}{lll} (0) & (i < d),\\ (R/I)^{\oplus t} & (i = d),\\ (R/I)^{\oplus (t^{2}-1){\cdot }t^{i - (d + 1)} } & (i > d). \end {array}\right .\)

Here, I(i) denotes the ideal ofRgenerated by the entries of the matrixi, and βi = rankRFi.

Because Ulrich ideals are a very special kind of ideals, it seems natural to expect that, in the behavior of Ulrich ideals, there might be contained ample information on base rings, once they exist. As stated above, this is the case of two-dimensional rational singularities, and the present objects of study are rings of dimension one.

In what follows, unless otherwise specified, let \((R,\mathfrak m)\) be a Cohen-Macaulay local ring with \(\dim R = 1\). Our main targets are chains \(I_{n} \subsetneq I_{n-1} \subsetneq {\cdots } \subsetneq I_{1}\) (n ≥ 2) of Ulrich ideals in R. Let I be an Ulrich ideal of R with a reduction Q = (a). We set A = I : I in the total ring of fractions of R. Hence, A is a birational finite extension of R, and I = aA. Firstly, we study the close connection between the structure of the ideal I and the R-algebra A. Secondly, let J be an Ulrich ideal of R and assume that \(I \subsetneq J\). Then, we will show that μR(J) = μR(I), where μR(∗) denotes the number of elements in a minimal system of generators, and that J = (b) + I for some \(a,b \in \mathfrak m\) with I = abA. Consequently, we have the following, which is one of the main results of this paper.

Theorem 1.3

Let \((R,\mathfrak m)\) be a Cohen-Macaulay local ring with \(\dim R = 1\) . Then the following assertions hold true.

(1) Let I be an Ulrich ideal ofR andA = I : I.Let\(a_{1},a_{2}, \ldots , a_{n}\in \mathfrak m\) (n ≥ 2) andassume thatI = a1a2anA.For 1 ≤ in, letIi = (a1a2ai) + I. Then eachIiis an Ulrichideal ofR and

$$I=I_{n}\subsetneq I_{n-1}\subsetneq {\cdots} \subsetneq I_{1}.$$

(2) Conversely, letI1, I2,…, In (n ≥ 2) be Ulrich ideals ofRand suppose that

$$I_{n}\subsetneq I_{n-1}\subsetneq {\cdots} \subsetneq I_{1}.$$

We setI = In and A = I : I. Then there exist elements\(a_{1},a_{2}, \ldots , a_{n}\in \mathfrak m\)such thatI = a1a2anAandIi = (a1a2ai) + Ifor all 1 ≤ in − 1.

Let I and J be Ulrich ideals of R and assume that \(I \subsetneq J\). We set B = J : J. Let us write J = (b) + I for some \(b \in \mathfrak m\). We then have that J2 = bJ and that B is a local ring with the maximal ideal \(\mathfrak n = \mathfrak m + \frac {I}{b}\), where \(\frac {I}{b} = \left \{\frac {i}{b} \mid i \in I\right \}~(=b^{-1}I)\). We furthermore have the following.

Theorem 1.4

\(\frac {I}{b}\) is an Ulrich ideal of the Cohen-Macaulay local ring B of dimension one and there is a one-to-one correspondence \(\mathfrak a \mapsto \frac {\mathfrak a}{b}\) between the Ulrich ideals \(\mathfrak a\) of R such that \(I \subseteq \mathfrak a \subsetneq J\) and the Ulrich ideals \(\mathfrak b\) of B such that \(\frac {I}{b} \subseteq \mathfrak b\) .

These two theorems convey to us that the behavior of chains of Ulrich ideals in a given one-dimensional Cohen-Macaulay local ring could be understood via the correspondence, and the relationship between the structure of Cohen-Macaulay local rings R and B could be grasped through the correspondence, which we shall closely discuss in this paper.

We now explain how this paper is organized. In Section 2, we will summarize some preliminaries, which we shall need later to prove the main results. The proof of Theorems 1.3 and 1.4 will be given in Section 3. In Section 4, we shall study the case where the base rings R are not regular but possess minimal multiplicity ([10]), and show that the set of Ulrich ideals of R are totally ordered with respect to inclusion. In Section 5, we explore the case where R is a GGL ring ([2]).

In what follows, let \((R,\mathfrak m)\) be a Cohen-Macaulay local ring with \(\dim R = 1\). Let Q(R) (resp. \(\mathcal X_{R}\)) stand for the total ring of fractions of R (resp. the set of all the Ulrich ideals in R). We denote by \(\overline {R}\), the integral closure of R in Q(R). For a finitely generated R-module M, let μR(M) (resp. R(M)) be the number of elements in a minimal system of generators (resp. the length) of M. For each \(\mathfrak m\)-primary ideal \(\mathfrak a\) of R, let

$$\mathrm{e}_{\mathfrak{a}}^{0}(R) = \underset{n \to \infty}{\lim}\frac{\ell_{R}(R/\mathfrak a^{n})}{n} $$

stand for the multiplicity of R with respect to \(\mathfrak a\). By v(R) (resp. e(R)), we denote the embedding dimension \(\mu _{R}(\mathfrak m)\) of R (resp. \(\mathrm {e}_{\mathfrak m}^{0}(R)\)). Let \(\widehat {R}\) denote the \(\mathfrak m\)-adic completion of R.

Preliminaries

Let us summarize preliminary facts on \(\mathfrak m\)-primary ideals of R, which we need throughout this paper.

In this section, let I be an \(\mathfrak m\)-primary ideal of R, for which we will assume Condition (C) in Definition 2.2 to be satisfied. This condition is a partial extraction from Definition 1.1 of Ulrich ideals; hence, every Ulrich ideal satisfies it (see Remark 2.3).

Firstly, we assume that I contains an element aI with I2 = aI. We set A = I : I and

$$\frac{I}{a} =\left\{\frac{x}{a} \mid x \in I\right\}=a^{-1}I $$

in Q(R). Therefore, A is a birational finite extension of R such that \(R \subseteq A\subseteq \overline {R}\), and \(A = \frac {I}{a}\), because I2 = aI; hence I = aA. We then have the following.

Proposition 2.1

IfI = (a) : RI,thenA = R : IandI = R : A,whenceR : (R : I) = I.

Proof

Notice that I = (a) : RI = (a) : I = a[R : I] and we have A = R : I, because I = aA. We get R : A = I, since \(R:A= R:\frac {I}{a}= a[R:I] = aA\). □

Let us now give the following.

Definition 2.2

Let I be an \(\mathfrak m\)-primary ideal of R and set A = I : I. We say that I satisfies Condition (C), if

  • (i) A/R≅(R/I)t as an R-module for some t > 0, and

  • (ii) A = R : I.

Consequently, I = R : A by Condition (i), when I satisfies Condition (C).

Remark 2.3

Let \(I \in \mathcal X_{R}\). Then I satisfies Condition (C). In fact, choose aI so that I2 = aI. Then, I/(a)≅(R/I)t as an R/I-module, where t = μR(I) − 1 > 0 ([4, Lemma 2.3]). Therefore, I = (a) : RI, so that I satisfies the hypothesis in Proposition 2.1, whence A = R : I. Notice that A/RI/(a)≅(R/I)t, because I = aA.

We assume, throughout this section, that our \(\mathfrak m\)-primary ideal I satisfies Condition (C). We choose elements {fi}1≤it of A so that

$$A=R+\sum\limits_{i = 1}^{t}Rf_{i}.$$

Therefore, the images \(\{\overline {f_{i}}\}_{1 \le i \le t}\) of {fi}1≤it in A/R form a free basis of the R/I-module A/R. We then have the following.

