Star operations on overrings of Noetherian domains

Journal of Pure and Applied Algebra 220 (2016) 810–821

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Journal of Pure and Applied Algebra www.elsevier.com/locate/jpaa

Star operations on overrings of Noetherian domains Evan Houston a,∗,1 , Abdeslam Mimouni b , Mi Hee Park c,2 a

Department of Mathematics and Statistics, University of North Carolina at Charlotte, Charlotte, NC 28223, USA b Department of Mathematics and Statistics, King Fahd University of Petroleum & Minerals, P. O. Box 278, Dhahran 31261, Saudi Arabia c Department of Mathematics, Chung-Ang University, Seoul 156-756, Republic of Korea

a r t i c l e

i n f o

Article history: Received 5 August 2014 Received in revised form 1 June 2015 Available online 12 August 2015 Communicated by D. Nakano/G. Rosolini MSC: 13A15; 13E05

a b s t r a c t Let R be a Noetherian domain, and let Star(R) denote the set of star operations on R. Call R star regular if |Star(T )| ≤ |Star(R)| for each overring T of R. In the case where Star(R) is finite we show that star regularity becomes a local property, and, further assuming that R is one-dimensional and local with infinite residue field, we prove that R is star regular. We also study the question of whether finiteness of Star(T ) for each proper overring of a one-dimensional Noetherian domain R implies finiteness of Star(R). © 2015 Elsevier B.V. All rights reserved.

0. Introduction Let R be an integral domain with quotient field K, and let F(R) be the set of nonzero fractional ideals of R. A mapping ∗ : F(R) → F(R), E → E ∗ , is called a star operation on R if the following conditions hold for all a ∈ K \ {0} and E, F ∈ F(R): (I) (aE)∗ = aE ∗ and R∗ = R; (II) E ⊆ E ∗ ; if E ⊆ F , then E ∗ ⊆ F ∗ ; and (III) (E ∗ )∗ = E ∗ . The simplest star operations are the d-operation, defined by E d = E for every E ∈ F(R), and the v-operation defined by E v = (R : (R : E)) for every E ∈ F(R). We denote by Star(R) the set of all star operations on R, and we say that R is a divisorial domain if d = v, equivalently, Star(R) = {d}, equivalently, |Star(R)| = 1. * Corresponding author. E-mail addresses: [email protected] (E. Houston), [email protected] (A. Mimouni), [email protected] (M.H. Park). The first-named author was supported by a visiting grant from GNSAGA of INdAM (Istituto Nazionale di Alta Matematica). 2 The third-named author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2010-0021883). 1

http://dx.doi.org/10.1016/j.jpaa.2015.07.018 0022-4049/© 2015 Elsevier B.V. All rights reserved.

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For several years, motivated by well-known characterizations of integrally closed and Noetherian divisorial domains [3,8], we have been interested in domains admitting only finitely many star operations [4–7]. Our investigations have often involved examination of the star operations on overrings of the base ring (that is, rings between the base ring and its quotient field), and the goal of this paper is to study, for a Noetherian domain R, how |Star(R)| affects |Star(T )| for each overring T of R, and conversely, with emphasis on the case where Star(R) is finite. Call a domain R star regular if |Star(R)| ≥ |Star(T )| for each overring T of R. Since for (non-field) Noetherian domains R, finiteness of Star(R) implies that the (Krull) dimension of R is one [5, Theorem 2.1], we usually impose this restriction. In Section 1 we give a simple example of a one-dimensional local Noetherian domain R that fails miserably to be star regular; indeed, for this R, we have |Star(R)| = 1 but |Star(T )| = ∞ for some overring T . We show, for a one-dimensional Noetherian domain R, that if R is locally star regular, then it is star regular, and we prove the converse in the case |Star(R)| < ∞ (and give an example showing that the converse does not hold without the finiteness assumption). We conjecture that if R is a local Noetherian domain with 1 < |Star(R)| < ∞, then R is star regular. The main result of the paper is that the conjecture holds when R has infinite residue field (Corollary 1.18). The proof relies heavily on the characterization in [7] of local Noetherian domains that have infinite residue field and only finitely many star operations. In Section 2, we consider the question of whether finiteness of Star(T ) for each proper overring of a Noetherian domain R implies finiteness of Star(R). We show in Theorem 2.2 that this does occur when R is non-local, and in Theorems 2.3 and 2.4, we show that if (R, M ) is a local Noetherian domain for which each proper overring of R admits only finitely many star operations, then dim R = 1, and, further assuming that R/M is infinite, that (R : M ) is either a PID with at most four maximal ideals or (R : M ) has two maximal ideals, both of whose residue fields are isomorphic to R/M . This is as close to a characterization of the situation as we come, and we close with several classes of examples of local Noetherian domains (R, M ) (with infinite residue field) for which |Star(R)| = ∞ but |Star(T )| < ∞ for each proper overring T of R. 1. Star regularity We recall the definition: Definition 1.1. A domain R is star regular if |Star(R)| ≥ |Star(T )| for each overring T of R. Since we shall refer to it several times, we restate [5, Theorem 2.3]. Lemma 1.2. For a Noetherian domain R, |Star(R)| =

 M ∈Max(R)

|Star(RM )|.

We remark that the proof of this result in [5] did not distinguish cardinalities in the infinite case. However, [2, Theorem 2.3] is a generalization of Lemma 1.2, and there the proof is valid for all cardinalities. We begin with an example of a local Noetherian divisorial domain R that not only fails to be star regular but that has an overring T satisfying |Star(T )| = ∞. Example 1.3. Let k be a field, D = k[X 3 , X 7 ], and P = (X 3 , X 7 )D. Then P −1 = k[X 3 , X 7 , X 11 ] = D + DX 11 . Now let R = DP and M = P DP . Then M −1 = P −1 DP , and dimk M −1 /M = 2. Thus |Star(R)| = 1. If N = (X 3 , X 7 , X 11 )M −1 , then S0 := (M −1 : N ) = (N : N ) = k[X 3 , X 4 ](X 3 ,X 4 )k[X 3 ,X 4 ] = M −1 + M −1 X 4 + M −1 X 8 . It follows that S0 is singly generated as an M −1 -algebra, and it is easy to see that dimk S0 /N = 3. If Q denotes the maximal ideal of S0 , then S1 := (S0 : Q) = (Q : Q) = k[X 3 , X 4 , X 5 ](X 3 ,X 4 ,X 5 )k[X 3 ,X 4 ,X 5 ] . Then, since S1 is not a PID and dimk S1 /N S1 = 3, [7, Proposition 2.8] yields |Star(M −1 )| ≥ |k| + 2. In particular, |Star(M −1 )| = ∞ if k is infinite. (It is easy to see that we also have |Star(D)| = 1 and |Star(P −1 )| = ∞ if |k| = ∞. Since we do not need this, we omit the details.)

