Abstract
This paper studies Ulrich ideals in onedimensional CohenMacaulay 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 onedimensional CohenMacaulay 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 twodimensional rational singularities are closely studied with a concrete classification. However, in contrast to the existing research on Ulrich ideals, the theory pertaining to the onedimensional 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 onedimensional 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 onedimensional CohenMacaulay 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 onedimensional case, we would like to state the general definition, in the case of any arbitrary dimension. Let \((R,\mathfrak m)\) be a CohenMacaulay 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 = (a_{1}, a_{2},…, a_{d}) of R as a reduction. We say that I is an Ulrichideal of R, if the following conditions are satisfied.

(1)
I ≠ Q,

(2)
I^{2} = QI, and

(3)
I/I^{2} is a free R/Imodule.
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 CohenMacaulay ring and a(gr_{I}(R)) = 1 − d, where a(gr_{I}(R)) denotes the ainvariant of gr_{I}(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/Imodule ([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 CohenMacaulay type of R, and let \(\text {Syz}_{R}^{i}(M)\) denote, for each integer i ≥ 0 and for each finitely generated Rmodule M, the ith syzygy module of M in its minimal free resolution.
Theorem 1.2
([4, 7]) Let I be an Ulrich ideal of a CohenMacaulay local ringR of dimensiond ≥ 0 andsett = n − d (> 0),where n denotes the number of elements in a minimal system of generators of I.Let
be a minimal free resolution ofR/I. Then r(R) = t⋅r(R/I) and the following assertions hold true.

(1)
I(∂_{i}) = I for i ≥ 1.

(2)
For i ≥ 0, \(\beta _{i} = \left \{\begin {array}{lll} t^{id}{\cdot }(t + 1)^{d} & (i \ge d),\\ \binom {d}{i}+ t{\cdot }\beta _{i1} & (1 \le i \le d),\\ 1 & (i = 0). \end {array}\right .\)

(3)
\(\text {Syz}_{R}^{i + 1}(R/I) \cong [\text {Syz}_{R}^{i}(R/I)]^{\oplus t}\)fori ≥ d.