Lemma 2.4

\(aA\cap R\subseteq (a)+I\)forallaR.

Proof

Let xaAR and write x = ay with yA. We write \(y=c_{0} + {\sum }_{i = 1}^{t}c_{i}f_{i}\) with ciR. Then, aciI for 1 ≤ it, since \(x = ac_{0} + {\sum }_{i = 1}^{t} (ac_{i})f_{i} \in R\). Therefore, (aci)fiIA = I for all 1 ≤ it, so that x ∈ (a) + I as claimed. □

Corollary 2.5

Let J be an\(\mathfrak m\)-primaryideal ofR and assume that J contains an elementbJsuchthatJ2 = bJandJ = (b) : RJ.If\(I \subseteq J\),thenJ = (b) + I.

Proof

We set B = J : J. Then B = R : J and J = bB by Proposition 2.1, so that \(B=R:J \subseteq A = R:I\), since \(I\subseteq J\). Consequently, \(J=bB\subseteq bA\cap R\subseteq (b)+I\) by Lemma 2.4, whence J = (b) + I. □

In what follows, let J be an \(\mathfrak m\)-primary ideal of R and assume that J contains an element bJ such that J2 = bJ and J = (b) : RJ. We set B = J : J. Then \(B=R:J=\frac {J}{b}\) by Proposition 2.1. Throughout, suppose that \(I \subsetneq J\). Therefore, since J = (b) + I by Corollary 2.5, we get

$$B=\frac{J}{b} = R+\frac{I}{b}. $$

Let \(\mathfrak a=\frac {I}{b}\). Therefore, \(\mathfrak a\) is an ideal of A containing I, so that \(\mathfrak a\) is also an ideal of B with

$$R/(\mathfrak a \cap R) \cong B/\mathfrak a. $$

With this setting, we have the following.

Lemma 2.6

The following assertions hold true.

  1. (1)

    \(A/B\cong (B/\mathfrak a)^{\oplus t}\)asa B-module.

  2. (2)

    \(\mathfrak a\cap R=I:_{R}J\) .

  3. (3)

    R([I : RJ]/I) = R(R/J).

  4. (4)

    I = [b⋅(I : RJ)]A.

Proof

(1) Since \(A=R+{\sum }_{i = 1}^{t}Rf_{i}\), we get \(A/B={\sum }_{i = 1}^{t}B\overline {f_{i}}\) where \(\overline {f_{i}}\) denotes the image of fi in A/B. Let {bi}1≤it be elements of \(B=\frac {J}{b}\) and assume that \({\sum }_{i = 1}^{t}b_{i}f_{i} \in B\). Then, since \({\sum }_{i = 1}^{t}(bb_{i})f_{i} \in R\) and bbiR for all 1 ≤ it, we have bbiI, so that \(b_{i}\in \frac {I}{b}=\mathfrak a\). Hence \(A/B\cong (B/\mathfrak a)^{\oplus t}\) as a B-module.

(2) This is standard, because J = (b) + I and \(\mathfrak a = \frac {I}{b}\).

(3) Since J/I = [(b) + I]/IR/[I : RJ], we get

$$\ell_{R}([I:_{R}J]/I)=\ell_{R}(R/I)-\ell_{R}(R/[I:_{R}J])=\ell_{R}(R/I)-\ell_{R}(J/I)=\ell_{R}(R/J). $$

(4) We have \([b{\cdot }(I:_{R}J)]A\subseteq I\), since \(b{\cdot }(I:_{R}J)\subseteq I\) and IA = I. To see the reverse inclusion, let xI. Then \(x \in J=bB \subseteq bA\). We write \(x=b[c_{0}+{\sum }_{i = 1}^{t}c_{i}f_{i}]\) with ciR. Then bciI for 1 ≤ it since xR, so that (bci)fiI for all 1 ≤ it, because I is an ideal of A. Therefore, bc0I, since \(x=bc_{0}+{\sum }_{i = 1}^{t}(bc_{i})f_{i} \in I\). Consequently, ciI : Rb = I : RJ for all 0 ≤ it, so that x ∈ [b⋅(I : RJ)]A as wanted.

Corollary 2.7

J/(b)≅([I : RJ]/I)tasanR-module. HenceR(J/(b)) = tR(R/J).

Proof

We consider the exact sequence

$$0 \to B/R \to A/R \to A/B \to 0 $$

of R-modules. By Lemma 2.6 (1), A/B is a free \(B/\mathfrak a\)-module of rank t, possessing the images of {fi}1≤it in A/B as a free basis. Because A/R is a free R/I-module of rank t, also possessing the images of {fi}1≤it in A/R as a free basis, we naturally get an isomorphism between the following two canonical exact sequences;

figurea

Since \(B/R=\frac {J}{b}/R \cong J/(b)\) and \(\mathfrak a \cap R=I:_{R}J\) by Lemma 2.6 (2), we get

$$J/(b)\cong([I:_{R}J]/I)^{\oplus t}. $$

The second assertion now follows from Lemma 2.6 (3). □

The following is the heart of this section.

Proposition 2.8

The following conditions are equivalent.

  1. (1)

    \(J\in \mathcal {X}_{R}\) .

  2. (2)

    μR([I : RJ]/I) = 1.

  3. (3)

    [I : RJ]/IR/JasanR-module.

When this is the case,μR(J) = t + 1.

Proof

The implication (3) ⇒ (2) is clear, and the reverse implication follows from the equality R([I : RJ]/I) = R(R/J) of Lemma 2.6 (3).

(1) ⇒ (3) Suppose that \(J\in \mathcal {X}_{R}\). Then J/(b) is R/J-free, so that by Corollary 2.7, [I : RJ]/I is a free R/J-module, whence [I : RJ]/IR/J by Lemma 2.6 (3).

(3) ⇒ (1) We have J/(b)≅([I : RJ]/I)t≅(R/J)t by Corollary 2.7, so that by Definition 1.1, \(J\in \mathcal {X}_{R}\) with μR(J) = t + 1.

We now come to the main result of this section, which plays a key role in Section 5.

Theorem 2.9

The following assertions hold true.

(1) Suppose that\(J\in \mathcal {X}_{R}\).Then there exists an element\(c \in \mathfrak m\)suchthatI = bcA.Consequently,\(I\in \mathcal {X}_{R}\)andμR(I) = μR(J) = t + 1.

(2) Suppose thatt ≥ 2.Then\(I\in \mathcal {X}_{R}\)ifand only if\(J\in \mathcal {X}_{R}\).

Proof

(1) Since \(J\in \mathcal {X}_{R}\), by Proposition 2.8, we get an element \(c \in \mathfrak m\) such that I : RJ = (c) + I. Therefore, by Lemma 2.6 (4) we have

$$I=[b{\cdot}(I:_{R}J)]A=[b{\cdot}((c)+I)]A=bcA+bIA = bcA + bI, $$

whence I = bcA by Nakayama’s lemma. Let a = bc. Then I2 = (aA)2 = aaA = aI, so that (a) is a reduction of I; hence \(A = \frac {I}{a}\). Consequently, I/(a)≅A/R≅(R/I)t, so that \(I\in \mathcal {X}_{R}\) with μR(I) = t + 1. Therefore, μR(I) = μR(J), because μR(J) = t + 1 by Proposition 2.8.