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Remark 1.4. (1) The reader may have noticed that, in Example 1.3, we could have taken R = kJX 3 , X 7 K, in which case the details are somewhat simpler. However, we will make use of the construction in Example 1.3 again in Example 1.7 below, and there we need the polynomial version. (2) For a simple example of a local Noetherian divisorial domain that is star regular, let k be a field, and take R = kJX 2 , X 3 K. It is well known that R is a divisorial domain (|Star(R)| = 1). Let T be a nonfield proper overring of R. Then since kJXK = R ⊆ T and kJXK is a 1-dimensional rank-one discrete valuation domain (DVR), T = kJXK and hence T ⊆ kJXK. Since there is no intermediate ring properly between R and kJXK, this implies that T = kJXK, and hence |Star(T )| = 1. (3) If k is finite in Example 1.3, then R is still not star regular, but it is easy to see via [5, Theorem 2.5] that each overring T of R satisfies |Star(T )| < ∞. We proceed to show that this can occur even when the residue field is infinite. Let R = kJX 3 , X 4 K, where k is a field, and let M be the maximal ideal of R. Since M −1 = k + X 3 kJXK = R + RX 5 , |Star(R)| = 1 by [5, Theorem 2.4]. Let N be the maximal ideal of M −1 . Then S := (M −1 : N ) = (N : N ) = kJXK. Since S is a 1-dimensional DVR with maximal ideal different from N and dimk S/N = 3, |Star(M −1 )| = 3 by [5, Theorem 3.1(4)]. Now let T be a nonfield proper overring of R. Since kJXK = R ⊆ T , we have T = kJXK and hence T ⊆ kJXK. Since R  T ⊆ kJXK, T contains an element of the form aX +bX 2 +cX 5 , with a, b, c ∈ k and not all zero. If a = 0, then T contains an element of the form X + b X 2 + c X 5 . Since T is a ring, (X + b X 2 + c X 5 )2 ∈ T . Also, since T contains R, this implies that X 2 ∈ T . Consequently, we have X 5 = X 2 · X 3 ∈ T and X ∈ T . Therefore, in this case, T = kJXK and |Star(T )| = 1. Assume that a = 0 but b = 0. Then T contains an element of the form X 2 + c X 5 . Since X 3 (X 2 + c X 5 ) ∈ T , we have X 5 ∈ T and consequently, X 2 ∈ T . Therefore, in this case, T ⊇ k + X 2 kJXK (in fact, T = k + X 2 kJXK) and hence |Star(T )| = 1. Finally, assume that a = b = 0 but c = 0. Then since T contains the element X 5 , T ⊇ M −1 (in fact, T = M −1 ) and hence |Star(T )| = 3. The next result shows that for the class of one-dimensional Noetherian domains admitting only finitely many star operations, star regularity is a local property: Theorem 1.5. Let R be a one-dimensional Noetherian domain. Then (1) if RN is star regular for each maximal ideal N of R, then R is star regular, and (2) if R is star regular and |Star(R)| < ∞, then RN is star regular for each maximal ideal N of R. K be an overring Proof. (1) Assume that RN is star regular for each maximal ideal N of R. Let T = of R, and let M denote the set of maximal ideals of R that are contracted from T . Since T is also a one-dimensional Noetherian domain, we may use Lemma 1.2 to obtain |Star(T )| =



|Star(TP )|

P ∈Max(T )

=





|Star(TP )|

M ∈M P ∩R=M

=



|Star(TR\M )|

M ∈M

≤



|Star(RM )|

M ∈M

≤ |Star(R)|. (2) Assume that |Star(R)| < ∞. We may assume that |Max(R)| > 1. Suppose that RN is not star regular for some maximal ideal N of R. Then RN has an overring T with |Star(T )| > |Star(RN )|. Set

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 U := {RM | M ∈ Max(R) \ {N }} and S = T ∩ U . We first note that T ⊇ RN , and, since RN is a PID with only finitely many maximal ideals, the same is true of T . It follows that T has only finitely many maximal ideals, say Q1 , . . . , Qn . Since R is a 1-dimensional Noetherian domain, it is h-local, and hence UR\N = K by [9, Theorem 22]. This yields SR\N = TR\N ∩ UR\N = T ∩ K = T . It follows that Qi ∩ S = Qj ∩ S for i = j. It is clear that if P is a maximal ideal of S different from the Qi ∩ S, then P ∩ R = M for some maximal ideal M of R distinct from N and hence P = M RM ∩ S and SP = RM . We then have Max(S) = {Qi ∩ S | i = 1, . . . , n} ∪ {M RM ∩ S | M = N }. Therefore, |Star(S)| =

