One-dimensional stable rings

arXiv:1603.02173v1 [] 7 Mar 2016

Bruce Olberding
Department of Mathematical Sciences, New Mexico State University, Las Cruces, NM
88003-8001

Abstract
A commutative ring R is stable provided every ideal of R containing a nonzerodivisor is projective as a module over its ring of endomorphisms. The class
of stable rings includes the one-dimensional local Cohen-Macaulay rings of multiplicity at most 2, as well as certain rings of higher multiplicity, necessarily
analytically ramified. The former are important in the study of modules over
Gorenstein rings, while the latter arise in a natural way from generic formal
fibers and derivations.
We characterize one-dimensional stable local rings in several ways. The
characterizations involve the integral closure R of R and the completion of R in
a relevant ideal-adic topology. For example, we show: If R is a reduced stable
ring, then there are exactly two possibilities for R: (1) R is a Bass ring, that
is, R is a reduced Noetherian local ring such that R is finitely generated over
R and every ideal of R is generated by two elements; or (2) R is a bad stable
domain, that is, R is a one-dimensional stable local domain such that R is not
a finitely generated R-module.
Keywords: local ring, Cohen-Macaulay ring, stable ideal, two-generated ideal
2010 MSC: 13H10, 13E15, 13A15

1. Introduction
The class of stable rings has a long history dating back at least to the “Ubiquity” paper of Bass, where he showed that rings for which every ideal can be
generated by two elements are stable [6, Corollary 7.3]. Following Lipman [21]
and Sally and Vasconcelos [39, 40], we define an ideal I of a commutative ring
R to be stable if I is projective over its ring of endomorphisms. A ring R is
stable if every regular ideal of R (that is, every ideal containing a nonzerodivisor) is stable1 . Independently of Bass, Lipman [21] studied stable ideals in
one-dimensional semilocal Cohen-Macaulay rings and showed how stability was
Email address: olberdin@nmsu.edu (Bruce Olberding)
definition of a stable ring differs slightly from Sally and Vasconcelos [39, 40] in that
we require only that every regular ideal is stable, not that every ideal is stable. However, for
a reduced one-dimensional Cohen-Macaulay ring, the two definitions agree [35, p. 260].
1 Our

Preprint submitted to Elsevier

March 8, 2016

reflected in terms of invariants of the ring such as multiplicity, embedding dimension and the Hilbert function. (We recall the definitions of these terms in
Section 2.1.) The terminology of “stable” ideal originates with Lipman; it reflects the stabilization of a certain chain of infinitely near local rings. Lipman
in turn was motivated to introduce these ideals as a way to unify ideas from Arf
and Zariski involving singularities of plane curves, and to produce the largest
ring between a one-dimensional local Cohen-Macaulay ring R and its integral
closure having the same multiplicity sequence as R [21, Corollary 3.10].
Sally and Vasconcelos, motivated by the work of Bass and Lipman, studied
stable Noetherian local rings in detail in [39, 40], and showed that a reduced
local Cohen-Macaulay ring with finite normalization is stable if and only if every
ideal of R can be generated by two elements, thus substantiating a conjecture of
Bass and proving a partial converse to his theorem mentioned above. Reduced
Noetherian local rings having finite normalization and every ideal generated
by two elements are known in the literature as Bass rings because of their
importance in Bass’s article [6]. Sally and Vasconcelos [40, Example 5.4] gave
the first example of a Noetherian stable domain R of multiplicity > 2 (and
hence without finite normalization). They used a construction of Ferrand and
Raynaud that exhibits R as the preimage of a derivation over a specific field of
characteristic 2. Heinzer, Lantz and Shah [15, (3.12)] showed that this technique
can be modified to produce over this same field of characteristic 2 for each
e > 2 an analytically ramified Noetherian local stable domain of multiplicity e.
More recently, analytically ramified Noetherian stable local rings have proved
to be useful theoretical tools for classifying the rank one discrete valuation
rings (DVRs) that arise as the normalization of an analytically ramified onedimensional Noetherian local domain, as well as for describing properties of the
generic formal fiber of a Noetherian local domain [31, 32]; see Remark 2.15 and
Theorem 2.16.2
The main goal of this article is to tie all these results together. The Noetherian assumption is not essential to most of our arguments, and thus we use general ideal-theoretic methods to work in a setting in which R is a one-dimensional
local ring with regular maximal ideal.3 For example, we show in Theorem 4.2
that R is stable if and only if the integral closure R is a Dedekind ring with
the property that every R-submodule containing R is a ring. This generalizes
a result of Rush [36, Theorem 2.4] to rings that are not necessarily Noetherian.
A finer classification is possible by distinguishing when the integral closure of
R is a finitely generated R-module: We show in Theorems 4.5 and 4.7 that R is
e of R
stable and has finite integral closure if and only if a suitable completion R
is a Bass ring, while R is stable without finite integral closure if and only if there
e such that P 2 = 0 and R/P
e
is a nonzero prime ideal P of R
is a DVR. From
2 The definitions of normalization, finite normalization and analytically ramified rings appear at the end of this section and in Section 2.1.
3 We use the terminology of “local ring” for a ring with a unique maximal ideal. In particular, we do not require a local ring to be Noetherian.

2

this we deduce in Corollary 4.9 that R is reduced and stable if and only if R is
a Bass ring or R is a stable domain without finite integral closure. The rings
in the first class, the Bass rings, are classical, while the second class, the stable
domains without finite integral closure, have a rather transparent structure, as
indicated by Theorem 2.13.
We also revisit the two-generator property in Section 5. For R a onedimensional local ring with regular maximal ideal M , we show that M n is
two-generated (i.e., can be generated by two elements), for some n ≥ 2, if and
only if every M -primary ideal of R is two-generated. From this we deduce:
Theorem 5.6. Let R be a Cohen-Macaulay ring with maximal ideal M . Then
these statements are equivalent.
(1) M n is two-generated, for some n ≥ 2.
(2) R has Krull dimension 1 and multiplicity at most 2.
(3) R is one of the following:
(3a) a Bass ring;
(3b) a one-dimensional analytically ramified stable domain; or
(3c) a ring containing a nonzero principal prime ideal P such that P 2 = 0
and R/P is a DVR.
We also generalize a theorem of Greither to rings that are not a priori
Noetherian: If R is a local ring with regular maximal ideal and the integral
closure of R is a Dedekind ring generated by two elements as an R-module,
then R is a Bass ring (Theorem 5.8).
Conventions. All rings are commutative with 1. The Jacobson radical of
the ring R is denoted Jac R. The total ring of quotients of R is denoted Q(R).
The integral closure of R is the integral closure of R in Q(R) and is denoted R.
We say that R has finite integral closure if R is a finitely generated R-module.
When R is reduced, R is the normalization of R, and R has finite normalization
if R is a finitely generated R-module. We write (R, M ) for a local ring R with
maximal ideal M .
2. Preliminaries and background
2.1. One-dimensional Cohen-Macaulay rings
We collect in this subsection properties of one-dimensional Noetherian local
rings that are needed in later sections. Recall that a local ring (R, M ) is anb has no nontrivial
alytically unramified provided that the M -adic completion R
nilpotent elements.
Proposition 2.1. ([11, Corollary 1.21] and [24, pp. 263-264]) Let (R, M ) be a
local ring with maximal ideal M and integral closure R. Then

3

(1) If R is reduced and one-dimensional, and there exists t > 0 such that M t
is finitely generated, then R is Noetherian.
(2) If R is Noetherian, M is regular and R is finitely generated as an Rmodule, then R is reduced.
(3) If R is Noetherian and analytically unramified, then R is finitely generated
as an R-module.
The embedding dimension of a Noetherian local ring R is the minimal number
of elements needed to generate its maximal ideal M . For sufficiently large values
of n, the length of R/M n is a polynomial in n of degree d = dim R. The product
of d! and the leading coefficient of this polynomial is the multiplicity of R. See
[24] for more information on multiplicity.
Proposition 2.2. ([41, Propositions 11.1.10 and 11.2.1], [38, Theorem 1.1,
p. 49] and [23, Theorem 10.2, p. 90]) Let (R, M ) be a one-dimensional CohenMacaulay ring with maximal ideal M and integral closure R. Then
(1) If M 2 = mM for some m ∈ M , then the embedding dimension and multiplicity of R agree. In particular, R has minimal mulitplicity.
(2) Every ideal of R can be generated by e or fewer elements, where e is the
multiplicity of R.
(3) R has finite integral closure if and only if R is analytically unramified.
2.2. Completions of a one-dimensional local ring
For a one-dimensional local ring (R, M ) that is not Noetherian, the M -adic
b of R can be too coarse to contain useful information about R. For
completion R
b = R/M . For this reason we work with another
example, if M = M 2 , then R
completion that agrees with the M -adic completion of R if R is one-dimensional
and Noetherian.
Notation 2.3. Let R be a one-dimensional local ring with regular maximal
e the mR-adic
ideal M , and let m be a nonzerodivisor in M . We denote by R
b the M -adic completion of R. Since R has Krull
completion of R and by R
dimension 1 and m is a nonzerodivisor, mR is M -primary, and hence the ring
e is independent of the choice of nonzerodivisor m ∈ M .
R