(4)
Fori ∈ ℤ, \(\text {Ext}_{R}^{i}(R/I, R)\cong \left \{\begin {array}{lll} (0) & (i < d),\\ (R/I)^{\oplus t} & (i = d),\\ (R/I)^{\oplus (t^{2}1){\cdot }t^{i  (d + 1)} } & (i > d). \end {array}\right .\)
Here, I(∂_{i}) denotes the ideal ofRgenerated by the entries of the matrix∂_{i}, and β_{i} = rank_{R}F_{i}.
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 twodimensional 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 CohenMacaulay local ring with \(\dim R = 1\). Our main targets are chains \(I_{n} \subsetneq I_{n1} \subsetneq {\cdots } \subsetneq I_{1}\) (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 Ralgebra A. Secondly, let J be an Ulrich ideal of R and assume that \(I \subsetneq J\). Then, we will show that μ_{R}(J) = μ_{R}(I), where μ_{R}(∗) 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 CohenMacaulay 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 = a_{1}a_{2}⋯a_{n}A.For 1 ≤ i ≤ n, letI_{i} = (a_{1}a_{2}⋯a_{i}) + I. Then eachI_{i}is an Ulrichideal ofR and
(2) Conversely, letI_{1}, I_{2},…, I_{n} (n ≥ 2) be Ulrich ideals ofRand suppose that
We setI = I_{n} and A = I : I. Then there exist elements\(a_{1},a_{2}, \ldots , a_{n}\in \mathfrak m\)such thatI = a_{1}a_{2}⋯a_{n}AandI_{i} = (a_{1}a_{2}⋯a_{i}) + Ifor all 1 ≤ i ≤ n − 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 J^{2} = 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 CohenMacaulay local ring B of dimension one and there is a onetoone 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 onedimensional CohenMacaulay local ring could be understood via the correspondence, and the relationship between the structure of CohenMacaulay 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 CohenMacaulay 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 Rmodule M, let μ_{R}(M) (resp. ℓ_{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
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 a ∈ I with I^{2} = aI. We set A = I : I and
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 I^{2} = aI; hence I = aA. We then have the following.
Proposition 2.1
IfI = (a) : _{R}I,thenA = R : IandI = R : A,whenceR : (R : I) = I.
Proof
Notice that I = (a) : _{R}I = (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 Rmodule 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 a ∈ I so that I^{2} = aI. Then, I/(a)≅(R/I)^{⊕t} as an R/Imodule, where t = μ_{R}(I) − 1 > 0 ([4, Lemma 2.3]). Therefore, I = (a) : _{R}I, so that I satisfies the hypothesis in Proposition 2.1, whence A = R : I. Notice that A/R≅I/(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 {f_{i}}_{1≤i≤t} of A so that