n 

|Star(SQi ∩S )| ·

n 

|Star(RM )|

M =N

i=1

=





|Star(TQi )| ·

i=1

= |Star(T )| ·



|Star(RM )|

M =N

|Star(RM )|

M =N

> |Star(RN )| ·



|Star(RM )|

M =N

= |Star(R)|. Therefore, R is not star regular. 2 In the next example, we show that a domain with infinitely many star operations may or not be star regular and that star regularity does not necessarily localize. We need the following lemma, the proof of which is a straightforward modification of the proof of [5, Theorem 2.5]. Lemma 1.6. Let (R, M ) be a local Noetherian domain such that k := R/M is countable and R a finite R-module. Then |Star(R)| ≤ 2ℵ0 . Example 1.7. (a) Let k be a countable field and {Xi | i = 0, 1, 2, . . .} a set of indeterminates. Then set ∞ D = k[X03 , X07 , {Xi3 , Xi4 , Xi5 | i > 0}], P0 = (X03 , X07 )D, Pi = (Xi3 , Xi4 , Xi5 )D for i > 0, and S = D \ i=0 Pi .  Suppose that I is an ideal of D with I  Pi for each i. We claim that I  i Pi . To verify this, let 0 = f ∈ I. Then f involves powers of only finitely many of the Xi ; in particular, f is contained in only finitely many Pi , n n n say f ∈ j=1 Pij , f ∈ / Pi for i = i1 , . . . , in . Then I  j=1 Pij ; choose g ∈ I \ j=1 Pij . Now let r be larger than any subscript of any Xi occurring in the representation of f as a sum of monomials. It is then easy ∞ to check that f + Xr3 g ∈ I \ i=0 Pi . From this it follows that R := DS is a one-dimensional Noetherian domain with Max(R) = {Mi | i ≥ 0}, where Mi = Pi R. Next, note that D = k[X0 , X1 , . . .] and that X0 D ∩D = (X03 , X07 )D and Xi D ∩D = (Xi3 , Xi4 , Xi5 )D for i > 0. Hence R = DS = D D\ Xi D is a PID with i Max(R) = {Xi R}. Now set ki = k({Xj | j = i}), and observe that RM0 = k0 [X03 , X07 ](X03 ,X07 )k0 [X03 ,X07 ] = k0 + (X03 , X07 )k0 [X03 , X07 , X011 ](X03 ,X07 ,X011 )k0 [X03 ,X07 ,X011 ] and RMi = ki [Xi3 , Xi4 , Xi5 ](Xi3 ,Xi4 ,Xi5 )ki [Xi3 ,Xi4 ,Xi5 ] = ki + Xi3 RXi R . Moreover, (RM0 : M0 RM0 ) = k0 [X03 , X07 , X011 ](X03 ,X07 ,X011 ) , and (RMi : Mi RMi ) = RXi R for i > 0, so that dimk0 (RM0 : M0 RM0 )/M0 RM0 = 2, while dimki (RMi : Mi RMi )/Mi RMi = 3 for i > 0. Hence |Star(RM0 )| = 1, and, by [5, Theorem 3.8], |Star(RMi )| = 3 for i > 0. Using Lemma 1.2, we then  have |Star(R)| = i |Star(RMi )| = 3ℵ0 . Now let T ( = K) be an overring of R. If P is a maximal ideal of T and P ∩ R = Mi , i > 0, then RMi ⊆ TP , and hence RMi ⊆ TP . Since RMi is a 1-dimensional DVR, RMi = TP and so TP ⊆ RMi . Then, according to the proof of [5, Theorem 3.8], TP must be one of RMi , RMi , or RMi + Xi2 RMi . The latter two are divisorial domains, and hence |Star(TP )| ≤ 3. (Hence RMi is star regular for i > 0.) Suppose P ∩ R = M0 . The residue field of TP is k0 , which is countable. Hence,

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by Lemma 1.6, TP has at most 2ℵ0 star operations. Therefore, since TR\M0 is semilocal (in fact, it is the local ring TP ), |Star(TR\M0 )| ≤ 2ℵ0 also (by Lemma 1.2). Thus, as in the proof of Theorem 1.5, we have  |Star(T )| = |Star(TR\M0 )| · P ∩R=M0 |Star(TP )| ≤ 2ℵ0 · 3ℵ0 = |Star(R)|. Therefore, R is star regular. Of course, RM0 is not star regular by Example 1.3. (b) Let everything be as above, but assume that |k| > 3ℵ0 . According to Example 1.3, the ring T := k0 [X03 , X07 , X011 ](X03 ,X07 ,X011 )k0 [X03 ,X07 ,X011 ] satisfies |Star(T )| ≥ |k0 | = |k| > |Star(R)|. Hence R is not star regular. (c) Modify the example in (a) by deleting X0 from consideration, and make the obvious changes in notation. We have |Star(R)| = 3ℵ0 . Then, as before, each RMi is star regular, and, by Theorem 1.5, so is R. Remark 1.8. The example in (a) also shows that star finiteness is not a local property, since Star(R) is infinite while |Star(RM )| < ∞ for each maximal ideal M of R. (This is not surprising.) In view of Theorem 1.5 (and Example 1.3), we proceed to consider star regularity for one-dimensional local domains (R, M ) for which 1 < |Star(R)| < ∞. Under these assumptions, we suspect–but cannot prove–that R must be star regular. However, if we further assume that the residue field is infinite, then we are able to show that R is star regular. The proof involves a case-by-case analysis of the characterization in [7]. We begin with a case that does not depend on the residue field (or on the characterization in [7]). It is convenient to insert a preparatory result. Proposition 1.9. Let D be a Noetherian domain, and assume that D is a finitely generated D-module. Then D is divisorial as a fractional ideal of D. Proof. Since D is finitely generated over D, J := (D : D) is a common nonzero ideal of D and D. Hence J −1 JD = J −1 J ⊆ D, and we have J −1 J ⊆ (D : D) = J. Thus J −1 = (J : J). Since D is Noetherian, (J : J) is integral over D, and we must have J −1 = (J : J) = D. Therefore, D is a divisorial fractional ideal of D. 2 Proposition 1.10. Let (R, M ) be a one-dimensional local Noetherian domain with quotient field K, and assume that M −1 is a (necessarily rank-one discrete) valuation domain. Then R is star regular. Proof. Let T = K be a proper overring of R. Then T ⊇ R = M −1 , and we must have T = M −1 . In particular, T ⊆ M −1 . Now let ∗ be a star operation on T , and define δ(∗) on R by E δ(∗) = (ET )∗ ∩ E v . Then δ(∗) is a star operation on R by [1, Theorem 2]. Let A be a fractional ideal of T . Then A is also a fractional ideal of R, and by [5, Lemma 3.7], we may assume that R  A ⊆ M −1 . Since A is a fractional ideal of T , we must then have T ⊆ A. By Proposition 1.9, A∗ ⊆ (T )∗ ⊆ (T )vT = T = M −1 . We also have Av = M −1 . Hence Aδ(∗) = A∗ ∩ Av = A∗ . It follows that the map δ : Star(T ) → Star(R) is 1–1. 2 Next, we give an example showing that |Star(T )| = |Star(R)| is possible. Example 1.11. Let (R, M ) and S be as in [7, Proposition 2.13 and its proof]. Then |Star(R)| = 5. We have that T = R + M S is a local ring with maximal ideal M S, and S = (T : M S) is a PID with two maximal ideals P1 , P2 and M S = P1 P2 . Therefore, we also have |Star(T )| = 5 by [5, Theorem 3.13]. (For a specific example, see [7, Example 2.14].) It is convenient to state the following results from [7,2]. Theorem 1.12. Let (R, M ) be a one-dimensional local Noetherian domain with residue field k. Assume that M is not principal.