(2) We have only to show the only if part. Suppose that \(I\in \mathcal {X}_{R}\) and choose aI so that I2 = aI; hence \(A=\frac {I}{a}\). We then have μR(I) = t + 1, since I/(a)≅A/R≅(R/I)t. Consequently, since J = (b) + I, we get

$$\mu_{R}(J/(b))=\mu_{R}([(b)+I]/(b)) \leq \mu_{R}(I)=t + 1. $$

On the other hand, we have μR(J/(b)) = tμR([I : RJ]/I), because J/(b)≅([I : RJ]/I)t by Corollary 2.7. Hence

$$t{\cdot}(\mu_{R}([I:_{R}J]/I)-1)\leq 1, $$

so that μR([I : RJ]/I) = 1 because t ≥ 2. Thus by Proposition 2.8, \(J \in \mathcal {X}_{R}\) as claimed.

Chains of Ulrich Ideals

In this section, we study the structure of chains of Ulrich ideals in R. First of all, remember that all the Ulrich ideals of R satisfy Condition (C) stated in Definition 2.2 (see Remark 2.3), and summarizing the arguments in Section 2, we readily get the following.

Theorem 3.1

Let\(I, J \in \mathcal {X}_{R}\)andsuppose that\(I\subsetneq J\).ChoosebJsothatJ2 = bJ.Then the following assertions hold true.

(1) J = (b) + I.

(2) μR(J) = μR(I).

(3) There exists an element\(c\in \mathfrak m\)suchthatI = bcA,so that (bc) isa reduction of I, whereA = I : I.

We begin with the following, which shows that Ulrich ideals behave well, if R possesses minimal multiplicity. We shall discuss this phenomenon more closely in Section 4.

Corollary 3.2

Suppose that v(R) = e(R) > 1 andlet\(I \in \mathcal X_{R}\).ThenμR(I) = v(R) andR/Iisa Gorenstein ring.

Proof

We have \(\mathfrak m \in \mathcal X_{R}\) and r(R) = v(R) − 1, because v(R) = e(R) > 1. Hence, by Theorem 3.1 (2), \(\mu _{R}(I) = \mu _{R}(\mathfrak m)= \mathrm {v}(R)\). The second assertion follows from the equality r(R) = [μR(I) − 1]⋅r(R/I) (see [7, Theorem 2.5]). □

For each \(I \in \mathcal X_{R}\), Assertion (3) in Theorem 3.1 characterizes those ideals \(J \in \mathcal X_{R}\) such that \(I \subsetneq J\). Namely, we have the following.

Corollary 3.3

Let \(I\in \mathcal {X}_{R}\) . Then

$$\{ J\in \mathcal{X}_{R} | I\subsetneq J \} =\left\{(b)+I | b \in \mathfrak m~\text{such~that}~(bc)~\text{is~a~reduction~of}~I~\text{for~some}~c \in \mathfrak m \right\}. $$

Proof

Let \(b, c \in \mathfrak m\) and suppose that (bc) is a reduction of I. We set J = (b) + I. We shall show that \(J \in \mathcal X_{R}\) and \(I \subsetneq J\). Because \(bc \not \in \mathfrak m I\), we have b, cI, whence \(I \subsetneq J\). If J = (b), we then have \(I=bcA \subseteq J = (b)\) where A = I : I, so that \(cA \subseteq R\). This is impossible, because cR : A = I (see Lemma 2.1). Hence \((b)\subsetneq J\). Because I2 = bcI, we have J2 = bJ + I2 = bJ + bcI = bJ. Let us check that J/(b) is a free R/J-module. Let {fi}1≤it (t = μR(I) − 1 > 0) be elements of A such that \(A= R + {\sum }_{i = 1}^{t}Rf_{i}\), so that their images \(\{\overline {f_{i}}\}_{1 \le i \le t}\) in A/R form a free basis of the R/I-module A/R (remember that I satisfies Condition (C) of Definition 2.2). We then have

$$J=(b)+I=(b)+bcA=(b) + \sum\limits_{i = 1}^{t}R{\cdot}(bc)f_{i}. $$

Let {ci}1≤it be elements of R and assume that \({\sum }_{i = 1}^{t} c_{i}{\cdot }(bcf_{i}) \in (b)\). Then, since \({\sum }_{i = 1}^{t}c_{i}c{\cdot }f_{i} \in R\), we have cicI = bcA, so that cibAR for all 1 ≤ it. Therefore, because \(bA\cap R\subseteq (b)+I=J\) by Lemma 2.4, we get ciJ, whence J/(b)≅(R/J)t. Thus, \(J = (b)+I \in \mathcal X_{R}\). □

The equality μR(I) = μR(J) does not hold true in general, if I and J are incomparable, as we show in the following.

Example 3.4

Let S = k[[X1, X2, X3, X4]] be the formal power series ring over a field k and consider the matrix \({\Bbb M}= \left (\begin {array}{lll} X_{1} & X_{2} & X_{3} \\ X_{2} & X_{3} & X_{1} \end {array}\right ) \). We set \(R=S/[\mathfrak a + ({X_{4}^{2}})]\), where \(\mathfrak a\) denotes the ideal of S generated by the 2 × 2 minors of \(\mathbb M\). Let xi denote the image of Xi in R for each i = 1,2,3,4. Then, (x1, x2, x3) and (x1, x4) are Ulrich ideals of R with different numbers of generators, and they are incomparable with respect to inclusion.

We are now ready to prove Theorem 1.3.

Proof of Theorem 1.3

(1) This is a direct consequence of Corollary 3.3.

(2) By Theorem 3.1, we may assume that n > 2 and that our assertion holds true for n − 1. Therefore, there exist elements \(a_{1}, a_{2}, \ldots , a_{n-1} \in \mathfrak m\) such that (a1a2an− 1) is a reduction of In− 1 and Ii = (a1a2ai) + In− 1 for all 1 ≤ in − 2. Now apply Theorem 3.1 to the chain \(I_{n} \subsetneq I_{n-1}\). We then have In− 1 = (a1a2an− 1) + In together with one more element \(a_{n}\in \mathfrak m\) so that (a1a2an− 1)⋅anA = In. Hence

$$I_{i}=(a_{1}a_{2}{\cdots} a_{i})+I_{n-1}=(a_{1}a_{2}{\cdots} a_{i})+I_{n} $$

for all 1 ≤ in − 1.

In order to prove Theorem 1.4, we need more preliminaries. Let us begin with the following.

Theorem 3.5

Suppose that\(I, J\in \mathcal {X}_{R}\)and\(I\subsetneq J\).LetbJsuchthatJ2 = bJandB = J : J.Then the following assertions hold true.

(1) \(B=R+\frac {I}{b}\)and\(\frac {I}{b}=I:J\).

(2) B is a Cohen-Macaulay local ring with\(\dim B = 1\)and\(\mathfrak n=\mathfrak m+\frac {I}{b}\)themaximal ideal. Hence\(R/\mathfrak m \cong B/\mathfrak n\).

(3) \(\frac {I}{b}\in \mathcal {X}_{B}\)and\(\mu _{B}(\frac {I}{b})=\mu _{R}(I)\).(4) r(B) = r(R) and e(B) = e(R).Therefore, v(B) = e(B) ifand only if v(R) = e(R).