Therefore, the images \(\{\overline {f_{i}}\}_{1 \le i \le t}\) of {f_{i}}_{1≤i≤t} in A/R form a free basis of the R/Imodule A/R. We then have the following.
Lemma 2.4
\(aA\cap R\subseteq (a)+I\)foralla ∈ R.
Proof
Let x ∈ aA ∩ R and write x = ay with y ∈ A. We write \(y=c_{0} + {\sum }_{i = 1}^{t}c_{i}f_{i}\) with c_{i} ∈ R. Then, ac_{i} ∈ I for 1 ≤ i ≤ t, since \(x = ac_{0} + {\sum }_{i = 1}^{t} (ac_{i})f_{i} \in R\). Therefore, (ac_{i})f_{i} ∈ IA = I for all 1 ≤ i ≤ t, so that x ∈ (a) + I as claimed. □
Corollary 2.5
Let J be an\(\mathfrak m\)primaryideal ofR and assume that J contains an elementb ∈ JsuchthatJ^{2} = bJandJ = (b) : _{R}J.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 b ∈ J such that J^{2} = bJ and J = (b) : _{R}J. 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
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
With this setting, we have the following.
Lemma 2.6
The following assertions hold true.

(1)
\(A/B\cong (B/\mathfrak a)^{\oplus t}\)asa Bmodule.

(2)
\(\mathfrak a\cap R=I:_{R}J\) .

(3)
ℓ_{R}([I : _{R}J]/I) = ℓ_{R}(R/J).

(4)
I = [b⋅(I : _{R}J)]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 f_{i} in A/B. Let {b_{i}}_{1≤i≤t} 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 bb_{i} ∈ R for all 1 ≤ i ≤ t, we have bb_{i} ∈ I, so that \(b_{i}\in \frac {I}{b}=\mathfrak a\). Hence \(A/B\cong (B/\mathfrak a)^{\oplus t}\) as a Bmodule.
(2) This is standard, because J = (b) + I and \(\mathfrak a = \frac {I}{b}\).
(3) Since J/I = [(b) + I]/I≅R/[I : _{R}J], we get
(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 x ∈ I. Then \(x \in J=bB \subseteq bA\). We write \(x=b[c_{0}+{\sum }_{i = 1}^{t}c_{i}f_{i}]\) with c_{i} ∈ R. Then bc_{i} ∈ I for 1 ≤ i ≤ t since x ∈ R, so that (bc_{i})f_{i} ∈ I for all 1 ≤ i ≤ t, because I is an ideal of A. Therefore, bc_{0} ∈ I, since \(x=bc_{0}+{\sum }_{i = 1}^{t}(bc_{i})f_{i} \in I\). Consequently, c_{i} ∈ I : _{R}b = I : _{R}J for all 0 ≤ i ≤ t, so that x ∈ [b⋅(I : _{R}J)]A as wanted.
□
Corollary 2.7
J/(b)≅([I : _{R}J]/I)^{⊕t}asanRmodule. Henceℓ_{R}(J/(b)) = t⋅ℓ_{R}(R/J).
Proof
We consider the exact sequence
of Rmodules. By Lemma 2.6 (1), A/B is a free \(B/\mathfrak a\)module of rank t, possessing the images of {f_{i}}_{1≤i≤t} in A/B as a free basis. Because A/R is a free R/Imodule of rank t, also possessing the images of {f_{i}}_{1≤i≤t} in A/R as a free basis, we naturally get an isomorphism between the following two canonical exact sequences;
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
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)
\(J\in \mathcal {X}_{R}\) .