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(1) If dimk M −1 /M = 2, then |Star(R)| = 1. (2) If dimk M −1 /M = 3 and M −1 is local with maximal ideal M , then M −1 is a PID and |Star(R)| = 2. (3) Assume that dimk M −1 /M = 3 and that M −1 is local with maximal ideal N = M . Let S = (N : N ). (a) If M −1 is a PID, then |Star(R)| = 3. (b) If M −1 is not a PID, M −1 is singly generated as an R-algebra, and S is a local PID with dimk S/M S = 3, then |Star(R)| = 4. (c) If M −1 is not a PID, M −1 is singly generated as an R-algebra, and S is nonlocal with dimk S/M S = 3, then S is a PID and |Star(R)| = 5. (4) Assume that dimk M −1 /M = 3 and that M −1 has three maximal ideals. Then M −1 is a PID and |Star(R)| = 9. (5) Assume that dimk M −1 /M = 3 and that M −1 is a PID with two maximal ideals. If M is the Jacobson radical of M −1 , then |Star(R)| = 3; otherwise, |Star(R)| = 5. (6) Assume that dimk M −1 /M = 3, and that M −1 has two maximal ideals, exactly one of which is principal. Set S−1 = M −1 , and for i ≥ 0, set Si = (Ji−1 : Ji−1 ), where Ji−1 is the Jacobson radical of Si−1 . Further assume that Sn is a PID for some n > −1 and that dimk Si /M Si = 3 for −1 ≤ i ≤ n. Then Sn has two or three maximal ideals. (a) If Sn has two maximal ideals and M Sn = Jn , then |Star(R)| = 4n + 7. (b) If Sn has two maximal ideals and M Sn = Jn , then |Star(R)| = 4n + 9. (c) If Sn has three maximal ideals, then (M Sn = Jn and) |Star(R)| = 4n + 13. Theorem 1.13. Let (R, M ) be a local Noetherian domain with infinite residue field such that |Star(R)| < ∞. Then R has Krull dimension one and is either a DVR or falls into one of the categories in Theorem 1.12. Lemma 1.14. Let (W, Q) be a one-dimensional local Noetherian domain, and let (V, P ) be a local overring that is finitely generated as a W -module. Assume that the inclusion W/Q ⊆ V /P is an isomorphism and that P = QV . Then V = W . Proof. Since W/Q ∼ = V /P , V = W + P . Hence Q(V /W ) = (W + QV )/W = (W + P )/W = V /W , and the conclusion follows from Nakayama’s lemma. 2 The next lemma is undoubtedly well known, but we include a brief proof for the sake of completeness. Lemma 1.15. Let (V, P ) be a DVR with residue field k, and let W be a local subring with maximal ideal P n , also with residue field k. Then W is Noetherian, dimk V /P n = n, and V is an n-generated W -module. Proof. Write P = V x. Then V = W + P = W + V x = W + (W + V x)x = W + W x + V x2 = · · · = W + W x + · · · + W xn−1 . Thus V is an n-generated W -module. It is routine to show that 1, x, . . . , xn−1 are linearly independent modulo P n and hence that dimk V /P n = n. That W is Noetherian then follows from Eakin’s theorem. 2 The next lemma is given in [7, Lemma 3.17] with the extra condition |k| > 2. We proceed to show that this condition is unnecessary. Lemma 1.16. Let (R, M ) be a local Noetherian domain with residue field S and dimk M −1 /M = 3. Assume that M −1 is a PID, or that R is as in Theorem 1.12(6). Then M is principal in M −1 . Proof. Let n ≥ −1 be the smallest n for which Sn is a PID. If n = −1, i.e., M −1 is a PID, then M = M M −1 is principal in M −1 . Assume n > −1. Let N1 , N2 be maximal ideals of M −1 with N2 principal. By [7, Lemma 3.10(2)], each Si , i < n, has exactly two maximal ideals, one of which is principal. By induction

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(with (R + N1 N2 , N1 N2 ) replacing (R, M )) N1 N2 is principal in S0 . Hence N1 = N1 S0 is also principal in S0 , and, since S0 is a 3-generated R-module, so is N1 . Therefore, dimk N1 /M N1 = 3, and, since M N1 = N12 N2  M  N1 N2  N1 by [7, Lemmas 3.2(4) and 3.10(1)], there can be no R-modules and hence −1 no M −1 -modules properly between M N1 and M . This implies that there are no MN -modules properly 1 −1 −1 −1 −1 between M N1 MN1 and M MN1 . Then, by Nakayama’s lemma, M MN1 is principal. Since MN is a PID, 2 −1 −1 M MN2 is also principal, and it follows that M is principal in M . 2 Theorem 1.17. Let (R, M ) be a local Noetherian domain that falls into one of the categories (2)–(6) of Theorem 1.12. Then R is star regular. Proof. Let W be a proper overring of R. We consider each case of Theorem 1.12 in turn. (2) Since M −1 is a local PID, we can apply Proposition 1.10. (3a) This also follows from Proposition 1.10. (3b) Since W ⊇ R = S, we have W = S. In particular, W ⊆ S. Referring to the proof of [7, Proposition 2.10], W = R + M S, W = M −1 , or W = S. In the first case, the maximal ideal P of S satisfies P = M S by [7, Lemma 2.3], and it is easy to see that (W : M S) = (M S : M S) = S. Hence by applying Theorem 1.12(3a) (W in place of R), we obtain |Star(W )| = 3 < 4 = |Star(R)|. In the second case, M  M S  N  P by [7, Lemmas 2.3 and 2.5], and hence M −1 /N ∼ = k and dimk S/N = 2. Thus |Star(W )| = 1 by Theorem 1.12(1). In the third case, obviously, |Star(W )| = 1. (3c) Refer to [7, Proposition 2.13] and its proof. In particular, |Star(R)| = 5. Let P1 , P2 be maximal ideals of S. Then M S = P12 P2 or P1 P22 . Assume that M S = P12 P2 . If W ⊆ S, then W = R + M S, (R + M S) + P12 = R + P12 , M −1 (= R + N = R + P1 P2 ), or S. In the first case, |Star(W )| = 5 as in Example 1.11. If W = M −1 , then S = (W : N ) = (N : N ), and it is clear that dimk S/N = 2. Hence, in this case, |Star(W )| = 1. Suppose W = R + P12 . Then W has two maximal ideals: P12 and Q := P2 ∩ W . We have P12 WP12 = P12 SW \P12 = (P1 SP1 )2 and hence SP1 is a 2-generated WP12 -module by Lemma 1.15. We also have QSP2 ⊇ M SP2 = P2 SP2 , whence WQ = SP2 by Lemma 1.14. Therefore, |Star(W )| = 1 by [8, Theorem 3.8]. The case W = S is trivial. Finally, suppose that W  S. Then W  S. We may write S = R[u1 , . . . , ul ] for elements ui ∈ S (actually, we can take l ≤ 2). Hence S  W [u1 , . . . , ul ] ⊆ W . Since S is a PID, W [u1 , . . . , ul ] is integrally closed and must therefore be equal to W . Thus W is a finitely generated W -module. Also, since S  W , W is one of the valuation domains SP1 or SP2 . In particular, W is local with maximal ideal (say) Q. Note that QW ⊇ M S = P12 P2 . Suppose that W = SP2 . Then QW = P2 W , and we have W = W by Lemma 1.14. Now suppose W = SP1 . In this case, QW ⊇ P12 W , and we may as well assume equality (otherwise, we obtain W = SP1 as above). Since W is a finitely generated W -module, W and W share an ideal, necessarily equal to P1m W for some m ≥ 2. In particular, Q ⊇ P1m . Now R + P1m is a ring properly containing R, and upon considering the possibilities for such a ring (listed above), we must have R + P1m = U := R + P12 . Then, since W is local and Q ⊇ P12 , we must in fact have UP12 ⊆ W . Now observe that, since U = S, UP12 = SU \P12 = W , and, since P12 is an ideal of S, P12 UP12 is an ideal of W , necessarily equal to P12 W . By Lemma 1.15, W is a 2-generated UP12 -module, whence W = UP12 (and |Star(W )| = 1). (4) Refer to the proof of [5, Theorem 3.14]. In particular, denote the maximal ideals of M −1 by N1 , N2 , N3 , and note that all residue fields are equal to k = R/M . There are three rings properly between R and M −1 : T = R + N1 N2 , R + N1 N3 , and R + N2 N3 . By symmetry, we need only consider T , which has two maximal ideals, N1 N2 and P := N3 ∩T . It is easy to see that dimk M −1 /N1 N2 = 2, from which it follows that −1 −1 MN = MT−1 = N3 . \N1 N2 is a 2-generated TN1 N2 -module. Hence |Star(TN1 N2 )| = 1. We also have P M 1 ∪N2 (Write 1 = y + z with y ∈ N1 N2 and z ∈ P . Then for x ∈ N3 , we have x = xy + xz ∈ M + P N3 ⊆ P M −1 .) −1 Now apply Lemma 1.14 to obtain TP = MN . It follows that |Star(T )| = 1. 3 −1 If W = M , then, obviously, |Star(W )| = 1. Now suppose that W  M −1 . As above, W  M −1 , and −1 −1 we may assume by symmetry that W = MN or W = MN . Also as above, W is a finitely generated 1 1 ∪N2