Proof

We set A = I : I. Hence \(R \subsetneq B \subsetneq A\) by Proposition 2.1. Let t = μR(I) − 1. (1) Because J = (b) + I and \(B = \frac {J}{b}\), we get \(B=R+\frac {I}{b}\). We have \(I:J \subseteq \frac {I}{b}\), since bJ. Therefore, \(\frac {I}{b}=I:J\), because

$$J\cdot \frac{I}{b}=I\cdot \frac{J}{b}=IB\subseteq IA=I. $$

(2) It suffices to show that B is a local ring with maximal ideal \(\mathfrak n=\mathfrak m + \frac {I}{b}\). Let \(\mathfrak a = \frac {I}{b}\). Choose \(c \in \mathfrak m\) so that I = bcA. We then have \(\mathfrak a =cA \subseteq \mathfrak m A \subseteq \mathrm {J}(A)\), where J(A) denotes the Jacobson radical of A. Therefore, \(\mathfrak n = \mathfrak m + cA\) is an ideal of B = R + cA, and \(\mathfrak n \subseteq \mathrm {J}(B)\), because A is a finite extension of B. On the other hand, because \(R/\mathfrak m \cong B/\mathfrak n\), \(\mathfrak n\) is a maximal ideal of B, so that \((B,\mathfrak n)\) is a local ring.(3) We have \(\mathfrak a^{2}=c\mathfrak a\), since \(\mathfrak a = cA\). Notice that \(\mathfrak a \ne cB\), since AB. Then, because \(\mathfrak a/cB\cong A/B\cong (B/\mathfrak a)^{\oplus t}\) by Lemma 2.6 (1), we get \(\mathfrak a\in \mathcal {X}_{B}\) and \(\mu _{B}(\mathfrak a)=t + 1=\mu _{R}(I)\).(4) We set L = (c) + I. Then, since bcA = I, \(L\in \mathcal {X}_{R}\) and μR(L) = μR(I) = t + 1 by Corollary 3.3 and Theorem 3.1 (2). Therefore, r(R) = t⋅r(R/L) by [7, Theorem 2.5], while \(\mathrm {r}(B)=t{\cdot }\mathrm {r}(B/\mathfrak a)\) for the same reason, because \(\mathfrak a \in \mathcal X_{B}\) by Assertion (3). Remember that the element c is chosen so that I : RJ = (c) + I (see the proof of Theorem 2.9 (1)). We then have \(\mathrm {r}(B/\mathfrak a)=\mathrm {r}(R/[I:_{R}J])\), because \(B = R + \mathfrak a\) and

$$R/L=R/[\mathfrak a \cap R] \cong B/\mathfrak a $$

where the first equality follows from Lemma 2.6 (2). Thus

$$\mathrm{r}(B)=t{\cdot}\mathrm{r}(B/\mathfrak a)=t{\cdot}\mathrm{r}(R/L)=\mathrm{r}(R), $$

as is claimed. To see the equality e(B) = e(R), enlarging the residue class field of R, we may assume that \(R/\mathfrak m\) is infinite. Choose an element \(\alpha \in \mathfrak m\) so that (α) is a reduction of \(\mathfrak m\). Hence, αB is a reduction of \(\mathfrak m B\), while \(\mathfrak m B\) is a reduction of \(\mathfrak n\), because

$$\mathfrak n A=(\mathfrak m + cA)A=\mathfrak m A = (\mathfrak m B)A. $$

Therefore, αB is a reduction of \(\mathfrak n\), so that

$$\mathrm{e}(B)=\ell_{B}(B/\alpha B)=\ell_{R}(B/\alpha B)=\mathrm{e}_{\alpha R}^{0}(B)=\mathrm{e}_{\alpha R}^{0}(R)=\mathrm{e}(R), $$

where the second equality follows from the fact that \(R/\mathfrak m \cong B/\mathfrak n\) and the fourth equality follows from the fact that \(\ell _{R}(B/R)<\infty \). Hence, e(B) = e(R) and r(B) = r(R). Because v(R) = e(R) > 1 if and only if r(R) = e(R) − 1, the assertion that v(B) = e(B) if and only if v(R) = e(R) now follows. □

We need one more lemma.

Lemma 3.6

Suppose that\(I, J\in \mathcal {X}_{R}\)and\(I\subsetneq J\).LetαJ.ThenJ = (α) + Iifand only ifJ2 = αJ.

Proof

It suffices to show the only if part. Suppose J = (α) + I. We set A = I : I, B = J : J, and choose bJ so that J2 = bJ. Then J = bB and \(B \subseteq A\), whence JA = bA, while JA = [(α) + I]A = αA + I. We now choose \(c\in \mathfrak m\) so that I = bcA (see Theorem 3.1 (3)). We then have bA = JA = αA + bcA, whence bA = αA by Nakayama’s lemma. Therefore, JA = αA, whence (α) is a reduction of J, so that J2 = αJ. □

We are now ready to prove Theorem 1.4.

Proof of Theorem 1.4

Let \(I, J \in \mathcal X_{R}\) such that \(I \subsetneq J\). We set A = I : I and B = J : J. Let bJ such that J = (b) + I. Then J2 = bJ by Lemma 3.6 and B is a local ring with \(\mathfrak n = \mathfrak m + \frac {I}{b}\) the maximal ideal by Theorem 3.5. □

Let \(\mathfrak a \in \mathcal X_{R}\) such that \(I \subseteq \mathfrak a \subsetneq J\). First of all, let us check the following.

Claim 1

\(\frac {\mathfrak a}{b} \in \mathcal X_{B}\) and \(\frac {\mathfrak a}{b} = \mathfrak a :J\) .

Proof of Claim 1

Since bJ, \(\mathfrak a:J \subseteq \frac {\mathfrak a}{b}\). On the other hand, since

$$B=R:J \subseteq R:\mathfrak a=\mathfrak a:\mathfrak a $$

by Lemma 2.1, we get

$$J\cdot \frac{\mathfrak a}{b}=\mathfrak a\cdot \frac{J}{b}=\mathfrak a B\subseteq \mathfrak a {\cdot}(\mathfrak a:\mathfrak a)=\mathfrak a, $$

so that \(\frac {\mathfrak a}{b}\) is an ideal of \(B = \frac {J}{b}\) and \(\mathfrak a:J=\frac {\mathfrak a}{b}\). Since \(\frac {I}{b}\in \mathcal {X}_{B}\) by Theorem 3.5 (3), to show that \(\frac {\mathfrak a}{b} \in \mathcal X_{B}\), we may assume \(I \subsetneq \mathfrak a\). We then have, by Theorem 1.3 (2), elements \(a_{1}, a_{2}\in \mathfrak m\) such that I = ba1a2A and \(\mathfrak a = (ba_{1})+I\); hence \(\frac {\mathfrak a}{b}=a_{1}R+\frac {I}{b}\). We get \(\frac {\mathfrak a}{b}=a_{1}B+\frac {I}{b}\), since \(\frac {\mathfrak a}{b}\) is an ideal of B. Therefore, \(\frac {\mathfrak a}{b}\in \mathcal {X}_{B}\) by Corollary 3.3, because a1a2B is a reduction of \(\frac {I}{b}=a_{1}a_{2}A\).

We now have the correspondence φ defined by \(\mathfrak a \mapsto \frac {\mathfrak a}{b}\), and it is certainly injective. Suppose that \(\mathfrak b\in \mathcal {X}_{B}\) and \(\frac {I}{b}\subsetneq \mathfrak b\). We take \(\alpha \in \mathfrak b\) so that \(\mathfrak b^{2}=\alpha \mathfrak b\). Then, since B is a Cohen-Macaulay local ring with maximal ideal \(\mathfrak m + \frac {I}{b}\), we have \(\mathfrak b=\alpha B+\frac {I}{b}\) by Theorem 3.1. Let us write α = a + x with \(a\in \mathfrak m\) and \(x\in \frac {I}{b}\). We then have \(\mathfrak b=aB+\frac {I}{b}\), so that \(\mathfrak b^{2}=a\mathfrak b\) by Lemma 3.6. Set \(L=\frac {I}{b}\). Then, since A = I : I = L : L, by Theorem 3.1 we have an element \(\beta \in \mathfrak n=\mathfrak m+L\) such that L = aβA; hence aβL. Let us write β = c + y with \(c\in \mathfrak m\) and yL. We then have ac = aβayL and \(yA \subseteq L\), so that because

$$L=a\beta A\subseteq acA+a{\cdot}yA\subseteq acA+\mathfrak m L, $$

we get L = acA by Nakayama’s lemma. Therefore, I = abcA. On the other hand, since \(aB=aR+a{\cdot }\frac {I}{b}\), we get \(\mathfrak b=aB + \frac {I}{b} = aR+\frac {I}{b}\). Hence, because \(b\mathfrak b=(ab)+I\) and I = (ab)cA, we finally have that \(b \mathfrak b \in \mathcal {X}_{R}\) and

$$I=abcA \subsetneq b\mathfrak b = (ab)+ I \subsetneq J $$

by Theorem 1.3 (1). Thus, the correspondence φ is bijective, which completes the proof of Theorem 1.4. □