(2)
μ_{R}([I : _{R}J]/I) = 1.

(3)
[I : _{R}J]/I≅R/JasanRmodule.
When this is the case,μ_{R}(J) = t + 1.
Proof
The implication (3) ⇒ (2) is clear, and the reverse implication follows from the equality ℓ_{R}([I : _{R}J]/I) = ℓ_{R}(R/J) of Lemma 2.6 (3).
(1) ⇒ (3) Suppose that \(J\in \mathcal {X}_{R}\). Then J/(b) is R/Jfree, so that by Corollary 2.7, [I : _{R}J]/I is a free R/Jmodule, whence [I : _{R}J]/I≅R/J by Lemma 2.6 (3).
(3) ⇒ (1) We have J/(b)≅([I : _{R}J]/I)^{⊕t}≅(R/J)^{⊕t} by Corollary 2.7, so that by Definition 1.1, \(J\in \mathcal {X}_{R}\) with μ_{R}(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,\(I\in \mathcal {X}_{R}\)andμ_{R}(I) = μ_{R}(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 : _{R}J = (c) + I. Therefore, by Lemma 2.6 (4) we have
whence I = bcA by Nakayama’s lemma. Let a = bc. Then I^{2} = (aA)^{2} = a⋅aA = 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 μ_{R}(I) = t + 1. Therefore, μ_{R}(I) = μ_{R}(J), because μ_{R}(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 a ∈ I so that I^{2} = aI; hence \(A=\frac {I}{a}\). We then have μ_{R}(I) = t + 1, since I/(a)≅A/R≅(R/I)^{⊕t}. Consequently, since J = (b) + I, we get
On the other hand, we have μ_{R}(J/(b)) = t⋅μ_{R}([I : _{R}J]/I), because J/(b)≅([I : _{R}J]/I)^{⊕t} by Corollary 2.7. Hence
so that μ_{R}([I : _{R}J]/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\(I\subsetneq J\).Chooseb ∈ JsothatJ^{2} = bJ.Then the following assertions hold true.
(1) J = (b) + I.
(2) μ_{R}(J) = μ_{R}(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\(I \in \mathcal X_{R}\).Thenμ_{R}(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) = [μ_{R}(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
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, c∉I, 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 c∉R : A = I (see Lemma 2.1). Hence \((b)\subsetneq J\). Because I^{2} = bcI, we have J^{2} = bJ + I^{2} = bJ + bcI = bJ. Let us check that J/(b) is a free R/Jmodule. Let {f_{i}}_{1≤i≤t} (t = μ_{R}(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/Imodule A/R (remember that I satisfies Condition (C) of Definition 2.2). We then have
Let {c_{i}}_{1≤i≤t} 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 c_{i}c ∈ I = bcA, so that c_{i} ∈ bA ∩ R for all 1 ≤ i ≤ t. Therefore, because \(bA\cap R\subseteq (b)+I=J\) by Lemma 2.4, we get c_{i} ∈ J, whence J/(b)≅(R/J)^{⊕t}. Thus, \(J = (b)+I \in \mathcal X_{R}\). □
The equality μ_{R}(I) = μ_{R}(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[[X_{1}, X_{2}, X_{3}, X_{4}]] 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 x_{i} denote the image of X_{i} in R for each i = 1,2,3,4. Then, (x_{1}, x_{2}, x_{3}) and (x_{1}, x_{4}) 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 \(a_{1}, a_{2}, \ldots , a_{n1} \in \mathfrak m\) such that (a_{1}a_{2}⋯a_{n− 1}) is a reduction of I_{n− 1} and I_{i} = (a_{1}a_{2}⋯a_{i}) + I_{n− 1} for all 1 ≤ i ≤ n − 2. Now apply Theorem 3.1 to the chain \(I_{n} \subsetneq I_{n1}\). We then have I_{n− 1} = (a_{1}a_{2}⋯a_{n− 1}) + I_{n} together with one more element \(a_{n}\in \mathfrak m\) so that (a_{1}a_{2}⋯a_{n− 1})⋅a_{n}A = I_{n}. Hence
for all 1 ≤ i ≤ n − 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\(I\subsetneq J\).Letb ∈ JsuchthatJ^{2} = 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 CohenMacaulay 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 = μ_{R}(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 b ∈ J. Therefore, \(\frac {I}{b}=I:J\), because
(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 A ≠ B. 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 μ_{R}(L) = μ_{R}(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 : _{R}J = (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
where the first equality follows from Lemma 2.6 (2). Thus
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
Therefore, αB is a reduction of \(\mathfrak n\), so that
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\(I\subsetneq J\).Letα ∈ J.ThenJ = (α) + Iifand only ifJ^{2} = αJ.
Proof
It suffices to show the only if part. Suppose J = (α) + I. We set A = I : I, B = J : J, and choose b ∈ J so that J^{2} = 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 J^{2} = 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 b ∈ J such that J = (b) + I. Then J^{2} = 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 b ∈ J, \(\mathfrak a:J \subseteq \frac {\mathfrak a}{b}\). On the other hand, since
by Lemma 2.1, we get
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 = ba_{1}a_{2}A 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 a_{1}a_{2}B 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 CohenMacaulay 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 aβ ∈ L. Let us write β = c + y with \(c\in \mathfrak m\) and y ∈ L. We then have ac = aβ − ay ∈ L and \(yA \subseteq L\), so that because
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
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 μ_{R}(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 b ∈ J with J^{2} = 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)
ℓ_{R}(J/I) = 1.