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−1 W -module. If W = MN , then W is local; call its maximal ideal Q. Since Q ⊇ M = N1 N2 N3 , we have 1 −1 QW = N1 W and hence W = W (so that |Star(W )| = 1) by Lemma 1.14. Now suppose that W = MN . 1 ∪N2 Then W is either local with maximal ideal Q or has two maximal ideals, Q1 and Q2 . In the first case, since W is a finitely generated W -module, so that W and W share an ideal, from which it follows that Q ⊇ (N1 N2 )m for some m ≥ 1. Then R + (N1 N2 )m is a ring properly between R and M −1 , and the only −1 possibility is R + (N1 N2 )m = R + N1 N2 . It follows that TN1 N2 ⊆ W . Then, since W = MN is a 1 ∪N2 2-generated TN1 N2 -module, we must have W = TN1 N2 or W . If W is nonlocal, then W has two maximal ideals, Q1 := N1 W ∩ W and Q2 := N2 W ∩ W . For i = 1, 2, we have Qi W Ni W ⊇ M W Ni W = Ni W Ni W , and hence WQi = W Ni W (Lemma 1.14). It follows that W = W . In both cases, we have |Star(W )| = 1. (5) First, suppose that M is the Jacobson radical of M −1 . Let N1 , N2 be maximal ideals of M −1 . Referring to [7, Lemma 3.1], we may assume that [M −1 /N1 : R/M ] = 2 and M −1 /N2 = k = R/M . Then U := R + N1 = R + N12 is the only ring properly between R and M −1 , and U has two maximal ideals, −1 −1 −1 −1 N1 and P := N2 ∩ U . Note that P MN ⊇ M MN = N2 MN . Hence UP = MN by Lemma 1.14. Also, 2 2 2 2 −1 −1 −1 −1 N1 UN1 is the maximal ideal of MU \N1 = MN1 . Since dimk MN1 /N1 MN1 = 2, UN1 is a divisorial domain, and hence U is a divisorial domain (and |Star(U )| = 1). We may now assume that W is a proper overring −1 of R and that W  M −1 . As above W is a finitely generated W -module and is therefore equal to MN 1 −1 −1 or MN2 . In particular, W is local with maximal ideal Q. If W = MN2 , then, using the fact that Q ⊇ M , −1 −1 it is easy to see via Lemma 1.14 that W = MN . If W = MN , then much as in (4) we may argue that 2 1 Q ⊇ N1m for some integer m. Also, as before, we then have U = R + N1m , whence, in fact, Q ⊇ N1 . We then −1 −1 have UN1 ⊆ W ⊆ MN , and hence W = UN1 or MN (whence |Star(W )| = 1). 1 1 Now suppose that M is not the Jacobson radical of M −1 . We refer to [5, Theorem 3.13] and its proof. We have |Star(R)| = 5. Denote the maximal ideals of M −1 by N1 , N2 , and note that both residue fields are isomorphic to k = R/M . We may assume that M = N12 N2 . There are then two rings properly between R and M −1 : T = R + N1 N2 and U = R + N12 . It is clear that T is local with maximal ideal N1 N2 , and, since dimk M −1 /M = 3, M −1 = (T : N1 N2 ) is a 2-generated T -module. Hence |Star(T )| = 1. The −1 ring U has two maximal ideals N12 and P := N2 ∩ U . We have UP = MU−1 \P = MN2 with P UP = N2 UP −1 (since P ⊇ M = N12 N2 ). We may then apply Lemma 1.14 to obtain UP = MN2 . It is easy to see that (U : N12 ) = (N12 : N12 ) = M −1 . Also, since dimk M −1 /M = 3, we must have dimk M −1 /N12 = 2. It follows −1 that MN = MU−1 is a 2-generated UN12 -module. Therefore, U is a divisorial domain by [8, Theorem 3.8]. \N 2 1 1