The Case Where R Possesses Minimal Multiplicity

In this section, we focus our attention on the case where R possesses minimal multiplicity. Throughout, we assume that v(R) = e(R) > 1. Hence, \(\mathfrak m \in \mathcal X_{R}\) and μR(I) = v for all \(I \in \mathcal X_{R}\) by Corollary 3.2, where v = v(R). We choose an element \(\alpha \in \mathfrak m\) so that \(\mathfrak m^{2}=\alpha \mathfrak m\).

Let \(I, J\in \mathcal {X}_{R}\) such that \(I\subsetneq J\) and assume that there are no Ulrich ideals contained strictly between I and J. Let bJ with J2 = bJ and set B = J : J. Hence \(B = \frac {J}{b}\), and J = (b) + I by Theorem 3.1. Remember that by Theorem 3.5, B is a local ring and v(B) = e(B) = e(R) > 1. We have \(\mathfrak n^{2}=\alpha \mathfrak n\) by the proof of Theorem 3.5 (4), where \(\mathfrak n\) denotes the maximal ideal of B.

We furthermore have the following.

Lemma 4.1

The following assertions hold true.

  1. (1)

    R(J/I) = 1.

  2. (2)

    \(I=b\mathfrak n=J\mathfrak n\).Hence, the ideal I is uniquely determined by J,and\(I:I=\mathfrak n:\mathfrak n\).

  3. (3)

    (bα) isa reduction of I. IfI = (bα) + (x2, x3,…, xv),thenJ = (b, x2, x3,…, xv).

Proof

By Theorem 1.4, we have the one-to-one correspondence

$$\left\{\mathfrak a\in\mathcal{X}_{R} | I\subseteq \mathfrak a\subsetneq J\right\} \overset{\varphi}{\longrightarrow} \left\{\mathfrak b\in\mathcal{X}_{B} | \frac{I}{b}\subseteq \mathfrak b\right\},\enskip \mathfrak a\mapsto \frac{\mathfrak a}{b}, $$

where the set of the left hand side is a singleton consisting of I, and the set of the right hand side contains \(\mathfrak n\). Hence, \(\mathfrak n=\frac {I}{b}\), that is \(I=b\mathfrak n=J\mathfrak n\), because J = bB. Therefore, \(I^{2}=b^{2}\mathfrak n^{2}=b\alpha {\cdot }b\mathfrak n=b\alpha {\cdot }I\), so that (bα) is a reduction of I. Because

$$J/I=bB/b\mathfrak n \cong B/\mathfrak n $$

and \(R/\mathfrak m \cong B/\mathfrak n\) by Theorem 3.5 (2), we get R(J/I) = 1. Assertion (3) is clear, since J = (b) + I. □

Let I, J be ideals of R such that \(I \subsetneq J\) and \(\ell _{R}(J/I) < \infty \). Then we say that a chain \(I = I_{\ell } \subsetneq I_{\ell -1} \subsetneq {\ldots } \subsetneq I_{1} = J\) of ideals in R is a composition series which connects I with J, if R(Ii/Ii+ 1) = 1 for all 1 ≤ i − 1, where = R(J/I) + 1. With this terminology, since \(\ell _{R}(R/I) < \infty \) for all \(I \in \mathcal X_{R}\), we have the following.

Corollary 4.2

Suppose that\(I, J\in \mathcal {X}_{R}\)and\(I\subsetneq J\).Then there exists a composition series\(I=I_{\ell }\subsetneq I_{\ell -1}\subsetneq \cdots \subsetneq I_{1}=J\)connecting I with J such that\(I_{i}\in \mathcal {X}_{R}\)forall 1 ≤ i.

The following is the heart of this section.

Theorem 4.3

The set \(\mathcal {X}_{R}\) is totally ordered with respect to inclusion.

Proof

Suppose that there exist \(I, J\in \mathcal {X}_{R}\) such that \(I\nsubseteq J\) and \(J\nsubseteq I\). Since \(I\subsetneq \mathfrak m\) and \(J\subsetneq \mathfrak m\), thanks to Corollary 4.2, we get composition series

$$I=I_{\ell}\subsetneq I_{\ell-1}\subsetneq\cdots\subsetneq I_{1}=\mathfrak m\quad \text{and}\quad J=J_{n}\subsetneq J_{n-1}\subsetneq\cdots\subsetneq J_{1}=\mathfrak m $$

connecting I with \(\mathfrak m\) and J with \(\mathfrak m\), respectively, such that \(I_{i}, J_{j}\in \mathcal {X}_{R}\) for all 1 ≤ i and 1 ≤ jn. We may assume n. Then Lemma 4.1 (2) shows that Ii = Ji for all 1 ≤ i, whence \(J \subseteq J_{\ell }=I_{\ell }\subseteq I\). This is a contradiction. □

Remark 4.4

Theorem 4.3 is no longer true, unless R possesses minimal multiplicity. For example, let k be a field and consider R = k[[t3, t7]] in the formal power series ring k[[t]]. Then, \(\mathcal X_{R}= \{(t^{6}-ct^{7}, t^{10}) \mid 0 \ne c \in k\}\), which is not totally ordered, if k > 2. See Example 5.7 (3) also.

Let us now summarize the results in the case where R possesses minimal multiplicity.

Theorem 4.5

Let\(I\in \mathcal {X}_{R}\)and take a compositionseries

$$(E)\ \ \ I=I_{\ell}\subsetneq I_{\ell-1}\subsetneq\cdots\subsetneq I_{1}=\mathfrak m $$

connectingIwith\(\mathfrak m\)such that\(I_{i}\in \mathcal {X}_{R}\)for every 1 ≤ i = R(R/I). We setB0 = RandBi = Ii : Ii for 1 ≤ iand let\(\mathfrak n_{i}=\mathrm {J}(B_{i})\)denote the Jacobson radical ofBifor each 0 ≤ i. Then we obtain a tower

$$R=B_{0} \subsetneq B_{1} \subsetneq {\cdots} \subsetneq B_{\ell -1} \subsetneq B_{\ell} \subseteq \overline{R} $$

of birational finite extensions ofRand furthermore have the following.

(1) (αi) is a reduction ofIifor every 1 ≤ i.

(2) \(B_{i}=\mathfrak n_{i-1}:\mathfrak n_{i-1}\)for every 1 ≤ i.

(3) For 0 ≤ i − 1, \((B_{i},\mathfrak n_{i})\)is a local ring with v(Bi) = e(Bi) = e(R) > 1 and\(\mathfrak {n_{i}^{2}}=\alpha \mathfrak n_{i}\).

(4) Choosex2, x3,…, xvIso thatI = (α, x2,…, xv). ThenIi = (αi, x2, x3,…, xv) for every 1 ≤ i. In particular, \(\mathfrak m = (\alpha , x_{2}, x_{3}, \ldots , x_{v})\), so that theseries (E) is a unique composition series of ideals inRwhich connectsIwith\(\mathfrak m\).