(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)
(bα) isa reduction of I. IfI = (bα) + (x_{2}, x_{3},…, x_{v}),thenJ = (b, x_{2}, x_{3},…, x_{v}).
Proof
By Theorem 1.4, we have the onetoone correspondence
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
and \(R/\mathfrak m \cong B/\mathfrak n\) by Theorem 3.5 (2), we get ℓ_{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 ℓ_{R}(I_{i}/I_{i+ 1}) = 1 for all 1 ≤ i ≤ ℓ − 1, where ℓ = ℓ_{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 ≤ ℓ.
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
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 ≤ ℓ and 1 ≤ j ≤ n. We may assume ℓ ≤ n. Then Lemma 4.1 (2) shows that I_{i} = J_{i} for all 1 ≤ i ≤ ℓ, 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[[t^{3}, t^{7}]] 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
connectingIwith\(\mathfrak m\)such that\(I_{i}\in \mathcal {X}_{R}\)for every 1 ≤ i ≤ ℓ = ℓ_{R}(R/I). We setB_{0} = RandB_{i} = I_{i} : I_{i} for 1 ≤ i ≤ ℓand let\(\mathfrak n_{i}=\mathrm {J}(B_{i})\)denote the Jacobson radical ofB_{i}for each 0 ≤ i ≤ ℓ. Then we obtain a tower
of birational finite extensions ofRand furthermore have the following.
(1) (α^{i}) is a reduction ofI_{i}for every 1 ≤ i ≤ ℓ.
(2) \(B_{i}=\mathfrak n_{i1}:\mathfrak n_{i1}\)for every 1 ≤ i ≤ ℓ.
(3) For 0 ≤ i ≤ ℓ − 1, \((B_{i},\mathfrak n_{i})\)is a local ring with v(B_{i}) = e(B_{i}) = e(R) > 1 and\(\mathfrak {n_{i}^{2}}=\alpha \mathfrak n_{i}\).
(4) Choosex_{2}, x_{3},…, x_{v} ∈ Iso thatI = (α^{ℓ}, x_{2},…, x_{v}). ThenI_{i} = (α^{i}, x_{2}, x_{3},…, x_{v}) for every 1 ≤ i ≤ ℓ. 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 = I_{i}for some 1 ≤ i ≤ ℓ.
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 ℓ. □
Corollary 4.6
Suppose that there exists a minimal element I in\(\mathcal {X}_{R}\).Then\(\sharp \mathcal {X}_{R}=\ell <\infty \)withℓ = ℓ_{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 I_{i}’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 twodimensional regular local ring. Let \(\mathfrak n=(X, Y)\) and consider the ring A = S/(Y^{2}). Then v(A) = e(A) = 2 and
where x, y denote the images of X, Y in A, respectively. Hence \(\sharp \mathcal {X}_{A}=\infty \).
Proof
Let I_{n} = (x^{n}, 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 I_{n} = (x^{n}) : _{A}y, we get I_{n}/(x^{n})≅A/I_{n}. Therefore, \(I_{n}\in \mathcal {X}_{A}\) for all n ≥ 1. To see that \(\mathcal X_{A}\) consists of these ideals I_{n}’s, let \(I\in \mathcal {X}_{A}\) and set ℓ = ℓ_{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_{ℓ}, because ℓ_{A}(A/I_{ℓ}) = ℓ. 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≅K_{R} as an Rmodule 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≅K_{R}asanRmodule. 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 μ_{R}(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 onetoone 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 twodimensional 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) = (Y^{2}) 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 = V^{⊕n} (n ≥ 2). Let t be a regular parameter of V . Then for each n ≥ 1, I_{n} = (t^{n}) × 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 twodimensional CohenMacaulay 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[[X_{0}, X_{1},…, X_{n}]] (n ≥ 3) be the formal power series ring over a field k. Let ℓ ≥ 1 be an integer and consider the 2 × n matrix
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 CohenMacaulay local ring of dimension two, possessing minimal multiplicity. For this ring, we have
where x_{i} denotes the image of X_{i} in R for each 0 ≤ i ≤ n. 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 = X − Y.) We consider the ring R = S/(Z^{2} − FGH). Then R is a twodimensional CohenMacaulay 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 CohenMacaulay local ring with \(d= \dim R \ge 0\), possessing the canonical module K_{R}. We say that R is a generalized Gorenstein local (GGL for short) ring, if one of the following conditions is satisfied.