−1 −1 Now let W be an overring of R with W  M −1 . Then W = MN or MN . As above, we can show that 1 2 −1 W is a finitely generated W -module. Also as above, we can then show that, if W = MN , then W = UN12 1 −1 −1 −1 or MN1 , and, if W = MN2 , then W = MN2 . (6) Let n > −1 be the smallest integer for which Sn is a PID. We induct on n. Let N1 , N2 be maximal ideals of M −1 with N2 = M −1 y. Then the ring U = R + N12 also has two maximal ideals, which are easily seen to be Q1 := M + N12 and Q2 := N2 ∩ U . By Lemma 1.16 and its proof, N1 = S0 x for some x ∈ N1 , and M is principal in M −1 , say M = M −1 c. By [7, Lemma 3.10], M S0 = N1 N2 = S0 xy. Hence c = uxy for some unit u of S0 . By replacing x by ux, we may therefore assume that M = M −1 xy. We then have Q1 = M + N12 = M −1 xy + S0 x2 = (M −1 y + S0 x)x = M −1 x. Hence (U : Q1 ) = (Q1 : Q1 ) = M −1 . Since dimk M −1 /M = 3, there are no (R-modules and hence no) U -modules properly between U and M −1 . Since −1 −1 −1 Q1 = M −1 x, x ∈ / N2 and hence UQ2 = Q1 UQ2 = xMU−1 \Q2 = xMN2 = MN2 . Therefore, UQ1  MU \Q1 and there are no UQ1 -modules properly between them. Since (UQ1 : Q1 UQ1 ) = (U : Q1 )UQ1 = MU−1 \Q1 , it follows −1 −1 that UQ1 is a divisorial domain. Also, since UQ2 = MN2 is a PID (with principal maximal ideal N2 MN ), 2 it follows that U is a divisorial domain (|Star(U )| = 1). Now let T0 = R + N1 N2 , which is local with maximal ideal N1 N2 . It is clear that (T0 : N1 N2 ) = (N1 N2 : N1 N2 ) = S0 . By [7, Lemma 3.10(2)], we can check that T0 falls into one of the categories (4)–(6) of Theorem 1.12, and it is easy to verify that we have |Star(T0 )| = |Star(R)| −4 in each case. Then, by cases (4), (5) and, if T0 is in category (6), by induction, we obtain |Star(Z)| ≤ |Star(T0 )| = |Star(R)| − 4 < |Star(R)|

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for every overring Z of T0 . Now let W be a proper overring of R. We may assume that W  T0 . If W ⊆ Sn , then by [7, Lemma 3.12], we must have W = U . Thus we may assume W  Sn . The proof now splits into two cases: (6a,b) We can handle these cases together. By [7, Lemma 3.10(3)], we have W = (Sn )Pn or (Sn )Sn y , where, Pn is the maximal ideal of Sn that contracts to N1 and y is as above. (In case (a), one has that Pn is also the nonprincipal maximal ideal of Sn−1 ; in case (b), the nonprincipal maximal ideal of Sn−1 is Pn2 .) Then W is local with maximal ideal Q. As usual, we have that W is a finitely generated W -module and QW ⊇ M W . If W = (Sn )Sn y , then M W is the maximal ideal of (Sn )Sn y , and we have W = (Sn )Sn y by Lemma 1.14. (Technically, this is a contradiction, since we have assumed that W  T0 .) Suppose that W = (Sn )Pn . Then QW ⊇ M W ⊇ S0 x = N1 , and with the usual argument, we have W ⊇ U and hence UQ1 ⊆ W ⊆ W = (Sn )Pn = (Sn )U \Q1 . Let w ∈ W . Then sw ∈ Sn ∩ W for some s ∈ U \ Q1 . By what has already been shown, Sn ∩ W = U , and hence w = s−1 (sw) ∈ UQ1 . Thus W = UQ1 , a divisorial domain. (6c) Now Sn has three maximal ideals, Pn and Pn , both lying over N1 in M −1 (in fact Pn Pn = N1 Sn = Sn x), and N2 Sn = Sn y, lying over N2 . There are quite a few possibilities for W : (Sn )Pn , (Sn )Pn , (Sn )N2 Sn , (Sn )Pn ∪Pn , (Sn )Pn ∪N2 Sn and (Sn )Pn ∪N2 Sn . Suppose W = (Sn )Pn . Then W is local with maximal ideal Q, and QW ⊇ M W = Pn W ⊇ N1 . As above, we obtain W ⊇ UQ1 and then that W = UQ1 . The situation with W = (Sn )Pn is identical, and if W = (Sn )N2 Sn , we again obtain the contradiction that W = (Sn )N2 Sn ⊇ T0 . Suppose W = (Sn )Pn ∪Pn . If W is local with maximal ideal Q, then QW ⊇ M W = Pn Pn W ⊇ N1 . In this case, as before, we obtain W = UQ1 . If W has two maximal ideals, they must be L = Pn W ∩ W and L = Pn W ∩ W . In this case, we obtain WL = W Pn W = (Sn )Pn and WL = (Sn )Pn , from which it follows that W = W , a contradiction. By symmetry, the only other case to consider is W = (Sn )Pn ∪N2 Sn . We shall show that this cannot occur. First, suppose that W is local with maximal ideal Q. We then have QW ⊇ M W = Pn N2 W . Via the usual argument, we obtain Q ⊇ (Pn N2 )m for some m and hence that W ⊇ R + (Pn N2 )m , a proper overring of R contained in Sn . According to [7, Lemma 3.12], such a ring must contain T0 = R + N1 N2 , or must be contained in M −1 . Since we have assumed that W  T0 , R+(Pn N2 )m  T0 . Suppose R+(Pn N2 )m ⊆ M −1 . Then (Pn N2 )m ⊆ Pn ∩M −1 = N1 ⊆ Pn , a contradiction. This completes the proof. 2 From Theorems 1.13 and 1.17, we have: Corollary 1.18. If R is a local Noetherian domain with infinite residue field, and 1 < |Star(R)| < ∞, then R is star regular. 2 Corollary 1.18 does not globalize. To see this let D = k[X03 , X07 , X13 , X14 , X15 ], where k is an infinite field, and set R = D(X03 ,X07 )D∪(X13 ,X14 ,X15 )D . Denoting the maximal ideals of R by M, N , with M = (X03 , X07 )R, we have, as in Example 1.7 (but with much simpler details!), that |Star(RM )| = 1, |Star(RN )| = 3, both residue fields are infinite, and RM is not star regular. Hence |Star(R)| = 3 by Lemma 1.2, but R is not star regular by Theorem 1.5. (Moreover, there exists an overring T of RM (and hence of R) with |Star(T )| = ∞.) 2. A converse question In this section, we consider the following: Question 2.1. If R is a one-dimensional Noetherian domain each of whose proper overrings admits only finitely many star operations, must Star(R) also be finite? As phrased, the question is naive, and we give many (classes of) counterexamples at the end of the section. However, in the nonlocal case, the answer is always “yes”:

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Theorem 2.2. Let R be a nonlocal Noetherian domain such that |Star(T )| < ∞ for each proper overring T of R. Then |Star(R)| < ∞. Proof. Let M be a maximal ideal of R. Since RM is a proper overring of R, |Star(RM )| < ∞. This implies that ht M = 1 by [5, Theorem 2.1] and hence that (M RM )v = (M RM )t = M RM = RM .  Thus since M −1 = R, the ring T = N ∈Max(R)\{M } RN is a proper overring of R, and hence |Star(T )| < ∞. Since T is an overring of the 1-dimensional Noetherian domain R, T is also a 1-dimensional Noetherian domain. Thus, for each maximal ideal N of R distinct from M , N RN ∩ T is a maxi mal ideal of T and TN RN ∩T = RN . Therefore, by Lemma 1.2, |Star(T )| = N  ∈Max(T ) |Star(TN  )| ≥   |Star(RN )|, and hence N ∈Max(R)\{M } |Star(RN )| < ∞. Then, by Lemma 1.2, |Star(R)| = N ∈Max(R)\{M } |Star(RM )| · ( N ∈Max(R)\{M } |Star(RN )|) < ∞. 2 Theorem 2.3. Let (R, M ) be a local Noetherian domain. If |Star(T )| < ∞ for each proper overring T of R, then dim R = 1. Proof. Proceeding contrapositively, assume that dim R > 1. Then R is a Krull domain of dimension greater than one and must therefore admit infinitely many star operations by [4, Corollary 2.11]. Hence we may assume that R = R. Then there exist elements a, b ∈ M with (a, b)−1 = M −1 = R. It is obvious that R[a/b] is a proper Noetherian overring of R, and therefore, since R[a/b] ∼ = R[x]/(a−bx) and (a −bx)  M [x]  M +(x) is a chain of distinct prime ideals in the polynomial ring R[x], dimR[a/b] ≥ 2, and so |Star(R[a/b])| = ∞ [5, Theorem 2.1]. 2 Theorem 2.4. Let (R, M ) be a local Noetherian domain with M nonprincipal and infinite residue field k. Assume that |Star(T )| < ∞ for each proper overring T of R. Then M −1 is a PID with at most four maximal ideals, or M −1 has at most two maximal ideals, each of whose residue fields is isomorphic to k. Proof. Let N1 , · · · , Nr be the maximal ideals of M −1 . Then the ring R + N1 · · · Nr is a local ring with maximal ideal N1 · · · Nr and residue field k. Case 1. R = R + N1 · · · Nr . Since M = N1 · · · Nr , M −1 is a PID by [5, Lemma 3.2]. Assume that r ≥ 2. Let S = M −1 \N1 ∪· · ·∪Nr−1 . Then the ring R + M MS−1 is a proper overring of R with maximal ideal M MS−1 and residue field k, and (R + M MS−1 : M MS−1 ) = (M MS−1 : M MS−1 ) = MS−1 . Since |Star(R + M MS−1 )| < ∞, dimk MS−1 /M MS−1 = dimk M −1 /N1 ⊕ · · · ⊕ M −1 /Nr−1 ≤ 3 by [5, Corollary 2.8]. Therefore, r ≤ 4. Case 2. R + N1 · · · Nr = M −1 . Then r = 1 and M −1 /N1 ∼ = k. Case 3. R  R + N1 · · · Nr  M −1 . Let R1 = R + N1 · · · Nr and M1 = N1 · · · Nr . Then M −1 ⊆ (N1 : N1 ) ⊆ (N1 · · · Nr : N1 · · · Nr ) = (M1 : M1 ) = (R1 : M1 ). Since |Star(R1 )| < ∞, dimk (R1 : M1 )/M1 ≤ 3. This implies that dimk M −1 /M1 ≤ 3 and hence that r ≤ 3. If M −1 = (R1 : M1 ), then M −1 is a PID [5, Lemma 3.2]. Assume that M −1  (R1 : M1 ). Then dimk (R1 : M1 )/M1 = 3 and dimk M −1 /M1 = 2. Therefore, r ≤ 2. If r = 1, then M −1 is a local ring with maximal ideal M1 . However, by the proof of [7, Lemma 2.2] and by [7, Lemmas 3.1 and 3.2], there is no such intermediate ring properly between R1 and (R1 : M1 ). Therefore, r = 2 and M −1 /N1 ∼ =k∼ = M −1 /N2 . 2 Although Theorems 2.2–2.4 yield a good deal of information, we have not been able to characterize when Question 2.1 has a positive answer. We close with several categories of domains where the answer is negative. Lemma 2.5. Let (R, P ) ⊆ (T, Q) be local rings with R/P ∼ = T /Q. Then there are no R-modules properly between Q and T . 2