(5) LetJbe an ideal ofRand assume that\(I \subseteq J \subseteq \mathfrak m\). ThenJ = Iifor some 1 ≤ i.

Proof

The uniqueness of composition series in Assertion (4) follows from the fact that the maximal ideal \(\mathfrak m/I\) of R/I is cyclic, and then, Assertion (5) readily follows from the uniqueness. Assertions (1), (2), (3), and the first part of Assertion (4) follow by standard induction on . □

Corollary 4.6

Suppose that there exists a minimal element I in\(\mathcal {X}_{R}\).Then\(\sharp \mathcal {X}_{R}=\ell <\infty \)with = R(R/I).

Proof

Since \(\mathcal X_{R}\) is totally ordered by Theorem 4.3, I is the smallest element in \(\mathcal X_{R}\), so that \(I \subseteq J\) for all \(J \in \mathcal X_{R}\). Therefore, by Theorem 4.5 (5), J is one of the Ii’s in the compoosition series \(I=I_{\ell }\subsetneq I_{\ell -1}\subsetneq \cdots \subsetneq I_{1}=\mathfrak m\). □

Corollary 4.7

If \(\widehat {R}\) is a reduced ring, then \(\mathcal X_{R}\) is a finite set.

Proof

Since by Theorem 4.5 \(\ell _{R}(R/I) \le \ell _{R}(\overline {R}/R)<\infty \) for every \(I \in \mathcal X_{R}\), the set \(\mathcal X_{R}\) contains a minimal element, so that \(\mathcal X_{R}\) is a finite set. □

Here let us note the following.

Example 4.8

Let \((S, \mathfrak n)\) be a two-dimensional regular local ring. Let \(\mathfrak n=(X, Y)\) and consider the ring A = S/(Y2). Then v(A) = e(A) = 2 and

$$\mathcal{X}_{A}=\left\{(x^{n}, y) \mid n \geq 1\right\} $$

where x, y denote the images of X, Y in A, respectively. Hence \(\sharp \mathcal {X}_{A}=\infty \).

Proof

Let In = (xn, y) for each n ≥ 1. Then \((x^{n})\subsetneq I_{n}\) and \({I_{n}^{2}}=x^{n}I_{n}\). Let J(A) = (x, y) be the maximal ideal of A. We then have J(A)2 = xJ(A), whence v(A) = e(A) = 2. Because In = (xn) : Ay, we get In/(xn)≅A/In. Therefore, \(I_{n}\in \mathcal {X}_{A}\) for all n ≥ 1. To see that \(\mathcal X_{A}\) consists of these ideals In’s, let \(I\in \mathcal {X}_{A}\) and set = A(A/I). Then \(I\subseteq I_{\ell }\) or \(I\supseteq I_{\ell }\), since \(\mathcal {X}_{A}\) is totally ordered. In any case, I = I, because A(A/I) = . Hence \(\mathcal {X}_{A}=\left \{(x^{n}, y) \mid n\geq 1\right \}\). □

We close this section with the following. Here, the hypothesis about the existence of a fractional canonical ideal K is equivalent to saying that R contains an \(\mathfrak m\)-primary ideal I such that I≅KR as an R-module and such that I possesses a reduction Q = (a) generated by a single element a of R ([3, Corollary 2.8]). The latter condition is satisfied, once \(\mathrm {Q}(\widehat {R})\) is a Gorenstein ring and the field \(R/\mathfrak m\) is infinite.

Theorem 4.9

Suppose that there exists a fractional ideal K ofR suchthat\(R\subseteq K\subseteq \overline {R}\)andK≅KRasanR-module. Then the following conditions are equivalent.

(1) \(\sharp \mathcal {X}_{R}=\infty \).

(2) e(R) = 2 and\(\widehat {R}\)isnot a reduced ring.

(3) The ring\(\widehat {R}\)hasthe form\(\widehat {R} \cong S/(Y^{2})\)forsome regular local ring\((S,\mathfrak n)\)ofdimension two with\(Y \in \mathfrak n \setminus \mathfrak n^{2}\).

Proof

(1) ⇒ (2) The ring \(\widehat {R}\) is not reduced by Corollary 4.7. Suppose R is not a Gorenstein ring; hence \(R \subsetneq K\) and e(R) > 2. We set \(\mathfrak a = R:K\). Let \(I \in \mathcal X_{R}\). Then, since μR(I) = v = e(R) > 2 by Corollary 3.2, we have \(\mathfrak a \subseteq I\) by [7, Corollary 2.12], so that \(\ell _{R}(R/I)\leq \ell _{R}(R/\mathfrak a)<\infty \). Therefore, the set \(\mathcal X_{R}\) contains a minimal element, which is a contradiction.

(3) ⇒ (1) See Example 4.8 and use the fact that there is a one-to-one correspondence \(I \mapsto I\widehat {R}\) between Ulrich ideals of R and \(\widehat {R}\), respectively.

(2) ⇒ (3) Since v(R) = e(R) = 2, the completion \(\widehat {R}\) has the form \(\widehat {R} = S/I\), where \((S,\mathfrak n)\) is a two-dimensional regular local ring and I = (f) a principal ideal of S. Notice that e(S/(f)) = 2 and \(\sqrt {(f)} \ne (f)\). We then have (f) = (Y2) for some \(Y \in \mathfrak n \setminus \mathfrak n^{2}\), because \(f \in \mathfrak n^{2} \setminus \mathfrak n^{3}\).

Remark 4.10

In Theorem 4.9, the hypothesis on the existence of fractional canonical ideals K is not superfluous. In fact, let V denote a discrete valuation ring and consider the idealization \(R = V \ltimes F\) of the free V -module F = Vn (n ≥ 2). Let t be a regular parameter of V . Then for each n ≥ 1, In = (tn) × F is an Ulrich ideal of R ([4, Example 2.2]). Hence \(\mathcal X_{R}\) is infinite, but v(R) = e(R) = n + 1 ≥ 3.

Higher dimensional cases are much wilder. Even though \((R,\mathfrak m)\) is a two-dimensional Cohen-Macaulay local ring possessing minimal multiplicity, the set \(\mathcal X_{R}\) is not necessarily totally ordered. Before closing this section, let us note examples.

Example 4.11

We consider two examples.

(1) Let S = k[[X0, X1,…, Xn]] (n ≥ 3) be the formal power series ring over a field k. Let ≥ 1 be an integer and consider the 2 × n matrix

$$\mathbb M = \left( \begin{array}{llll} X_{1}&X_{2}&\cdots&X_{n}\\ X_{0}^{\ell}&X_{1}&\cdots&X_{n-1} \end{array}\right). $$

We set \(R = S/\mathbb I_{2}(\mathbb M)\), where \(\mathbb I_{2}(\mathbb M)\) denotes the ideal of S generated by the 2 × 2 minors of the matrix \(\mathbb M\). Then, R is a Cohen-Macaulay local ring of dimension two, possessing minimal multiplicity. For this ring, we have

$$\mathcal X_{R}=\{({x_{0}^{i}}, x_{1}, x_{2}, \ldots, x_{n}) \mid 1 \le i \le \ell\}, $$

where xi denotes the image of Xi in R for each 0 ≤ in. Therefore, the set \(\mathcal X_{R}\) is totally ordered with respect to inclusion.

(2) Let \((S,\mathfrak n)\) be a regular local ring of dimension three. Let \(F, G, H, Z \in \mathfrak n\) and assume that \(\mathfrak n =(F,G, Z)=(G,H,Z)=(H,F,Z)\). (For instance, let S = k[[X, Y, Z]] be the formal power series ring over a field k with chk ≠  2, and choose F = X, G = X + Y, H = XY.) We consider the ring R = S/(Z2FGH). Then R is a two-dimensional Cohen-Macaulay local ring of minimal multiplicity two. Let f, g, h, z denote, respectively, the images of F, G, H, Z in R. Then, (f, gh, z), (g, fh, z), (h, fg, z) are Ulrich ideals of R, but any two of them are incomparable.