(1)
R is a Gorenstein ring.

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

(i)
C is an Ulrich Rmodule with respect to \(\mathfrak a\) and

(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 onedimensional case, here let us summarize a few results on GGL rings of dimension one. Suppose that \((R,\mathfrak m)\) is a CohenMacaulay local ring of dimension one, admitting a fractional canonical ideal K. Hence, K is an Rsubmodule of \(\overline {R}\) such that K≅K_{R} as an Rmodule 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 = K^{n} 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)
R is aGGL ringwith respect to some\(\mathfrak m\)primaryideal\(\mathfrak a\)ofR.

(2)
K/Risa free\(R/\mathfrak c\)module.

(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.

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

(ii)
\(S/R\cong (R/\mathfrak c)^{\oplus \mathrm {r}(R)}\) as an Rmodule.
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)
If \(I\subseteq \mathfrak c\) , then \(I=\mathfrak c\) .

(2)
Ifμ_{R}(I) ≠ 2,then\(\mathfrak c\subseteq I\).

(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 a ∈ I so that I^{2} = aI. We then have I ≠ (a) and I/I^{2} is a free R/Imodule. 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
□
Claim 2
A is a Gorenstein ring andA/Kisthe canonical module ofR/I.
Proof of Claim 2
Taking the Kdual of the canonical exact sequence 0 → I → R → R/I → 0, we get the exact sequence
where ι : K → K : I denotes the embedding. On the other hand, K : I = A, because
(remember that \(K \subseteq A\)). Therefore, since I = K : A is a canonical ideal of A ([11, Korollar 5.14]) and I = aA≅A, A is a Gorenstein ring, and \(A/K \cong \text {Ext}_{R}^{1}(R/I,K)\).
We consider the exact sequence 0 → (a)/aI → I/aI → I/(a) → 0 of R/Imodules. Then, because I = aA, we get the canonical isomorphism between the exact sequences
of R/Imodules, 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/K≅R/I by Claim 2. Consequently, the exact sequence
of R/Imodules is split, whence K/R is a nonzero free R/Imodule, 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\(\{ I \in \mathcal {X}_{R}  \mathfrak c\subsetneq I \}\)isnonempty.(2) μ_{R}(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 nonempty, 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 μ_{R}(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 setI_{i} = (a_{1}a_{2}⋯a_{i}) \(+\mathfrak c\)foreach 1 ≤ i ≤ n.Then\(\mathfrak c \in \mathcal X_{R}\)and\(I_{i} \in \mathcal {X}_{R}\)forall 1 ≤ i ≤ n,forming a chain\(\mathfrak c \subsetneq I_{n} \subsetneq I_{n1} \subsetneq {\ldots } \subsetneq I_{1}\)in\(\mathcal X_{R}\).(2) Conversely, let\(I_{1}, I_{2}, \ldots , I_{n} \in \mathcal {X}_{R}\)(n ≥ 1) andassume that\(\mathfrak c \subsetneq I_{n} \subsetneq I_{n1} \subsetneq {\ldots } \subsetneq I_{1}\).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 ≤ i ≤ n.
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[[t^{5}, t^{7}, t^{9}, t^{13}]] the semigroup ring of H over k. Then, R is a GGL ring, possessing S = k[[t^{3}, t^{5}, t^{7}]] 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[[t^{4}, t^{9}, t^{15}]]. Then, R is a GGL ring, possessing S = k[[t^{3}, t^{4}]] and \(\mathfrak c = (t^{9}, t^{12}, t^{15})\) = t^{9}S. For this ring R, \(\mathcal X_{R}=\{\mathfrak c\}\).(3) Let R = k[[t^{6}, t^{13}, t^{28}]]. Then, R is a GGL ring, possessing S = k[[t^{2}, t^{13}]] 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 a ∈ k, we have a maximal chain
in \(\mathcal X_{R}\). On the other hand, for a, b ∈ k such that a≠ 0,
is also a maximal chain in \(\mathcal X_{R}\).(4) Let \(H = \left <6, 13, 28\right >\). Choose integers 0 < α ∈ H and 1 < β ∈ ℤ so that α∉{6,13,28} and GCD(α, β) = 1. We consider R = k[[t^{α}, t^{6β}, t^{13β}, t^{28β}]]. Then, R is a GGL ring with v(R) = 4 and r(R) = 2. For this ring R, S = k[[t^{α}, t^{2β}, t^{13β}]], 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 GrantinAid 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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Goto, S., Isobe, R. & Kumashiro, S. The Structure of Chains of Ulrich Ideals in CohenMacaulay Local Rings of Dimension One. Acta Math Vietnam 44, 65–82 (2019). https://doi.org/10.1007/s403060180283y
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Keywords
 CohenMacaulay ring
 Gorenstein ring
 Generalized Gorenstein ring
 Canonical ideal
 Ulrich ideal
 Minimal multiplicity
Mathematics Subject Classification (2010)
 13H10
 13H15