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Theorem 2.6. Let (R, M ) be a local Noetherian domain with infinite residue field k. Then |Star(R)| = ∞, while |Star(T )| < ∞ for every proper overring T of R, in each of the following cases. (1) M −1 is a DVR with maximal ideal M , and dimk M −1 /M = p for some prime p ≥ 5. (2) M −1 is a DVR and dimk M −1 /M = 4. (3) M −1 is local with nonprincipal maximal ideal N , dimk M −1 /M = 3, M −1 is not singly-generated as an R-algebra, S := (N : N ) is a DVR, and dimk S/M S = 3. Proof. (1) We have |Star(R)| = ∞ by [10, Theorem 2.10] or [5, Theorem 2.7]. The result follows, since the only overrings of R are M −1 and K. (2) Suppose that M −1 is as stated. Again, we have |Star(R)| = ∞ by [5, Theorem 2.7]. Let T be a non-field overring of R. Then we may assume that R  T  M −1 . Clearly, T is local with maximal ideal, say, Q. Suppose that (T : Q) = (Q : Q)  M −1 . We then have the chain M  R  T  (Q : Q)  M −1 . This implies that there are no (R-modules and hence no) T -modules properly between T and (Q : Q), from which it follows that |Star(T )| = 1. Hence we may as well assume that (T : Q) = (Q : Q) = M −1 . It is clear that dimT /Q M −1 /Q < dimk M −1 /M = 4. If dimT /Q M −1 /Q = 2, then |Star(T )| = 1. If dimT /Q M −1 /Q = 3, then |Star(T )| = 2 or 3 by Theorem 1.12(2), (3a). (Actually, |Star(T )| = 2 forces Q to be the maximal ideal of M −1 , and it is not difficult to show that this yields dimk M −1 /M > 4, a contradiction.) (3) Assume the situation stated, and let P denote the maximal ideal of S. Since M −1 is not a singly generated R-algebra, N 2 ⊆ M by [7, Lemma 2.1]. There are three cases to consider. Case 3a. Suppose that M S = P . Write P = Sx, where (we may assume that) x ∈ M . Note that N = P also. Since dimk S/M S = 3, this implies that [S/M S : k] = 3. Since k ⊆ M −1 /N ⊆ S/M S, it must be the case that k = M −1 /N (otherwise, M −1 /N = S/M S = S/N , yielding the contradiction that M −1 = S). Note that |Star(M −1 )| = 2 by Theorem 1.12(2). Now suppose that T is a ring properly between R and M −1 , and let Q denote the maximal ideal of T . We have N Q ⊆ N 2 ⊆ M ⊆ Q. Then M −1 Q = (R + N )Q ⊆ Q, and Q is an ideal of M −1 , that is, (T : Q) = (Q : Q) ⊇ M −1 . Since QS = N = Q, we must have (Q : Q) = M −1 . As there are no R-modules properly between T and M −1 , this yields dimk (T : Q)/Q = 2 and hence T is a divisorial domain. Finally, suppose, by way of contradiction, that U is a ring with R  U  S and U  M −1 . Denote the maximal ideal of U by L. We have N L = N LS = N 2 ⊆ M ⊆ L, from which it follows (as above) that (U : L) = (L : L) ⊇ M −1 . Since U  M −1 , this forces (L : L) = S. However, this implies that L is an ideal of S, whence L = LS = N . We then have M −1 = R + N = R + L ⊆ U , a contradiction (since there are no rings properly between M −1 and S). Case 3b. Suppose that M S = P , but M S = N . We claim that k = S/P . Otherwise, we have the chain S  R + P  P  (R + P )x  P 2 ⊇ M S, contradicting that dimk S/M S = 3. Next, by Lemma 2.5, there are no R-modules properly between S and P and hence none between P = Sx and P 2 = Sx2 . This yields dimk S/P 2 = 2. Hence M S  P 2 , and we must have M S = P 3 . We now examine the overrings of R. By [5, Theorem 3.8] and its proof, |Star(M −1 )| = 3, and there is only one ring, namely U := M −1 + P 2 , properly between M −1 and S. It is clear that (U : P 2 ) = (P 2 : P 2 ) = S and that dimU/P 2 S/P 2 = 2. Hence U is a divisorial domain. Now let (T, Q) be a ring properly between R and M −1 . We have N Q = N SQ = N 2 ⊆ M ⊆ Q, from which it follows easily (since M −1 = R + N ) that (T : Q) = (Q : Q) ⊇ M −1 . Since QS = N  Q, this forces (Q : Q) = M −1 or (Q : Q) = U . In the first case, T is a divisorial domain. In the second case, note that Q ⊇ M U  M . Since there are no R-modules between Q and M , this forces Q = M U . The usual arguments then show that U  M −1  N  Q = M U is a saturated chain of R-modules, whence dimk U/Q = 3. Next, note that since S is a 3-generated R-module, so is N , whence dimk N/M N = 3. Hence there are no (R-modules and hence no) T -modules properly between M and M N . Then dimk M/N M = 1, that is, M is principal in M −1 , and we must have M = M −1 ux3 for some unit u ∈ S. Suppose we have Sx4 = (P 2 )2 ⊆ Q = M U = U ux3 . Then P = Sx ⊆ U u, whence P = P u−1 ⊆ U , a contradiction. Hence (P 2 )2  Q, whence U is a singly-generated T -module by [7, Lemma 2.1]. Since QS = P 3 , we also have

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dimk S/QS = 3. Therefore, |Star(T )| = 4 by Theorem 1.12(3b). To conclude, it suffices to show that every ring between R and S lies between R and M −1 or between M −1 and S. To this end, let (W, L) be a ring between R and S. If LS = P , then W = S by Lemma 1.14. If LS = P 3 , then L ⊆ P 3 = N , whence W = R + L ⊆ R + N = M −1 . Finally, suppose that LS = P 2 . Then vx2 ∈ L for some unit v ∈ S. We also have ux3 ∈ M ⊆ L (with u a unit of S), whence uvx5 ∈ L. Write uv = r + sx for some unit r ∈ R and element s ∈ S. This yields rx5 + sx6 ∈ L. However, Sx6 = N 2 ⊆ M ⊆ L, and we have x5 ∈ L. In turn, this gives Sx5 = (R + Sx)x5 = Rx5 + Sx6 ⊆ L. We can apply a similar argument to v 2 x4 ∈ L to obtain Sx4 ⊆ L and another similar argument (to ux3 ∈ L) to get N = Sx3 ⊆ L. It then follows that M −1 ⊆ W , as desired. Case 3c. Suppose that M S = N , so that M S ⊆ P 2 . If N = P , then k ⊆ M −1 /N  S/P , and we have the chain S  R + P  P  P 2 ⊇ M S. However, this yields M S = P 2 = N 2 ⊆ M , a contradiction (since M is not an ideal of S). Therefore, N = P , and we must have N = P 2 and M S = P 3 . We also must have k = S/P . We then have Sx4 = N 2  M  Sx3 , whence P = Sx  M x−3  S, contradicting Lemma 2.5. Thus Case 3c is impossible. 2 The following examples illustrate the various parts of Theorem 2.6. Examples 2.7. (1) Examples are easy to produce. √ √ (2a) Let Q be the field of rational numbers and X an indeterminate over Q. Let V = Q( 2, 3 )JXK = √ √ Q( 2, 3 ) + M and set R = Q + M , where M is the maximal ideal of V . Clearly, M is the maximal ideal of R, V = M −1 , and dimR/M V /M = 4. Hence R is as in Theorem 2.6(2). (2b) Let k be an infinite field, X an indeterminate, and R = kJX 4 , X 5 , X 6 , X 7 K. Then M −1 = kJXK, and dimk M −1 /M = 4. Again R illustrates Theorem 2.6(2). (3a) Let S = F JXK, where F is an infinite field, and assume that F has a subfield k with [F : k] = 3. Let Z = k + XF JXK, M = ZX, and R = k + M . Note that S = Z + Zu + Zu2 for some u ∈ F . We have M −1 = (M : M ) = Z (since M is principal in Z). Moreover, it is easy to check that M −1 = R+RuX +Ru2 X and that S = R + Ru + Ru2 , whence dimk M −1 /M = dimk S/M S = 3. Also, since N = SX ⊆ M −1 , N 2 = SX 2 ⊆ M −1 X = M and M −1 is not a singly-generated R-algebra by [7, Lemma 2.1]. This illustrates Case 3a in the proof of Theorem 2.6. (3b) Let k be an infinite field, X an indeterminate, and R = kJX 3 , X 7 , X 8 K. Denote the maximal ideal of R by M . Then M −1 = kJX 3 , X 4 , X 5 K. It is easy to check that dimk M −1 /M = 3. It is also easy to see that if N denotes the (clearly nonprincipal) maximal ideal of M −1 , then N 2 ⊆ M . As above, M −1 is not a singly-generated R-algebra. Finally, S = (N : N ) = kJXK, and it is easy to check that dimk S/M S = 3. Hence R is as in Case 3b of the proof of Theorem 2.6. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]

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