Remark 2.4. Let R be a one-dimensional local ring with regular maximal ideal
e can be identified
M . If (i) M is finitely generated, or (ii) M is stable, then R
b
with R. In case (i) some power of M is contained in a principal regular ideal
mR, and so the mR-adic and M -adic topologies agree on R. For case (ii),
Lemma 3.4(6) implies that M 2 = mM for some m ∈ M . Thus M 2 ⊆ mR and
so again the two topologies coincide.
Proposition 2.5. ([23, Theorem 2.1(3), p. 11], [23, Theorem 2.1(4), p. 11] and
[23, Theorem 2.8(1), p. 19]) Let R be a one-dimensional local ring with regular
maximal ideal M .
4

e is a regular ideal of R.
e
(1) If I is a regular ideal of R, then I R

e denotes the canonical mapping, then R
e = λ(R) + xR
e for all
(2) If λ : R → R
nonzerodivisors x ∈ R.
(3) If I is a regular ideal of R, then the mapping λ in (2) induces an isomore R
e and I = λ−1 (I R).
e
phism R/I ∼
= R/I
Proof. Matlis proves the statements in the proposition for what he calls the “Rcompletion” H of a commutative ring R; that is, H = lim R/I, where I ranges
←−
over the regular ideals of R. Since in our case R is local of Krull dimension one
e
and the maximal ideal M of R is regular, the R-completion of R is simply R.
e
Thus the cited references apply also to R.
2.3. Quadratic extensions
Rush [36, Proposition 2.1] has shown that, if R is a stable ring, every Rsubmodule of R that contains R is a ring. In this section we review some
properties of extensions that share this property.
Definition 2.6. An extension R ⊆ S of rings is quadratic if xy ∈ xR + yR + R
for all x, y ∈ S; equivalently, every R-module between R and S is a ring.
Remarks 2.7. (1) Quadratic extensions were considered by Handelman [14]
and Rush [36], and more recently in [31, 32, 33].
(2) If an extension R ⊆ S of rings is quadratic, then, for every x ∈ S, we
have x2 ∈ xR + R; that is, every x ∈ S is a root of a monic polynomial of degree
at most 2 with coefficients in R. Thus every quadratic extension is an integral
extension.
(3) If R ⊆ S ⊆ T is an extension of rings and R ⊆ T is quadratic, then
R ⊆ S is quadratic. Also, if I is an ideal of S that is also an ideal of R, then
R/I ⊆ S/I is quadratic if and only if R ⊆ S is quadratic.
A key tool for analyzing quadratic rings is Handelman’s classification of
finite-dimensional algebras that are quadratic extensions of a base field.
Lemma 2.8. (Handelman [14, Lemma 5]) Let F be a field and let S be a finitedimensional F -algebra such that F ⊆ S is a quadratic extension. Then S is
isomorphic as an F -algebra to one of the following:
(i) F ,
(ii) a field extension of F of degree 2,
(iii) a local ring with square zero maximal ideal and residue field isomorphic to
F,
(iv) F × F , or
(v) F × F × F , in which case F = F2 .

5

Remark 2.9. If K/F is a finite field extension that is a quadratic extension
of rings in the sense of Definition 2.6, then from Lemma 2.8 it follows that
[K : F ] ≤ 2. Thus an extension of fields that is quadratic in the sense of
Definition 2.6 agrees with the usual notion of a quadratic field extension.
The following proposition is deduced in [34] from Handelman’s lemma.
Proposition 2.10. [34, Proposition 3.3] If R ⊆ S is a quadratic extension,
then there are at most three prime ideals of S lying over any prime ideal of R.
2.4. Stable rings
Noetherian stable rings have been studied in a number of contexts; cf. [2,
6, 9, 12, 14, 15, 21, 35, 39, 40, 42, 43]. Also, the related class of archimedean
stable domains have been investigated recently by Gabelli and Roitman [10]
and shown to generalize features of one-dimensional stable domains, and hence
Noetherian stable domains, to a wider setting.
The following theorem summarizes some of the results discussed in the introduction.
Theorem 2.11. (cf. [6, Corollary 7.3], [9], [35, Theorem 1.4] and [39, Theorem
2.4]) Let R be a reduced Cohen-Macaulay ring. If every ideal of R can be generated by 2 elements (equivalently, R has Krull dimension ≤ 1 and multiplicity
≤ 2), then R is a stable ring. Conversely, if R is a reduced stable ring with
finite normalization, then R is a Bass ring.
Thus in the case that R is an analytically unramified local Cohen-Macaulay
ring, R is stable if and only if every ideal of R can be generated by two elements.
The analytically ramified case is more complicated, as illustrated by results
throughout the paper. We single out in the following definition a specific class
of such rings.
Definition 2.12. A domain R is a bad stable domain if R is a one-dimensional
stable local domain without finite normalization.
Bad stable domains were studied in [31, 32]. The appellation “bad” here is
borrowed from the title of Nagata’s appendix in [25], “Examples of bad Noetherian rings,” and refers to the fact that the domain does not have finite normalization.
Theorem 2.13. [31, Proposition 2.1, Theorems 3.4 and 4.2 and Corollaries 3.5
and 4.3] The following are equivalent for a local domain R with quotient field
F.
(1) R is a bad stable domain.
(2) R is a DVR, R ⊆ R is a quadratic extension and R/R is a divisible Rmodule.
e as in Notation 2.3, there is a nonzero prime ideal P of R
e such
(3) With R
2
e
that P = 0 and R/P
is a DVR.
6

L
(4) R is a DVR and R/R ∼
= i∈I F/R for some index set I.
(5) R is a DVR and R/R is a direct sum of divisible Artinian uniserial Rmodules.
Moreover, R is a bad Noetherian stable domain of multiplicity e if and only if
the cardinality of the set I in (4) is e − 1.
Theorem 2.13(3) and a theorem of Lech’s can be used to guarantee the
existence of bad stable domains with prescribed completions:
Corollary 2.14. The following are equivalent for a Noetherian local ring (R, M )
that is complete in the M -adic topology.
(1) The ring R is the completion of a bad Noetherian stable domain.
(2) No nonzero integer of R is a zerodivisor and there is a prime ideal P of
R such that P 2 = 0 and R/P is a DVR.
Proof. That (1) implies (2) follows from Theorem 2.13(3). Conversely, assume
(2). By a theorem of Lech [20, Theorem 1], the fact that M 6= 0, M is not an associated prime of R, and no nonzero integer of R is a zerodivisor implies that the
ring R is the completion of a Noetherian local domain A. By Theorem 2.13(3),
A is a bad Noetherian stable domain.
Remark 2.15. The archetypal example of a ring R such as in Corollary 2.14
is produced by Nagata idealization: Let V be a complete DVR, and let L be
a nonzero finitely generated free V -module. Define V ∗ L as a V -module to be
V ⊕ L, and view V ∗ L as a ring with multiplication given by (v1 , ℓ1 )(v2 , ℓ2 ) =
(v1 v2 , v1 ℓ2 + v2 ℓ1 ) for all v1 , v2 ∈ V and ℓ1 , ℓ2 ∈ L. With P = 0 ∗ L, the ring
R = V ∗ L is a complete local ring meeting the hypotheses of Corollary 2.14, and
hence R is the completion of a bad Noetherian stable domain A. Moreover, since
the multiplicity of A is the same as that of its completion R, the multiplicity of
A is e = 1 + rank L.
To further motivate the class of bad stable domains, we mention three other
contexts in which these rings arise. An extension U ⊆ V of DVRs is immediate
if U and V have the same residue field and the maximal ideal of U extends
to the maximal ideal of V . The m-adic completion of a DVR is an immediate
extension.
Theorem 2.16. ([31, Corollary 5.7 and Theorem 5.9] and [32, Theorem 4.1])
(1) If k is a field and A is an affine k-domain with Krull dimension d > 1
and quotient field F , then there exists a bad Noetherian stable domain R
between A and F having multiplicity d and residue field finite over k.
(2) If (A, M ) is a Noetherian local domain and there exists a DVR V of A that
birationally dominates A and satisfies V = A + M V , then the kernel of
the exterior differential dV /A : V → ΩV /A is a bad stable domain between
A and V .
7

(3) If U ⊆ V is an immediate extension of DVRs having quotient fields Q and
F , respectively, then there is a one-to-one correspondence between bad
stable domains R containing U and having normalization V and proper
full V -submodules of the module ΩF/Q of Kähler differentials.
Remarks 2.17. (1) The ring R in Theorem 2.16(1) is constructed along the
following lines: For an appropriate local A-algebra B contained in F such that B
b (2) ),
is essentially of finite type over k, teh ring R is defined to be R = F ∩ (B/P
b is the m-adic completion of B, P is a certain prime ideal of B
b and
where B
P (2) is the second symbolic power of P . That R has multiplicity d depends on
the fact that B is an excellent local ring [31, Corollary 5.6].
(2) Every bad Noetherian stable domain of positive characteristic must arise
as in Theorem 2.16(2) from a derivation [32, Theorem 6.4].
(3) The one-to-one correspondence in Theorem 2.16(3) is used in [32, Theorem 7.6] to describe complete DVRs that are the normalization of an analytically
ramified Noetherian local domain. For example, every equicharacteristic complete DVR is the normalization of both Noetherian and non-Noetherian bad
stable domains.
3. Quadratic rings
In this section we introduce the notion of a quadratic ring and show that
these rings exemplify some key technical features of stable local rings. We
require first some preliminaries on stable ideals.
Definitions 3.1. Let R be a ring, and let Q(R) be its total ring of quotients.
(1) An R-submodule I of Q(R) such that bI ⊆ R for some nonzerodivisor
b ∈ R is called a fractional ideal of R. The fractional ideal I is regular if
it contains a nonzerodivisor.
(2) If I is a regular fractional ideal of the ring R, then I is invertible if I is a
finitely generated fractional ideal such that IRM is a principal fractional
ideal of RM for each maximal ideal M of R.
(3) The ring R is finitely stable4 if every finitely generated regular ideal of R
is stable.
Remark 3.2. Let I be a regular fractional ideal of a ring R. Then I is invertible
if and only if IJ = R for some fractional ideal J [16, Proposition 2.3, p. 97].
If R has only finitely many maximal ideals, then an invertible ideal is principal
[17, Exercise 2.11B, p. 33.].
Notation 3.3. Let I be a regular fractional ideal of a ring R. Then, by [34,
(2.8)], End(I) = {q ∈ Q(R) : qI ⊆ I}. To simplify notation we write E(I) =
End(I).
4 The

rings we call “finitely stable” were called “stable” by Rush in [35, 36].

8

We collect in the next lemma some basic properties of a regular ideal I and
its endomorphism ring E(I).
Lemma 3.4. With the terminology of Definition 3.1, let I be a regular fractional
ideal of a ring R. Then
(1)
(2)
(3)
(4)

I is a regular fractional ideal of E(I).
I ⊆ Jac(E(Jac R))).
I is stable if and only if I is an invertible fractional ideal of E(I).
If I 2 = AI for some invertible fractional ideal A of R with A ⊆ I, then
I = AE(I) and I is stable.
(5) If there exists x ∈ I such that I 2 = xI, then I is stable.
(6) Let R be finitely stable and local. Then I is stable if and only if I 2 = xI
for some x ∈ I.