The Case Where R is a GGL Ring

In this section, we study the case where R is a GGL ring. The notion of GGL rings is given by [2]. Let us briefly review the definition.

Definition 5.1

([2]) Suppose that \((R,\mathfrak m)\) is a Cohen-Macaulay local ring with \(d= \dim R \ge 0\), possessing the canonical module KR. We say that R is a generalized Gorenstein local (GGL for short) ring, if one of the following conditions is satisfied.

  1. (1)

    R is a Gorenstein ring.

  2. (2)

    R is not a Gorenstein ring, but there exists an exact sequence

    $$0 \to R \xrightarrow{\varphi} \mathrm{K}_{R} \to C \to 0 $$

of R-modules and an \(\mathfrak m\)-primary ideal \(\mathfrak a\) of R such that

  1. (i)

    C is an Ulrich R-module with respect to \(\mathfrak a\) and

  2. (ii)

    the induced homomorphism \(R/\mathfrak a \otimes _{R} \varphi : R/\mathfrak a \to \mathrm {K}_{R}/\mathfrak a \mathrm {K}_{R}\) is injective.

When Case (2) occurs, we especially say that R is a GGL ring with respect to \(\mathfrak a\).

Since our attention is focused on the one-dimensional case, here let us summarize a few results on GGL rings of dimension one. Suppose that \((R,\mathfrak m)\) is a Cohen-Macaulay local ring of dimension one, admitting a fractional canonical ideal K. Hence, K is an R-submodule of \(\overline {R}\) such that K≅KR as an R-module and \(R \subseteq K \subseteq \overline {R}\). One can consult [3, Sections 2, 3] and [11, Vortrag 2] for basic properties of K. We set S = R[K] in Q(R). Therefore, S is a birational finite extension of R with S = Kn for all n ≫ 0, and the ring S = R[K] is independent of the choice of K ([1, Theorem 2.5]). We set \(\mathfrak c=R:S\). First of all, let us note the following.

Lemma 5.2

(cf. [3, Lemma 3.5]) \(\mathfrak c = K : S\)and\(S=\mathfrak c:\mathfrak c=R:\mathfrak c\).

Proof

Since R = K : K ([11, Bemerkung 2.5 a)]), we have \(\mathfrak c = (K:K):S = K : KS = K : S\), while \(R:\mathfrak c = (K:K):\mathfrak c = K: K\mathfrak c = K : \mathfrak c\). Hence \(R : \mathfrak c = K : \mathfrak c = K:(K:S) = S\) ([11, Definition 2.4]). Therefore, \(\mathfrak c:\mathfrak c = (K:S):\mathfrak c = K:S\mathfrak c = K:\mathfrak c = S\). □

We then have the characterization of GGL rings.

Theorem 5.3

([2]) Suppose thatR is not a Gorenstein ring. Then the following conditions areequivalent.

  1. (1)

    R is aGGL ringwith respect to some\(\mathfrak m\)-primaryideal\(\mathfrak a\)ofR.

  2. (2)

    K/Risa free\(R/\mathfrak c\)-module.

  3. (3)

    S/Risa free\(R/\mathfrak c\)-module.

When this is the case, one necessarily has\(\mathfrak a = \mathfrak c\),and the following assertions hold true.

  1. (i)

    \(R/\mathfrak c\) is a Gorenstein ring.

  2. (ii)

    \(S/R\cong (R/\mathfrak c)^{\oplus \mathrm {r}(R)}\) as an R-module.

The following result is due to [2, 7]. Let us include a brief proof of Assertion (1) for the sake of completeness.

Theorem 5.4

([2, 7]) Suppose thatR is not a Gorenstein ring.Let\(I\in \mathcal {X}_{R}\).Then the following assertions hold true.

  1. (1)

    If \(I\subseteq \mathfrak c\) , then \(I=\mathfrak c\) .

  2. (2)

    IfμR(I) ≠  2,then\(\mathfrak c\subseteq I\).

  3. (3)

    \(\mathfrak c\in \mathcal {X}_{R}\) if and only if R is a GGL ring and S is a Gorenstein ring.

Proof

(1) Let \(I \in \mathcal X_{R}\) and assume that \(I \subseteq \mathfrak c\). We choose an element aI so that I2 = aI. We then have I ≠ (a) and I/I2 is a free R/I-module. Let A = I : I; hence I = aA. On the other hand, because \(\mathfrak c \subseteq I\), by Lemmata 2.1 and 5.2 we have

$$A = R:I \supseteq R : \mathfrak c = S \supseteq K. $$

Claim 2

A is a Gorenstein ring andA/Kisthe canonical module ofR/I.

Proof of Claim 2

Taking the K-dual of the canonical exact sequence 0 → IRR/I → 0, we get the exact sequence

$$0 \to K \overset{\iota}{\to} K:I \to \text{Ext}_{R}^{1}(R/I,K) \to 0, $$

where ι : KK : I denotes the embedding. On the other hand, K : I = A, because

$$I = R:A=(K:K):A = K : KA = K: A $$

(remember that \(K \subseteq A\)). Therefore, since I = K : A is a canonical ideal of A ([11, Korollar 5.14]) and I = aAA, A is a Gorenstein ring, and \(A/K \cong \text {Ext}_{R}^{1}(R/I,K)\).

We consider the exact sequence 0 → (a)/aII/aII/(a) → 0 of R/I-modules. Then, because I = aA, we get the canonical isomorphism between the exact sequences

figureb

of R/I-modules, where A/I is a Gorenstein ring, since A is a Gorenstein ring and I = aA. Therefore, since A/I (≅I/aI) is a flat extension of R/I, R/I is a Gorenstein ring, so that A/KR/I by Claim 2. Consequently, the exact sequence

$$0 \to K/R \to A/R \to A/K \to 0 $$

of R/I-modules is split, whence K/R is a non-zero free R/I-module, because so is A/R (≅I/(a)). Hence, \(\mathfrak c = R:S \subseteq R:K =R:_{R}K = I\), so that \(I = \mathfrak c\). □

Thanks to Theorem 5.4, we get the following.

Theorem 5.5

LetR be aGGL ringand assume thatR is not a Gorenstein ring. Then the following assertions holdtrue.(1) \( \{ I \in \mathcal {X}_{R} | \mathfrak c\subsetneq I \} =\{(a)+\mathfrak c \mid a\in \mathfrak m \text { such that } \mathfrak c =abS \text { for~some }b \in \mathfrak m\}\).

In particular,\(\mathfrak c\in \mathcal {X}_{R}\),once the set\(\{ I \in \mathcal {X}_{R} | \mathfrak c\subsetneq I \}\)isnon-empty.(2) μR(I) = r(R) + 1 forall\(I \in \mathcal X_{R}\)suchthat\(\mathfrak c \subseteq I\).(3) \(\{I \in \mathcal {X}_{R} | \mathfrak c\subseteq I \}=\{I \in \mathcal {X_{R}} | \mu _{R}(I)\neq 2\}\).Therefore, ifR possesses minimal multiplicity, then theset\(\mathcal X_{R}\)istotally ordered, and\(\mathfrak c\)isthe smallest element of\(\mathcal X_{R}\).

Proof

(1) Let us show the first equality. First of all, assume that \(\mathfrak c\in \mathcal {X}_{R}\). Then since \(S=\mathfrak c:\mathfrak c\), for each \(\alpha \in \mathfrak c\), (α) is a reduction of \(\mathfrak c\) if and only if \(\mathfrak c=\alpha S\), so that the required equality follows from Corollary 3.3. Assume that \(\mathfrak c\notin \mathcal {X}_{R}\). Hence, by Theorem 5.4 (3), S is not a Gorenstein ring, because R is a GGL ring. Therefore, since \(\mathfrak c = K:S\) is a canonical module of S (Lemma 5.2 and [11, Korollar 5.14]), we have \(\mathfrak c \ne \alpha S\) for any \(\alpha \in \mathfrak c\), whence the set \(\{(a)+\mathfrak c \mid a \in \mathfrak m\text { such that } abS=\mathfrak c \text { for~some } b \in \mathfrak m\}\) is empty. On the other hand, since \(S=\mathfrak c :\mathfrak c = R:\mathfrak c\) and \(S/R\cong (R/\mathfrak c)^{\oplus \mathrm {r}(R)}\) (see Theorem 5.3 (ii)), the \(\mathfrak m\)-primary ideal \(\mathfrak c\) of R satisfies Condition (C) in Definition 2.2. Therefore, if the set \(\{ I\in \mathcal {X}_{R} \mid \mathfrak c\subsetneq I \}\) is non-empty, then \(\mathfrak c \in \mathcal {X}_{R}\) by Theorem 2.9 (2), because r(R) ≥ 2. Thus, \(\{ I \in \mathcal {X}_{R} \mid \mathfrak c\subsetneq I \}=\emptyset \).(2) By Assertion (1), we may assume \(\mathfrak c\in \mathcal {X}_{R}\). Then, \(\mathfrak c = \alpha S\) for some \(\alpha \in \mathfrak c\), and therefore, \(\mu _{R}(\mathfrak c)=\mathrm {r}(R)+ 1\), since \(\mathfrak c/(\alpha ) \cong S/R\cong (R/\mathfrak c)^{\oplus \mathrm {r}(R)}\). Thus, by Theorem 3.1, \(\mu _{R}(I)=\mu _{R}(\mathfrak c)=\mathrm {r}(R)+ 1\) for every \(I \in \mathcal {X}_{R}\) with \(\mathfrak c \subseteq I\).(3) The assertion follows from Assertion (2) and Theorem 5.4 (3).

The last assertion follows from Assertion (3), since μR(I) = v(R) > 2 for every \(I \in \mathcal X_{R}\) (see Corollary 3.2). □

Combining Theorems 1.3 and 5.5, we have the following.

Corollary 5.6

LetR be aGGL ringand assume thatR is not a Gorenstein ring. Then the following assertions holdtrue.(1) Let\(a_{1},a_{2}, \ldots , a_{n}, b\in \mathfrak m\)(n ≥ 1) andassume that\(\mathfrak c = a_{1}a_{2}{\cdots } a_{n}bS\).We setIi = (a1a2ai) \(+\mathfrak c\)foreach 1 ≤ in.Then\(\mathfrak c \in \mathcal X_{R}\)and\(I_{i} \in \mathcal {X}_{R}\)forall 1 ≤ in,forming a chain\(\mathfrak c \subsetneq I_{n} \subsetneq I_{n-1} \subsetneq {\ldots } \subsetneq I_{1}\)in\(\mathcal X_{R}\).(2) Conversely, let\(I_{1}, I_{2}, \ldots , I_{n} \in \mathcal {X}_{R}\)(n ≥ 1) andassume that\(\mathfrak c \subsetneq I_{n} \subsetneq I_{n-1} \subsetneq {\ldots } \subsetneq I_{1}\).Then\(\mathfrak c \in \mathcal X_{R}\)andthere exist elements\(a_{1},a_{2}, \ldots , a_{n}, b \in \mathfrak m\)suchthat\(\mathfrak c =a_{1}a_{2}{\cdots } a_{n}bS\)and\(I_{i}=(a_{1}a_{2}{\cdots } a_{i})+\mathfrak c\)forall 1 ≤ in.

Concluding this paper, let us note a few examples of GGL rings.

Example 5.7

Let k[[t]] be the formal power series ring over a field k.(1) Let \(H=\left <5,7,9, 13\right >\) denote the numerical semigroup generated by 5,7,9,13 and R = k[[t5, t7, t9, t13]] the semigroup ring of H over k. Then, R is a GGL ring, possessing S = k[[t3, t5, t7]] and \(\mathfrak c = (t^{7},t^{9}, t^{10}, t^{13})\). For this ring R, S is not a Gorenstein ring, and \(\mathcal X_{R}=\emptyset \).(2) Let R = k[[t4, t9, t15]]. Then, R is a GGL ring, possessing S = k[[t3, t4]] and \(\mathfrak c = (t^{9}, t^{12}, t^{15})\) = t9S. For this ring R, \(\mathcal X_{R}=\{\mathfrak c\}\).(3) Let R = k[[t6, t13, t28]]. Then, R is a GGL ring, possessing S = k[[t2, t13]] and \(\mathfrak c = (t^{24},t^{26}, t^{28})=t^{24}S\). For this ring R, the set \(\{I \in \mathcal X_{R} \mid \mathfrak c \subsetneq I\}\) consists of the following families.(i) \(\{(t^{6}+ at^{13}) + \mathfrak c \mid a \in k\}\),(ii) \(\{(t^{12} + at^{13} + bt^{19}) + \mathfrak c \mid a, b \in k\}\), and(iii) \(\{(t^{18}+ at^{25}) + \mathfrak c \mid a \in k\}\).

For each ak, we have a maximal chain

$$\mathfrak c \subsetneq (t^{18}+at^{25})+\mathfrak c \subsetneq (t^{12}+at^{19})+\mathfrak c \subsetneq (t^{6}+at^{13})+\mathfrak c $$

in \(\mathcal X_{R}\). On the other hand, for a, bk such that a≠ 0,

$$\mathfrak c \subsetneq (t^{12}+at^{13}+ bt^{19})+\mathfrak c $$

is also a maximal chain in \(\mathcal X_{R}\).(4) Let \(H = \left <6, 13, 28\right >\). Choose integers 0 < αH and 1 < β so that α∉{6,13,28} and GCD(α, β) = 1. We consider R = k[[tα, t6β, t13β, t28β]]. Then, R is a GGL ring with v(R) = 4 and r(R) = 2. For this ring R, S = k[[tα, t2β, t13β]], and \(\mathfrak c = t^{24\beta }S\). For instance, take α = 12 and β = 5n, where n > 0 and GCD(2, n) = GCD(3, n) = 1. Then, \(\mathfrak c = t^{120n}S =(t^{12})^{10n}S\), so that the set \(\{I \in \mathcal X_{R} \mid \mathfrak c \subsetneq I\}\) seems rather wild, containing chains of large length.

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Funding

The first author was partially supported by the JSPS Grant-in-Aid for Scientific Research (C) 16K05112. The second and third authors were partially supported by Birateral Programs (Joint Research) of JSPS and International Research Supporting Programs of Meiji University.

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Correspondence to Shinya Kumashiro.

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Goto, S., Isobe, R. & Kumashiro, S. The Structure of Chains of Ulrich Ideals in Cohen-Macaulay Local Rings of Dimension One. Acta Math Vietnam 44, 65–82 (2019). https://doi.org/10.1007/s40306-018-0283-y

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Keywords

  • Cohen-Macaulay ring
  • Gorenstein ring
  • Generalized Gorenstein ring
  • Canonical ideal
  • Ulrich ideal
  • Minimal multiplicity

Mathematics Subject Classification (2010)

  • 13H10
  • 13H15