Proof. (1) This follows from the fact that every nonzerodivisor in R is also a
nonzerodivisor in Q(R).
(2) Let x ∈ Jac R and w ∈ E(Jac R). Then wx ∈ Jac R, and so 1 − wx is a
unit of R, hence also of E(Jac R). Since w was arbitrary in E(Jac R), we have
x ∈ Jac E(Jac R).
(3) This is proved in [34, Proposition 2.11].
(4) This is proved in [34, Proposition 2.13].
(5) Since I is regular and I 2 ⊆ xR, the ideal xR is regular, hence invertible,
and so this follows from (4).
(6) Under the assumptions of (6), it is proved in [34, Corollary 5.7] that a
regular ideal J of R is stable if and only if J is a principal ideal of E(I). The
characterization in (6) now follows from (5).
Notation 3.5. Let R be a ring such that Jac R is a regular ideal. We associate
to R a tower of rings in Q(R) by defining
S∞
R0 := R,
Ri := E(Jac Ri−1 ) for every i ≥ 1,
R∞ := i=1 Ri .
Lemma 3.4(1) implies that for each i, Jac Ri is a regular ideal of Ri+1 =
E(Jac Ri ). In Lipman’s terminology [21, Section 2], if R is a local CohenMacaulay ring and the Jacobson radical of each Ri is stable, then the ring Ri+1
is the blow-up of Ri at its Jacobson radical and the localizations of the rings Ri
at maximal ideals are the local rings “infinitely near” R.
Definition 3.6. Let R be a ring such that Jac R is a regular ideal. With
Notation 3.5 we define R to be a quadratic ring if
(1) R is local,
(2) R ⊆ R∞ is a quadratic extension, and
(3) the maximal ideal of R is stable and regular.

9

As we show in Theorem 3.9, the relevance of this notion is that a finitely
stable local ring with stable regular maximal ideal is a quadratic ring. Some
technical properties of quadratic rings are elaborated in Lemmas 3.7 and 3.10
and Proposition 3.11. Parts of these results were proved for finitely stable local
domains with stable maximal ideal in [29, Section 4], but the arguments given
here are somewhat different and more general in that they permit zero divisors.
Lemma 3.7. Let R be a local ring with regular maximal ideal M and residue
field F = R/M . Let R1 be as in Notation 3.5. Suppose that R ⊆ R1 is a
quadratic extension with R 6= R1 .
(1) If R1 is a local ring such that Jac R1 = M , then R1 = R2 and R1 /M is a
field extension of F of degree 2.
(2) If R1 is a local ring such that Jac R1 6= M , then R1 = R + M1 and
R1 /M1 ∼
= F.
(3) If R1 is not a local ring, then Jac R1 = M , R1 = R2 and R1 /M is
isomorphic as an F -algebra to either
(i) F × F , or
(ii) F × F × F , in which case F = F2 .
(4) (Jac R1 )2 ⊆ M .
Proof. (1) Suppose that R1 is a local ring such that Jac R1 = M . Then R1 /M
is a field and Jac R = M = Jac R1 . Hence R1 = R2 . To see that R1 /M
has degree 2 over F , let x, y ∈ R1 \ R, and let S = xR + yR + R. Then S
is a ring since R ⊆ R1 is a quadratic extension. Hence R/M ⊆ S/M is a
quadratic extension, which by Remark 2.7 must be integral. Since R1 /M is a
field, the subring S/M is an integral domain, and hence since S/M is integral
over the field R/M , S/M is a field. Since R/M ( S/M , Lemma 2.8 implies
that S/M has degree 2 as a field extension of R/M . Since x, y 6∈ R, this forces
S/M = (xR + R)/M = (yR + R)/M . Hence x ∈ yR + R for all x, y ∈ R1 \ R,
and so for each x ∈ R1 \ R, we have R1 = xR + R. Therefore the degree of
R1 /M over R/M is 2.
(2) Suppose that R1 is a local ring such that JacR1 6= M . Let M1 denote the
maximal ideal of R1 . We claim that R1 = R + M1 . By Lemma 3.4(2), M ( M1 .
Let x ∈ M1 \M , let y ∈ R1 , and let S = R+xR+yR. Since x ∈ M1 ∩S, we have
M1 ∩ S 6= M . Because R ⊆ R1 is quadratic, S is a ring, and, by Remarks 2.7(2),
R ⊆ R1 is an integral extension. Since R1 is local, so is S. Also, S/M cannot
be a field since M ( M1 ∩ S ( S. Furthermore, since S is local, the ring S/M is
not isomorphic to F × F or F × F × F . By Lemma 2.8, S/M is a local ring with
square zero maximal ideal and residue field F . If N is the maximal ideal of S,
we have N 2 ⊆ M and S/N ∼
= F = R/M . Therefore y ∈ S = R + N ⊆ R + M1 .
The choice of y ∈ R1 was arbitrary, and so R1 = R + M1 , as desired. Moreover,
since M ⊆ M1 , this implies that R1 has residue field F .
(3) Suppose that R1 is not a local ring, and let k denote the number of
maximal ideals of R1 . By Lemma 3.4(2), M ⊆ Jac R1 , and so R1 /M has
k > 1 maximal ideals. Also, since R ⊆ R1 is a quadratic extension of rings,
10

R/M ⊆ R1 /M is a quadratic extension in which R1 /M has more than one
maximal ideal. By [34, Lemma 5.2], R1 /M is isomorphic as an F -algebra to
Qk
i=1 F , and F = F2 when k = 3. Also, since we have obtained a decomposition
of R1 /M as a finite product of fields, Jac R1 = M and hence R1 = R2 .
(4) In light of (1) and (3), the only case in which (4) is not immediate is
that of (2). Suppose that R1 is local and M 6= Jac R1 . By Lemma 3.4(2),
M ( Jac R1 . Let a, b ∈ Jac R1 with a, b 6∈ M . Define T = aR + bR + R. Then,
as in the proof (2), Lemma 2.8 forces the square of the maximal ideal of the ring
T to be contained in M . Thus ab ∈ M , and it follows that (Jac R1 )2 ⊆ M .
For the next lemma, we recall that a ring R is Prüfer if every finitely generated regular ideal is invertible.
Lemma 3.8. ([36, Proposition 2.1 and Theorem 2.3] and [34, Corollary 5.11])
A ring R is finitely stable if and only if
(a) R ⊆ R is a quadratic extension,
(b) R is a Prüfer ring, and
(c) there are at most two maximal ideals of R lying over each regular maximal
ideal of R.
The next theorem shows that the class of quadratic rings contains the finitely
stable local rings with stable maximal ideal.
Theorem 3.9. Let (R, M ) be a finitely stable local ring with stable regular
maximal ideal M , and let R∞ be defined as in Notation 3.5. Then R is a
quadratic ring for which R∞ has at most two maximal ideals.
Proof. With {Ri }∞
i=0 as in Notation 3.5, we prove the theorem by establishing
four claims.
Claim 1. For each i ≥ 0, if Ri is a local ring with maximal ideal Mi , then
Ri ⊆ Ri+1 is a quadratic extension.
We use the fact that Ri ⊆ Ri+1 is a quadratic extension if and only if every
finitely generated Ri -submodule of Ri+1 containing Ri is a stable fractional
ideal of Ri ; see [34, Proposition 3.5] or [35, proof of Lemma 2.1]. Thus we let
A be a finitely generated Ri -submodule of Ri+1 with Ri ⊆ A. Let m ∈ M be a
nonzerodivisor. Then m ∈ Mi and
m ∈ mRi ⊆ mA ⊆ mRi+1 ⊆ Ri ,
since Ri+1 = {q ∈ Q(R) : qMi ⊆ Mi }. Therefore mA is a finitely generated
regular Ri -submodule of Ri , hence a finitely generated regular ideal of Ri . Every
ring between a finitely stable ring and its total quotient ring is finitely stable
[34, Proposition 5.1]. Hence Ri is finitely stable, and so mA is a stable ideal
of Ri by Definition 3.1(3). Since mA ∼
= A, A is a stable fractional ideal of Ri .
This proves Claim 1.
11

Claim 2. For each i ≥ 0 with Ri ( Ri+1 , Ri is local.
For i = 0, the claim holds, since R = R0 is local. Suppose that i > 0 and
that Ri is local whenever Ri ( Ri+1 . If Ri+1 ( Ri+2 , then also Ri ( Ri+1 and
so Ri is local by induction. By Claim 1, Ri ( Ri+1 is a quadratic extension.
Since Ri+1 6= Ri+2 , Lemma 3.7(3) implies that Ri+1 is a local ring, as desired
for Claim 2.
Claim 3. (i) For each i ≥ 0, Ri ⊆ Ri+1 is an integral extension. (ii) The
extension R ⊆ R∞ is integral.
The first statement follows from Claims 1 and 2 and Remark 2.7(2). The
second statement follows from the first, since compositions of integral extensions
are again integral [7, Proposition 6, p. 307].
Claim 4. R∞ has at most two maximal ideals.
By Claim 3, R∞ ⊆ R, and by Lemma 3.8, R has at most two maximal ideals.
Since R is an integral extension of R∞ , R∞ also has at most two maximal ideals.
Claim 5. R is a quadratic ring.
By Lemma 3.8, R ⊆ R is a quadratic extension, and Claim 3 implies R∞ ⊆
R. By Remark 2.7(3), R ⊆ R∞ is a quadratic extension. By Definition 3.6, R
is a quadratic ring.
By adding to Lemma 3.7 the assumption that the maximal ideal of R is
stable, we obtain stronger results in the case where R1 is local. In particular,
R1 inherits the property of having a stable maximal ideal.
Lemma 3.10. Let R be a local ring with stable regular maximal ideal M . Suppose that R ⊆ R1 is a quadratic extension and R1 is a local ring with maximal
ideal M1 .
(1) If R1 6= R2 , then M1 = M R2 , M1 is a stable ideal of R1 and R1 = R+M1 .
(2) If R1 = R2 , then M1 is a principal ideal of R1 and R1 = R + xR for every
x ∈ R1 \ R.
Proof. By Lemma 3.4(3), M is an invertible ideal of R1 . Since R1 is a local
ring, M = mR1 for some nonzerodivisor m ∈ M . By Lemma 3.7(4), M12 ⊆
M = mR1 , so that M12 = mA for some ideal A of R1 .
Claim: (i) If A = R1 , then M1 is invertible in R1 and R1 = R2 . (ii) If
A ( R1 , then M12 = mM1 and M1 = mR2 = M R2 .
For (i), A = R1 =⇒ mR1 = M12 =⇒ M1 (m−1 M1 ) = R1 =⇒ M1 is invertible. We have
R2 = {q ∈ Q(R1 ) : qM1 ⊆ M1 } = {q ∈ Q(R1 ) : qM1 M1−1 ⊆ M1 M1−1 } = R1 ,
and so (i) holds.
For (ii), A ( R1 =⇒ M12 = mA ⊆ mM1 ⊆ M12 =⇒ M12 = mM1 . By
Lemma 3.4(4) we have
M1 = mR2 ⊆ M R2 ⊆ M1 R2 ⊆ mR2 =⇒ M1 = mR2 = M R2 ,
12

and so (ii) holds.
We return to the proof of Lemma 3.10. For (1), M1 = M R2 and M1 is
invertible in R2 by Claim (ii). Thus M1 is stable by Lemma 3.4(3). Since
E(M ) = R1 6= R2 = E(M1 ), we have M 6= M1 . By Lemma 3.7(2), R1 = R+M1 .
This verifies (1).
For (2), we have R1 = R2 . If M1 = M , we have M1 = mR1 is principal. By
Lemma 3.7(1), the field R1 /M has degree 2 over R/M , and hence R1 = R + xR
for all x ∈ R1 \ R, and so (2) holds if M1 = M .
Thus we assume that M1 6= M . We have M1 is a principal ideal of R1 in
either case A = R1 or A ( R1 , using either (i) of the claim (M1 is invertible and
so is a principal ideal of R1 ), or (ii) of the claim (M1 = mR2 = mR1 ). Thus
the first part of (2) holds.
It remains to show R1 = R + xR, for all x ∈ R1 , given that R1 = R2 ,
M1 = nR1 , for some n ∈ R1 , and M1 6= M . We have also R 6= R1 and n is a
nonzerodivisor.
By Lemma 3.7(4), M12 ⊆ M . If mM1 = M12 , then
mnR1 = n2 R1 =⇒ M = mR1 = nR1 = M1 ,
a contradiction. Thus mM1 ( M12 ⊆ M = mR1 . Hence there exist c, d ∈ M1
with cd ∈ mR1 \ mM1 ; say cd = um, where u is a unit of R1 . Then m =
u−1 cd ∈ M12 , and
mR1 ⊆ M12 ⊆ M = mR1 =⇒ M12 = M.
Therefore

M1 /M = M1 /M12 ∼
= R/M,
= R1 /M1 ∼

by Lemma 3.7(2). Since R/M is a field, we have two equations.
(3.10a) M1 = qR + M, ∀q ∈ M1 \ M.

(3.10b) R1 = R + M1 .

Let x ∈ R1 \ R. By Equation 3.10b, x = r + m1 , for some r ∈ R and
m1 ∈ M1 , where m1 6∈ M , since x 6∈ R. Then, using Equations 3.10b and 3.10a,
we have
R + xR ⊆ R1 = R + M1 ⊆ R + m1 R = R + (r + m1 )R = R + xR.
This completes the proof of the lemma.
We collect now some of the main technical properties of quadratic rings that
will be needed in the next section.
Proposition 3.11. Let R be a quadratic ring with maximal ideal M , let {Ri }∞
i=0
and {R∞ } be as in Notation 3.5, and let
n = min{k ∈ {0, 1, 2, . . . , ∞} : Rk = R∞ }.
Then the following statements hold for R.
13

(1) There exists a nonzerodivisor m ∈ M such that M = mR1 .
(2) For each i < n, Ri = R + Mi and Ri is a quadratic ring with stable
maximal ideal Mi = mRi+1 .
(3) If n = ∞, then R∞ is a local ring with maximal ideal mR∞ and R∞ =
R + mR∞ .
(4) R∞ has at most 3 maximal ideals, each of which is principal.
(5) If R∞ is not local, then Jac R∞ = mR∞ .
(6) The R-module R∞ is finitely generated if and only if n < ∞.
T∞
(7) Suppose n < ∞ and J := i=1 M i is a one-dimensional ideal of R. Then
J is the intersection of the one-dimensional prime ideals of R, the number
of which is not more than the number of maximal ideals of R∞ . Moreover
J is also the intersection of the one-dimensional prime ideals of R∞ , and
hence J is an ideal of R∞ .
Proof. (1) By Definition 3.6, M is stable. Since R ⊆ R1 is a quadratic extension, Proposition 2.10 implies that R1 has at most 3 maximal ideals. By
Lemma 3.4(3), M is invertible in R1 . Since R1 is local, M = mR1 , for some
nonzerodivisor m ∈ M .
(2) The proof is by induction. If n = 0 or i = 0, the statement holds by
(1). Suppose the result holds for some i with 0 < i < n − 1. Then Ri is a
quadratic ring with stable maximal ideal Mi = mRi+1 and Ri = R + Mi . Also
R ( Ri ( Ri+1 ( Ri+2 , by the minimality of n. By Lemma 3.7(3), Ri+1
is a local ring. By Lemma 3.7(1), the maximal ideal Mi+1 of Ri+1 properly
contains Mi . By Lemma 3.10(1), Mi+1 = Mi Ri+2 = mRi+1 Ri+2 = mRi+2 , and
Ri+1 = Ri + Mi+1 = R + Mi+1 . Since R ⊆ R∞ is a quadratic extension, it is
clear that Ri+1 ⊆ R∞ is also a quadratic extension, and so Ri+1 is a quadratic
ring.
(3) If n = ∞, then by (2), each Ri is local with maximal ideal mRi+1
and Ri = R + Mi . Hence R∞ is a local ring with maximal ideal mR∞ and
R∞ = R + mR∞ .
(4) By Proposition 2.10, Rn has at most three maximal ideals. If n = ∞,
then (4) follows from (3). If n = 0, then R = R0 is local and it has principal
maximal ideal by (1), so that (4) holds.
Suppose that 0 < n < ∞. Then Rn−1 6= Rn . By (2), Rn−1 is a quadratic
ring and the maximal ideal Mn−1 = mRn is stable. Also Rn = Rn+1 . If Rn is
local, then Lemma 3.10(2) implies that the maximal ideal Mn is principal, and
so we have established (4).
If Rn is not local, then Lemma 3.7(3) implies that Jac Rn = Mn−1 =
mRn . Since Rn has at most three maximal ideals,
denote them by
Tt we may Q
t
N1 , N2 , . . . , Nt , where 2 ≤ t ≤ 3. Then mRn = i=1 Ni = i=1 Ni , by [24,
Qt
Theorem 1.3]. Then each Ni is invertible; e.g., N1 has inverse m−1 i=2 Ni . By
Remark 3.2, each of the Ni is principal. Thus (4) holds.
(5) Since R∞ is not local, (3) implies that n < ∞. As noted in the proof of
(4), we have Jac Rn = mRn .
S∞
(6) If n = ∞, then clearly R∞ = i=1 Ri is not finitely generated.
14

For the converse, the case n = 0 is clear. Let 0 < n < ∞. We show first
that Rn is a finitely generated Rn−1 -module.
Case 1: If Rn is local, then Lemma 3.10(2) implies that Rn = Rn−1 +xRn−1 ,
for every x ∈ Rn \ Rn−1 , and so Rn is finitely generated as an Rn−1 -module if
Rn is local.
Case 2: If Rn is not local, then, by (3), n < ∞. By (2), Rn−1 is a quadratic
ring, and so by Lemma 3.7(3), Rn /Mn−1 has dimension at most three as a vector
space over Rn−1 /Mn−1 . Therefore Rn is a finitely generated Rn−1 -module.
If n = 1, then the proof is complete. If n > 1, the result will follow from the
following claim (using induction).
Claim. Suppose 0 ≤ k ≤ n − 2 and Rk+2 is finitely generated as an Rk+1 module. Then Rk+1 is finitely generated as an Rk -module.
Write Rk+2 = b1 Rk+1 + · · · + bt Rk+1 , for some b1 , . . . , bt ∈ Rk+2 . By (2), we
have Mk+1 = mRk+2 and
Rk+1

=

Rk + mRk+2 = Rk + mb1 Rk+1 + · · · + mbt Rk+1

=

Rk + mb1 (Rk + mRk+2 ) + · · · + mbt (Rk + mRk+2 ).

2
By Lemma 3.7(4), m(mRk+2 ) ⊆ Mk+1
⊆ Mk . Each mbi ∈ Rk+1 . Therefore

Rk+1 ⊆ Rk + mb1 Rk + · · · + mbt Rk + Mk ⊆ Rk+1 .
This proves the claim.
(7) By (4), R∞ has at most 3 maximal ideals, say N1 , . . . , Nr , all of which are
regular and principal. For eachTj = 1, . . . , r, since Nj is a principal regular ideal
of R∞ , it follows that Qj := k>0 Njk is a prime ideal and every prime ideal
of R∞ properly contained in Nj is contained in Qj [3, Theorem 2.2]. Therefore
each nonmaximal prime ideal of R∞ is contained in one of the T
Qj . Now R ⊆ R∞
is a quadratic, hence integral, extension. By assumption J = k>0 M k is a onedimensional ideal of R. Every prime ideal of R∞ containing J also has dimension
at most one. Since J ⊆ Q1 ∩ · · · ∩ Qr , each prime ideal Qi has dimension one.
Since also each nonmaximal prime ideal of R∞ is contained in one of the Qi ,
the set of one-dimensional prime ideals of R∞ is {Q1 , . . . , Qr }. Since R∞ is an
integral extension of R, it follows that {Q1 ∩ R, . . . , Qr ∩ R} is the set of onedimensional
T prime ideals of R. Now since JacR∞ = N1 · · · Nr = N1 ∩· · ·∩Nr , we
have that k>0 (JacR∞ )k = Q1 ∩ · · · ∩ Qr . In fact, since Rn = R∞ , an inductive
argument using Lemma 3.7(4)
T some power of the ideal Jac R∞ is
T shows that
contained in M . Hence J = k>0 M k = k>0 (Jac R∞ )k = Q1 ∩ · · · ∩ Qr .
Corollary 3.12.
Let R be a one-dimensional quadratic ring with maximal ideal
T
∞
i
M , let J = ∞
i=1 M , let R be the integral closure of R, and let {Ri }i=0 and
R∞ be as in Notation 3.5. Then
(1) R∞ = R.
(2) The R-module R is finitely generated if and only if R∞ = Rn for some
n ≥ 0.
15

(3) The ring R/J has integral closure R∞ /J and total quotient ring Q(R)/J.
Proof. (1) Since R ⊆ R∞ is a quadratic, hence integral, extension, and R has
Krull dimension one, R∞ also has Krull dimension one. Since there are only
finitely many maximal ideals of the one-dimensional ring R∞ , all of which are
principal and regular, it follows that every regular ideal of R∞ is principal,
and hence the integral extension R∞ of R is integrally closed in Q(R); cf. [19,
Theorem 10.18, p. 237].
(2) This follows from (1) and Proposition 3.11(6).
(3) By Proposition 3.11(1), there exists a nonzerodivisor m ∈ M such that
M 2 = mM . Since R has dimension one, Q(R) = R[1/m]. Also, since M 2 =
mM , it follows that J = (R :R R[1/m]), and so J is an ideal of Q(R). Let
m∗ denote the image of m in R/J. Then m∗ is a nonzerodivisor in the onedimensional ring R/J. Hence Q(R)/J = R[1/m]/J = (R/J)[1/m∗] = Q(R/J).
By (1), R = R∞ , and so it follows that R∞ /J is the integral closure of R/J in
Q(R/J) = Q(R)/J.
By Proposition 3.11(4), R∞ has at most three maximal ideals if R is a
quadratic ring. In the next theorem we describe the quadratic rings R for which
R∞ has exactly three maximal ideals.
Theorem 3.13. Let R be a local ring with a regular maximal ideal. Then the
following statements are equivalent.
(1) R is a quadratic ring such that R∞ has three maximal ideals.
(2) There is a ring S between R and Q(R) such that
(i) S has exactly 3 maximal ideals,
(ii) each maximal ideal of S is principal and regular with residue field
isomorphic to F2 , and
(iii) R = k + Jac S, where k is the prime subring of R.
Proof. (1) ⇒ (2) Suppose that R is a quadratic ring and R∞ has three maximal
ideals. Since R is local, R 6= R∞ , and hence R 6= R1 . If R1 = R∞ , then,
by Lemma 3.7(3), M = Jac R1 and the three maximal ideals of R1 each have
residue field isomorphic to F2 . Moreover, by Proposition 3.11(4), each maximal
ideal of R1 is regular and principal. Since (k + M )/M ⊆ R/M and R/M has
only 2 elements (because R/M is a subfield of R1 /M ∼
= F2 × F2 × F2 ), it must
be that k + M = R. Therefore to verify (2) it remains to show that R1 = R∞
and choose S = R1 .
Since R is a quadratic ring, each Ri with Ri 6= Ri+1 is a quadratic ring
by Proposition 3.11(2). Denote the maximal ideal of each such Ri by Mi .
Since R∞ has 3 maximal ideals, Proposition 3.11(3) implies that there is n > 0
such that Rn−1 ( Rn = R∞ . We show that n = 1. Suppose by way of
contradiction that n > 1. By Proposition 3.11(2), Rn−2 is a quadratic ring with
Rn−2 ( Rn−1 ( Rn = R∞ , and, by Lemma 3.7(2), Rn /Mn−1 ∼
= F2 ×F2 ×F2 . To
16

show that such a case is impossible, we may after relabeling assume that n = 2.
We show the assumption that R is a quadratic ring with R ( R1 ( R2 = R∞
and the fact that R2 /M1 ∼
= F2 × F2 × F2 lead to a contradiction of the fact that
R 6= R1 .
Claim: The dimension of the R/M -vector space R1 /R is 1.
Let A be an R-module such that R ⊆ A ( R1 . It suffices to show that R = A.
Since R ⊆ R1 is a quadratic extension, R ⊆ R1 is an integral extension and A is
a ring. Thus A ⊆ R1 is an integral extension. Since R1 is local, A is local with
maximal ideal, say N , lying over M and under M1 . By Proposition 3.11(1) and
(2), there is a nonzerodivisor m ∈ M such that M = mR1 and M1 = mR2 .
Then mR1 = M ⊆ N ⊆ M1 = mR2 , so that R1 ⊆ m−1 N ⊆ R2 . Since
R1 ⊆ R2 is a quadratic extension, m−1 N is a ring, and hence N = mA1 , where
A1 = E(N ) = m−1 N . Moreover R1 ⊆ A1 ⊆ R2 . To prove that R = A, we
show that R1 = A1 . Once this is established, we have that A has maximal
ideal N = mA1 = mR1 = M . Since by Lemma 3.10(1), R and R1 have the
same residue field R/M , so does A. Thus the fact that N = M implies that
A = R + N = R, as claimed.
For the claim it remains to show that with A and A1 as above, R1 = A1 .
Since R1 /M1 ⊆ A1 /M1 ⊆ R2 /M1 ∼
= F2 × F2 × F2 , and the last ring is reduced,
A1 /M1 is also reduced. Hence M1 = Jac A1 . Since every element of the ring
F2 × F2 × F2 is an idempotent, either R1 = A1 or A1 /M1 is a decomposable
ring. Suppose that R1 6= A1 . Then A1 /M1 is a decomposable ring, and hence
A1 has more than one maximal ideal. Since R ⊆ R∞ is a quadratic extension
and R ⊆ A ⊆ A1 ⊆ R∞ , we have that A ⊆ A1 is also a quadratic extension
with A1 not local. By Lemma 3.7(3) (applied to A ⊆ A1 ), N = Jac A1 . We
have established above that Jac A1 = M1 , and so N = M1 . Therefore, by
Lemma 3.7(1), R1 = R + M1 = R + N ⊆ A, contrary to the choice of A
with A ( R1 . This shows A1 /M1 cannot be a decomposable ring, which forces
R1 = A1 and proves that R1 /R has dimension 1 as an R/M -vector space. This
verifies the claim.
We show now that the claim implies a contradiction to our assumption that
R 6= R1 . Let t ∈ R1 \ R. Since R1 /R has dimension 1, we have R1 = R + tR,
and hence M = mR1 = mR + mtR, so that as an R/M -vector space, M/M 2
has dimension at most 2. By Lemma 3.7(4), M12 ⊆ M . Also, M12 /M 2 is a
proper subspace of M/M 2 , and as such has dimension at most 1. (We have
M12 ( M since M1 = mR2 , M = mR1 and m is a nonzerodivisor.) However,
M12 /M 2 = m2 R2 /m2 R1 ∼
= R2 /R1 , so that R2 /R1 has dimension 1 as an R/M vector space. There is an exact sequence of R/M -vector spaces,
0 → R1 /M1 → R2 /M1 → R2 /R1 → 0.
The left and right vector spaces have dimension 1, while the middle vector
space has dimension 3, since R2 /M1 ∼
= F2 × F2 × F2 . This contradiction implies
that the assumption R ( R1 ( R2 is impossible under the conditions in (1).
Therefore R1 = R2 , from which it follows that R1 = R∞ . This completes the
proof that (1) implies (2).
17

(2) ⇒ (1) By (2), M = Jac S is the product of the three principal maximal
ideals of S, so Jac S is a principal regular ideal of S. Hence Jac S is a stable
maximal ideal of R. Moreover S = E(M ) = R1 = R∞ and R1 /M ∼
= F2 ×F2 ×F2 .
By Lemma 2.8, R/M ⊆ R1 /M is a quadratic extension, and hence R ⊆ R1 =
R∞ is also a quadratic extension by Remark 2.7. Therefore R is a quadratic
ring and R∞ has three maximal ideals.
4. Characterizations of one-dimensional stable rings
In this section we work in the following setting.
Setting 4.1. Let R be a one-dimensional local ring having a regular maximal
ideal M , let R denote the integral closure of T
R in its total quotient ring Q(R),
and let m ∈ M be a nonzerodivisor. Let J = i>0 mi R. As in Notation 2.3, we
e and R
b the completions of R in the mR-adic and M -adic topologies,
denote by R
e are independent of the choice of nonzerodivisor m
respectively. Both J and R
in M . Moreover, since m is a nonzerdivisor, R/J is a one-dimensional ring with
regular maximal ideal M/J.
We first characterize a one-dimensional stable local ring in terms of its integral closure. Recall that a ring A is a Dedekind ring if every regular ideal I of
A is invertible.
Theorem 4.2. With Setting 4.1, the following statements are equivalent.
(1)
(2)
(3)
(4)

R is a stable ring.
R is a finitely stable ring with stable maximal ideal.
R is a quadratic ring such that R has at most two maximal ideals.
R ⊆ R is a quadratic extension and R is a Dedekind ring with at most
two maximal ideals.
(5) R/J is a stable ring.

Proof. (1) ⇒ (2) This is clear.
(2) ⇒ (3) This follows from Lemma 3.8 and Theorem 3.9.
(3) ⇒ (4) By Corollary 3.12(1), R = R∞ , and so R ⊆ R is a quadratic extension. By (3), R has at most 2 maximal ideals, and Proposition 3.11(4) implies
the maximal ideals of R are principal regular ideals. Since R has dimension one,
every regular ideal of R is a principal regular ideal, and hence R is a Dedekind
ring.
(4) ⇒ (1) Let I be a regular ideal of R. Since R has Krull dimension 1
and at most 2 maximal ideals, both of which are principal and regular (they are
regular because they lie over M ), it follows that every regular ideal of R contains
a power of Jac R. Since R is a Dedekind ring with only finitely many maximal
ideals, every regular ideal is principal. In particular, IR is a principal regular
ideal; that is, IR = xR for some nonzerodivisor x ∈ R. An argument of Rush
[36, proof of Theorem 2.2], which is also given explicitly in [34, Proposition 3.6],
shows that since R has at most two maximal ideals, x can be chosen an element
18

of I. Therefore R ⊆ x−1 I ⊆ R. Since R ⊆ R is a quadratic extension, x−1 I is a
ring, and hence I = xE(I). This shows that every regular ideal of R is stable.
(1) ⇒ (5) Let I be an ideal of R such that I/J is a regular ideal in R/J.
Then I is an M -primary ideal R and hence a regular ideal in R. Since R is
stable, Lemma 3.4(6) implies that I 2 = xI, for some x ∈ I. Since I 2 is M primary, we have J ⊆ I 2 ⊆ xR, and hence I 2 /J = (I/J)(xR/J). Moreover,
since I 2 ⊆ xR, the ideal xR is M -primary, hence regular in R. Therefore xR/J
is M/J-primary in R/J and hence regular in R/J. Lemma 3.4(4) implies then
that I/J is a stable ideal in R/J, which proves that R/J is a stable ring.
(5) ⇒ (1) Let I be a regular ideal of R. Then I is M -primary, and so there
is k > 0 such that mk ∈ I. Since the image of mk in R/J is a nonzerodivisor,
I/J is regular. By (5) and Lemma 3.4(6) there exists x ∈ I such that I 2 /J =
(xI + J)/J. Thus I 2 = xI + J, so that since J is an ideal of R[1/m], we have
R[1/m] = I 2 R[1/m] = xIR[1/m] + JR[1/m] = xR[1/m] + J. Therefore there
exist ℓ ≥ 0, r ∈ R and j ∈ J such that 1 = x(r/mℓ ) + j. Thus (1 − j)mℓ ∈ xR,
and, since 1 − j is a unit in R, mℓ ∈ xR. Therefore J ⊆ mk+ℓ R ⊆ xI, so that
I 2 = xI + J = xI. By Lemma 3.4(5), I is stable.
Remark 4.3. Rush [35, Theorem 2.4] has proved the equivalence of (1) and
(4) of Theorem 4.2 in the case where R is a Noetherian ring. Our proof that
(4) implies (1) follows Rush’s argument.
From the theorem we deduce a characterization of one-dimensional quadratic
rings that shows these are stable except in a very special case.
Corollary 4.4. A one-dimensional local ring R is a quadratic ring if and only
if either R is a stable ring or R = k + Jac S, where k is the prime subring of
R and S is a Dedekind ring between R and Q(R) with exactly three maximal
ideals, each with residue field isomorphic to F2 .
Proof. Suppose that R is a quadratic ring that is not a stable ring. By Theorem 4.2, R has more than two maximal ideals. By Corollary 3.12(1), R = R∞ ,
and so, by Proposition 3.11(4), R∞ has exactly three maximal ideals. By Theorem 3.13, there exists a ring S between R and Q(R) having exactly 3 maximal
ideals, all of which are principal and have residue field isomorphic to F2 , such
that R = k + Jac S. As noted in the proof of Theorem 3.13, we may take S
to be R∞ . Since the maximal ideals of the one-dimensional ring R = R∞ are
principal and regular, R is a Dedekind ring. The converse is a consequence of
Theorems 3.9 and 3.13.
In Theorem 4.5 we characterize the one-dimensional stable local rings with
finite integral closure; Theorem 4.7 similarly characterizes stable rings without
finite integral closure.
Theorem 4.5. With Setting 4.1, the following statements are equivalent.
(1) R is a stable ring with finite integral closure.
(2) R/J is a Bass ring.
19

e is a Bass ring.
(3) R
Proof. (1) ⇒ (2) By Theorem 4.2, R/J is a stable ring. Since also R has
finite integral closure, Corollary 3.12(2) implies that Rn = R∞ for some n.
By Proposition 3.11(7), J is the nilradical of R. Therefore R/J is a reduced
one-dimensional stable ring. To show that R/J is a Bass ring, it remains by
Theorem 2.11 to show that R/J is a Noetherian ring with finite integral closure.
By Proposition 3.11(7), since Rn = R∞ , J is the intersection of the minimal
prime ideals Q1 , . . . , Qr of R∞ , and, as noted in the proof of the proposition,
these prime ideals are comaximal. Now R∞ is a one-dimensional ring with
principal maximal ideals (Proposition 3.11(4)), and so for each i, R∞ /Qi is
a principal ideal domain. Since Q1 , . . . , Qr are pairwise comaximal, the ring
R∞ /J is the product of the principal ideal domains R∞ /Qi . Therefore R∞ /J
is a principal ideal ring. By Proposition 3.11(6), R∞ /J is a finitely generated
R/J-module, and so the Eakin-Nagata Theorem [24, Theorem 3.7, p. 18] implies
that R/J is a Noetherian ring. Moreover, by Corollary 3.12(3), R∞ /J is the
integral closure of R/J in its total quotient ring, and so R/J has finite integral
closure. Therefore R/J is a Bass ring.
(2) ⇒ (3) Let M denote the maximal ideal of R. Since S := R/J is reduced
and has finite normalization, Proposition 2.2(3) implies that the M/J-adic completion Sb of S is also reduced, and hence, again by Proposition 2.2(3), Sb has
b Thus Sb is a Bass
finite normalization. Since S has multiplicity ≤ 2, so does S.
e
e∼
b Since
ring. To show then that R is a Bass ring, we need only prove that R
= S.
mR/J contains a power of M/J, the (mR/J)-adic completion of S = R/J coincides with the (M/J)-adic completion Sb of S. For each i > 0, J ⊆ mi R, so
that S/mi S ∼
= R/mi R as R-algebras. Since these isomorphisms are natural,
e
R/mi R = R,
Sb = lim S/mi S ∼
= lim
←−
←−
which completes the proof of (3).
(3) ⇒ (2) As noted in the proof of (2) implies (3), the mR/J-adic completion
e so we may assume without loss of generality that J = 0
of R/J coincides with R,
and show that R is a Bass ring. Since J = 0, we may identify R with its image
e Also, since R
e is a reduced ring, so is R.
in R.
We claim that R is Noetherian. Since R is reduced and has Krull dimension
1, it suffices by Proposition 2.1(1) to show that M is finitely generated. To
e is a Noetherian ring, the R-module
e
e R
e is
this end, observe that since R
M R/m
e
e
e
e
finitely generated. By Proposition 2.5(2), R = R + mR, and so M R/mR is
e R
e
finitely generated as an R-module. By Proposition 2.5(3), M/mR ∼
= M R/m
as R-modules, and hence M/mR is a finitely generated R-module. We conclude
that M is a finitely generated ideal of R and hence that R is a Noetherian ring.
Since R is Noetherian, the M -adic and mR-adic topologies agree on R.
b of the Noetherian ring R is a Bass ring. In
Therefore the M -adic completion R
b has
particular, R is an analytically unramified Noetherian local ring. Since R
multiplicity ≤ 2, so does R, which proves that R is a Bass ring.

20

(2) ⇒ (1) By Theorem 4.2, R is stable ring, and by Corollary 3.12(3), R∞ /J
is the integral closure of R/J. Since R/J is a Bass ring, R∞ /J is a finitely
generated R-module. Hence R∞ is a finitely generated R-module, so that R has
finite integral closure by Corollary 3.12(1).
A simple example shows that the conditions of the theorem are not in general
equivalent to the condition that R is a Bass ring. The example, as well as later
examples in this section, are constructed using Nagata idealization, which was
discussed in Remark 2.15.
Example 4.6. Let V be a DVR with quotient field F . Then R = V ⋆ F is a
one-dimensional stable local ring with regular maximal ideal [31, Lemma 3.3].
The ring R has finite integral closure (it is in fact integrally closed), and R is
not a Bass ring because it is not reduced, nor even Noetherian.
We turn next to the characterization of stable rings without finite integral
closure.
Theorem 4.7. With Setting 4.1, the following statements are equivalent.
(1) R is a stable ring without finite integral closure.
(2) R/J is a stable ring without finite integral closure.
e such that P 2 = 0 and R/P
e
(3) There exists a nonzero prime ideal P of R
is
a DVR.
Proof. (1) ⇒ (2) Suppose that R is a stable ring without finite integral closure.
By Theorem 4.2, R/J is a stable ring. By Corollary 3.12(1) and (3), R/J is
the integral closure of R/J. Thus, since R does not have finite integral closure,
neither does R/J.
(2) ⇒ (3) As observed in the proof of (2) implies (3) of Theorem 4.5, the
e We assume without loss of
(mR/J)-adic completion of R/J coincides with R.
generality that J = 0 and R is a stable ring without finite integral closure.
Corollary 3.12(2) implies that Ri 6= R∞ , for all i > 0. By Proposition 3.11(3)
there exists a nonzerodivisor x ∈ M such that xR∞ is the maximal ideal of R∞
and R∞ = R + xR∞ . To simplify notation, let S = R∞ and Q = Q(R).
Claim 1: The inclusion mapping R → S lifts to a surjective ring homomore → S,
e where Se denotes the completion of S in the mR-adic completion
phism R
of S.
Consider the exact sequence
δ

HomR (Q/R, Q/R) → HomR (Q/R, Q/S) → Ext1R (Q/R, S/R).
Since x is a nonzerodivisor in R, the ideal xR is M -primary. Since R has
Krull dimension one, this implies that some power of x is contained in every
principal regular ideal of R. Consequently, since S = R + xS, it follows that
S = R + rS for all nonzerodivisors r ∈ R; that is, S/R is divisible with respect
to the set of nonzerodivisors of R. This property, along with the fact that Q =
21

R[1/m] is a countably generated R-module, implies that Ext1R (Q/R, S/R) = 0
[4, Theorem 1.1], and hence the map δ is a surjection. By [23, Theorem 2.2,
e while HomR (Q/R, Q/S) can
p. 13], HomR (Q/R, Q/R) can be identified with R,
e
e maps onto S.
e
be identified with S. Since these identifications are natural, R
As discussed in the proof of [23, Theorem 2.9, p. 21], this mapping is a ring
homomorphism.
e such that R/P
e
Claim 2: There is a prime ideal P of R
is a DVR.
Since x is a nonzerodivisor in S, Proposition 2.5(1) implies that x is also a
e Since S has principal maximal ideal xS, Se has principal
nonzerodivisor in S.
T k
e Thus, since x is a nonzerodivisor in S,
e the ideal L =
e
maximal ideal xS.
kx S
e
e
has residue ring S/L that is a DVR. Since Claim 1 implies that R maps onto
e
e such that R/P
e
S/L,
there is a prime ideal P of R
is a DVR.

Claim 3: The ideal P in Claim 2 has the property that P 2 = 0.
e
e → S/
e T xk S.
The ideal P in Claim 2 is the kernel of the induced map R
k
T ke T
ke
With m as in Setting 4.1, since mR is M -primary, k x S = k m S. Viewing
Q
i
i
e and Se as subrings of Q
R
i>0 S/m S, respectively, we have
i>0 R/m R and
\
e : h ri + mi S i ∈
e
P = {h ri + mi R i ∈ R
mk S}.
k

Let p, q ∈ P , and write p = h ai + mi R i and q = h bi + mi R i. Since p, q ∈ P ,
there exist ci , di , si , ti ∈ S such that ai = mi+1 ci + mi si and bi = mi+1 di + mi ti .
Thus ai , bi ∈ mi S ∩ R. Since S = R + mS, R ⊆ S ⊆ R[1/m] and R ⊆ S is a
quadratic extension, it follows that (mi S ∩ R)2 ⊆ mi R for all i > 0 [31, Lemma
3.2]. Therefore ai bi ∈ mi R for all i > 0, which shows that pq = 0, and hence
P 2 = 0.
e
e is a one-dimensional stable
(3) ⇒ (1) Since R/P
is a DVR and P 2 = 0, R
ring [31, Lemma 3.3]. We claim that this implies R is a stable ring. To this end,
e
we first make an observation about the canonical homomorphism λ : R → R.
e is such that λ−1 (z R)
e is a regular ideal of R, then the
Claim: If z ∈ R
e
e
principal ideal z R is generated by the image of an element of λ−1 (z R).
e be such that λ−1 (z R)
e is a regular ideal of R, and let x be a
Let z ∈ R
−1
e
e If w is a
nonzerodivisor in λ (z R).
Write λ(x) = zw for some w ∈ R.
e
e
e
e
unit in R, then xR = z R, as claimed. Suppose that w is not a unit in R.
e
e
Since x is a nonzerodivisor in R, Proposition 2.5(2) gives that R = λ(R) + xR.
e = λ(R) + zwR,
e and so, for some t ∈ R
e and y ∈ R,
Therefore z ∈ λ(R) + xR
e is a local ring, 1 − wt
z(1 − wt) = z − zwt = λ(y). Since w is not a unit and R
e Since z(1 − wt) = λ(y), we have z R
e = y R.
e This shows that in
is a unit in R.
e
all cases z R is generated as a principal ideal by the image of an element of R.
This proves the claim.
We use the claim now to verify that R is stable. Let I be a regular ideal of R.
e is a stable local ring, Lemma 3.4(6) implies that there exists z ∈ I R
e such
Since R
2
−1
2
2e
2e
e
e
e
that I R = zI R. Since I R ⊆ z R, we have I ⊆ λ (z R). Since I is a regular
22

e such that z R
e = xR.
e
ideal of R, the claim implies that there exists x ∈ λ−1 (z R)
e = I, and so x ∈ λ−1 (z R)
e ⊆ λ−1 (I R)
e = I.
By Proposition 2.5(3), λ−1 (I R)
e = xI R,
e so again by Proposition 2.5(3), I 2 = λ−1 (I 2 R)
e =
Furthermore, I 2 R
−1
e
λ (xI R) = xI. By Lemma 3.4(4), I is a stable ideal of R, which proves that
e is
R is stable. Finally, if R has finite integral closure, then, by Theorem 4.5, R
e is reduced, contrary to (3). Thus R does not have
a Bass ring. In particular R
finite integral closure.
In the case where R is separated in the mR-adic topology (e.g., if R is
Noetherian), we obtain the following classification of one-dimensional stable
local rings.
Corollary 4.8. With Setting 4.1 and J = 0, the ring R is stable if and only if
e such that R/P
e
R is a Bass ring or there is a nonzero prime ideal P of R
is a
DVR and P 2 = 0.
Proof. Apply Theorems 4.5 and 4.7.
In particular, if R is complete in the mR-adic topology, then there are only
two classes of (one-dimensional) stable rings: Bass rings and rings having a
square zero prime ideal whose residue ring is a DVR. Thus, in the case in which
R is mR-adically complete, R is stable and reduced if and only if R is a Bass
ring. More generally, we have the following corollary.
Corollary 4.9. With Setting 4.1, the ring R is a reduced stable ring if and only
if R is either a Bass ring or a bad stable domain.
Proof. Suppose that R is a reduced stable ring. If R has finite normalization,
then by Corollary 3.12(2), Ri = R∞ , for some i, and so, by Proposition 3.11(7),
J = 0. In this case Theorem 4.5 implies that R is a Bass ring. Otherwise,
suppose that R does not have finite normalization. By Corollary 3.12(2), Ri 6=
R∞ , for all i > 0, and so Proposition 3.11(3) implies R∞ is local. In this case
Proposition 3.11(7) implies that R has a unique minimal prime ideal, and hence,
since R is reduced, R is a domain. By Theorem 4.7 there exists a nonzero prime
e such that R/P
e
ideal P of R
is a DVR and P 2 = 0. By Theorem 2.13, R is a
bad stable domain. The converse is clear in light of Theorem 2.11.
The next two examples illustrate some complications with characterizing
the non-reduced, non-complete one-dimensional stable local rings R such that
e has a prime ideal P with P 2 = 0 and R/P
e
R
is a DVR. Such a ring R has a
2
nonzero prime ideal Q such that Q = 0. The first example shows that even if
R is Noetherian, R/Q need not be a DVR. The second example shows that the
maximal ideal of R can be two-generated but the ring R not be Noetherian.
Example 4.10. A local Cohen-Macaulay stable ring R having a nonzero prime
ideal P such that P 2 = 0 and R/P is a bad stable domain (and hence R/P is not
a DVR). Let (A, M ) be a bad Noetherian stable domain, let R = A ⋆ (A/M )
be the Nagata idealization of the A-module A/M (cf. Remark 2.15) and let
23

∼ A is a bad stable domain. Since
P = 0 ⋆ (A/M ). Then P 2 = 0 and R/P =
A/M is a finitely generated A-module, R is a one-dimensional local CohenMacaulay ring. We claim that R is stable. Let m be a nonzerodivisor in M .
e=A
e ⋆ (A/M
e A),
e where R
e is the mR-adic completion of R and A
e is the
Then R
mA-adic completion of A. Since A is a bad stable domain, Theorem 2.13 implies
e such that A/Q
e
there is a nonzero prime ideal Q of A
is a DVR and Q2 = 0.
e A)
e is a prime ideal of R,
e and L2 = 0 since Q2 = 0 and
Now L := Q ⋆ (A/M
∼
e
e
e
e
e
Q · (A/M A) = 0. Also, R/L = A/Q, so that R/L
is a DVR. By Corollary 4.8,
R is a stable ring. If A is chosen instead to be a non-Noetherian bad stable
domain, then R is a non-Noetherian stable ring.
Example 4.11. A non-Noetherian separated stable local ring (R, M ) of Krull
dimension one whose maximal ideal can be generated by two elements. Let k be
a field, let X be an indeterminate over k, and let V = k[X](X) . The (X)-adic
completion of V is Ve = k[[X]]. Define R = V ⋆ Ve . Since Ve is not a finitely
generated V -module, R is not a Noetherian ring. However, the maximal ideal
M = XV ⋆ Ve of R is generated by two elements, (X, 0) and (0, 1), the first a
nonzerodivisor in R and the second a zerodivisor. Since the ring R has a prime
ideal whose square is zero and
T whoseTresidue ring is a DVR, R is a stable ring
[31, Lemma 3.3]. Moreover i M i = i (mi V ⋆ mi Ve ) = 0, and so R is separated
in the M -adic topology.
5. The two-generator property
In this section we consider the two-generator property in our context. By a
proper power of the ideal I, we mean an ideal of the form I k , where k > 1. Sally
[37, Proposition 1] has shown that if a proper power of a regular ideal (which,
a priori, is not necessarily finitely generated) is two-generated, then so is every
power of the ideal, including the ideal itself. We use this in the next proposition
to connect the property of being two-generated with that of being stable.
Proposition 5.1. Let I be a regular ideal of the local ring R. Then, for some
n > 1, I n is two-generated if and only if I is two-generated with I 2 = aI for
some a ∈ I.
Proof. Suppose that a proper power of I is a two-generated ideal. By the
result of Sally discussed above, I is a two-generated ideal and we may write
I = (x, y)R and I 2 = (x2 , xy, y 2 )R. The argument that follows is based on
the proof of [40, Theorem 3.4]. Since I 2 is two-generated, Nakayama’s lemma
implies that I 2 = (x2 , xy)R = xI, I 2 = (xy, y 2 )R = yI or I 2 = (x2 , y 2 )R. In
the first two cases, we have that I 2 = aI for some a ∈ I. Consider the last case,
I 2 = (x2 , y 2 )R. Then xy = rx2 + sy 2 for some r, s ∈ R. If r is a unit, then
x2 ∈ (xy, y 2 )R = yI, so that I 2 = yI. Similarly, if s is a unit, then I 2 = xI.
If neither r nor s is a unit, then xy ∈ M I 2 , with M the maximal ideal of R.
Thus I 2 = (x2 , y 2 )R = (x2 − xy, xy − y 2 )R + M I 2 , and by Nakayama’s lemma,
I 2 = (x2 − xy, xy − y 2 )R = (x − y)I. This proves that in all cases I 2 = aI for
some a ∈ R. The converse of the proposition is clear.
24

Corollary 5.2. Let R be a reduced one-dimensional local ring with regular
maximal ideal M . If, for some n > 1, M n is two-generated, then every ideal of
R is two-generated.
Proof. Suppose M n is two-generated for some n > 1. By the result of Sally [37,
Proposition 1] discussed above, M is two-generated. Also, by Proposition 5.1,
M 2 = mM for some m ∈ M . Since R is reduced and one-dimensional, Proposition 2.1(1) implies that R is a Noetherian ring. Also, since M 2 = mM and M
is two-generated, Proposition 2.2(1) implies that R has multiplicity at most 2.
By Proposition 2.2(2) every ideal of R is two-generated.
Rush [36, Proposition 2.5] has shown that if R is a local ring for which
every finitely generated regular ideal is two-generated, then R is finitely stable.
Proposition 5.1 leads to a different proof of this result in slightly stronger form.
Corollary 5.3. If every finitely generated regular ideal of the local ring R has
a power that is two-generated, then R is a finitely stable ring.
Proof. Let I be a finitely generated regular ideal of R. By assumption I 2 has a
power that is two-generated. Hence I has a proper power that is two-generated.
By Proposition 5.1, I 2 = aI for some a ∈ I, and so I is stable by Lemma 3.4(5).
Theorem 5.4. With Setting 4.1, the following statements are equivalent.
(1) M n is two-generated, for some n ≥ 2.
e is a Noetherian local ring of multiplicity at most 2.
(2) R
(3) Every regular ideal of R is two-generated.
Proof. (1) ⇒ (2) By Proposition 5.1, M is two-generated and M 2 = mM for
e of R
e is two-generated and M 2 R
e=
some m ∈ M . Thus the maximal ideal M R
T
ib
2e
2
e
b
mM R. Since M = mM , Remark 2.4 implies R = R. Since i M R = 0 and
b is M R-adically
b
b is
R
complete with finitely generated maximal ideal, the ring R
b
Noetherian [8, Theorem 3]. By Proposition 2.5(1), m is a nonzerodivisor in R
2b
b
b
e
with M R = mM R, and so by Proposition 2.2(1) the multiplicity of R = R is
at most 2.
e is a
(2) ⇒ (3) Let I be a regular ideal of R. By Proposition 2.5(1), I R
e
e
regular ideal of R. Since R has Krull dimension 1 and multiplicity at most 2,
e is two-generated. Since I is
Proposition 2.2(2) implies that every ideal of R
k
e k+1 R
e ∼
regular, m ∈ I for some k > 0, and so I R/m
= I/mk+1 R by Proposie = λ(R) + mk+1 R,
e where λ : R → R
e is
tion 2.5(3). By Proposition 2.5(2), R
k+1 e
e
the canonical embedding. Since I R/m R can be generated by two elements
e
as an R-module,
I/mk+1 R can be generated by two elements as an R-module.
Therefore I = (x, y, mk+1 )R for some x, y ∈ I. Since mk+1 ∈ M I, Nakayama’s
lemma implies that I can be generated by two elements.
(3) ⇒ (1) This is clear.
25

Remark 5.5. Example 4.11 is a non-Noetherian ring satisfying the conditions
of Theorem 5.4.
Restricting to Noetherian rings, we have a stronger characterization.
Theorem 5.6. Let R be a Cohen-Macaulay ring with maximal ideal M . Then
these statements are equivalent.
(1) M n is two-generated, for some n ≥ 2.
(2) R has Krull dimension 1 and multiplicity at most 2.
(3) R is one of the following:
(3a) a Bass ring;
(3b) a bad stable domain; or
(3c) a ring containing a nonzero principal prime ideal P such that P 2 = 0
and R/P is a DVR.
Proof. (1) ⇒ (2) Proposition 5.1 implies that every power of M is two-generated,
and hence the Hilbert polynomial of M is constant. Since the degree of this polynomial is one less than the Krull dimension of R, the Hilbert-Samuel theorem
e is a stable
[38, p. 4] implies that R has Krull dimension one. By Theorem 5.4, R
ring of multiplicity at most 2. Since R is Noetherian and has Krull dimension
e = R.
b Therefore, since R
b has multiplicity 2, so does
one, Remark 2.4 implies R
R.
(2) ⇒ (3) By Proposition 2.2(2) every ideal of R is two-generated, and hence
R is a stable ring by Corollary 5.3. If R is a finitely generated R-module, then
R is a Bass ring by Theorem 4.5 and R satisfies (3a).
Suppose that R is not a finitely generated R-module. Since R is one-dimenb = R.
e By Theorem 4.7 there exists
sional and Noetherian, Remark 2.4 implies R
2
b
b
a nonzero prime ideal L of R such that L = 0 and R/L
is a DVR. Viewing R
b we consider two cases.
as a subring of R,
Case 1: L ∩ R = 0.
In this case R is a domain. By Theorem 2.13, R is a bad stable domain and
hence (3b) holds for R.

Case 2: P := L ∩ R is a nonzero prime ideal of R.
We show that (3c) holds. Since L2 = 0, we have that P 2 = 0. Since R is
stable, Lemma 3.4(6) implies there exists m ∈ M such that M 2 = mM . We
claim that P 6⊆ mR. Suppose to the contrary that P ⊆ T
mR. SinceTP is a prime
ideal and m 6∈ P , it follows that P = mP . Hence P ⊆ i mi R = i M i = 0, a
contradiction. Thus P 6⊆ mR.
Next we claim that M 2 ( mR. Indeed, if M 2 = mR, then, since m is a
nonzerodivisor, M is an invertible, hence principal, ideal of R. Thus R is a
regular local ring, hence a domain. However, P 2 = 0 and P is nonzero, so this
contradiction shows that M 2 ( mR.

26

Now since M 2 ( mR, P 6⊆ mR and M/M 2 is a vector space of dimension
at most 2 over R/M , it follows that M = mR + P . Thus R/P is a DVR.
Since M = mR + P and P 2 = 0, we have P/M P = P/(mP + P 2 ) = P/mP =
P/(mR ∩ P ) ∼
= (P + mR)/mR = M/mR. Since M/M 2 has dimension at most 2
as a vector space over R/M and M 2 ( mR, it follows that P/M P has dimension
at most one as an R/M -vector space. By Nakayama’s lemma, P is a principal
ideal, and so (3c) is satisfied by R.
(3) ⇒ (1) It is clear that if R is a Bass ring, then every power of M is twogenerated. Similarly, if R is a bad stable domain with multiplicity ≤ 2, then, by
Proposition 2.2(2), every power of M is two-generated. Finally, suppose there is
a nonzero principal prime ideal P of R with R/P a DVR and P 2 = 0. Since R/P
is a DVR and P 2 = 0, it follows that M 2 = mM for some m ∈ M [31, Lemma
3.3]. Since P is a principal ideal and R/P is a DVR, M is two-generated. Thus
M 2 = mM is also two-generated, which verifies (1).
Greither [13, Theorem 2.1] has shown using multiplicity theory that a reduced Noetherian local ring R is a Bass ring if and only if R can be generated
by two elements as an R-module. Theorem 5.8, which uses a different method
of proof, generalizes this result to rings that a priori need not be Noetherian.
Lemma 5.7. If R ⊆ S is an extension of rings such that S can be generated by
two elements as an R-module, then R ⊆ S is a quadratic extension.
Proof. By [34, Lemma 3.1], it suffices to show that RM ⊆ SM is a quadratic
extension for each maximal ideal M of R. Thus we may assume without loss
of generality that R is a local ring. Since S can be generated by two elements,
Nakayama’s lemma implies that S = R + xR for some x ∈ S. Let s, t ∈ S.
We claim that st ∈ sR + tR + R. Write s = a + xb and t = c + xd for some
a, b, c, d ∈ R. Since S = R + xR and S is a ring, x2 ∈ R + xR. Hence
st

= ac + bcx + adx + bdx2 ∈ R + bxR + dxR + dx2 R
⊆ R + bxR + dxR = R + sR + tR,

This proves R ⊆ S is a quadratic extension.
Recall that a generalizedTlocal ring is a local ring with finitely generated
maximal ideal M such that i M i = 0; cf. [8].
Theorem 5.8. The following are equivalent for a one-dimensional local ring R
with regular maximal ideal M .
(1) R is a Bass ring.
(2) R is a generalized local ring whose integral closure R can be generated by
two elements as an R-module.
(3) R is a Noetherian ring that can be generated by two elements as an Rmodule.

27

Proof. (1) ⇒ (2) Suppose R is a Bass ring. Then R is a generalized local ring.
Since R is a finitely generated R-submodule of Q(R), there is a nonzerodivisor
r ∈ R such that rR is an ideal of R. Since R is a Bass ring, rR is a two-generated
ideal of R. Since r is a nonzerodivisor, R and rR are isomorphic as R-modules.
Thus R can be generated by two elements as an R-module.
(2) ⇒ (3) Since R is a finitely generated R-submodule of Q(R) and M is
a regular ideal, the ideal (R : R) contains
a nonzerodivisor
m ∈ M . Also,
T
T
since R is a generalized local ring, i mi R = i M i = 0. We claim that R
is reduced. The argument is essentially that of [24, pp. 263-264]. Let r be a
nilpotent element of R. For each i >T0, r/mi ∈ R. Since m ∈ (R : R), we have
m(r/mi ) ∈ R for all i > 0. Thus r ∈ i>1 mi−1 R = 0, proving that r = 0 and R
is reduced. Since the maximal ideal of R is finitely generated, R is a Noetherian
ring by Proposition 2.1(1). Therefore, since R is a reduced one-dimensional
Noetherian local ring with regular maximal ideal, R is a Noetherian ring [1,
Theorem 3].
(3) ⇒ (1) Given (3), to prove that R is a Bass ring it suffices by Corollary 4.9
to show that R is a reduced stable ring. For this it is enough by Theorem 4.2
to prove that R ⊆ R is a quadratic extension and R is a reduced Dedekind ring
with at most two maximal ideals. Since the R-module R can be generated by
two elements, R/(JacR) has dimension at most 2 as an R/M -vector space. Thus
R has at most 2 maximal ideals. Since R is a one-dimensional local Noetherian
ring with regular maximal ideal and R is a Noetherian ring, R is a finite product
of Dedekind domains [1, Theorem 3]. Hence R is a reduced Dedekind ring. By
Lemma 5.7, R ⊆ R is a quadratic extension, and so the proof is complete.
Acknowledgment. I am grateful to the referee for a careful reading of the
paper and many suggestions that have improved the presentation.
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